All categories

3 years ago

A large dog with a mass of 30 kg chases a car at 2 m/s. What is the magnitude of its momentum?

Physics

1 answer:

AlexFokin [52]3 years ago

6 0

Answer: 60 kg-m/s

Explanation: I just answered that question and got it correct.

You might be interested in

A 2kg bowling ball rolls at a speed of 5 m/s on a roof of the building that is 40 meters tall. What is the kinetic energy

MrRa [10]

The kinetic energy is $\frac 1 2 m v^2$ and the height of the building doesn't matter at all.

$E = \frac 1 2 m v^2 = \frac 1 2 (2)(5)^2 = 25$ joules

8 0

3 years ago

Kids are pelting a window with snowballs. On average, two snowballs of roughly 300-g mass hit the window each second, moving hor

hram777 [196]

Answer:

The average force exerted on the window due to two snowballs is 6 N

Explanation:

Given:

Mass of snowballs $m = 300 \times 10^{-3}$ Kg

Velocity of snowball $v = 10 \frac{m}{s}$

For finding the average force,

Force is equal to the change in momentum,

$F = \frac{dP}{dt}$

Here, final velocity is zero so we write,

$F = \frac{mv}{1}$

Where $dt = 1$ sec

$F = 300 \times 10^{-30} \times 10$

$F = 3$ N

Above value of force is due to one ball, but here given in question there are two ball,

$F = 3 \times 2$

$F = 6$ N

Therefore, the average force exerted on the window due to two snowballs is 6 N

5 0

3 years ago

A train moving at a constant speed on a surface inclined upward at 10.0° with the horizontal travels a distance of 400 meters in

Amiraneli [1.4K]

If the velocity of the train is v=s/t, where s is the distance and t is time, then v=400/5=80m/s. To get the vertical component of the velocity we need to multiply the velocity v with a sin(α): Vv=v*sin(α), where Vv is the vertical component of the velocity and α is the angle with the horizontal. So:

Vv=80*sin(10)=80*0.1736=13.888 m/s.

So the vertical component of the velocity of the train is Vv=13.888 m/s.

7 0

3 years ago

A 0.050 kg bullet strikes a 5.0 kg wooden block with a velocity of 909 m/s and embeds itself in the block which fies off its sta

serg [7]

Answer:

The final velocity of the bullet is 9 m/s.

Explanation:

We have,

Mass of a bullet is, m = 0.05 kg

Mass of wooden block is, M = 5 kg

Initial speed of bullet, v = 909 m/s

The bullet embeds itself in the block which flies off its stand. Let V is the final velocity of the bullet. The this case, momentum of the system remains conserved. So,

$mv=(m+M)V\\\\V=\dfrac{mv}{m+M}\\\\V=\dfrac{0.05\times 909}{0.050+5}\\\\V=9\ m/s$