Ask question

Ask question

All categories

2 years ago

7

Bumper car A (331 kg) moving east at 3.87 m/s hit stationary car B (208 kg). After, the car A moves .88 m/s east. What is the ve

locity of car B?

1 answer:

Gre4nikov [31]2 years ago

4 0

Answer:

Ub = 221.99m/s

Explanation:

(Ma x Ua)+(Mb x Ub) = (Ma + Mb)V

(331 x 3.87)+(208 x Ub) = (331 + 208)88

(1257.8) + (208 x Ub) = 47432

(208 x Ub) = 47432 - 1257.8

(208 x Ub) = 46174.2

Ub = 46174.2/208 = 221.99m/s

Ub = 221.99m/s

You might be interested in

How do cells maintain homeostasis

Vikki [24]

Answer:

because they become home for them when they are in there for such a along time

5 0

2 years ago

Giving 100 points to the correct answers for all 8 questions! I need help. This assignment is about constant acceleration for Ph

solong [7]

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

8 0

2 years ago

Ordinary ocean waves are created by the interaction of the ____ and ____.

dmitriy555 [2]

Wind and amplitude creates the waves in an ordinary ocean

7 0

2 years ago

An engineer is designing a runway. She knows that a plane, starting at rest, needs to reach a speed of 180mph at take-off. If th

kvv77 [185]

Answer:

The plane would need to travel at least $8,\!580\; {\rm ft}$ ( $8.58 \times 10^{3}\; {\rm ft}$ .)

The $10,\!000\; {\rm ft}$ runway should be sufficient.

Explanation:

Convert unit of the the take-off velocity of this plane to $\rm ft\cdot s^{-1}$ :

$\begin{aligned}v &= 180\; {\rm mph} \\ &= 180\; {\rm mi \cdot hrs^{-1}} \times \frac{1\; {\rm hrs}}{3600\; {\rm s}} \times \frac{5280\; {\rm ft}}{1\; {\rm mi}} \\ &= 264\; {\rm ft \cdot s^{-1}}\end{aligned}$ .

Initial velocity of the plane: $u = 0\; {\rm ft \cdot s^{-1}}$ .

Take-off velocity of the plane $v =264\; {\rm ft\cdot s^{-1}}$ .

Let $x$ denote the distance that the plane travelled along the runway. Since acceleration is constant but unknown, make use of the SUVAT equation $x = ((u + v) / 2) \, t$ .

Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration $t$ times average velocity $(u + v) / 2$ .

The distance that the plane need to cover would be:

$\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \frac{0\; {\rm ft \cdot s^{-1}} + 264\; {\rm ft \cdot s^{-1}}}{2} \times 65.0\; {\rm s} \\ &= 8.58\times 10^{3}\; {\rm ft}\end{aligned}$ .

4 0

2 years ago

Earth exerts a 100 N gravitational force on a metal box. What is the magnitude of the gravitational force the metal box exerts o

jeka94

We have that the magnitude of the gravitational force is mathematically given as

f=6.377N

<h3>Force</h3>

Question Parameters:

Earth exerts a 100 N gravitational force on a metal box.

(Mass of the earth is 6e24 kg and radius of the earth is 6.4e6m.)

Generally the equation for the Gravitational mForce is mathematically given as

$F=\frac{GMm}{r^2}\\\\Therefore\\\\F=\frac{100/9.8* 6e116e24}{6.4e6^2}$

f=6.377N

For more information on Force visit

brainly.com/question/26115859

7 0

1 year ago