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sammy [17]
2 years ago
13

When an object is lifted 12 meters off the ground, it gains a certain amount of potential energy. If the same object is lifted 2

4 meters off the ground, its potential energy is
A.
the same.
B.
three times as much.
C.
twice as much.
D.
four times as much.
Physics
1 answer:
alexandr402 [8]2 years ago
8 0
I think the answer is twice as much
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A 66-kg fellow stands on a digital scale in an elevator that accelerates upwards from rest to 4.5 m/s in 2.00 s. show answer Inc
Nina [5.8K]

Answer:

Explanation:

mass of the fellow ( m ) = 66kg

acceleration of fellow a

v = u + at

4.5 = 0 + a x 2

a = 4.5 /2

= 2.25 m / s²

Net force acting on fellow in upward direction by the surface of  elevator

R - mg where R is reaction force of the surface of the elevator

Applying Newton's law of motion

R - mg = ma

R = m (g +a )

= 66 x ( 9.8 + 2 )

= 778.8 N

This will be the scale reading .

7 0
3 years ago
A 26.0 kg beam is attached to a wall with a hinge while its far end is supported by a cable such that the beam is horizontal. If
Nuetrik [128]

Answer:

Force exerted by the hinge on the beam = 109.24N

Explanation:

Weight = mg = 26 x 9.81 = 255.06 N

Vertical component = T sin θ

Horizontal component = Tcos θ

Now, there are 3 vertical forces acting on the beam. These are;

- The downward force which is the weight of the beam.

- The vertical components of the tension in the cable.

-The force that hinge exerts on the beam are the upward forces.

Hence, for the beam to remain horizontal, the sum of the upward forces must be equal to the weight of the beam.

For us to determine the vertical component of the tension in the cable, we will do a torque problem. Let the pivot point be at the hinge. Let’s assume that the length of the beam is L. The vertical component of the tension in the cable will produce clockwise torque while the weight of the beam will produce counter clockwise torque.

Tbus;

Clockwise torque = TL sin 61

Since the center of mass of beam is at the middle of the beam, the distance from the hinge to the weight of the beam is L/2.

Counter clockwise torque = WL/2

Thus;

TL sin 61 = WL/2

L will cancel out.

T sin 61 = 255.06/2

T x 0.8746 = 127.53

T = 127.53/0.8746 = 145.82 N

Now, the equation to determine the vertical component of the force that the hinge exerts on the beam is given as;

T + F = W

Thus;

145.82 + F = 255.06

F = 255.06 - 145.82 = 109.24 N

8 0
3 years ago
An experimental apparatus has two parallel horizontal metal rails separated by 1.0 m. A 3.0 Ω resistor is connected from the lef
Blizzard [7]

Answer:

The induced current and the power dissipated through the resistor are 0.5 mA and 7.5\times10^{-7}\ Watt.

Explanation:

Given that,

Distance = 1.0 m

Resistance = 3.0 Ω

Speed = 35 m/s

Angle = 53°

Magnetic field B=5.0\times10^{-5}\ T

(a). We need to calculate the induced emf

Using formula of emf

E = Blv\sin\theta

Where, B = magnetic field

l = length

v = velocity

Put the value into the formula

E=5.0\times10^{-5}\times1.0\times35\sin53^{\circ}

E=1.398\times10^{-3}\ V

We need to calculate the induced current

E =IR

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Put the value into the formula

I=\dfrac{1.398\times10^{-3}}{3.0}

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(b). We need to calculate the power dissipated through the resistor

Using formula of power

P=I^2 R

Put the value into the formula

P=(0.5\times10^{-3})^2\times3.0

P=7.5\times10^{-7}\ Watt

Hence, The induced current and the power dissipated through the resistor are 0.5 mA and 7.5\times10^{-7}\ Watt.

6 0
3 years ago
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One way to monitor global warming is to measure the average temperature of the ocean. Researchers are doing this by measuring th
Marat540 [252]

Answer:

0.07°C

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<u>solution:</u>

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t(5◦C) = ( 7600 × 103 m ) / (1484 m/s) = 5188.7 s

<u>one degree C corresponds to a ∆t of 14 s so temperature difference is:</u>

ΔT=1 s/14 s=0.07◦C

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Answer:

the glucose in air will solidify and it will stop the process.

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3 years ago
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