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laiz [17]
3 years ago
10

If the input work on a machine is equal to it's output work. the machine has _____ efficiency.

Physics
1 answer:
AleksAgata [21]3 years ago
7 0

Answer:

100% efficiency

Explanation:

the machine has _100%_ efficiency.

(fill in the blank)

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Solve the science problem
scZoUnD [109]

Answer:

What is the problem I cant help unless you have the problem.

Explanation:

5 0
2 years ago
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A 65.0-kg woman steps off a 10.0-m diving platform and drops straight down into the water. 1) If she reaches a depth of 3.20 m,
timurjin [86]

Answer:

F=2627.6N

Explanation:

The work done by this resistive force while traveling a distance <em>d</em> underwater would be:

W=F.d=-Fd

where the minus sign appears because the force is upwards and the displacement downwards.

This work is equal to the change of mechanical energy. At the diving plataform and underwater, when she stops moving, the woman has no kinetic energy, so all can be written in terms of her total change of gravitational potential energy:

W=\Delta E=U_f-U_i=mgh_f-mgh_i=mg(h_f-h_i)

Putting all together:

F=-\frac{W}{d}=-\frac{mg(h_f-h_i)}{d}=-\frac{(65kg)(9.8m/s^2)(-3.2m-10m)}{3.2m}=2627.6N

7 0
2 years ago
Objects with masses of 235 kg and a 535 kg are separated by 0.330 m. (a) find the net gravitational force exerted by these objec
saul85 [17]
The 61.0 kg object<span> ... F = (300kg)(6.673×10−11 </span>N m<span>^2 </span>kg<span>^−2)(61kg)/(.225m)^2. F = 2.412e-5 </span>N<span> towards the 495 </span>kg<span> block. </span>b. [195kg] ===.45m ... (b<span>) You cannot achieve this </span>position<span>. For the </span>net force<span> to become zero, one or both of the </span>masses<span> must ...</span>
7 0
3 years ago
A 1250 kg car, driving 7.39 m/s, runs into the back of a stationary 5380 kg truck. After the collision, the truck moves forward
Leno4ka [110]

Explanation:

Given that,

Mass of the car, m₁ = 1250 kg

Initial speed of the car, u₁ = 7.39 m/s

Mass of the truck, m₂ = 5380 kg

It is stationary, u₂ = 0

Final speed of the truck, v₂ = 2.3 m/s

Let v₁ is the final velocity of the car. Using the conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

1250\times 7.39+5380\times 0=1250\times v_1+5380\times 2.3

v_1=-2.5\ m/s

So, the final velocity of the car is 2.5 m/s but in opposite direction. Hence, this is the required solution.

3 0
3 years ago
As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an ax
AysviL [449]

Answer:

I=2.766\ kg.m^2

Explanation:

We have:

diameter of the wheel, d=0.88\ m

weight of the wheel, w_w=280\ N

mass of hanging object to the wheel, m_o=6.32\ kg

speed of the hanging mass after the descend, v_o=4\ m.s^{-1}

height of descend, h=2.5\ m

(a)

moment of inertia of wheel about its central axis:

I=\frac{1}{2} m.r^2

I=\frac{1}{2} \frac{w_w}{g}.r^2

I=\frac{1}{2} \times \frac{280}{9.8}\times 0.44^2

I=2.766\ kg.m^2

3 0
3 years ago
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