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netineya [11]
3 years ago
10

What happens to the amplitude of a sound wave when energy transfer is increased?

Physics
1 answer:
dezoksy [38]3 years ago
3 0

Answer:

Most likely A, as energy increases the amplitude increases too.

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If a car increases its velocity from 1m/s to 3.6km/hr in 5 seconds what’s its acceleration
EleoNora [17]

Answer:

applying 1st eq of motion vf=vi+at here we have to find a=vf-vi/t , a= 1-1/5 , a=0/5 then we got a=0 here(vf value 3.6km/h is converted in standard unit 3.6×1000/3600 so we get vf=1m/s²

7 0
3 years ago
The entropy of an isolated system must be conserved, so it never changes.a. Trueb. Fasle
Snowcat [4.5K]

Answer:

B: False

Explanation:

The second law of thermodynamics states that: the entropy of an isolated system will never decrease because isolated systems always tend to evolve towards thermodynamic equilibrium which is a state with maximum entropy.

Thus, it means that the entropy change will always be positive.

Therefore, the given statement in the question is false.

6 0
3 years ago
What is the acceleration of a 7 kg mass pushed by a 3.5 N force?
defon

Answer:

0.5m/s²

Explanation:

F =m x a

3.5 = 7 x a

a = 3.5 /7

a = 0.5m/s ²

4 0
3 years ago
The fulcrum is the
olga_2 [115]

Answer:

point of support on which a lever rotates.

Explanation:

The fulcrum is the point of support on which a lever rotates. Fulcrum is a pivotal part of simple machines.

The fulcrum provides the platform for a lever to torque.

  • The force that opposes motion by the applied force is termed the frictional force.
  • Friction is a force that opposes motion.
  • The stored energy of an object is its potential energy.
  • The potential energy is the energy due to the position of a body.
  • The distance an object moves when doing work is termed its displacement.
6 0
3 years ago
You are riding in a school bus. as the bus rounds a flat curve at constant speed, a lunch box with a mass of 0.470 kg suspended
valina [46]
Missing question: "<span> What is the speed v of the bus?"</span>

Solution:
Let's solve the problem by writing the equilibrium conditions on both x- and y- axis:
mg=T cos \alpha
m \frac{v^2}{r}=T sin \alpha
where mg is the weight of the lunch box, m \frac{v^2}{r} is the centripetal force, T is the tension of the string and \alpha=30^{\circ}.

The radius r is the one with respect to the vertical position of the string, therefore
r=L sin30^{\circ}=0.875 m.

From the first equation we find
T= \frac{mg}{cos \alpha}
and if we replace this into the second one, we find
m \frac{v^2}{r}=mg \tan \alpha
from which we can find the velocity:
v= \sqrt{rg \tan \aplha}= \sqrt{(0.875m)(9.81m/s^2)(\tan 35^{\circ})}=2.45 m/s
4 0
3 years ago
Read 2 more answers
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