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yuradex [85]
3 years ago
15

What is the equation used to calculate power?

Physics
2 answers:
shtirl [24]3 years ago
6 0

Answer:

Power = Work ÷ Time

( Watts ) = ( Joules ) ÷ ( Seconds )

Explanation:

Paladinen [302]3 years ago
3 0

Answer:

p =  \frac{w}{t}  \\    p = power(watts) \\ w = work(joules) \\ t = time(seconds)

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A 5000 kg open train car is rolling on frictionless rails at 22 m/s when it starts pouring rain. A few minutes later, the car’s
Phantasy [73]

Answer:

500 kg

Explanation:

It is given that,

The mass of a open train car, M = 5000 kg

Speed of open train car, V = 22 m/s

A few minutes later, the car’s speed is 20 m/s

We need to find the mass of water collected in the car. It is based on the conservation of momentum as follows :

initial momentum = final momentum

Let m is final mass

MV=mv

m=\dfrac{MV}{v}\\\\m=\dfrac{5000\times 22}{20}\\\\=5500\ kg

Water collected = After mass of train - before mass of train

= 5500 - 5000

= 500 kg

So, 500 kg of water has collected in the car.

3 0
3 years ago
A bar on a hinge starts from rest and rotates with an angular acceleration α = 12 + 5t, where α is in rad/s2 and t is in seconds
Rama09 [41]

Answer:

Ф = 239.73 rad

Explanation:

α = 12 + 15×t

W = ∫α×dt

   = ∫(12 + 5×t)×dt

   = 12×t + 2.5×t^2

then:

Ф = ∫W×dt

   = ∫(12×t + 2.5×t^2)dt

   = 6×t^2 + 5/6×t^3

therefore the angle at t = 4.88s is:

Ф = 6×(4.88)^2 + 5/6×(4.88)^3

   = 239.73 rad    

5 0
3 years ago
A man walking on a tightrope carries a long a pole which has heavy items attached to the two ends. If he were to walk the tight-
katen-ka-za [31]

Answer:

 I_weight = M L²

this value is much larger and with it it is easier to restore balance.I

Explanation:

When man walks a tightrope, he carries a linear velocity, this velocity is related to the angular velocity by

            v = w r

For man to maintain equilibrium needs the total moment to be zero

             ∑τ = I α

              S  τ = 0

The forces on the home are the weight of the masses, the weight of the man and the support on the rope, the latter two are zero taque the distance to the center of rotation is zero.

Therefore the moment of the masses and the open is the one that must be zero.

If the man carries only the bar, we could approximate it by two open one on each side of the axis of rotation formed by the free of the rope

              I = ⅓ m L² / 4

As the length of half the length of the bar and the mass of the bar is small, this moment is small, therefore at the moment if there is some imbalance it is difficult to recover.

If, in addition to the opening, each of them carries a specific weight, the moment of inertia of this weight is

             I_weight = M L²

this value is much larger and with it it is easier to restore balance.

5 0
3 years ago
If the acceleration due to gravity at the surface of planet x is double the value of earth's, how does planet x's mass compare t
love history [14]
There is not enough information given to answer with. The force of gravity at the planet's surface depends on the planet's radius as well as its mass. The planet could have exactly the same mass as Earth has. But if it's radius is only 71% of Earth's radius, then gravity on its surface will be twice as strong as gravity on Earth.
3 0
3 years ago
An airplane has a starting velocity of 300m/s. It then accelerates at a rate of 45m/s2 for a time of 10s. What is it's final vel
Olenka [21]
A = (v - u) / t

a = acceleration
v = final velocity
u = initial velocity
t = time

45 = (v - 300) / 10

45 × 10 = v - 300

450 + 300 = v

v = 750 m/s

Hope this helps!

P.S. Let me know if you need an explanation
8 0
2 years ago
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