Which is a characteristic of diatoms?

A 1000-kilogram car traveling due east at 15 meters per second hit from behind and receives a forward impulse of 6000 newton-sec

strojnjashka [21]

The change in momentum of the car is 6000 kg m/s

Explanation:

According to the impulse theorem, the change in momentum of an object is equal to the impulse exerted on the object, therefore:

$\Delta p = I$

where

$\Delta p$ is the change in momentum

I is the impulse exerted

For the car in this problem, the impulse received is

I = 6000 kg m/s (in the forward direction)

Therefore, the change in momentum of the car is equal to this value:

$\Delta p = I = 6000 kg m/s$ (in the forward direction)

We can also calculate what is the new momentum of the car. In fact, the initial momentum is

$p_i = mu = (1000 kg)(15 m/s)=15,000 kg m/s$

And so, the new momentum is

$p_f = p_i + \Delta p = 15,000 + 6,000 = 21,000 kg m/s$

Learn more about impulse and momentum:

brainly.com/question/9484203

#LearnwithBrainly

8 0

2 years ago

In mechanics, massless strings are often assumed. Why is that not a good assumption when discussing waves on strings?

Marizza181 [45]

In mechanics, massless strings are often assumed. but this is not a good assumption when discussing waves on strings because the speed of a wave on a massless string would be infinite.

<h3>How to explain the information?</h3>

It should be noted that waves simply means the dynamic disturbance of a quantity.

It should be noted that in mechanics, massless strings are often assumed. but this is not a good assumption when discussing waves on strings because the speed of a wave on a massless string would be infinite.

Learn more about waves in:

brainly.com/question/15663649

#SPJ4

6 0

1 year ago

A fan that is rotating at 960 rev/s is turned off. It makes 1500 revolutions before it comes to a stop. a) What was its angular

Evgesh-ka [11]

Answer:

α = 1930.2 rad/s²

Explanation:

The angular acceleration can be found by using the third equation of motion:

$2\alpha \theta=\omega_f^2-\omega_i^2$

where,

α = angular acceleration = ?

θ = angular displacement = (1500 rev)(2π rad/1 rev) = 9424.78 rad

ωf = final angular speed = 0 rad/s

ωi = initial angular speed = (960 rev/s)(2π rad/1 rev) = 6031.87 rad/s

Therefore,

$2\alpha(9424.78\ rad) = (0\ rad/s)^2-(6031.87\ rad/s)^2\\\\\alpha = -\frac{(6031.87\ rad/s)^2}{(2)(9424.78\ rad)}$

<u>negative sign shows deceleration</u>

5 0

3 years ago

Why are lunar calendars not more widely used today?

Nookie1986 [14]

The seasons affect the scheduling of far more events the moon phases do.

6 0

3 years ago

Read 2 more answers

A runner achieves a velocity of 11.1 m/s, 9 sec after he begins. What is his

Elena-2011 [213]

Answer:

1.23 m/s²

Explanation:

Given:

v₀ = 0 m/s

v = 11.1 m/s

t = 9 s

Find: a

Equation:

v = at + v₀

Plug in:

11.1 m/s = a (9 s) + 0 m/s

a = 1.23 m/s²

The runner's acceleration is 1.23 m/s².

7 0

3 years ago