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Elodia [21]
3 years ago
8

Question 15: At the edge of a lake you throw a stone with a velocity of 29 m/s at an angle of 45°. The stone is in the air for 4

.23 s. How far did you throw the stone? (What is the range?)
I think I have this question's right answer, but I would like to check.
Physics
1 answer:
Marianna [84]3 years ago
8 0
Get your numbers gathered up and solve the problem in the ordered step
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When AC Electric is producing 120 volts with 20 amps, What controls the the 120 volts at 60cycles a second?
lubasha [3.4K]

Answer:

Cycles per second is dependent on the construction of the alternator and the 120 volts is dependent upon the current and resistance in the circuit according to the ohms law.

Explanation:

We are given with AC of 120 volts, 20 amperes and 60 hertz frequency.

<u>According to the Ohm's law, we find its resistance:</u>

R=\frac{V}{I}

R=\frac{120}{20}

R=6\ \Omega

So, this 6 ohm resistance controls the current controls the magnitude of the AC current, while the frequency of the current remains constant and depends upon the construction and rotational speed of the armature of the alternator producing the current.

Here the value of frequency is the number of times the current changes its direction or the polarity in one second.

6 0
4 years ago
Your high beam headlights illuminate the road in front of you for __________ feet. A. 150 B. 450 C. 650
ASHA 777 [7]

Answer:

B. 450 feet

Explanation:

Due to the angle at which high beam headlights illuminate, they can illuminate the road for about 450 feet.

8 0
3 years ago
a body weights 28N at a height of 3200km from the earth surface.What will be the gravitational force on that body if its lies on
alekssr [168]

Answer:

The object would weight 63 N on the Earth surface

Explanation:

We can use the general expression for the gravitational force between two objects to solve this problem, considering that in both cases, the mass of the Earth is the same. Notice as well that we know the gravitational force (weight) of the object at 3200 km from the Earth surface, which is (3200 + 6400 = 9600 km) from the center of the Earth:

F_G=G\,\frac{M_E\,m}{d^2} \\28\,\,N=G\,\frac{M_E\,m}{9600000^2}

Now, if the body is on the surface of the Earth, its weight (w) would be:

F_G=G\,\frac{M_E\,m}{d^2} \\w=G\,\frac{M_E\,m}{6400000^2}

Now we can divide term by term the two equations above, to cancel out common factors and end up with a simple proportion:

\frac{w}{28} =\frac{9600000^2}{6400000^2} \\\frac{w}{28} =\frac{9}{4} \\\\ \\w=\frac{9\,*\,28}{4}\,\,\,N\\w=63\,\,N \\

4 0
3 years ago
For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the e
Sati [7]

Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg  is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

5 0
4 years ago
Help hurry help pls it’s timed
Jobisdone [24]

Answer:

its the teal/second from the left one

Explanation:

4 0
2 years ago
Read 2 more answers
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