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rjkz [21]
3 years ago
13

Consider a 100 g object dropped from a height of 1 m. Assuming no air friction (drag), when will the object hit the ground and a

t what speed? Take g 9.81 m/s m/'s.
Physics
1 answer:
Katyanochek1 [597]3 years ago
3 0

Answer:

speed and time are Vf = 4.43 m/s and  t = 0.45 s

Explanation:

This is a problem of free fall, we have the equations of kinematics

      Vf² = Vo² + 2g x

As the object is released the initial velocity is zero, let's look at the final velocity with the equation

      Vf = √( 2 g X)

      Vf = √(2 9.8  1)

      Vf = 4.43 m/s

This is the speed with which it reaches the ground

 

Having the final speed we can find the time

      Vf = Vo + g t

       t = Vf / g

       t = 4.43 / 9.8

       t = 0.45 s

This is the time of fall of the body to touch the ground

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A scientist testing to see how a cat “always” lands on their feet drops a cat upside down out of a third story window at a heigh
makkiz [27]

<u>Answer: </u>

Cat has 2.02 seconds to right itself.

<u>Explanation: </u>

Initial height of cat from ground = 20 meter.

We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

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Substituting

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7 0
3 years ago
Help please, I need it
love history [14]

Answer:

1.97×10⁻²¹ J

Explanation:

Use ideal gas law to find temperature.

PV = nRT

(9 atm) (9 L) = (83.3 mol) (0.0821 L·atm/mol/K) T

T = 11.9 K

The average kinetic energy per atom is:

KE = 3/2 kT

KE = 3/2 (1.38×10⁻²³ J/K) (11.9 K)

KE = 2.46×10⁻²² J

For a mass of 5.34×10⁻²⁶ kg, the kinetic energy is:

KE = (5.34×10⁻²⁶ kg) (1 mol / 0.004 kg) (6.02×10²³ atom/mol) (2.46×10⁻²² J)

KE = 1.97×10⁻²¹ J

5 0
3 years ago
The winch takes in cable at the constant rate of 130 mm/s. if the cylinder mass is 115 kg, determine the tension in cable 1. neg
nikitadnepr [17]
By applying Newton's second law of motion;

ma = mg - T

Where,
m = mass; a = downward accelerations (+ve value) or upward acceleration (-ve value); g = gravitational acceleration; T = tension.

For the current case, the velocity is constant therefore,
a = 0

Then,
0 = mg - T
T = mg = 115*9.81 = 1128.15 N

Tension in the cable is 1128.15 N.
8 0
3 years ago
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