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3 years ago

1. A car starting from rest accelerates uniformly at 3

Physics

1 answer:

amid [387]3 years ago

5 0

The car's final speed in m/s after the acceleration is 30.

Given the following data:

Initial velocity, U = 0 m/s (since the cars starts from rest)

Time, t = 10 seconds

Acceleration, a = 3 meter per seconds square.

To find the car's final speed in m/s after the acceleration, we would use the first equation of motion;

Mathematically, the first equation of motion is given by the formula;

$V = U + at\\\\V = 0 + 3(10)$

Final speed, V = 30 m/s

Therefore, the car's final speed in m/s after the acceleration is 30.

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A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used

zloy xaker [14]

Answer:

The compression is $\sqrt{2} \ d$ .

Explanation:

A Hooke's law spring compressed has a potential energy

$E_{potential} = \frac{1}{2} k (\Delta x)^2$

where k is the spring constant and $\Delta x$ the distance to the equilibrium position.

A mass m moving at speed v has a kinetic energy

$E_{kinetic} = \frac{1}{2} m v^2$ .

So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity $v_1$ . Knowing that the energy is constant.

$\frac{1}{2} m v_1^2 = \frac{1}{2} k d^2$

If we want to double the kinetic energy, then, the knew kinetic energy for a obtained by compressing the spring a distance D, implies:

$2 * (\frac{1}{2} m v_1^2) = \frac{1}{2} k D^2$

But, in the left side we can use the previous equation to obtain:

$2 * (\frac{1}{2} k d^2) = \frac{1}{2} k D^2$

$D^2 = \frac{2 \ (\frac{1}{2} k d^2)}{\frac{1}{2} k}$

$D^2 = 2 \ d^2$

$D = \sqrt{2 \ d^2}$

$D = \sqrt{2} \ d$

And this is the compression we are looking for

3 0

4 years ago

Read 2 more answers

A solid conducting sphere has net positive charge and radiusR = 0.600 m . At a point 1.20 m from the center of the sphere, the e

gayaneshka [121]

Answer:

V_inside = 36 V

Explanation:

Given

We are given a sphere with a positive charge q with radius R = 0.400 m Also, the potential due to this charge at distance r = 1.20 m is V = 24.0 V.

Required

We are asked to calculate the potential at the centre of the sphere

Solution

The potential energy due to the sphere is given by equation

V = (1/4*π*∈o) × (q/r) (1)

Where r is the distance where the potential is measured, it may be inside the sphere or outside the sphere. As shown by equation (1) the potential inversely proportional to the distance V

V ∝ 1/r

The potential at the centre of the sphere depends on the radius R where the potential is the same for the entire sphere. As the charge q is the same and the term (1/4*π*∈o) is constant we could express a relation between the states , e inside the sphere and outside the sphere as next

V_1/V_2=r_2/r_1

V_inside/V_outside = r/R

V_inside = (r/R)*V_outside (2)

Now we can plug our values for r, R and V_outside into equation (2) to get V_inside

V_inside = (1.2 m )/(0.600)*18

= 36 V

V_inside = 36 V

7 0

3 years ago

Read 2 more answers

What happens to a light ray if it is incident on a reflective surface along the normal?

Bad White [126]

Answer: The incident ray and the reflected ray and the normal will be parallel to each other.

Explanation:

The normal is perpendicular to the surface of the mirror or the reflective surface.

According to the law of reflection which state that:

The angle of incidence is always equal to the angle of reflection on a smooth surface.

If a light ray is incident on a reflective surface along the normal. The angle of incidence will be at 90 degrees which will be perpendicular to the surface of the mirror, the reflected ray will bounce back likewise at the same angle which will be perpendicular to the reflective surface.

Both the incident ray and the reflected ray and the normal will be parallel to each other.

4 0

3 years ago

In the nucleus ______________ tell cells what to do and how to change.

anyanavicka [17]

Answer:

i believe it is dna

Explanation:

if i remember correctly if the dna tells the cell what to do and how to do it

7 0

3 years ago

A physical therapy exercise has a person shaking a 5.00 kg weight up and down rapidly. if the barbell is moving at 4.50 m/s, wha

Ivanshal [37]

The magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.

<h3>Magnitude of required force to stop the weight</h3>

The magnitude of the force required to stop the weight in 0.333 seconds is calculated by applying Newton's second law of motion as shown below;

F = ma

F = m(v/t)

F = (mv)/t

F = (5 x 4.5)/0.333

F = 67.6 N

Thus, the magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.

Learn more about force here: brainly.com/question/12970081

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7 0

2 years ago