Answer:
sultan's displacement is
[tex]20 \sqrt{13} [/tex] km
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.
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)
If an atom of oxygen has an atomic number of eight that means...…
E. there are 8 protons
F. there are 8 neutrons
G. it weighs 8 amu
H. it is in group 8
How does the water cycle influence weather in central Wisconsin?
Answer:
The water cycle describes how water evaporates from the surface of the earth, rises into the atmosphere, cools and condenses into rain or snow in clouds, and falls again to the surface as precipitation.The water cycle involves the exchange of energy, which leads to temperature changes.
Answer:
so be more specific please
Many scientific studies have found that colds are caused by viruses. What is this? *
Fact
Interpretation
Analysis
Opinion
A 5 cm diameter orfice discharges fluid from a tank with a head of 5 meters. The discharge rate, Q, is measured at 0.015 m^3/s. The actual velocity at the vena contracta is 9 m/s. The coefficient of discharge is nearest to:________
Answer:
[tex]\mathbf{ C_d = 0.86}[/tex]
Explanation:
Given that:
diameter of the orifice D = 5 cm = 5 × 10⁻² m
discharge rate Q = 0015 m³/s
Actual velocity V = 9 m/s
By using the formula
[tex]Q = C_d \times A \times \sqrt{2gh}[/tex]
where;
[tex]v = \sqrt{2gh}[/tex] = 9 m/s
[tex]Q = C_d \times A \times v}[/tex]
where;
[tex]A = \dfrac{\pi D^2}{4}[/tex]
[tex]A = \dfrac{\pi (5 \times 10^{-2} ) ^2}{4}[/tex]
[tex]A = 0.00196 \ m^2[/tex]
Again,
[tex]Q = C_d \times A \times v}[/tex]
[tex]0.015 = C_d \times 0.00196 \times 9[/tex]
[tex]C_d = \dfrac{0.015 }{ 0.00196 \times 9}[/tex]
[tex]\mathbf{ C_d = 0.86}[/tex]
Hence, the coefficient of discharge [tex]\mathbf{ C_d = 0.86}[/tex]
You have now seen examples during lecture on how to calculate the electric field for a line of charge and a ring of charge - both uniformly distributed. This activity will ask you to solve for the electric field, on-axis, of a uniformly-charged disk sitting in the yz plane. Below is a picture of the situation of interest. Note: Treat it as a totally flat disk and ignore its thickness in the x direction. Also, let x be the distance between the center of the disc and point P.
Answer:
1/4πε₀[Hx/ (√x² + b²)^3/2]i.
Explanation:
So, without mincing words let's dive straight into the solution to the question above. There is need to determine the electric field on-axis of a uniformly-charged disk sitting.
The electric field in the x-component, dεₓ = 1/4πε₀[H/ x² + b²] cos .
Thus, the total electric field in the x-component, εₓ = 1/4πε₀ [ xdH/ (x^2 + b^2)^3/2.
Therefore, the electric field = 1/4πε₀ [ xdH/ (x^2 + b^2)^3/2i.
Where x=0
1.How much work does it take to get a 2Kg ball moving 15m/s if it starts from rest?
2. If a force of 235N was added to the ball, through what distance would this force have to act to give the ball a velocity of 15m/s
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?
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 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.
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).
b. What would the momentum be if the mass of the bowling ball were doubled and its velocity still was 3 m/s?
Answer:
Twice.
Explanation:
The momentum of an object is given by :
p = mv
Where
m is mass and v is the velocity
If the mass of the ball were doubled, m'=2m and v'=v=3 m/s
New momentum,
p'=m'v'
p'=2m × v
p'=2mv
or
p'=2p
So, the new momentum becomes twice the initial momentum.
i’ll mark you brainlist :)
a student lifts a 15N mass through a distance of 1.5m. whats the works done ?
Answer:
10N/m
Explanation:
Calculating workdone=Force/Distance
Therefore=15N/1.5m
=10N/m
A swift moving hawk is moving due west with a speed of 30 m/s; 5.0 s later it is moving due north with a speed of 20 m/s.
(A) What are the magnitude and direction of Δvav vector during this 5 s interval? and
(B) What are the magnitude and direction of vector aav during this 5 s interval?
Answer:
See explanation
Explanation:
Now we have;
vi = (-30, 0) m/s
vf = (0, 20) m/s
Δvav= vf - vi= (30, 20) m/s
magnitude of Δvav= √30^2 + 20^2 = 36.0 m/s
Direction = tan-1(20/30) = 33.69°
For aav
aav= Δvav/t = (30/5, 20/5) = (6,4) m/s^2
magnitude of aav = √6^2 + 4^2 = 7.2 m/s^2
direction of aav = tan-1(4/6) = 33.69°
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?
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).
PLS HELP ME!
A motorist is traveling 40ms-¹ and applies brakes and slow down at a rate of 2ms-² the available distance for the the motorist to stop is 400m will the motorist be able to stop?
Answer:
[tex] \underline{ \boxed{ yes}}\\[/tex]
Explanation:
[tex]given : initial \: velocity \: (u )= 40 {ms}^{ - 1} \\ given : final \: velocity \: (u )= 0 {ms}^{ - 1} \\ given : - (acceleration) \: (a_r) = 2 {ms}^{ - 2} \\ given : distance \: (s) \: = \: ? : \\ but \: {v}^{2} = {u}^{2} + 2( a)s\\ {0}^{2} = {40}^{2} + 2( - 2)s \\ - {40}^{2} = - 4s \\ s = \frac{ - {40}^{2} }{ - 4} \\ s = \frac{1600}{4} \\s = 400 \: m[/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
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].
Which type of heat transfer causes air movement between land and ocean?
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.
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?
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
A girl walks 20.0 m east then 70.0 meters north. What is the girl’s displacement (mag. And direction)? What is the girl’s distance?
Explanation:
Given that,
A girl walks 20.0 m east then 70.0 meters north.
Displacement is the shortest path covered by an object. Let it is d. It is calculated as :
[tex]d=\sqrt{20^2+70^2} \\\\=72.80\ m[/tex]
For direction,
[tex]\theta=\tan^{-1}(\dfrac{70}{20})\\\\\theta=74.05^{\circ}[/tex]
Girl's distance = 20 m + 70 m
= 90 m
Hence, this is the required solution.
In the equation for the gravitational force between two objects, which quantity must be squared?
•mi
•m2
•G
•d
Answer:
d
Explanation:
The quantity that must be squared in the equation of gravitational force is distance d.
According to the universal gravitational law, the square of the distance between two objects is inversely proportional to the force of gravity.
Therefore, the quantity to be squared is dThe formula is given as:
Fg = [tex]\frac{G m_{1} m_{2} }{d^{2} }[/tex]
So d is the quantity that must be squared
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
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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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:______
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 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?
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
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
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.
some waves can only travel through matter. What is this matter called?
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.
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.
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)
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%
In the figure, if Q = 52 µC q =10 µC and d = 55 cm, what is the magnitude of the electrostatic force on q?
Answer:
F = 15.47 N
Explanation:
Given that,
Q = 52 µC
q = 10 µC
d = 55 cm = 0.55 m
We need to find the magnitude of the electrostatic force on q. The formula for the electrostatic force is given by :
[tex]F=k\dfrac{q_1q_2}{d^2}\\\\F=9\times 10^9\times \dfrac{52\times 10^{-6}\times 10\times 10^{-6}}{(0.55)^2}\\\\F=15.47\ N[/tex]
So, the magnitude of the electrostatic force is 15.47 N.
The magnitude of electrostatic force will be "15.47 N".
Electrostatic forceAccording to the question,
Charges, Q = 52 μC
q = 10 μC
Distance, d = 55 cm, or
= 0.55 m
Constant, k = 9 × 10⁹
We know the relation,
→ Electrostatic force, F = k [tex]\frac{q_1 q_2}{d^2}[/tex]
By substituting the values, we get
= 9 × 10⁹ × [tex]\frac{10\times 10^{-6}}{(0.55)^2}[/tex]
= 15.47 N
Thus the above answer is correct.
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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
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.
Many scientific studies have found that colds are caused by viruses. What is this? *
Fact
Interpretation
Analysis
Opinion
Answer:
Analysis
Explanation:
Because you must Analysis each and every cold too find out which virus caused this.
It’s weird because Interpretation and Analysis have the meaning of examination