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Natasha2012 [34]
2 years ago
15

The density of aluminum is 2.7 × 103 kg/m3 . the speed of longitudinal waves in an aluminum rod is measured to be 5.1 × 103 m/s.

what is the value of young's modulus for this aluminum?
Physics
1 answer:
andrey2020 [161]2 years ago
5 0
<span>The speed of longitudinal waves, S, in a thin rod = âšYoung modulus / density , where Y is in N/m^2. So, S = âšYoung modulus/ density. Squaring both sides, we have, S^2 = Young Modulus/ density. So, Young Modulus = S^2 * density; where S is the speed of the longitudinal wave. Then Substiting into the eqn we have (5.1 *10^3)^2 * 2.7 * 10^3 = 26.01 * 10^6 * 2.7 *10^6 = 26.01 * 2.7 * 10^ (6+3) = 70.227 * 10 ^9</span>
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A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio
dlinn [17]

Answer:

a = 4.72 m/s²  

Explanation:

given,

mass of the box (m)= 6 Kg

angle of inclination (θ) = 39°

coefficient of kinetic friction (μ) = 0.19

magnitude of acceleration = ?

box is sliding downward so,

F - f = m a                        

f is the friction force

m g sinθ - μ N = ma                        

m g sinθ - μ m g cos θ = ma            

a = g sinθ - μ g cos θ                    

a = 9.8 x sin 39° - 0.19 x 9.8 x cos 39°

a = 4.72 m/s²                                      

the magnitude of acceleration of the box down the slope is a = 4.72 m/s²  

3 0
2 years ago
A boy throws a 15 kg ball at 4.7 m/s to a 65 kg girl who is stationary and standing on a skateboard. After catching the ball, th
jek_recluse [69]

Answer:

a)v_{f}=0.88m/s

Explanation:

To solve this problem we use the Momentum's conservation Law, before and after the girl catch the ball:

\\ p_{1}=p_{2}\\m_{ball}*v_{o.ball}+m_{girl}*v_{o.girl} = m_{ball}*v_{f.ball} + m_{girl}*v_{f.girl}        (1)

At the beginning the girl is  stationary:

v_{o.girl}=0m/s       (2)

If the girl catch the ball, both have the same speed:

v_{f.girl}=v_{f.ball}=v_{f}       (3)

We replace (2) and (3) in (1):

m_{ball}*v_{o.ball} = (m_{ball}+m_{girl})*v_{f} \\

We can now solve the equation for v_{f}:

v_{f}=\frac{m_{ball}*v_{o.ball}}{(m_{ball}+m_{girl})}=\frac{15*4.7}{15+65}=0.88m/s

4 0
2 years ago
If you wished to warm 100 kg of water by 15 degrees celsius for your bath, how much heat would be required? (give your answer in
Anit [1.1K]
For the answer to the question above, 
<span>Q = amount of heat (kJ) </span>
<span>cp = specific heat capacity (kJ/kg.K) = 4.187 kJ/kgK </span>
<span>m = mass (kg) </span>
<span>dT = temperature difference between hot and cold side (K). Note: dt in °C = dt in Kelvin </span>

<span>Q = 100kg * (4.187 kJ/kgK) * 15 K </span>
<span>Q = 6,280.5 KJ = 6,280,500 J = 1,501,075.5 cal</span>
6 0
2 years ago
Why do most objects tend to contain nearly equal numbers of positive and negative charges?
mart [117]
<span>Most objects tend to contain the same numbers of positive and negative charge because this is the most stable situation. In fact, if an object has an excess of positive charge, it tends to attract an equal number of negative charges to balance this effect and restore neutrality: the attracted negative charges combine with the excess of positive charges, leaving the object electrically neutral.</span>
3 0
2 years ago
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A man pushes a drum of oil 10kg up an incline OA. If OA is 5m and AB is 3m, what is the potential energy of the drum at A.​
Paladinen [302]

Answer:

294 Joules

Explanation:

From the question;

  • Mass of the drum is 10 kg
  • The length of inclined surface, OA is 5 m
  • The final height of the drum on the plane AB is 3 m

We are required to determine the potential energy of the drum at A

We know that potential energy is given by the formula;

Potential energy = mgh

where m is the mass, g is the gravitational pull and h is the height

Taking g as 9.8 N/kg

Then;

Potential energy = 10 kg × 9.8 N/kg × 3 m

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Thus, the potential energy of the drum at A is 294 Joules

6 0
2 years ago
Read 2 more answers
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