A lorry pulls forward after initially being stationary, it takes the lorry 80m to reach a speed of

A 1000-kg whale swims horizontally to the right at a speed of 6.0 m/s. It suddenly collides directly with a stationary seal of m

anzhelika [568]

Answer:

Momentum after collision will be 6000 kgm/sec

Explanation:

We have given mass of the whale = 1000

Initial velocity v = 6 m/sec

It collides with other mass of 200 kg which is at stationary

Initial momentum of the whale = 1000×6 = 6000 kgm/sec

We have to find the momentum after collision

From conservation of momentum

Initial momentum = final momentum

So final momentum = 6000 kgm/sec

5 0

3 years ago

Select the statement that correctly lists the forces from strongest to weakest. (2 points)

Vilka [71]

Strong Nuclear force: it is the short range force and strongest fundamental force in all type of forces.

Electromagnetism: this is the force due to magnetic and electric behavior of the particles. It is moderate type of force and its range is more than Nuclear force.

Weak Nuclear Force: This force is also short range force which act between the nucleoside. But this force is also moderate type of force

Gravitational force: this force is between two point masses and least order of force. also the range of this force is upto infinite.

so the correct order of this fundamental force is

<em>strong nuclear, electromagnetism, weak nuclear, gravitational</em>

5 0

3 years ago

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

An example of an atom that has no change is one that has A. 1 proton, 2 electrons, and 3 neutrons B. 3 protons,2 electrons, and

Vikki [24]

Your question asks which is an example of an atom that has no change, or charge.

Your answer would be C). 2 protons, 2 electrons, and 1 neutron

These atoms would be known as a neutral atom.

In order for an atom to be neutral, it must have the same amount of protons and electrons.

We know that Protons are positive

We also know that Electrons are negative

In order for the atom to be neutral, they must have an equal amount of protons and neutrons to "cancel out"

In answer choice C, you would see that there are 2 protons and 2 electrons.

2- 0 = 0, in which allows this atom to be neutral.

Therefore, answer choice C. 2 protons, 2 electrons, and 1 neutron would be your answer.

4 0

3 years ago

Read 2 more answers

A 40 kg person sits on top of a 400 kg rock. What is the person’s weight? 390 N

algol13

The weight of the person is given by:

W = mg

W = weight, m = mass, g = gravitational acceleration

Given values:

m = 40kg, g = 9.81m/s²

Plug in and solve for W:

W = 40(9.81)

W = 390N

7 0

3 years ago