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hjlf
3 years ago
7

The picture above shows a football player kicking a football. This is known as two dimensional motion. In which direction does t

he football move?
A) vertical only (y)
B) horizontal only (x)
C) horizontal and vertical (x, y)
D) horizontal, vertical, and side to side (x, y, and z)

The diagram shows the motion of a tennis ball that has just been hit with a racket (air resistance is neglected). Which of these is true of the horizontal and vertical components of the ball’s velocity?
A) Both the horizontal and the vertical components are constant.
B) Both the horizontal and the vertical components are accelerated.
C) The horizontal component is constant but the vertical component is accelerated.
D) The horizontal component is accelerated but the vertical component is constant.

Physics
1 answer:
fgiga [73]3 years ago
8 0

1. C) horizontal and vertical (x, y)

The picture shows the motion of a projectile, which consists of two separate motions along two different directions:

- horizontal (x): along this direction, the football has a uniform motion, with constant horizontal speed v_0 cos \theta,where v_0 is the magnitude of the initial velocity of the ball and \theta the angle at which it has been thrown

- vertical (y): along this direction, the football has an accelerated motion, with initial vertical velocity v_0 sin \theta upward and constant acceleration g=9.8 m/s^2 downward (acceleration due to gravity)


2. C) The horizontal component is constant but the vertical component is accelerated.

As described in the previous part of the exercise:

- along the horizontal direction there are no forces exerted on the ball, so it is a uniform motion, therefore the acceleration is zero and the horizontal component of the velocity is constant

- along the vertical direction there is one force acting on the ball (the force of gravity, which pushes downward), so there is an acceleration (downward) equals to g=9.8 m/s^2 and therefore the vertical component of the velocity is not constant.

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5 0
3 years ago
if the current is 0.4 A for the 20 ohm resistor, what is the value for the potential difference? ii) Would it be different for t
Genrish500 [490]

Answer:potential difference is more or less like voltage. Using ohms, V=IR

Where V is Voltage

I is Current =0.4A

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V=0.4*20

V=8V

Hence the potential difference will be 8V.

ii) V=0.4*30

V=12V

Explanation:

The voltage of potential difference is directly proportional to the current and the resistance. So if one increase or decrease, it will have impact on the other.

From the calculations, when the resistance increase, the voltage will increase to appreciate the change.

8 0
3 years ago
A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o
iVinArrow [24]
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s
- radius of the hill: r=100 m

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, m \frac{v^2}{r}, so we can write:
mg-N=m \frac{v^2}{r} (1)
By rearranging the equation and substituting the numbers, we find N:
N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
mg=m \frac{v^2}{r}
from which we find
v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s
7 0
3 years ago
A sound wave leaves the loudspeaker. As it travels, it experiences a temporary increase in wavelength and then returns to its or
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A high-pressure and a low-pressure zone are created in the medium as a result of this constant back and forth action. Compressions and rarefactions, respectively, are terms used to describe these high- and low-pressure zones. The sound waves go from one medium to another as a result of these regions being transmitted to the surrounding media.

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4 0
2 years ago
What is an RC circuit? 
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Hope this helps!
8 0
3 years ago
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