A runaway train car that has a mass of 15,000 kg travels at a speed of 5.4 m/s down a track. Compute the time required for a force of 1500 N to bring it to rest.
(a) 0.019 s
(b) 54s
(c) 4.2Ã106s
(d) 2.4Ã10â7s

Answers

Answer 1

Answer:

54 s

Explanation:

Given that,

Mass of a car, m = 15000 kg

Initial speed, u = 5.4 m/s

Final speed, v = 0 (it comes to rest)

Force, F = 1500 N

We need to find time when it comes to rest. Let it is t. Net force is given by :

F = ma, a is acceleration of the car

[tex]F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{15000\times (0-5.4)}{1500}\\\\t=54\ s[/tex]

So, it will take 54 seconds to come to rest.

Answer 2

The time required to bring the car to rest is 54 s and the right option is (b) 54 s.

To calculate the time required for the force to bring the car to rest, we use the formula below.

Formula:

F = m(v-u)/t............ Equation 1

Where:

F = Force applied to the carm = mass of the carv = final velocity of the caru = initial velocity of the cart = time.

Make t the subject of the equation

t = m(v-u)/F............. Equation 2

From the question,

Given:

F = -1500 N (stopping force)m = 15000 kgv = 0 m/s (to rest)u = 5.4 m/s

Substitute these values into equation 2

t = [1500(0-5.4)]/-1500t = -81000/-1500t =  54 secondst ≈ 54 s

Hence, The time required to bring the car to rest is 54 s and the right option is (b) 54 s.

Learn more about time here: https://brainly.com/question/4931057


Related Questions

A 8.45μC particle with a mass of 6.15 x 10^-5 kg moves perpendicular to a 0.493-T magnetic field in a circular path of radius 34.1 m. How much time will it take for the particle to complete one orbit?

a. 92.7 s
b. 0.0927 s
c. 9.27 s
d. 927 s

Answers

This question is incomplete, the complete question is;

A 8.45μC particle with a mass of 6.15 x 10⁻⁵ kg moves perpendicular to a 0.493-T magnetic field in a circular path of radius 34.1 m.

How much time will it take for the particle to complete one orbit?

a. 92.7 s

b. 0.0927 s

c. 9.27 s

d. 927 s

Answer:

it will take 92.7 seconds for the particle to complete one orbit.

Option a) 92.7 s is the correct option

Explanation:

Given that;

mass m =  6.15 x 10⁻⁵ kg

q = 8.45μC = 8.45 × 10⁻⁶ C

B = 0.493

we know that

Time period T = 2πr / V

where r = mv/qB

so T = 2πm/qB

we substitute

T = (2 × 3.14 × 6.15 x 10⁻⁵) / ( 8.45 × 10⁻⁶ × 0.493)

T = 0.0003862 / 0.000004165

T = 92.7 sec

Therefore it will take 92.7 seconds for the particle to complete one orbit.

Option a) 92.7 s is the correct option

''The Evolution Of Sakura And Hinata''

Who looks better?! Who do you think has improved a lot in the Naruto Series?

Answers

Answer:

hinata

Explanation:

unlike sakura she has not change at all

In my opinion, Sakura looks prettier and she developed way more than Hinata. Personally, I dislike the both of them because of their Tsundere side, but there's no doubt Sakura developed way more than Hinata. In the original Naruto series, Sakura was honestly a burden. There were some missions where Team 7 had to rescue Sakura. However, after the training with Tsunade, she gained monster strength and healing abilities. Additionally, throughout Naruto Shippuden she learned new jutsu like Yin Seal: Release.    

     However with Hinata, it's a bit more different. In the original Naruto series, I feel believe she was stronger than Sakura. She relied on her infamous visual prowess--the Byakugan. However in Naruto Shippuden, she improved slightly compared to Sakura. In my opinion, she would have been the most useless shinobi out of the eleven characters (YES, more useless than Tenten) if it weren't for her visual prowess. She's fortunate that she was born into the Hyuga clan. Every jutsu that's in her arsenal relies heavily on the Byakugan. And if Sakura and Hinata were to battle each other, it's most likely that Sakura would win. It's true that Hinata can block Sakura's chakra points, but I think she could dodge her attacks. And even if her chakra points are blocked, Sakura can rely on Taijutsu and raw strength to win.

Though both kunoichi are honorable, I believe Sakura is more prettier and stronger than Hinata.

F = 5 Newtons
W = 75 Joules
d = ?

ANSWER

Answers

Substitution: d = 75 J/5 N
Answer with unit of measure: d = 15 m

When a mass of a cart is 10 kg, and an applied force is 5 N, The acceleration of the cart is

5.0 m/s2
2.0 m/s2
0.5 m/s2
0.2 m/s2

Answers

Answer:

0.5 m/s2

Explanation:

F = ma

5 = 10a

a = 5/10

a =0.5

The sound intensity at 4 m from a source is 100 W/me. What is the intensity of the sound at 12 m away from the source ?

Answers

Answer:

Intensity at 12 meters will be 11.11 W/m^2

Explanation:

Recall that the intensity of sound is inversely proportional to the square of the distance from the source. Therefore, if at 4 m the intensity is 100 W/m^2

we have: 100 W/m^2 = k/16 and therefore, k = 1600 W

Then the intensity (I) at 12 m will be:

I = k/12^2 = 1600/144  W/m^2 = 11.11 W/m^2

A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force) acts on her. The student has a mass of 71.0 kg, and the height of the water slide is 12.7 m. If the kinetic frictional force does -5.10 103 J of work, how fast is the student going at the bottom of the slide

Answers

Answer:

[tex]10.27m/s[/tex]

Explanation:

Given data

work W= -5.10 10^3 J

mass m= 71kg

final height of slide h2= 12.7m

initial height of slide h1=0m

initial velocity v1= 0m/s

final velocity v2=?

Step two:

required

Final velocity

The work-energy theorem is expressed as'

[tex]W=1/mv_2^2 +mgh_2-(1/mv_1^2+mgh_1)[/tex]

make V2 subject of formula we have final speed

[tex]v_2=\sqrt{\frac{2W}{m}+v_1^2-2g(h_1-h_2) } \\\\[/tex]

substitute our given data we have

[tex]v_2=\sqrt{\frac{2*(-5.1*10^3)}{71}+0^2-2*9.81(12.7) } \\\\v_2=\sqrt{143.66-249.174 } \\\\v_2=\sqrt{105.514 } \\\\v_2=10.27m/s[/tex]

The student going at  10.27m/s

This is a short question can anyone help me please

Thank you

Picture Above

Answers

Answer:

I thinks it's

deficit spending

Explanation:

cause When a government spends more than it collects in taxes, it is said to have a budget deficit.

A bullet of mass 4.2 g strikes a ballistic pendulum of mass 1.0 kg. The center of mass of the pendulum rises a vertical distance of 18 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

Answers

Answer:

449.020m/s

Explanation:

We have been provided with the following information:

Mass of bullet = Mb = 4.2g

Mass of pendulum = 1.0 = Md

Vertical distance = h = 18cm

Gravitational force = g = 9.8

We have kinetic energy converted to potential energy for the entire system

1/2(Mb+Md)V² = (Mb + Md)gh

We have V as the speed of system during collision

Mbv = (Mb+Md)V

We divide through by Mb

v = (Mb+Md/Mb)√2gh

4.2x10^-3 = 0.0042

18cm = 0.18m

(0.0042+1.0/0.0042)√2x9.8x0.18

= 1.0042/0.0042√3.528

= 239.095x1.878

= 449.020m/s

This is the answer to the question

Thank you!

A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) (4.00 m/s2 ) .At that instant and in unit-vector notation, what is the acceleration of the wallet

Answers

Complete Question:

A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.

They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet

Answer:

aw = 3 i + 6 j m/s2

Explanation:

Since both objects travel in uniform circular motion, the only acceleration that they suffer is the centripetal one, that keeps them rotating.It can be showed that the centripetal acceleration is directly proportional to the square of the angular velocity, as follows:

       [tex]a_{c} = \omega^{2} * r (1)[/tex]

Since both objects are located on the same radial line, and they travel in uniform circular motion, by definition of angular velocity, both have the same angular velocity ω.

       ∴ ωp = ωw (2)

           ⇒ [tex]a_{p} = \omega_{p} ^{2} * r_{p} (3)[/tex]

               [tex]a_{w} = \omega_{w}^{2} * r_{w} (4)[/tex]

Dividing (4) by (3), from (2), we have:

        [tex]\frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}[/tex]

Solving for aw, we get:

        [tex]a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2[/tex]

Objects falling through air are slowed by the force of air resistance. Which objects were slowed the most by air resistance?

Answers

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, it should be noted that when objects of different sizes fall in absence of air resistance, the objects will get to the ground at the same time. But with the presence of air resistance, the heaviest object gets to the ground first; meaning it has the least air resistance while the lightest object will arrive at the ground last because it has the greatest air resistance and is slowed down the most by the air resistance.

Thus, the lightest object in the completed question is the answer.

Approximating Venus's atmosphere as a layer of gas 50 km thick, with uniform density 21 kg/m3, calculate the total mass of the atmosphere.Express your answer using two significant figures.m venus atmosphere = ____ kg

Answers

Answer:

m = 4.9 10⁸ kg

Explanation:

The expression for the density is

           ρ = m / V

           m = ρ V

the volume of the atmosphere is the volume of the sphere of the outer layer of the atmosphere minus the volume of the plant

          V = V_atmosphere - V_planet

           V = 4/3 π R_atmosphere³ - 4/3 π R_venus³

           V = 4/3 π (R_atmosphere³ - R_venus³

)

the radius of the planet is R_venus = 6.06 10⁶ m.

The radius of the outermost layer of the atmosphere

          R_atmosphere = 50 10³ + R_ venus = 50 10³ + 6.06 10⁶

           R_atmosphere = 6.11 10⁶ m

let's find the volume

           V = 4/3 pi [(6,11 10⁶)³ - (6,06 10⁶)³]

            V = 23,265 10⁶ m³

let's calculate the mass

          m = 21  23,265 10⁶

          m = 4.89   10⁸ kg

with two significant figurars is

          m = 4.9 10⁸ kg

when you stir a cup of tea the floating clip collect at the centre of the Cup rather than in the outer Rim why​

Answers

Answer:

hsvshxansjusjsnwjwisks

Explanation:

When we stir a cup of tea we create a force in the center which pulls out all the particles towards it this is the basic reason for collection of tea leaves at the center of the cup rather than at the rim of the cup, it is similar to the the case of tornado where it takes all the particles present on it way to its ..

Are the refractive index and the speed of light in a vacuum direct propotional or inversley​

Answers

The refractive index of the medium is inversely proportional to the velocity of light in it. As the refractive index of a medium increases, the speed of light going through that medium decreases.

A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?

Answers

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

A parallel-plate capacitor with circular plates of radius 40 mm is being discharged by a current of 6.0 A. At what radius (a) inside and (b) outside the capacitor gap is the magnitude of the induced magnetic field equal to 75% of its maximum value?(c) What is that maximum value?

Answers

Answer:

A) r = 0.03 m

B) r = 0.0533 m

C) B_max = 0.00003 T

Explanation:

Formula for magnetic field inside the capacitor when it is parallel to the length element is;

B_in = (μ_o•I•r/(2πR²)

Formula for maximum magnetic field is;

B_max = (μ_o•I/(2πR)

Formula for magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)

A) Magnetic field inside the capacitor is gotten from our first equation above;

B_in = (μ_o•I•r/R²)

Since we want to find the radius at which the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value.

Thus;

B_in = 0.75B_max

(μ_o•I•r/(2πR²) = 0.75((μ_o•I/(2πR))

μ_o•I and 2πR will cancel out to give;

r/R = 0.75

r = 0.75R

We are given R = 40 mm = 0.04 m

r = 0.75 × 0.04

r = 0.03 m

B) magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)

Thus for the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value:

B_out = 0.75B_max

(μ_o•I/(2πr) = 0.75((μ_o•I/(2πR))

μ_o•I and 2π will cancel out to give;

1/r = 0.75/R

r = R/0.75

r = 0.04/0.75

r = 0.0533 m

C) B_max = μ_o•I/(2πR)

μ_o is a constant known as vacuum of permeability with a value of 4π × 10^(-7) T.m/A

Thus;

B_max = (4π × 10^(-7) × 6)/(2π × 0.04)

B_max = 0.00003 T

An electron is accelerated from rest through a potential difference of 300V. it then passes through a uniform 0.001-T magnetic field, oriented perpendicular to the electrons velocity. what is the magnitude of the magnetic force on the electron?

Answers

Given :

Potential difference, V = 300 V.

Magnetic Field, B = 0.001 T.

To Find :

The magnitude of the magnetic force on the electron.

Solution :

We know, for perpendicular orientation, force is given by :

[tex]F = qVB\\\\F = 1.6 \times 10^{-19} \times 300\times 0.001\ N\\\\F = 4.8\times 10^{-20}\ N[/tex]

Hence, this is the required solution.


9. In the graph below, what is the force being exerted on
the 16-kg cart?
A. 4N
C. 16N
B. 8N
D. 32 N

Answers

Answer:

D

Explanation:

What is the average power consumption in watts of an appliance that uses 4.69 kW · h of energy per day?

Answers

Answer:

The average power is  [tex]P = 195 .42 \ W[/tex]

Explanation:

From the question we are told that

   The  energy of the appliance is  [tex]E = 4.69 \ kWh = 4.69 *10^{3} \ \ Wh[/tex]

    The time considered is   [tex]t = 1 \ day = 24 \ hours[/tex]

Generally the average power consumption is mathematically represented as

      [tex]P = \frac{E}{t}[/tex]

=>   [tex]P = \frac{4.69 *10^{3}}{24 }[/tex]

=>   [tex]P = 195 .42 \ W[/tex]

A crossbow is fired horizontally off a cliff with an initial velocity of 15 m/s. If the arrow takes 4s to hit the ground, what is the range of the projectile?

Answers

Answer:

The range of the projectile is 60 m

Explanation:

Horizontal Motion

When an object is thrown horizontally with a speed vo from a height h, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.

The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:

[tex]v_x=v_o[/tex]

The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:

[tex]v_y=g.t[/tex]

The horizontal distance is calculated as a constant speed motion:

[tex]x = v_x.t[/tex]

Knowing the crossbow is fired horizontally at vo=vx=15 m/s and it takes t=4 s to hit the ground, thus the range of the projectile is:

x = 15*4 = 60

The range of the projectile is 60 m

Mr. Jones starts from rest and begins to accelerate straight to the bathroom at a rate of 0.5 m/s for 10 seconds. What kind of motion is this
A. linear
b centripetal
c.free fall
d projectile​

Answers

Answer:

A. linear

Explanation:

because they are going in a straight line

A 500-N box is at rest on the floor. Dennis Elbo makes several
attempts to move the box, pushing against the box with varying
amounts of horizontal force. Yet the box never does move. In this
situation, the amount of static friction force experienced by the box
Select all that apply.
-
0 is 500 N
O is equal to the force with which Dennis exerts on the
box
has an upper limit and Dennis O has not yet exceeded the upper limit
Ois always the coefficient of friction multiplied by the normal force value

Answers

Answer:

Select the second and the third options you listed.

Explanation:

Select the answer options:

"is equal to the force with which Dennis exerts on the  box."

and

"has an upper limit and Dennis has not yet exceeded the upper limit."

In fact, this upper limit of the static friction force is the product of the coefficient of static friction ([tex]\mu[/tex]) times the weight of the box.

What physical property does the symbol I_enclosed in problem 5 represent? a. The current along the path in the same direction as the magnetic field b. The current in the path in the opposite direction from the magnetic field c. The total current passing through the loop in either direction d. The net current through the loop

Answers

Answer:

C

Explanation:

Current passing through the loop in either direction

Net force causes motion

Answers

Answer:

yes

Explanation:

If an object has a net force acting on it, it will accelerate. The object will speed up, slow down or change direction. An unbalanced force (net force) acting on an object changes its speed and/or direction of motion.

In the concluding paragraph, how does the author make the conflicting point of view that vegetarianism is strange and unnatural seem less reasonable?

Answers

Answer:

By pointing out that vegetarianism isn't contagious

Explanation:

I got this right on iready!

Answer:

by point out how food isn't contagious

Explanation:

got it right

How do unbalanced forces acting on an object affect its motion when the object is at rest? What if it is moving?

Answers

Answer:

It pushes it because an unbalanced force is pushing more newtons than something that isn't even moving. Even if it is moving, it depends which side is pushing/pulling the most force.

Explanation:

Answer:

It pushes it because an unbalanced force is pushing more newtons than something that isn't even moving. Even if it is moving, it depends which side is pushing/pulling the most force.

Which of the following is NOT a type of variable?

A. Burette
B. Ordered
C. Continuous
D. Categoric

Answers

Answer:

A

Explanation:

because its not trust me :)

Elevator is accelerating upward 3.5 M/S2 and has a mass of 300 KG. The force of gravity is 2940 N. What is the tension force pulling elevator up?

Answers

Answer:

T = 3990 N

Explanation:

The free body diagram for the elevator consists of a tension force pointing up, and its weight pointing down. So the elevator's net force is:

F = T - 2940N

ad at the same time, using Newton's second law, we have that this net force should equal the elevator's mass (300 kg) times its acceleration (a):

T - 2940N = 300kg (3.5m/s^2)

then

T = 2940 N + 1050 N

T = 3990 N

A football player runs down the field at a speed of 8 m/s how long will it take him to run 20 m?

Answers

The answer is simple. Since the football player runs at 8m/s and you are trying to find out how long it will take him to run 20 m, you simply divide 20/8.

The answer to this would be 2.5 seconds for the football player to run 20 meters.
I hope I did this problem correct and I hope that this helped you :)))

A rack of seven spherical bolwing balls (each 7.00 kg, radius of 9.50 cm) is positioned along a line located a distance =0.850 m from a point ,

Calculate the gravitational force
the bowling balls exert on a ping-pong ball of mass 2.70 g, centered at point .

Answers

Answer:

8.72*[tex]10^{-12}[/tex] N

Explanation:

force of attraction f = G m1m2/ r^2 = [tex]\frac{6.67*10^{-11}*7*5*2.70 }{.85*.85*1000}[/tex] = 8.72*[tex]10^{-12}[/tex] N

The gravitational force the bowling balls exert on a ping-pong is given as  8.72*[tex]10^{-12}[/tex] N.

What is force?

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

force of attraction

f = G [tex]m_1m_2[/tex]/r²

 = [tex]6.67*10^{-11}[/tex]*7*5*2.70/.85²*1000

 = 8.72*[tex]10^{-12}[/tex] N

The gravitational force the bowling balls exert on a ping-pong is given as  8.72*[tex]10^{-12}[/tex] N.

To learn more about force refer to the link:

brainly.com/question/13191643

#SPJ3

find the fundamental units involved in derived units
newton
watt
joule
pascal
cubic meter ​

Answers

9ycy8c8t 7f fixfuozofuxt8lsrupsurpaurae6pUeoUe6eoUeFipzuroz6d0, 7d0z6e0z7e0zurpz6e0z

Explanation:

force newton N - m·kg·s-2

pressure, stress pascal Pa N/m2 m-1·kg·s-2

energy, work, quantity of heat joule J N·m m2·kg·s-2

power, radiant flux watt W J/s m2·kg·s-3

volume cubic meter m3

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