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Step2247 [10]
3 years ago
14

A 76-kg man standing on a scale in an elevator notes that as the elevator rises, the scale reads 840 N. What is the acceleration

of the elevator? 1.25 Correct: Your answer is correct. m/s2 upward
Physics
1 answer:
vovikov84 [41]3 years ago
8 0

Answer:

1.25 m/s^2

Explanation:

m = 76 kg, R = 840 N

Let a be the acceleration of teh elevator in upward direction.

By use of Newton's second law

R - mg = m a

R = m ( g + a)

840 = 76 ( 9.8 + a)

a = 1.25 m/s^2

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A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. how long does h
noname [10]
The shot putter should get out of the way before the ball returns to the launch position.

Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.

The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 =  0.45 s

t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.

Answer: 0.45 s
7 0
3 years ago
Determine W (or fuel energy) required to launch a satellite of mass m at rest from a launching pad placed at the surface earth,
jeka57 [31]

Answer:

5. 9GmM/(10R)

Explanation:

m is the mass of the satellite

M is the mass of the earth

W is the energy required to launch the satellite

Energy at earth surface = Potential energy (PE) + W

W = Energy at earth surface - Potential energy (PE)

But PE = -\frac{GMm}{R}

Therefore: W = Energy at earth surface - \frac{GMm}{R}

Energy at earth surface (E) at an altitude of 5R = -\frac{GMm}{5r} +\frac{1}{2}mV^2

But V=\sqrt{\frac{GM}{5R} }

Therefore: E=-\frac{GMm}{5R}+\frac{1}{2}m(\sqrt{\frac{GM}{5R} } )^2=  -\frac{GMm}{5R}+\frac{GMm}{10R}  = -\frac{GMm}{10R}

W = E - PE

W=-\frac{GMm}{10R}-(-\frac{GMm}{R})=-\frac{GMm}{10R}+\frac{GMm}{R}=\frac{9GMm}{10R} \\W=\frac{9GMm}{10R}

7 0
3 years ago
Please help asap! Thank you.
olganol [36]

Answer:

direct current

Explanation:

it has a direct path to go down to reach the specific point

4 0
3 years ago
Read 2 more answers
A steel drill making 180 rpm is used to drill a hole in a block of steel. The mass of the steel block and the drill is 180 g. If
SVETLANKA909090 [29]

Answer:

a) 37.8 W

b) 2 Nm

Explanation:

180 g = 0.18 kg

We can also convert 180 revolution per minute to standard angular velocity unit knowing that each revolution is 2π and 1 minute equals to 60 seconds

180 rpm = 180*2π/60 = 18.85 rad/s

We can use the heat specific equation to find the rate of heat exchange of the steel drill and block:

\dot{E} = mc\Delta \dot{t} = 0.18*420*0.5 = 37.8 J/s

Since the entire mechanical work is used up in producing heat, we can conclude that the rate of work is also 37.8 J/s, or 37.8 W

The torque T required to drill can be calculated using the work equation

E = T\theta

\dot{E} = T\dot{\theta} = T\omega

T = \frac{\dot{E}}{\omega} = \frac{37.8}{18.85} = 2 Nm

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3 years ago
Io experiences tidal heating primarily because __________. hints io experiences tidal heating primarily because __________. io i
maxonik [38]
Lo experiences tidal heating primarily because lo’s elliptical orbit causes the tidal force on lo to vary as it orbits the Jupiter. Thus, lo’s elliptical orbit is essential to its tidal heating. This elliptical orbit, in turn, is an end result of the orbital resonance among lo, Europa and ganymade. This orbital resonance origin lo to have a more elliptical orbit than it would because lo intermittently passes Europa and ganymade in the same orbital position. We cannot perceive tidal forces of tidal heating in lo but rather we foresee that they must occur based on the orbital characteristic of the moons and active volcanoes on lo is the observational evidence that tidal heating is significant in lo.
8 0
3 years ago
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