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Sedbober [7]
3 years ago
15

Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon

has an arm length (hand to shoulder) of 0.60 m. We can model its motion as that of a point mass swinging at the end of a 0.60-m-long, massless rod. At the lowest point of its swing, the gibbon is moving at 3.2 m/s .
What upward force must a branch provide to support the swinging gibbon?
Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
aleksklad [387]3 years ago
5 0

Answer:

230 N

Explanation:

At the lowest position , the velocity is maximum hence at this point, maximum support force  T  is given by the branch.

The swinging motion of the ape on a vertical circular path , will require

a centripetal force  in upward direction . This is related to weight as follows

T - mg = m v² / R

R is radius of circular path . m is mass of the ape and velocity is 3.2 m/s

T =  mg -  mv² / R

T = 8.5 X 9.8 + 8.5 X 3.2² / .60  { R is length of hand of ape. }

T = 83.3 + 145.06

= 228.36

= 230 N ( approximately )

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A police car moving at 36.0 m/s is chasing a speeding motorist traveling at 30.0 m/s. The police car has a siren that emits soun
maxonik [38]

Answer:

The frequency heard by the motorist is 4313.2 Hz.

Explanation:

let f1 be the frequency emited by the police car and f2 be the frequency heard by the motorist, let v1 be the speed of the police car and v2 be the speed of the motorist and v = 343 m/s be the speed of sound.

because the police car is moving towards the motorist at a higher speed, then the motorist will hear a increasing frequency and according to Dopper effect, that frequency is given by:

f1 =  [(v + v2/(v - v1))]×(f2)

   = [( 343 + 30)/(343 - 36)]×(3550)

   = 4313.2 Hz

Therefore, the frequency heard by the motorist is 4313.2 Hz.

5 0
3 years ago
A ship was traveling across the Atlantic Ocean. It traveled 2,000 kilometers at a rate of 40 kilometers per hour. How long did i
8_murik_8 [283]
The correct answer is D
6 0
3 years ago
Charge X has twice as much charge as particle Y. The two charges are placed near each other. Compared to the force on particle X
Art [367]

The force on charge Y is the same as the force on charge X

Explanation:

We can answer this problem by applying Newton's third law of motion, which states that:

"When an object A exerts a force on object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"

In this problem, we can identify object A as charge X and object B as charge Y. The magnitude of the electrostatic force between them is given by

F=k\frac{q_x q_y}{r^2}  (1)

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_x, q_y are the two charges

r is the separation between the two charges

According to Newton's third law, therefore, the magnitude of the force exerted by charge X on charge Y is the  same as the force exerted by charge Y on charge X (and it is given by eq.(1)), however their directions are opposite.

Learn more about Newton's third law:

brainly.com/question/11411375

#LearnwithBrainly

6 0
3 years ago
What is the frequency of a wave if the speed is 24 m/s and the wave is 2 meters
vekshin1
Velocity=frequency(wavelength)
24m/s=f(2m)
24/2=f(2)/2
12Hz=f
4 0
4 years ago
In one scene in the movie The Godfather II, a solid gold phone is passed around a large table for everyone to see. Suppose the v
dangina [55]
Volume of gold in the phone = 10 cm^3
                                              = 0.<span>00001 m^3 </span>
Density of gold = 19300 kg/m^3
1 kg mass = 2.2 pounds
Mass of 10 cm^3 of gold = 0<span>.00001 m^3 * (19300 kg/m^3)
                                        = 0.193 kg 
So
0.193 kg = 0.193 * 2.2 pounds
               = 0.43 pounds
I think there is something wrong with the options given in the question.</span>
7 0
4 years ago
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