Which has a greater momentum and greater kinetic energy-a truck with
a mass of 3530 kg moving at a speed of 21 m/s or a car with a mass of
1620 kg moving at a speed of 54 m/s? Answers below are given in the
form: greater momentum, greater kinetic energy.

Answers

Answer 1

Answer:

the car has greater momentum.

the car has greater kinetic energy.

Explanation:

FOR MOMENTUM:

Momentum is given as the product of mass and velocity of an object:

[tex]P = mv[/tex]

where,

P = momentum

m = mass

v = velocity

For Truck:

[tex]P_{truck} = (3530\ kg)(21\ m/s)\\P_{truck} = 74130 Ns[/tex]

For Car:

[tex]P_{car} = (1620\ kg)(54\ m/s)\\P_{car} = 87480\ Ns[/tex]

Therefore, car has greater momentum.

FOR KINETIC ENERGY:

Kinetic Energy is given as:

[tex]K.E = \frac{1}{2} mv^{2}[/tex]

where,

K.E = Kinetic Energy

m = mass

v = velocity

For Truck:

[tex]K.E_{truck} = \frac{1}{2} (3530\ kg)(21\ m/s)^{2}\\K.E_{truck} = 778365\ J = 778.36\ KJ[/tex]

For Car:

[tex]K.E_{truck} = \frac{1}{2} (1620\ kg)(54\ m/s)^{2}\\K.E_{truck} = 2361960\ J = 2361.96\ KJ[/tex]

Therefore, car has greater kinetic energy.


Related Questions

Can someone please help meee .

Answers

Answer:                  

32 amu is the right choice because both protons and neutrons have a mass of 1 amu. Electrons have no mass so go with the last choice

I’m pretty sure it’s the last one 32amu

An Egyptian pyramid contains approximately 1.95 million stone blocks. The average weight of each block is 2.55 tons. What is the weight of the pyramid in pounds?

Answers

Answer:

More than 2,300,000 limestone and granite blocks were pushed, pulled, and dragged into place on the Great Pyramid. The average weight of a block is about 2.3 metric tons (2.5 tons).

A ball is thrown directly downward with an initial speed of 7.70 m/s, from a height of 30.2 m. After what time interval does it strike the ground?

Answers

Answer:

[tex]t = 1.82[/tex]

Explanation:

Given

[tex]u = 7.70m/s[/tex] -- initial velocity

[tex]s = 30.2m[/tex] --- height

Required

Determine the time to hit the ground

This will be solved using the following motion equation.

[tex]s = ut + \frac{1}{2}gt^2[/tex]

Where

[tex]g = 9,8m/s^2[/tex]

So, we have:

[tex]30.2 = 7.70t + \frac{1}{2} * 9.8 * t^2[/tex]

[tex]30.2 = 7.70t + 4.9 * t^2[/tex]

Subtract 30.2 from both sides

[tex]30.2 -30.2 = 7.70t + 4.9 * t^2 - 30.2[/tex]

[tex]0 = 7.70t + 4.9 * t^2 - 30.2[/tex]

[tex]0 = 7.70t + 4.9t^2 - 30.2[/tex]

[tex]7.70t + 4.9t^2 - 30.2 = 0[/tex]

[tex]4.9t^2 + 7.70t - 30.2 = 0[/tex]

Solve using quadratic formula:

[tex]t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex]

Where

[tex]a = 4.9;\ b = 7.70;\ c = -30.2[/tex]

[tex]t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}[/tex]

[tex]t = \frac{-7.70\±\sqrt{651.21}}{9.8}[/tex]

[tex]t = \frac{-7.70\±25.52}{9.8}[/tex]

Split the expression

[tex]t = \frac{-7.70+25.52}{9.8}[/tex] or [tex]t = \frac{-7.70-25.52}{9.8}[/tex]

[tex]t = \frac{17.82}{9.8}[/tex] or [tex]t = -\frac{33.22}{9.8}[/tex]

Time can't be negative;  So, we have:

[tex]t = \frac{17.82}{9.8}[/tex]

[tex]t = 1.82[/tex]

Hence, the time to hit the ground is 1.82 seconds

*PLEASE HELP*
When an object is placed in front
of a convex lens, it creates a virtual
image at -12.8 cm with a
magnification of 2.85. What is the
focal length of the lens?
(Mind your minus signs.)
(Unit = cm)

Answers

Answer: The focal length of the lens = 3.32 cm.

Explanation:

Lens formula : [tex]\dfrac1f=\dfrac1v-\dfrac1u[/tex]    (i)

f= focal length , v=image distance , u =object distance.

magnification: m = [tex]\dfrac{v}{u}[/tex]   (ii )

Given: v= -12.8 cm , m =2.85

Put values in (ii), we get

[tex]2.85=\dfrac{-12.8}{u}\\\\\Rightarrow\ u=\dfrac{-12.8}{2.85}\\\\\Rightarrow\ u=-4.49\ cm[/tex]

substitute values of u , v in (i)

[tex]\dfrac1f=\dfrac1{-12.8}-\dfrac{1}{4.49}\\\\\Rightarrow\dfrac1f=-0.30084214922\\\\\Rightarrow\ f=\dfrac{-1}{-0.30084214922}\approx3.32\ cm[/tex]

Hence, the focal length of the lens = 3.32 cm.

Answer:

6.92

Explanation:

2.85=-(-12.8/x)

do=4.49

1/f= 1/4.49 + 1/-12.8

f=6.92

A motor is used to lift a 10 kg mass 2 m above the ground in 4 s. If the power input to the motor is 100 W. what is the efficiency of the motor?
(Show Work)

Answers

Answer:

98%

Explanation:

Given parameters

Mass of motor = 10kg

Height = 2m

Time = 2s

Power input = 100w

Unknown

Efficiency = ?

Solution

Efficiency is the percentage of the power output to the power input.

Power is the rate at which work is done.

Power output = mass x g x height / time

g is the acceleration due to gravity

Power output = 10x 2 x 9.8 / 2 = 98W

Efficiency = power output/ power input x 100

Efficiency = 98/100 x 100 = 98%

A basketball of mass 0.23kg is thrown horizontally against a rigid vertical wall with a velocity of 20m/s. It rebounds with a velocity of 15m/s. Calculate the impulse of the force of the wall on the basketball.

Answers

Answer:

[tex]8.1\:\mathrm{Ns}[/tex]

Explanation:

The impulse-momentum theorem gives the impulse on an object to be equal to the change in momentum of that object. Since mass is maintained, the change in momentum of the basketball is:

[tex]\Delta p = m\Delta v[/tex], where [tex]m[/tex] is the mass of the basketball and [tex]\Delta v[/tex] is the change in velocity.

Since the basketball is changing direction, its total change in velocity is:

[tex]\Delta v = 20-(-15)=35\:\mathrm{m/s}[/tex].

Therefore, the basketball's change in momentum is:

[tex]\Delta p = m\Delta v = 0.23\cdot 35= 8.05=8.1\:\mathrm{kg\cdot m/s}[/tex].

Thus, the impulse on the basketball is [tex]\fbox{$8.1\:\mathrm{Ns}$}[/tex] (two significant figures).

You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90. The true value is 9.81. Calculate the standard error of your results to ONE significant digit.

Answers

Answer:

0.01

Explanation:

Given the data:

10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, 9.90

True value = 9.81

Mean value :

Σx / n

Sample size, n = 9

(10.1 + 9.87 + 9.76 + 9.91 + 9.75 + 9.88 + 9.69 + 9.83 + 9.90) / 9

= 88.69 / 9

= 9.854

Standard deviation (σ) :

Sqrt (Σ(X - m)² / n)

[(10.1 - 9.854)^2 + (9.87 - 9.854)^2 + (9.76 - 9.854)^2 + (9.91 - 9.854)^2 + (9.75 - 9.854)^2 + (9.88 - 9.854)^2 + (9.69 - 9.854)^2 + (9.83 - 9.854)^2 + (9.90 - 9.854)^2] / 9

Sqrt(0.113824 / 9)

Sqrt(0.0126471)

σ = 0.1124593

Standard Error = σ / sqrt(n)

Standard Error = 0.1124593 / 9

Standard Error = 0.0124954

Standard Error = 0.01 ( 1 significant digit)

a toy car has a 2.0 A current, and its resistance is 1.75 ohms. How much voltage does the car require

Answers

Answer:

the answer will be 24.40 ohms law

Explanation:

Austin invested $11,000 in an account paying an interest rate of 5.7% compounded quarterly. Assuming no deposits or withdrawals are made, how much money, to the nearest dollar, would be in the account after 6 years?

Answers

Answer:

15448

Explanation:

A=11000(1.01425)^{24}

A=11000(1.01425)  

24

Austin invested $11,000 in an account paying an interest rate of 5.7% compounded quarterly. Assuming no deposits or withdrawals are made, the money to the nearest dollar, would be in the account after 6 years is 15448.

What is Compound interest?

The compound interest occurs when the interest is reinvested rather than paying it out. It's basically earning interest over interest.

The formula is:

Compound interest, [tex]A = P ( 1 +\frac{r}{n} )^{nt}[/tex]

Where:

A = final Amount

P = initial principal balance

r = interest rate

n = number of times interest applied per time period

t = number of time periods elapsed

Austin invested P=$11000 in an account with an interest rate of r=5.7% = 0.057 (decimal) during t=6 years compounded quarterly. Since there are 4 quarters in a year, n=4.

Thus, Substituting all the values in the given formula,

A = 11000 ( 1 + [tex]\frac{0.057}{4} )^{6*4}[/tex]

 = 11000 × 1.4043662796

 = 15448.0290

The money to the nearest dollar, would be in the account after 6 years is 15448.

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What are the three concepts of The Cell Theory?

Answers

Answer:

These findings led to the formation of the modern cell theory, which has three main additions: first, that DNA is passed between cells during cell division; second, that the cells of all organisms within a similar species are mostly the same, both structurally and chemically; and finally, that energy flow occurs within.

Explanation:

hope this helped! <3

can i have the crown ?

Find the value of F1 + F2 + F3.

Answers

Answer:

F = 0.78[N]

Explanation:

The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.

For F₁

[tex]F_{y}=2[N][/tex]

For F₂

[tex]F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N][/tex]

For F₃

[tex]F_{x}=-1*sin(60)\\F_{x}=-0.866[N]\\F_{y}=1*cos(60)\\F_{y}=0.5 [N][/tex]

Now we can sum each one of the forces in the given axes:

[tex]F_{x}=1-0.866=0.134[N]\\F_{y}=2-1.73+0.5\\F_{y}=0.77[N][/tex]

Now using the Pythagorean theorem we can find the total force.

[tex]F=\sqrt{(0.134)^{2} +(0.77)^{2}}\\F= 0.78[N][/tex]

. Dan drags a box across the floor. He uses 95 N of force and moves the box 12 m. How much work does he do? 0.13 J 1,140 J 7.9 J 107 J

Answers

Answer:

[tex]1140\:\mathrm{J}[/tex]

Explanation:

Work is given by [tex]W=F\Delta x[/tex], where [tex]F[/tex] is force and [tex]\Delta x[/tex] is displacement.

Plugging in given values, we get:

[tex]W=95\:\mathrm{N}\cdot12\:\mathrm{m}=\fbox{$1140\:\mathrm{J}$}[/tex].

A honeybee leaves the hive, flies in a straight line to a flower 6 km away in 15 min, and then takes 15 minutes to return (also in a straight line). a.) Please find the distance travelled and displacement for the entire trip: distance travelled: 1 12 km displacement: 2 0 km b.) Please find the average speed and average velocity for the entire trip: average speed: 3 0.4 km/min average velocity: 4 0.4 km/min c.) If the bee had not flown in a straight line-- but instead with an unknown motion-- to the flower and back, which could not be determined: the average speed or average velocity

Answers

Answer:

a)   d = 12 km, d = 0,  b)  v = 0.4 km / min, v = 0,  c) v = 0

Explanation:

This problem asks to find the distance and the displacement. We must emphasize that the distance is a scalar and the displacement is a vector quantity

In the exercise, the bold letters indicate vectors

a) Let's find the total distance of the trip

           d = d₁ + d₂

where d₁ is the distance to get to the flowers and d₂ is the distance to return to the hive from the flowers

             d₁ = d₂ = 6 km

             d = 12 km

   

we look for the displacement that is a vector quantity

          d = d₁ - d₂

since the second displacement is in the opposite direction of the first

          d = 0

b) the average speed of the trip, again this magnitude is a scalar

          v = d / t

distance is d = 12 km t total time is t = 15 + 15 = 30 min

          v = 12/30

          v = 0.4 km / min

average velocity (vector) of the entire travel, in that case the displacement is zero

          v = 0

c) If the path is not a straight line but the time is the same, the average speed cannot be calculated since it lacks the distance value

In the same case to calculate the average velocity that is a vector, it will be zero since the net displacement is zero

            v = 0

I need help with number 3!!!!!!!!!!!!​

Answers

Answer:

3. The frequency of the wave is 3 Hz.

Explanation:

3. Determination of the frequency of the wave.

Frequency is simply defined as the number of complete circle or oscillation made in 1 seconds. Mathematically, it can be expressed as:

f = n / t

Where:

f => is the frequency.

n => is the number of circle.

t => is the time.

With the above formula, we can obtain the frequency of the wave as follow:

Number of complete circle (n) = 3

Time (t) = 1 s

Frequency (f) =?

f = n / t

f = 3 / 1

f = 3 /s = 3 Hz

Therefore, the frequency of the wave is 3 Hz

John runs 3 km north then walks 2 km south. What is his total distance traveled and displacement?

Answers

Answer:

the total distance is 5km and the displacement is 1km

Explanation:

The total distance would be the addition of John running both ways so 3 km, 2 km.

However since he only walked back from a distance of 3 km to 2 km, he would be displaced 1 km because displacement is more like the position from the original point.

Think about 2 km as a positive value for the first part of the question and a negative value for the second part.

5.
What is the apparent colour of a red shirt when viewed in pure green light.?
Red
(b)- Green
Yellow (d) Black) (e) Blue​

Answers

Answer: black

Explanation: When green light is shone on a red object, it absorbs all of the green light and not reflecting anything. Hence, it appears black.

An air-filled parallel-plate capacitor is charged and then disconnected from the battery. The plates are then pulled apart to twice their original separation. Which of the following statements about this capacitor is true?
A. The capacitance has doubled.
B. The energy stored in the capacitor has doubled.
C. The potential difference across the plates has decreased.
D. The electric field between the plates has increased.

Answers

Answer:

B.

Explanation:

From the given information, since the capacitors are disconnected from the battery, the electric field between the plate does not change due to the fact that there is no difference in charge density.

Hence, the potential difference after separating the plates is:

V' = Ed

V = e (2d')

V' = 2V

For the energy stored in the capacitor;

[tex]U' = \dfrac{1}{2}QV'[/tex]

where;

V' = 2V

[tex]U' = \dfrac{1}{2} Q (2V)}[/tex]

[tex]U' =2( \dfrac{1}{2} Q V)[/tex]

U' = 2U

Thus, in the capacitor, the energy that is being stored is doubled.

Question 1 of 10
What might happen to personal information when it is transferred using
digital signals?
A. Some information might be changed when the data are copied.
B. It might be accessed by someone who was not the intended
recipient.
C. The information might change while being transmitted because of
noise.
D. The information might change to analog, making it less reliable.


Answers

Answer:

its b for sure

Explanation:

Answer:

B. It might be accessed by someone who was not the intended

recipient

In the laboratory, a ball is dropped onto a force-sensing platform several times, each time hitting a different surface (foam, feathers, clay, etc.). The momentum of the ball changes by the same amount in each trial; in each trial, the average scale reading is F, and the time of collision t are measured. What quantities would need to be graphed to exhibit a straight-line relationship

Answers

Answer:

Graphing the momentum against the change in moment yields a linear relationship.

Explanation:

This is an impulse experiment,

          I = ∫ F .dt

where the force and time of the collision are measured, therefore if we assume an average force the integral reduces to

           I = F t

Furthermore, the momentum is equal to the change in moment of the ball, this change in moment can be found using the energy relations measuring the height of the ball and calculating its speed, in the two intervals for the descent and for the exit, possibly the heights are different so the moment change is different from zero.

Starting point. Higher

          Em₀ = U = mgh

Lower end point, just before hitting the scale

          [tex]Em_{f}[/tex] = K = ½ m v²

in the path in the air there is no friction

          Em₀ = Em_{f}

          m g h = ½ m v²

          v = [tex]\sqrt{2gh}[/tex]

this height is different for the descent and ascent of the ball, so we have two moments

         Δp = [tex]p_{f}[/tex] - p₀

         Δp = m (v_{f} -v₀)

         

therefore we have the relationship

         

         I = Δp

Graphing the momentum against the change in moment yields a linear relationship.

A bat at rest sends out ultrasonic sound waves at 46.2 kHz and receives them returned from an object moving directly away from it at 21.8 m/s, what is the received sound frequency?
f= ? Hz

Answers

Answer:

f" = 40779.61 Hz

Explanation:

From the question, we see that the bat is the source of the sound wave and is initially at rest and the object is in motion as the observer, thus;

from the Doppler effect equation, we can calculate the initial observed frequency as:

f' = f(1 - (v_o/v))

We are given;

f = 46.2 kHz = 46200 Hz

v_o = 21.8 m/s

v is speed of sound = 343 m/s

Thus;

f' = 46200(1 - (21/343))

f' = 43371.4285 Hz

In the second stage, we see that the bat is now a stationary observer while the object is now the moving source;

Thus, from doppler effect again but this time with the source going away from the obsever, the new observed frequency is;

f" = f'/(1 + (v_o/v))

f" = 43371.4285/(1 + (21.8/343))

f" = 40779.61 Hz

1. An engine absorbs 600 J of heat while doing 650 J of work. What is the change
in internal energy of the enginge? *
1250 J
-50 J
-1250 J
50 J

Answers

J-473 is what it would be

The change in internal energy of the engine is -50 joule. Hence, option (B) is correct.

What is law of conservation of energy?

Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.

The absorb energy: Q = 600 Joule

Work done: W = 650 Joule.

Let, the change in internal energy of the engine= dU.

According to conservation of energy:

The absorb energy =  change in internal energy  + Work done

Q = dU + W

dU = Q - W

= 600 joule - 650 joule

= - 50 joule.

Hence, the change in internal energy of the engine is -50 joule.

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Which type of heat transfer causes air movement between land and ocean?

Answers

There are three methods of heat travel:

CONDUCTION -- The transfer of heat through a medium. This is how we cook food on top of a stove. The heat from the stove burner is conducted through a medium (a metal pot) to the food.

CONVECTION -- The transfer of heat due to the physical movement of an object. We can observe convection by looking at a pot of boiling water. Have you ever noticed that when a pot of water is boiling, the water seems to follow a vertical circular motion? This is convection. The parcel of heated water at the bottom of the pot rises, as it rises it gives off some of its heat. Because it loses some heat, the parcel is cooler than the surrounding water. It then sinks to the bottom of the pot and the process is started again. The path of the rising water followed by the sinking water traces out a circle.

RADIATION -- The transfer of heat by means of waves. This is the most difficult method of heat transfer to understand. Yet, we experience it every day. We feel the effects of radiation whenever we stand near a stove or oven which is being used. We feel the heat radiating from the stove or oven to our skin. Similarly, we have all been outside on a sunny, hot Summer's day. If we look up to the sky we can feel the rays of the Sun hitting our faces. The Sun is radiating its heat to the Earth.

It is through one of the above processes of heat transfer that causes the air temperature at deep-ocean station 41001 to be warmer than that of land station CLKN7 during the winter months. Which process do you believe to be the cause of the air temperature differences between these two stations? I'll give you a hint, it has something to do with the temperature of the ocean water. Lets look at a graph of both the average air and water temperatures from Station 41001.

As you can see from the graph, the January (month 1) and February (month 2) water temperatures are about 20 degrees while the respective air temperatures are about 15 degrees. This is a 5 degree difference in temperature between the air and the water at the same geographical location!!

We can figure out what heat transfer process is influencing the air temperature at station 41001 by applying the three methods to our situation and then we can choose the one that seems most logical.

First, lets look at conduction. This process involves the transfer of heat through a conductive medium. Well, nothing exists between the air and the water surface. In our situation, the heat is going directly from the water to the air without passing through a conductive medium. Therefore, this is not the applicable process that is causing the warm winter-time air temperatures at station 41001.

Convection involves the movement of heated objects. The physical movement must be a result of the heating, such as with the pot of boiling water where the vertical movement is caused by the intense heat applied to the bottom of the pot. Because the ocean water isn't moving into or through the atmosphere as a result of the sun's heating of the water, convection isn't the process influencing air and water temperature difference. Ocean water is moving through the lower few feet of the air as ocean surface waves, but this doesn't occur because of the sun's heat.

The final process, radiation, is causing the winter-time air temperatures over water to be warmer than the winter-time air temperatures over land. The heat of the ocean is being given off (radiated) into the air, thus making the air substantially warmer.

Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s as it leaves the hose nozzle. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation of the hose until the water takes 3.00 to reach a building 41.0m away. You can ignore air resistance; assume that the end of the hose is at ground level.

Required:
a. Find the angle of elevation of the hose.
b. Find the speed in m/s of the water at the highest point in its trajectory.
c. Find the acceleration in m/s^2 of the water at the highest point in its trajectory.
d. How high above the ground in m does the water strike the building?
e. How fast is it moving in m/s just before it hits the building?

Answers

Answer:

a) θ = 58.3º

b) vfh = 13.7 m/s

c) g = -9.8 m/s2

d) h = 22.2 m

e) vfb = 15.5 m/s

Explanation:

a)

Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:vₓ₀ = v * cos θ (1)where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:

        [tex]x_{f} = v_{ox} * t = v_{o} * cos \theta * t (2)[/tex]

Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:

        [tex]cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)[/tex]

⇒θ = cos⁻¹ (0.526) = 58.3º (4)

b)

At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)

c)

At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)

d)

Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:

       [tex]\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)[/tex]

Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)Replacing (7) in (6), we get:

       [tex]\Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)[/tex]

e)

When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.The horizontal component, since it keeps constant, is just v₀x:v₀ₓ = 13.7 m/sThe vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:

       [tex]v_{fy} = v_{oy} - g*t (9)[/tex]

Replacing by the time t (a given), g, and  v₀y from (7), we can solve (9) as follows:

       [tex]v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s (10)[/tex]

Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:

       [tex]v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)[/tex]

A charge of 7.1 x 10-4 C is placed at the origin of a Cartesian coordinate system. A second charge of 6.5 x 10-4 C lies 20 cm above the origin, and a third charge of 8.9 x 10-4 C lies 20 cm to the right of the origin. Determine the direction of the total force on the first charge at the origin. Express your answer as a positive angle in degrees measured counter clockwise from the positive x-axis.

Answers

Answer:

α = 36.21 °

β = 143.79°

Explanation:

To do this, we need to know the expression to calculate the angle.

In this case:

α₁ = tan⁻¹ (Fy₁/Fx₁)   (1)

Now, let's analize the given data.

We have a charge q₁ at the origin of the cartesian coordinate system, so, it's at the 0. The charge q₂ is 20 cm above q₁, meaning is on the y-axis. Finally q₃ it's 20 cm to the right, meaning it's on the x-axis.

Knowing this,we can calculate the force that q₂ and q₃ are exerting over q₁. As these forces are in the x and y-axis respectively, we also are calculating the value of the forces in the x and y axis, that are needed to calculate the direction.

The expression to calculate the force would be Coulomb's law so:

F = K q₁q₂ / r²    (2)

The value of K is 9x10⁹ N m² / C². Let's calculate the forces:

F₁₂ = Fy = 9x10⁹ * (7.1x10⁻⁴) * (6.5x10⁻⁴) / (0.020)²

Fy = 1.04x10⁷ N

F₁₃ = Fx = 9x10⁹ * (7.1x10⁻⁴) * (8.9x10⁻⁴) / (0.020)²

Fx = 1.42x10⁷ N

Now that we have both forces, we can calculate the magnitude of the force:

F = √(Fx)² + (Fy)²

F = √(1.04x10⁷)² + (1.42x10⁷)²

F = 1.76x10⁷ N

Finally, the direction would be applying (1):

α = tan⁻¹ (1.04x10⁷/1.42x10⁷)

α = 36.21 °

And counter clockwise it would be:

β = 180 - 36.21 = 143.79°

Hope this helps

A 71-kg swimmer dives horizontally off a 500-kg raft. If the diver's speed immediately after leaving the raft is 6m/s, what is the corresponding raft speed?

Answers

Answer:

The answer is below

Explanation:

Momentum is used to measure the quantity of motion in an object. Momentum is the product of mass and velocity.

Momentum = mass * velocity

The principle of conservation of momentum states that momentum cannot be created or destroyed but  can be transferred. Therefore the momentum before and after an action is equal.

Initial momentum = Final momentum

Let m be the mass of the diver, M be the mass of the raft, u be the initial velocity of the diver, U be the initial velocity of the raft, v be the final velocity of the diver and V be the final velocity of the raft.

m = 71 kg, M = 500 kg, v = 6 m/s

Initial both the raft and diver are at rest, hence u and U is zero, hence:

mu + MU = mv + MV

71(0) + 500(0) = 71(6) + 500(V)

0 = 426 + 500(V)

500(V) = -426

V = -426/500

V = -0.852 m/s

28. Which of the following correctly shows the order of highest amount of friction to the lowest amount of
friction?
a. Static, Rolling, Sliding
b. Sliding, Rolling, Static
c. Rolling, Static, Sliding
d. Static, Sliding, Rolling

Answers

Answer:

[tex]\mathrm{d.\:Static,\: Sliding,\:Rolling}[/tex]

Explanation:

Static friction occurs when an object initially starts at rest. When the surfaces of the materials touch, the microscopic unevenness interlock greatest with each other, causing the most friction out of the three.

During sliding friction, an object is already moving or in motion. The microscopic surfaces still interlock, but because the object is in motion, it has a momentum. Therefore, the magnitude of sliding friction is less than that of static friction.

Rolling friction occurs when an object rolls across some surface. Rather than surfaces interlocking, rolling friction is caused by the constant distortion of surfaces. As it rolls, the surfaces of the object are constantly wrapping and changing. This distortion causes the rolling friction. However, it is much less in magnitude when compared to static or sliding friction.

Two spherical objects with a mass of 3.17 kg each are placed at a distance of 2.96 m apart. How many electrons need to leave each object so that the net force between them becomes zero

Answers

Answer:

1.704 × 10¹⁹ electrons

Explanation:

The gravitational force due to the identical masses with mass, m = 3.17 kg at a distance r = 2.96 m from each other is F = Gm²/r².

Since equal number of electrons have to leave both masses, we have a charge q on each mass acting to oppose each other at distance r.

So, the electrical force of repulsion is thus

F'= kq²/r²

For the net force to be zero, the gravitational force of attraction must balance the electrical force of repulsion.

So, F = F'

So,  Gm²/r² = kq²/r²

Gm² = kq²

q² = Gm²/k

taking square root of both sides, we have

q = m√(G/k)

So, substituting the values of the variables into the equation, we have

q = m√(G/k)

= 3.17 kg√(6.67 × 10⁻¹¹ Nm²/kg²/9 × 10⁹ Nm²/C²)

= 3.17 kg√(0.741  × 10⁻²⁰ C²/kg²)

= 3.17 kg × 0.861 C/kg

= 2.73 C

Now, q = ne where n = number of electrons and e = electron charge = 1.602 × 10⁻¹⁹ C

n = q/e

= 2.73 C ÷ 1.602 × 10⁻¹⁹ C

= 1.704 × 10¹⁹ electrons

Two 0.60-kilogram objects are connected by a thread that passes over a light, frictionless pulley. The objects are initially held at rest. If a third object with a mass of 0.30 kilogram is added on top of one of the 0.60-kilogram objects and the objects are released, the magnitude of the acceleration of the 0.30-kilogram object is most nearly:______

Answers

Answer:

2 m/s²

Explanation:

From the given information:

The first mass m_1 = 0.6 kg

The second mass m_2 = 0.3 kg

The magnitude for the acceleration of 0.3 kg is:

a = net force/ effective mass

Mathematically, it can be computed as follows:

[tex]a = \dfrac{F}{m}[/tex]

[tex]a = \dfrac{(m_2 +m_1 -m_1) }{(m_2+m_1+m_1)}(g)[/tex]

[tex]a = \dfrac{0.3 +0.6 -0.6}{(0.3 +0.6+0.6)}(9.8)[/tex]

a ≅ 2 m/s²

A receiver catches a football on the 50.0 yard line and is tackled 5.42 seconds later on the 12 yard line. What
was the runner's average speed?

Answers

Answer:

7.01yard/sec

Explanation:

Given parameters:

Initial position  = 50yard

Final position = 12yard

Time  = 5.42s

Unknown:

Average speed of runner  = ?

Solution:

To solve this problem;

        Speed  = [tex]\frac{distance}{time}[/tex]  

Distance covered  = Initial position  - final position  = 50 - 12  = 38yards

So;

       Speed  = [tex]\frac{38}{5.42}[/tex]   = 7.01yard/sec

a student lifts a 15N mass through a distance of 1.5m. whats the works done ?

Answers

Answer:

10N/m

Explanation:

Calculating workdone=Force/Distance

Therefore=15N/1.5m

=10N/m

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