Ask question

Ask question

All categories

bagirrra123 [75]

2 years ago

8

Cart A and cart B are traveling in the same direction on a straight track. Cart A is moving at 9.25 m/s and cart B is moving at

7.15 m/s. The mass of cart A is 72.0 kg and the mass of cart B is 55.0 kg. Cart A catches up to and collides with cart B. After they collide, cart A slows down to 6.15 m/s. Find the final speed of cart B.

1 answer:

Liula [17]2 years ago

7 0

Answer:

The speed of cart B is 11.21 m/s.

Explanation:

Given that,

Speed of cart A = 9.25 m/s

Speed of cart B = 7.15 m/s

Mass of cart A = 72.0 kg

Mass of cart B = 55.0 kg

Speed of card A after collision = 6.15 m/s

We need to calculate the speed of cart B

Using conservation of momentum

$m_{A}v_{A}+m_{B}v_{B}=m_{A}v_{A}+m_{B}v_{B}$

Put the value into the formula

$72.0\times9.25+55.0\times7.15=72.0\times6.15+55.0\times v_{B}$

$v_{B}=\dfrac{72.0\times9.25+55.0\times7.15-72.0\times6.15}{55.0}$

$v_{B}=11.21\ m/s$

Hence, The speed of cart B is 11.21 m/s.

You might be interested in

A centrifuge is a device that rotates an object to produce an acceleration many times that of gravity Pilots are trained in such

Leviafan [203]

Answer:

ω = 0.571 rad/s

Explanation:

given data

radius = 30 m

solution

we take here g = 9.8 m/s²

and g is express as

g = r × ω² ....................1

put here value and we get

9.8 = 30 × ω²

solve it we get

ω = 0.571 rad/s

5 0

2 years ago

A 99.1-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =

Stels [109]

Answer:

628.022466 N

8.61 m/s

Explanation:

m = Mass

$\mu$ = Coefficient of friction

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$F_f=\mu mg\\\Rightarrow F_f=0.646\times 99.1\times 9.81\\\Rightarrow F_f=628.022466\ N$

Magnitude of frictional force is 628.022466 N

$F=ma\\\Rightarrow a=\frac{F_f}{m}\\\Rightarrow a=\frac{628.022466}{99.1}\\\Rightarrow a=6.33726\ m/s^2$

$v=u+at\\\Rightarrow 0=u-6.33726\times 1.36\\\Rightarrow u=8.61\ m/s$

Initial speed of the player is 8.61 m/s

4 0

2 years ago

Which sphere of Earth is associated with plate tectonics?

Marrrta [24]

When looking at this question, we can easily start by eliminating certain answers. In the selections you've provided, you've shown atmosphere. We can easily eliminate letter A, as that makes absolutely no sense. Moving on, you also eliminate letter B, as that deals with ecosystems and whatnot. And finally, you can eliminate hydrosphere, letter C - as that's not the same. That deals with water, like oceans or rivers.

That leaves you with D) Lithosphere for your answer. The Lithosphere is the rigid part of the earth, the outermost layer, I would say. The crust / mantle. That's why it would be letter D - plate tectonics seem to have relations with the Lithosphere. The lithosphere is affected.

5 0

3 years ago

Read 2 more answers

As the mass of an object increases, it’s inertia will

Scrat [10]

Answer:

Inertia is the property of mass that resists change. Therefore, it is safe to say that as the mass of an object increases so does its inertia.

Explanation:

7 0

2 years ago

Read 2 more answers

Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on

Natasha2012 [34]

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

6 0

3 years ago