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Alekssandra [29.7K]

1 year ago

11

A block of mass 5 kg slides down an inclined plane that has an angle of 10. If the inclined plane has no friction and the block

starts at a height of 0.8 m, how much kinetic energy does the block have when it reaches the bottom? Acceleration due to gravity is g = 9.8 m/s2

1 answer:

Westkost [7]1 year ago

4 0

The kinetic energy of the block when it reaches the bottom is 39.2 J.

<h3>Kinetic energy of the block at the bottom</h3>

Apply the principle of conservation of energy.

K.E(bottom) = P.E(top)

P.E(top) = mgh

where;

m is mass of the block
g is acceleration due to gravity
h is the vertical height of fall

P.E(top) = 5 x 9.8 x 0.8

P.E(top) = K.E(bottom) = 39.2 J

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

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Which statement is true?

Jet001 [13]

Answer:

A. Speed is a scalar quantity and velocity is a vector quantity.

Explanation:

A scalar quantity is one that consists of only a numerical value.

Speed is a scalar quantity because only the instantaneous value is indicated, for example the speedometer of a car that tells you your speed at the moment but not where you are going or in what direction are you going.

On the other hand, velocity is a vector quantity. Because it is composed of a <u>magnitude and a direction</u>, for example 10m/s to the south is a velocity, and 10m/s is a speed.

6 0

2 years ago

Consider an oil droplet of mass m and charge q. We want to determine the charge on the droplet in a Millikan-type experiment. We

Katen [24]

Answer:

$q=\frac{mg}{E_o}$

Explanation:

Given:

Charge = <em>q</em>

Electric field strength = $E_o$

weight of the droplet = <em>mg</em>

The charge is suspended motionless. This is because the electric force on the charge is balanced by the weight of the droplet.

electric force on charged droplet, $F=qE_o$

This is balanced by the weight, $mg$

Equating the two:

$qE_o=mg\\\Rightarrow q=\frac{mg}{E_o}$

4 0

3 years ago

A beam of white light enters a prism of crown glass (n = 1.5) from air (n = 1.00). Once inside, the colors in the light disperse

eduard

Answer:

= 2.33

Explanation:

.According to snell's law:

n1sin i = n2sin r ,

where n1 is refractive index of the medium in which incident ray is travelling, n2 is the refractive index of the medium in which refracted ray is travelling,

i is angle of incidence,

r is angle of refraction.

Given that,

n1 = 1,

i = 51 degrees,

r = 19.5 degrees. ,

n2= ?

So,

1*sin 51 = n2 sin 19.5

=> n2 = sin51 / sin19.5

= 2.33

7 0

2 years ago

1. How are each of Newton’s Laws of motion demonstrated

stepan [7]

Answer:

Newton's law of inertia - His first law states that every object remains at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. ... This is the first part cited in Newton's first law; "there is no net force on the airplane and it travels at a constant velocity in a straight line."

Newton's law of acceleration - "a net external force changes the velocity of the object. The drag of the aircraft depends on the square of the velocity. So the drag increases with increased velocity."

Newton's law of Action/Reaction - "As a plane flies, the force of the air hitting the plane is always equal and opposite to the force of the plane pushing against the air. The force generated by the engine pushes against air while the air pushes back with an equal and opposite force."

Hope this helps! god bless :)

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5 0

3 years ago

In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu

Arte-miy333 [17]

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

In a nuclear physics experiment, a proton (mass $1.67 \times 10^(-27)$ kg, charge +e = $+1.60 \times 10^(-19)$ C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed $2.50 \times 10^6$ m/s. The proton comes momentarily to rest at a distance $5.31 \times 10^(-13)$ m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5.31 \times 10^{-13}$ m apart?

Explanation:

The given data is as follows.

Mass of proton = $1.67 \times 10^{-27}$ kg

Charge of proton = $1.6 \times 10^{-19} C$

Speed of proton = $2.50 \times 10^{6} m/s$

Distance traveled = $5.31 \times 10^{-13} m$

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

$(K.E + P.E)_{initial}$ = $(K.E + P.E)_{final}$

$(\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)$

where, $\frac{kq_{p}q_{t}}{r} = U = Electric potential energy$

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

Putting the given values into the above formula as follows.

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

= $(\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})$

= $5.218 \times 10^{-15} J$

Therefore, we can conclude that the electric potential energy of the proton and nucleus is $5.218 \times 10^{-15} J$ .

4 0

3 years ago