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1 year ago

11

when you drop a pebble from height h, it reaches the ground with kinetic energy k if there is no air resistance. from what heigh

t

1 answer:

marysya [2.9K]1 year ago

6 0

Answer:

From the initial height h

Explanation:

When a material or substance is drop from a height h, it possesses potential energy, immediately it is dropped from that height, the potential energy is gradually converted to kinetic energy, it gets to a point where the potential energy equals the kinetic energy, as the material touches the ground, all potential energy has been converted to kinetic energy already

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Ugonna stands at the top of an incline and pushes a 100−kg crate to get it started sliding down the incline. The crate slows to

Anna [14]

Answer:(a)891.64 N

(b)0.7

Explanation:

Mass of crate $m=100 kg$

Crate slows down in $s=1.5 m$

initial speed $u=1.77 m/s$

inclination $\theta =30^{\circ}$

From Work-Energy Principle

Work done by all the Forces is equal to change in Kinetic Energy

$W_{friction}+W_{gravity}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2$

$W_{gravity}=mg(0-h)=mgs\sin \theta$

$W_{gravity}=-mgs\sin \theta$

$W_{gravity}=-100\times 9.8\times 1.5\sin 30=-735 N$

change in kinetic energy $=\frac{1}{2}\times 100\times 1.77^2=156.64 J$

$W_{friction}=156.64+735=891.645$

(b)Coefficient of sliding friction

$f_r\cdot s=W_{friciton}$

$891.645=f_r\times 1.5$

$f_r=594.43 N$

and $f_r=\mu mg\cos \theta$

$\mu 100\times 9.8\times \cos 30=594.43$

$\mu =0.7$

5 0

2 years ago

A shear wave (S wave) is a type of seismic _________ that shakes the ground back and forth perpendicular to the direction the wa

Zepler [3.9K]

Body waves

Explanation:

A shear wave(S-wave) is a type of seismic body waves that shakes the ground back and forth perpendicular to the direction the wave is moving.

Seismic waves are elastic waves usually generated when there is a disturbance within the earth.
There are two types of seismic waves:

Surface waves

Body waves

Body waves travel within the earth and they cause disturbances there. P and S waves are the two types of body waves that we have.
Surface waves travels on the earth surface. They are the love and rayleigh waves. They are the ones that cause destruction on the earth surface during an earthquake.

Learn more:

Earthquake brainly.com/question/6520403

#learnwithBrainly

7 0

1 year ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin

Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is $14,000 + 10,000x − 26,000x^{2}$ , where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

W = $\int_{0}^{0.54} F dx$

= $\int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx$

= $[14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}$

= 7560 + 1458 - 1364.69

= 7653.31 J

or, = 7.65 kJ (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0

1 year ago

The bowling ball is whizzing down the bowling lane at 4 m/s. If the mass of the bowling ball is 7 kg, what is its kinetic energy

Lisa [10]

Kinetic Energy = 1/2xmassx(velocity)^2
Input values;
K.E=1/2x7kgx(4m/s)^2
K.E.=56J

3 0

2 years ago

Read 2 more answers

Two chargedparticles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged part

erica [24]

Answer:

Two possible points

<em>x= 0.67 cm to the right of q1</em>

<em>x= 2 cm to the left of q1</em>

Explanation:

<u>Electrostatic Forces</u>

If two point charges q1 and q2 are at a distance d, there is an electrostatic force between them with magnitude

$\displaystyle f=k\frac{q_1\ q_2}{d^2}$

We need to place a charge q3 someplace between q1 and q2 so the net force on it is zero, thus the force from 1 to 3 (F13) equals to the force from 2 to 3 (F23). The charge q3 is assumed to be placed at a distance x to the right of q1, and (2 cm - x) to the left of q2. Let's compute both forces recalling that q1=1, q2=4q and q3=q.

$\displaystyle F_{13}=k\frac{q_1\ q_3}{d_{13}^2}$

$\displaystyle F_{13}=k\frac{(q)\ (q)}{x^2}$

$\displaystyle F_{23}=k\frac{q_2\ q_3}{d_{23}^2}$

$\displaystyle F_{23}=k\frac{(q)(4q)}{(0.02-x)^2}$

$\displaystyle F_{23}=\frac{4k\ q^2}{(0.02-x)^2}$

Equating

$\displaystyle F_{13}=F_{23}$

$\displaystyle \frac{K\ q^2}{x^2}=\frac{4K\ q^2}{(0.02-x)^2}$

Operating and simplifying

$\displaystyle (0.02-x)^2=4x^2$

To solve for x, we must take square roots in boths sides of the equation. It's very important to recall the square root has two possible signs, because it will lead us to 2 possible answer to the problem.

$\displaystyle 0.02-x=\pm 2x$

Assuming the positive sign :

$\displaystyle 0.02-x= 2x$

$\displaystyle 3x=0.02$

$\displaystyle x=0.00667\ m$

$x=0.67\ cm$

Since x is positive, the charge q3 has zero net force between charges q1 and q2. Now, we set the square root as negative

$\displaystyle 0.02-x=-2x$

$\displaystyle x=-0.02\ m$

$\displaystyle x=-2\ cm$

The negative sign of x means q3 is located to the left of q1 (assumed in the origin).

5 0

1 year ago