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A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The snowball falls straight back down in to a 6 inches of snow. The snowball feels a deceleration of 100 m/s2 upon impact with the snow bank before coming to rest. (a) When does the snowball hit the top of the snow bank? (b) How far from the ground does the snowball come a rest?

Answers

Answer 1

Answer:

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

$v=u+at\n\Rightarrow 0=4.5-32.1* t\n\Rightarrow (-4.5)/(-32.1)=t\n\Rightarrow t=0.14 \s$

Time taken by the ball to reach the highest point is 0.14 seconds

$s=ut+(1)/(2)at^2\n\Rightarrow s=4.5* 0.14+(1)/(2)* -32.1* 0.14^2\n\Rightarrow s=0.315\ ft$

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

$s=ut+(1)/(2)at^2\n\Rightarrow 4.315=0t+(1)/(2)* 32.1* t^2\n\Rightarrow t=\sqrt{(4.315* 2)/(32.1)}\n\Rightarrow t=0.518\ s$

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

$v=u+at\n\Rightarrow v=0+32.1* 0.518\n\Rightarrow v=16.62\ ft/s$

$v^2-u^2=2as\n\Rightarrow s=(v^2-u^2)/(2a)\n\Rightarrow s=(0^2-16.62^2)/(2* -100* 3.28)\n\Rightarrow s=0.42\ ft$

The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches

2.4

Explanation:

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Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 25 kg and the larger bottom crate has a mass of m2 = 91 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is µs = 0.79 and the coefficient of kinetic friction between the two crates is µk = 0.62. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem).1)The rope is pulled with a tension T = 234 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate?2)In the previous situation, what is the magnitude of the frictional force the lower crate exerts on the upper crate?3)What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide?4)The tension is increased in the rope to 1187 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?m/s25)As the upper crate slides, what is the acceleration of the lower crate?

Answers

Answer:

Explanation:

Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.

1 ) Common acceleration a = force / total mass

= 234 / ( 25 +91 )

= 2.017 m s⁻².

2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂

= mass x acceleration

= 25 x 2.017

= 50.425 N

The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.

3 ) Maximum friction force that is possible between m₁ and m₂

= μ_s m₁g

= .79 x 25 x 9.8

= 193.55 N

Acceleration of m₁

= 193 .55 / 25

= 7.742 m s⁻²

This is the common acceleration in case of maximum tension required

So tension in rope

= ( 25 +91 ) x 7.742

= 898 N

4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁

= μ_k m₁g

= .62 x 25 x 9.8

= 151.9 N

Acceleration of m₁

= 151.9 / 25

= 6.076 m s⁻².

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 V. If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answers

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 $\Omega$ . It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 $\Omega$ . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer:

The additional resistance is $R_z = 4.4 \Omega$

The rate at which internal energy increase at the supply is $Z_1 = 32 W$

The rate at which internal energy increase in the battery is $Z_1 = 32 W$

The rate at which internal energy increase in the added series resistance is $Z_3 = 70.4 W$

the increase rate of the chemically energy in the battery is $C = 48 W$

Explanation:

From the question we are told that

The open circuit voltage is $V = 40.0V$

The internal resistance is $R = 2 \Omega$

The emf of each battery is $e = 6.00 V$

The internal resistance of the battery is $r = 0.300V$

The charging current is $I = 4.00 \ A$

Let assume the the additional resistance to to added to the circuit is $R_z$

So this implies that

The total resistance in the circuit is

$R_T = R + 2r +R_z$

Substituting values

$R_T = 2.6 +R_z$

And the difference in potential in the circuit is

$E = V -2e$

=> $E = 40 - (2 * 6)$

$E = 28 V$

Now according to ohm's law

$I = (E)/(R_T)$

Substituting values

$4 = (28)/(R_z + 2.6)$

Making $R_z$ the subject of the formula

So $R_z = (28 - 10.4)/(4)$

$R_z = 4.4 \Omega$

The increase rate of internal energy at the supply is mathematically represented as

$Z_1 = I^2 R$

Substituting values

$Z_1 = 4^2 * 2$

$Z_1 = 32 W$

The increase rate of internal energy at the batteries is mathematically represented as

$Z_2 = I^2 r$

Substituting values

$Z_2 = 4^2 * 2 * 0.3$

$Z_2 = 9.6 \ W$

The increase rate of internal energy at the added series resistance is mathematically represented as

$Z_3 = I^2 R_z$

Substituting values

$Z_3 = 4^2 * 4.4$

$Z_3 = 70.4 W$

Generally the increase rate of the chemically energy in the battery is mathematically represented as

$C = 2 * e * I$

Substituting values

$C = 2 * 6 * 4$

$C = 48 W$

What is the angular width of a person's thumb viewed at arm's length? Assume that the width of the thumb is 17.3 mm and that the distance between the eyes and the thumb is 71.9 cm. Use the small-angle approximation and then convert the answer to degrees.

Answers

Answer:

$\theta_(degrees) =0.024\°=0\°1'26.62''$

Explanation:

To solve the problem it is necessary to take into account the concepts related to arc length and the radius that make up the measurements of an angle.

An angle is given by the length of arc displaced as a function of the radius, that is

$\theta = (Arc_(length))/(Radius)$

$\theta = (17.3*10^(-3))/(71.9*10^(-2))$

$\theta = 0.02406rad$

360° is equal to do $2\pi$ rad, therefore:

$\theta_(degrees) = 0.02406rad *((180\°)/(2\pi rad))$

$\theta_(degrees) =0.024\°=0\°1'26.62''$

Carol is farsighted ( presbyopia) and cannot see objects clearly that are closer to her eyes than about meter. She sees objects clearly with a relaxed eye when they are distant. What is the refractive power of reading glasses that would allow her to read a book 50 cm away with a relaxed eye

Answers

Answer:

1.0 dioptres

Explanation:

Farsightedness is an eye defect in which a person can see far objects clearly but not near objects. That implies that the patients' near point is farther than 25cm which is the normal least distance of distinct vision.

Farsightedness results from the eyeball being too long or the crystalline lens not being sufficiently converging.

Carol is farsighted with a near point of about a meter (100cm). We desire to make a lens to enable her near point be reduced to about 50cm. The focal length and power of this lens is calculated in the image attached.

The power of a lens is the inverse of its focal length in meters hence the 100 in the formula for power of the lens.

Answer:

+1.00 diopter

Explanation:

The power of a lens can be described simply as the reciprocal of the focal length of the lens measured in meters.

But f is unknown, hence we look for the focal length with the formula

1/f = 1/u + 1/v

where u is former near point = 100cm

v is the new intended near point = 50cm

1/f = 1/50 - 1/100

1/f = 1/100

f = 100 cm

Hence we get Power (D) = 1/f

where f = focal length of the lens in meter

From the question, the focal length of the lens = 100cm = 1m

Hence D = 1/1

D = +1.00

Hence the refractive power of the reading glasses that would allow Carol to read a book 50cm away from the relaxed eye will be +1.00 diopters.

If Earth were completely blanketed with clouds and we couldn’t see the sky, could we learn about the realm beyond the clouds? What forms of radiation might penetrate the clouds and reach the ground?

Answers

The definition of waves that propagate through electric fields is called electromagnetic waves. The earth, despite being covered with clouds, can be 'affected' because waves such as sunlight or the moon have the ability to penetrate and be visible to the inhabitants of the earth. Microwaves and radio waves would be less affected by the clouds that cover the Earth.

Through these waves, you can know that there is beyond the clouds.

Ultraviolet light, microwaves and radio waves are the radiations that penetrate through the clouds and reach the Earth's surface.

Therefore, the answer is Yes, ultraviolet light, microwaves and radio waves are the forms of radiation that penetrate and reach the ground.

Final answer:

It is indeed possible to learn about the universe beyond the clouds due to other non-visual forms of radiation, mainly radio waves and gamma rays, which can penetrate through the clouds and reach the earth's surface.

Explanation:

Yes, even if Earth were completely blanketed with clouds and we could not see the sky, we could still learn about the universe beyond the clouds. This is because, in addition to visible light which would be blocked by the clouds, the universe also emits various other forms of radiation that can penetrate the clouds and reach the ground.

Two major types of radiation that could penetrate the dense clouds are radio waves and gamma rays. Radio waves are a form of electromagnetic radiation used in many areas of science and technology, while gamma rays are highly energetic forms of radiation and are used in fields such as astronomy to get valuable information about distant celestial bodies.