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DaniilM [7]
2 years ago
6

A force of 15 N is applied to a spring, causing it to stretch 0. 3 m. What is the spring constant for this particular spring? N/

m.
Physics
1 answer:
puteri [66]2 years ago
4 0
Get tyy guy hhh guy ggh fggg
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A ball with an initial velocity of 12 m/s is rolling up a hill. The ball accelerates at a rate of -2.6 m/s/s.
Anon25 [30]

Answer:

4.61 seconds

Explanation:

Given data

Initial velocity= 12m/s

acceleration= -2.6m/s^2

From the given data

we can find the time t

we know that

Acceleration= velocity/time

time= velocity/acceleration

time= 12/2.6

time= 4.61 seconds

4 0
3 years ago
Tessa wonders how many different ways she can light a lightbulb. Circle all the ways you think will work.
adell [148]
It’s just E because ethe positiv and negative current are supposed to flow thorough the bulb in opppsote sides at a equel level.In some them negerive/postive is absent and some of them are connected to the same side
4 0
3 years ago
Read 2 more answers
A bobsledder pushes her sled across horizontal snow to get it going, then jumps in. After she jumps in, the sled gradually slows
anastassius [24]

Answer:

In the vertical direction the acting forces are the normal force and the weight of the bobsleder plus the sled. In the horizontal direction the acting force is the friciton force.

Explanation:

Hi there!

Please, see the attached figure for a graphic representation of the forces acting on the sled after the bobsleder jumped in.

In the vertical direction, the acting forces are the normal force (N) and the weight of the sled plus the bobsledder (W).

Since the sled is not being accelerated in the vertical direction, the sum of forces in that direction is zero:

∑Fy = W + N = 0 ⇒ W = N

The weight is calculated as follows:

W = (mb + ms) · g

Where:

mb = mass of the bobsleder.

ms = mass of the sled.

g = acceleration due to gravity.

In the horizontal direction the only acting force is the friction force (Fr). The friction force is calculated a follows:

Fr = N · μ

Where:

N = normal force.

μ = kinetic friction coefficient.

Since N = W = (mb + ms) · g

Fr = (mb + ms) · g · μ

If we want to find the acceleration of the sled after the bobsleder jumps in, we can apply Newton's second law:

∑F = m · a

Where "a" is the acceleration and "m" is the mass of the object (in this case, the mass of bobsleder plus the mass of the sled).

∑F = Fr =  (mb + ms) · g · μ =  (mb + ms) · a

(mb + ms) · g · μ =  (mb + ms) · a

Solving for "a":

g · μ = a

3 0
3 years ago
Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as far
marta [7]

a) Their angular speeds are the same

b) Andrea's tangential speed is twice the value of Chuck's tangential speed

Explanation:

a)

The angular speed of Andrea and Chuck is the same.

Let's call \omega the angular speed at which the merry-go-round is rotating. We know that the angular speed is defined as:

\omega= \frac{2\pi}{T}

where

2 \pi is the angular displacement covered in one revolution

T is the period of revolution

The merry go round is a rigid body, so all its point cover the same angular displacement in the same time: this means that it doesn't matter where Andrea and Chuck are located along the merry-go-round, their angular speed will still be the same.

b)

For an object in circular motion, the tangential speed is given by

v=\omega r

where

\omega is the angular speed

r is the distance from the centre of rotation

Here let's call r_c the distance at which Chuck is rotating, so his tangential speed is

v_c = \omega r_c

Now we know that Andrea is rotating twice as far from the centre, so at a distance of

r_a = 2 r_c

So his tangential speed is

v_a = \omega r_a = \omega (2 r_c) = 2(\omega r_c) = 2 v_c

So, Andrea's tangential speed is twice the value of Chuck's tangential speed.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

5 0
3 years ago
An air-filled parallel-plate capacitor is connected to a battery and allowed to charge up. Now a slab of dielectric material is
natita [175]

Answer:

The charge on the capacitor had increased

Explanation:

The expression for the capacitance of an air-filled parallel-plate capacitor is as follows as;

C=\frac{\epsilon _{0}A}{d}  

Here, C is the capacitance, \epsilon _{0} is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.

When a slab of dielectric material is placed between the plates of the capacitor then the expression for the capacitance is as follows;

C=\frac{K\epsilon _{0}A}{d}

Here, K is the dielectric constant.

In the given problem, a slab is inserted between the plates of the capacitor then the capacitance of the capacitor will increase in this case. Therefore, the option (a) is true.

The expression for the charge stored in the capacitor is as;

Q= CV

Here, Q is the charge and V is the potential.

The charge will also increase in this case as the charge stored in the capacitor is directly proportional to the capacitance. Therefore, the option (d) is not true.

The expression for the energy stored in the capacitor is as follows;

E=\frac{CV^{2}}{2}

The voltage is constant in the given problem but the capacitance increases then the energy stored in the capacitor will increase. Therefore, the option (b) is not true.

The voltage across the capacitor will remain same as the capacitor is still connected to the battery. Therefore, the option (c) is not true.

Therefore, only option (a) is true.

3 0
3 years ago
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