All categories

2 years ago

Help with this physics question please :)

Physics

1 answer:

irina [24]2 years ago

4 0

Answer:

is that rm in you profile pic

You might be interested in

Can someone complete this assignment will give brainliest!<br> The assignment is attached

Musya8 [376]

... I can’t see the attached assignment that you put on here.

3 0

2 years ago

The power used by an electric tool can be determined by multiplying the _____.

posledela

Power = (current) x (voltage) .

3 0

3 years ago

Read 2 more answers

Calculate the total distance traveled (in m)

nlexa [21]

Answer:

Distance is 13m and Displacement is 9m

Explanation:

3 0

3 years ago

Read 2 more answers

A motorcycle has a mass of 350kg. A force of 1295N is applied to the cycle what is the motorcycles acceleration

Maru [420]

Answer: $3.7 m/s^2$

Explanation:

The acceleration of the motorcycle is given by Newton's second law:

$a=\frac{F}{m}$

where

F = 1295 N is the force applied to the motorcycle

m = 350 kg is the mass of the motorbike

a is the acceleration

By substituting the numbers into the formula, we find

$a=\frac{F}{m}=\frac{1295 N}{350 kg}=3.7 m/s^2$

4 0

2 years ago

0.180-kg stone rets on a frictionless, horizontal surface. A bullet of mass 7.50 g, traveling horizontally at 390 m/s, strikes t

Musya8 [376]

Answer:

The magnitud of the velocity is

$8.46m/s$

and the direccion:

$-28.3$ degrees from the horizontal.

Explanation:

Fist we define our variables:

$m_{s}=0.18kg\\m_{b}=7.5g = 0.0075kg\\v_{1b}= 390m/s -i\\v_{1s}=0m/s\\v_{2b}=210m/s -j\\v_{2s}=?$

The letters i and j represent the direction of the movement, i in this case is the horizontal direction, and j is perpendicular to i.

velocities with sub-index 1 are the speeds before the crash, and with sub-index 2 are the velocities after the crash.

Using conservation of momentum:

$m_{s}v_{1s}+m_{b}v_{1b}=m_{s}v_{2s}+m_{b}v_{2b}\\v_{1s}=0, so\\m_{b}v_{1b}=m_{s}v_{2s}+m_{b}v_{2b}$

Clearing for the velocity of the stone after the crash:

$v_{2s}=\frac{m_{b}v_{1b}-m_{b}v_{2b}}{m_{s}}$

Substituting known values:

$v_{2s}=\frac{0.0075kg(((390m/s-i)-210m/s-j)}{0.18kg}\\v_{2s}=(16.25m/s-i) - (8.75m/s-j)$

The magnitud of the velocity is :

$|v_{2s}|=\sqrt{(16.25m/s-i)^2 + (8.75m/s-j)^2}\\|v_{2s}|=18.46m/s$

and the direction:

$tan^{-1}(-8.75/16.25)=-28.3$

this is -28.3 degrees from the +i direction or the horizontal direcction.

Note: i and j can also be seen as x and y axis.

3 0

3 years ago