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rjkz [21]
2 years ago
15

If an engine does 660 J of work in 10 seconds, its average power is ...

Physics
1 answer:
givi [52]2 years ago
4 0

Answer:

66w

Explanation:

p=w/t

p=660/10

p=66

prolly a bad explanation but hope it helps...

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Giving quadrilateral a(2,-1 ) b ( 1,3) c(6,5) d(7,1) you want to prove that it is a parallelogram by showing opposite sides are
solong [7]

Answer:

Opposite sides are congruent (AB = DC).

Opposite angels are congruent (D = B).

Consecutive angles are supplementary (A + D = 180°).

If one angle is right, then all angles are right.

The diagonals of a parallelogram bisect each other.

Each diagonal of a parallelogram separates it into two congruent triangles.

Explanation: #if you need any queshtions answered within secs/mins hit me up and I gotchu.

8 0
3 years ago
Protons<br> Neutrons<br> Electrons<br> Location<br> Charge<br> Size
igor_vitrenko [27]
Protons are positive, and neutrons are negative, electrons are neutral. I’m not sure about the rest but I hope that helps for now
6 0
2 years ago
A force F=0.12N is aplied on spring and spring elongates by 3cm . specific constant of spring ​
PilotLPTM [1.2K]

The spring constant is 4 N/m

Explanation:

When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law:

F=kx

where

F is the magnitude of the spring applied

k is the spring constant

x is the elongation of the spring, relative to its equilibrium position

For the spring in this problem, we have:

F = 0.12 N (force applied)

x = 3 cm = 0.03 m (elongation of the spring)

Therefore, we can solve the formula for k to find the spring constant:

k=\frac{F}{x}=\frac{0.12}{0.03}=4 N/m

Learn more about forces:

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4 0
3 years ago
Two tuning forks, 254 Hz. and 260 Hz., are struck simultaneously. How many beats will be heard?
Lisa [10]

The longer you continue to listen, the more beats will be heard.

They'll occur at the rate of (260Hz - 254Hz) = 6 Hz .
 
7 0
3 years ago
At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There
katen-ka-za [31]

a) Constant

b) Constant

Explanation:

a)

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

\tau_{net} = I \alpha (1)

where

\tau_{net} is the net torque acting on the object in rotation

I is the moment of inertia of the object

\alpha is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

\alpha = \frac{\Delta \omega}{\Delta t}

where

\Delta \omega is the change in angular velocity

\Delta t is the time interval

So we can rewrite eq.(1) as

\tau_{net}=I\frac{\Delta \omega}{\Delta t}

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

\tau_{net}=0

From the previous equation, this implies that

\Delta \omega =0

Which means that the angular velocity at that instant does not change anymore.

b)

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

\tau_{net}=0

Therefore, this means that

\Delta \omega =0

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

8 0
3 years ago
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