All categories

3 years ago

Object B has a mass of 10 kg. Object B collides with another

Physics

1 answer:

vladimir2022 [97]3 years ago

5 0

Answer:

1.8m/s

Explanation:

We know that the momentum of an impact of a body is the mass times he velocity

M= mv

Given data

Mass m= 10kg

Momentum M= 18 kg.m/s

Velocity v=?

We can solve for velocity by submitting our given data in the momentum equation

18 = 10*v

v= 18/10

v= 1.8m/s

You might be interested in

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago

WILL GIVE BRAINLIEST ASAP

Sedaia [141]

Explanation:

1. Height Relatives to reference point, Mass, and strength of the gravitational field it's in

2. Distance in the magnetic field

8 0

2 years ago

A car travels for 5.5 h at an average speed<br> of 75 km/h. How far did it travel?<br> T TE

ANEK [815]

Answer:

$\boxed {\boxed {\sf 412.5 \ kilometers }}$

Explanation:

Distance is the product of speed and time.

$d=s*t$

The speed of the car is 75 kilometers per hour. It traveled for 5.5 hours.

$s= 75 \ km/hr \\t= 5.5 \ hr$

Substitute the values into the formula.

$d= 75 \ km/hr * 5.5 \ hr$

Multiply. Note that the hours will cancel each other out.

$d= 75 km * 5.5 \\d= 412.5 \ km$

The car travelled <u>412.5 kilometers.</u>

5 0

3 years ago

Partially correct answer iconYour answer is partially correct. A proton initially has and then 2.30 s later has (in meters per s

morpeh [17]

Answer: The question has some missing details. The initial velocity given as u = -6.5i + 17j + 13k and the final velocity v = -2.8i + 17j -9.3k.

a) = (1.82i - 9.69k)m/s2

b) magnitude = 9.85m/s2

c) direction = 280.64 degree

Explanation:

The detailed and step is shown in the attachment.

7 0

3 years ago

In the aftermath of an intense earthquake, the earth as a whole "rings" with a period of 54 minutes.

lawyer [7]

The frequency of the oscillation in hertz is calculated to be 0.00031 Hz.

The frequency of a wave is defined as the number of cycles completed per second while the period refers to the time taken to complete a cycle. The frequency is the inverse of period.

So;

Period(T) = 54 minutes or 3240 seconds

Frequency (f) = T-1 = 1/T = 1/3240 seconds = 0.00031 Hz

Learn more: brainly.com/question/14588679

6 0

2 years ago