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rodikova [14]
2 years ago
8

Ten-story building has identical stairs throughout its stairways. A student calculates the amount of work required to climb one

set of stairs and the amount of work to climb the entire ten sets.
Which quantity will remain constant in this experiment?

Physics
1 answer:
BabaBlast [244]2 years ago
8 0

Answer:

c

Explanation:

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A fan connected to a 120-volt electrical system by an extension cord was worn through and exposed the bare, energized conductor,
kari74 [83]

Answer: the ladder

Explanation: since the energized conductor is already in contact with the ladder there by making electric current to flow. The base of the ladder is on the ground there by making the circuit to be complete and causing electrocution.

8 0
2 years ago
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Dr. John Paul Stapp was U.S. Air Force officer whostudied the effects of extreme deceleration on thehuman body. On December 10,
Ira Lisetskai [31]

Answer:

(a) a = 56.4 m/s², his acceleration a, in multiples of gravity g, is 5.76 g

(b) a = -201.43 m/s², his deceleration -a, in multiples of gravity g, is -20.56 g

Explanation:

(a)

When moving upwards, the initial velocity, u = 0 (he accelerated from rest)

When moving upwards, the final velocity, v = 282 m/s

time of  motion during this acceleration, t = 5 s

His acceleration is calculated as;

v = u + at

282 = 0 + 5a

a = 282 / 5

a = 56.4 m/s²

Ratio of his acceleration, a to gravity, g = a/g = 56.4 / 9.8 = 5.76

a = 5.76 g

(b)

When moving downwards, the initial velocity, u = 282 m/s

When moving downwards, the final velocity, v = 0 (he was brought to rest)

time of  motion during this deceleration, t = 1.4 s

His deceleration is calculated as;

v = u + at

0 = 282 + 1.4a

1.4a = -282

a = -282 / 1.4

a = -201.43 m/s²

Ratio of his deceleration, -a to gravity, g = -a/g = 201.43 / 9.8 = 20.56

a = -20.56 g

3 0
3 years ago
A person wearing rollerblades starts from rest and accelerates at a constant rate. She reaches a speed of 13 mph (5.18 m/s) exac
kvv77 [185]

Answer: a. ) What is the magnitude of their average acceleration (in m/s2)

5 0
2 years ago
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A vertical spring gun is used to launch balls into the air. If the spring is compressed by 4.9 cm, the ball of mass 5.5 g is lau
AleksandrR [38]

We know, by conservation of energy :

\dfrac{kx^2}{2}=mgh

Therefore,

\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}

Putting given values, we get :

\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}\\\\\dfrac{4.9^2}{x_2^2}=\dfrac{50.2}{2\times 50.2}\\\\x_2^2=2\times 4.9^2\\\\x_2 = 4.9\times \sqrt{2}\\\\x_2=6.93\ cm

Therefore, the spring be compressed to 6.93 cm to send the ball twice as high.

Hence, this is the required solution.

6 0
2 years ago
A model train engine was moving at a constant speed on a straight horizontal track. As the engine moved​ along, a marble was fir
bagirrra123 [75]

Answer:

The marble was moving in a projectile and the speed of the engine was 2.716 m/s

Explanation:

The vertical component of the marble's flight path relative to the train

is given by the equation y(t) = v*t - (4.9)*t^2,

where, v is the initial upward velocity of the marble relative to the train.  

So with y(1) = v - 4.9 = 0 we have  

v = 4.9 m/s.

The marble will reach maximum height after 0.5 seconds, at which the

height will be y(0.5) = (4.9)*(0.5) - (4.9)*(0.5)^2 = (4.9)*(0.25) = 1.225 m.

Now,  the marble has a vertical velocity component of 4.9 m/s and a horizontal velocity component

of V m/s such that tan(61) = 4.9 / V

V = 4.9 / tan(61) = 2.716 m/s

This horizontal velocity component of the marble is the same as the

speed of the train i.e. 2.716 m/s.

3 0
3 years ago
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