a net horizontal force of 5000N is applied to a stalled car whose mass is 1500kg. what will be the car's speed after 10s?

A 1234 kg freight car moving at 6 m/s runs into a 2468 kg freight car at rest. They stick together upon collision. What was the

kirill [66]

Answer:

2 m/s

Explanation:

Applying,

The law of conservation of momentum

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m')............... Equation 1

Where m = mass of the first freight car, m' = mass of the second freight car, u = initial velocity of the first freight car, u' = initial velocity of the second freight car, V = final combined velocity/ speed.

make V the subject of the equation

V = (mu+m'u')/(m+m')........... Equation 2

From the question,

Given: m = 1234 kg, m' = 2468 kg, u = 6 m/s, u' = 0 m/s (at rest)

Substitute these values into equation 2

V = [(1234×6)+(2468×0)]/(1234+2468)

V = 7404/3702

V = 2 m/s

5 0

3 years ago

Which of the following is NOT a type of model used in scientific inquiry? interactive, mathematical, conceptual, convergent

katrin2010 [14]

A conceptual model. Hope this helps.

5 0

3 years ago

Read 2 more answers

An object stays at rest until what happens to it?

bixtya [17]

D. an outside or unbalanced force acts upon the object.

4 0

3 years ago

Considerable scientific work is currently under way to determine whether weak oscillating magnetic fields such as those found ne

slega [8]

Answer:

$\epsilon = 2.96 \times 10^{-11} \ V$

Explanation:

given,

magnetic field strength = 1.40 ✕ 10⁻³ T

frequency of oscillation = 60 Hz

diameter of RBC = 7.5 μm

EMF = ?

$\epsilon = NBA\omega$

$\epsilon = NB(\pi\ r^2)\ (2\pi f)$

$\epsilon = NB(\pi\ (\dfrac{d}{2})^2)\ (2\pi f)$

$\epsilon = (1)\ 1.4 \times 10^{-3}(\pi\ (\dfrac{7.5 \times 10^{-6}}{2})^2)\ (2\pi\times 60)$

$\epsilon = 2.96 \times 10^{-11} \ V$

maximum emf that can generate around the perimeter of the cell $\epsilon = 2.96 \times 10^{-11} \ V$

5 0

3 years ago

An object 1 of mass m1 is separated by some distance d from an object 2 of mass 2m1 . An object 3 of mass m3 is to be placed bet

Harrizon [31]

If the potential energy of the three-object system is to be a maximum (closest to zero), should object 3 be placed closer to object 1, closer to object 2, or halfway between them?

Object 3 should be placed closer to object 1.

Object 3 should be placed on a halfway between object 2 and object 1.

Object 3 should be placed closer to object 2.

Solution

I think that Object 3 should be placed closer to object 2.

6 0

3 years ago

a net horizontal force of 5000N is applied to a stalled car whose mass is 1500kg. what will be the car's speed after 10s?​

a net horizontal force of 5000N is applied to a stalled car whose mass is 1500kg. what will be the car's speed after 10s?