All categories

3 years ago

What force is necessary to accelerate a 70kg object at a rate of 4.2m/s squared

1 answer:

6 0

Newton's 2nd law of motion:

Force = (mass) · (acceleration)

   = (70 kg) · (4.2 m/s²)

   = (70 · 4.2) kg·m/s²

   = 294 newtons

You might be interested in

Skater begins to spend with arms held out at shoulder height. The skater wants to match the speed of the spin to the beat of the

Aleksandr [31]

Answer:

the moment of inertia with the arms extended is Io and when the arms are lowered the moment

I₀/I > 1 ⇒ w > w₀

Explanation:

The angular momentum is conserved if the external torques in the system are zero, this is achieved because the friction with the ice is very small,

L₀ = L_f

I₀ w₀ = I w

w = $\frac{I_o}{I}$ w₀

where we see that the angular velocity changes according to the relation of the angular moments, if we approximate the body as a cylinder with two point charges, weight of the arms

I₀ = I_cylinder + 2 m r²

where r is the distance from the center of mass of the arms to the axis of rotation, the moment of inertia of the cylinder does not change, therefore changing the distance of the arms changes the moment of inertia.

If we say that the moment of inertia with the arms extended is Io and when the arms are lowered the moment will be

I <I₀

I₀/I > 1 ⇒ w > w₀

therefore the angular velocity (rotations) must increase

in this way the skater can adjust his spin speed to the musician.

7 0

3 years ago

In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a fur

Margaret [11]

Answer:

D = -4/7 = - 0.57

C = 17/7 = 2.43

Explanation:

We have the following two equations:

$3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)$

First, we isolate C from equation (2):

$2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)$

using this value of C from equation (3) in equation (1):

$3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}$

Put this value in equation (3), we get:

$C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C = \frac{17}{7}\\$

5 0

3 years ago

The sun is 150,000,000 km from the earth.

aniked [119]

Answer:

<h3>What is the angular speed of the earth around the sun? </h3>

It takes the Earth approximately 23 hours, 56 minutes and 4.09 seconds to make one complete revolution (360 degrees). This length of time is known as a sidereal day. The Earth rotates at a moderate angular velocity of

$7.2921159 × 10 {}^{ - 5} \: \: \frac{radians}{second}$

<h3>What is the tangential speed of the earth? </h3>

The earth rotates once every 23 hours, 56 minutes and 4.09053 seconds, called the sidereal period, and its circumference is roughly 40,075 kilometers. Thus, the surface of the earth at the equator moves at a speed of 460 meters per second--or roughly 1,000 miles per hour.

8 0

3 years ago

A proton in a particle accelerator is traveling at a speed of 0.99c

Gwar [14]

A proton in a particle accelerator is traveling at a speed of 0.99c has a speed magnitude of 2.97 x 10⁸ m/s.

<h3>What is speed of proton?</h3>

The speed of a proton is the rate at which a proton is moving through a given space.

The given speed of the proton is 0.99c

where;

c is speed of light

<h3>What is speed of light?</h3>

The speed of light in vacuum, commonly denoted c, is a universal physical constant that is important in many areas of physics.

The value of speed of light in a vacuum is given as 3 x 10⁸ m/s.

The speed of the proton is calculated as follows;

v = 0.99 x 3 x 10⁸ m/s.

v = 2.97 x 10⁸ m/s.

Thus, a proton in a particle accelerator is traveling at a speed of 0.99c has a speed magnitude of 2.97 x 10⁸ m/s.

Learn more about speed of proton here: brainly.com/question/14663642

#SPJ1

8 0

1 year ago

Kinetic energy varies jointly as the mass and the square of the velocity. A mass of 1515 grams and velocity of 77 centimeters pe

Alexus [3.1K]

Answer:

The second kinetic energy is 162 J.

Explanation:

Given that,

Mass, $m_1=15\ g$

Velocity, $v_1=7\ cm/s$

Kinetic energy, $K_1=147\ ergs$

Mass, $m_2=10\ g$

Velocity, $v_2=9\ cm/s$

We need to find kinetic energy $K_2$ . Kinetic energy is given by :

$K=\dfrac{1}{2}mv^2$

So,

$\dfrac{K_1}{K_2}=\dfrac{m_1}{m_2}\times \dfrac{v_1^2}{v_2^2}\\\\K_2=\dfrac{K_1}{\dfrac{m_1}{m_2}\times \dfrac{v_1^2}{v_2^2}}\\\\K_2=\dfrac{147}{\dfrac{15}{10}\times \dfrac{7^2}{9^2}}\\\\K_2=162\ J$

So, the second kinetic energy is 162 J.

4 0

3 years ago