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mihalych1998 [28]
3 years ago
7

In this example we will analyze the forces acting on your body as you move in an elevator. Specifically, we will consider the ca

se where the elevator is accelerating. Suppose that you stand on a bathroom scale that rests on the floor of an elevator. (Don’t ask why!) Standing on the scale compresses its internal springs and activates a dial that indicates your weight in newtons. When the elevator is at rest, the scale reads 600 N. Suppose that the elevator is accelerating downward at 2.50 m/s2 . What does the scale read during the acceleration?
Physics
1 answer:
dexar [7]3 years ago
4 0

Answer:

The reading of the scale during the acceleration is 446.94 N

Explanation:

Given;

the reading of the scale when the elevator is at rest = your weight, w = 600 N

downward acceleration the elevator, a = 2.5 m/s²

The reading of the scale can be found by applying Newton's second law of motion;

the reading of the scale  = net force acting on your body

R = mg + m(-a)

The negative sign indicates downward acceleration

R = m(g - a)

where;

R is the reading of the scale which is your apparent weight

m is the mass of your body

g is acceleration due to gravity, = 9.8 m/s²

m = w/g

m = 600 / 9.8

m = 61.225 kg

The reading of the scale is now calculated as;

R = m(g-a)

R = 61.225(9.8 - 2.5)

R = 446.94 N

Therefore, the reading of the scale during the acceleration is 446.94 N

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A bullet with kinetic energy of 400J strikes a wooden block where a 8.00x10N resistive force stops the bullet what is the penetr
tia_tia [17]

The "penetration of the bullet" is 5 m

<u>Explanation</u>:

A "bullet" with "kinetic energy" of = 400J

A resistive force stops the bullet  = 8.00 x 10 N

Work = change in energy  

Work = ∆ Kinetic Energy   (equation 1)

Work = F\times d   (equation 2)

From equations 1 and 2 we have,

F\times d = ∆ Kinetic Energy

Where ,

Kinetic Energy = 400 J

F = 8.00 x 10 N

(8.00 x 10 N) d = 400 J

(80 N) d = 400 J

d=\frac{400}{80}

d = 5 m

The penetration of the bullet is 5 m

5 0
3 years ago
A 153 g mass is attached to the end of an unstressed vertical spring (of constant 24.7 N/m) and then dropped. The acceleration o
Arte-miy333 [17]

Answer:

The answer to the question is

Its maximum speed is 1.54 m/s

Explanation:

Work done = Kinetic energy

0.5·m·v² = 0.5·k·x²

Where

m = mass

v = velocity

k =  spring constant

x = extension of the spring

We note that Force F is given by

F = m·a

Where

a = acceleration due to gravity

= 0.153×9.8 = 1.4994 N

Equating the work done by the force to the work done on the spring gives

Work done = Force × Distance = 1.4994×x = 0.5×k÷x² = 0.5×24.7×x²

x = 1.4994÷12.35 = 0.121 m

Substituting the value of x into the equation below gives

0.5·m·v² = 0.5·k·x²

0.5×0.153×v² = 12.35×0.121²

v² = 0.182÷0.0765 = 2.379

v = 1.54 m/s

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3 years ago
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Two satellites A and B of the same mass are orbiting Earth in concentric orbits. The distance of satellite B from Earth’s center
riadik2000 [5.3K]

Answer:

ratio of tangential velocity of satellite b and a will be 0.707

Explanation:

We have given distance of satellite B from satellite A is twice

So r_b=2r_a

Tangential speed of the satellite is given by

v=\sqrt{\frac{GM}{r}}, G is gravitational constant. M is mass of satellite and r is distance from the earth

We have to find the ratio of tangential velocities of b and a

From the relation we can see that tangential velocity is inversely proportional to square root of distance from earth

So \frac{v_b}{v_a}=\sqrt{\frac{r_a}{r_b}}

\frac{v_b}{v_a}=\sqrt{\frac{r_a}{2r_a}}

\frac{v_b}{v_a}=\sqrt{\frac{1}{2}}

\frac{v_b}{v_a}=0.707

So ratio of tangential velocity of satellite b and a will be 0.707

5 0
3 years ago
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