Ask question

Ask question

All categories

3 years ago

7

A 4.0 kilogram train going right at 6.0 meters per second bumps into a toy wagon moving the same direction at 2.0 meters per sec

ond. The model train has a final speed of 4.8 meters per second, and the wagon has a final speed of 5.2 meters per second. WHAT IS THE MASS OF THE WAGON? round answer to two significant digits.

1 answer:

aliya0001 [1]3 years ago

5 0

Answer: 1.5kg

Explanation:

You might be interested in

Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.

KIM [24]

Answer:

The velocity is $v = 4.76 \ m/s$

Explanation:

From the question we are told that

The first distance is $d_1 = 4.0 \ km = 4000 \ m$

The first speed is $v_1 = 5.0 \ m/s$

The second distance is $d_2 = 1.0 \ km = 1000 \ m$

The second speed is $v_2 = 4.0 \ m/s$

Generally the time taken for first distance is

$t_1 = \frac{d_1 }{v_1 }$

$t_1 = \frac{4000}{5}$

$t_1 = 800 \ s$

The time taken for second distance is

$t_1 = \frac{d_2 }{v_2 }$

$t_1 = \frac{1000}{4}$

$t_1 = 250 \ s$

The total time is mathematically represented as

$t = t_1 + t_2$

=> $t = 800 + 250$

=> $t = 1050 \ s$

Generally the constant velocity that would let her finish at the same time is mathematically represented as

$v = \frac{d_1 + d_2}{t }$

=> $v = \frac{4000 + 1000}{1050 }$

=> $v = 4.76 \ m/s$

7 0

3 years ago

. Energy can neither be created nor be destroyed, but it can be changed from one form to another", this law is known as kinetic

Fynjy0 [20]

Answer:

The law of conservation of energy

3 0

3 years ago

After a displacement of 17 m, a train on a straight track is at the position xf = –2.5 m

EastWind [94]

-19.5m

-19.5+17=-2.5m

5 0

3 years ago

Read 2 more answers

4. The value of the before and after-collision momentum of two colliding objects is shown in the

ExtremeBDS [4]

Answer:A) WHICH is 0kgm/s

Explanation:

3 0

3 years ago

A mass-spring system has k = 56.8 N/m and m = 0.46 kg.

andriy [413]

Answer:

A. $\omega=11.1121\ rad.s^{-1}$

B. $f=1.7685\ Hz$

C. $T=0.5654\ s$

Explanation:

Given:

spring constant, $k=56.8\ N.m^{-1}$

mass attached, $m=0.46\ kg$

A)

for a spring-mass system the frequency is given as:

$\omega=\sqrt{\frac{k}{m} }$

$\omega=\sqrt{\frac{56.8}{0.46}}$

$\omega=11.1121\ rad.s^{-1}$

B)

frequency is given as:

$f=\frac{\omega}{2\pi}$

$f=\frac{11.1121}{2\pi}$

$f=1.7685\ Hz$

C)

Time period of a simple harmonic motion is given as:

$T=\frac{1}{f}$

$T=0.5654\ s$

7 0

3 years ago