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Stels [109]
3 years ago
7

In an engine governor, the two spheres (total mass of 1.0kg) are at 0.05m and rotating at 37rad/s If the engine increases the an

gular velocity to 58rad/s what is the new location of the spheres?
Physics
1 answer:
kirill115 [55]3 years ago
8 0

Here we can say that there is no external torque on this system

So here we can say that angular momentum is conserved

so here we will have

I_1\omega_1 = I_2\omega_2

now we have

I_1 = mr^2

I_1 = (1kg)(0.05^2)

I_1 = 25\times 10^{-4} kg m^2

similarly let the final distance is "r"

so now we have

I_2 = mr^2

I_2 = 1r^2

now from above equation we have

(25\times 10^{-4})37 = (r^2)(58)

r = 0.04 m

so final distance is 0.04 m between them

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The__ of friction is a number that represents the resistance to sliding
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Answer:

B. coefficient

Explanation:

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3 years ago
Which of the following is an example of an incline plane? a) lever b)gear c)pulley d)screw
slega [8]
Gear because each spike is a wedge shape that adds a mechanical advatage
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3 years ago
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Suppose a tank filled with water has a liquid column with a height of 10 meters. If the area is 2 square meters (2 m²), what's t
Vesna [10]

Answer:

196000 N

Explanation:

The following data were obtained from the question:

Height (h) = 10 m

Area (A) = 2 m²

Force (F) =.?

Next, we shall determine the pressure in the tank.

This can be obtained as follow:

P = dgh

Where

P is the pressure.

d is the density of the liquid.

g is acceleration due to gravity

h is the height.

Height (h) = 10 m

Density (d) of water = 1000 kg/m³

Acceleration due to gravity (g) = 9.8 m/s²

Pressure (P) =...?

P = dgh

P = 1000 × 9.8 × 10

P = 98000 N/m²

Therefore, the pressure acting on the tank is 98000 N/m²

Finally, we shall determine the force of gravity acting on the column of water as follow:

Area (A) = 2 m²

Pressure (P) = 98000 N/m²

Force (F) =.?

Pressure (P) = Force (F) /Area (A)

P = F /A

98000 = F/ 2

Cross multiply

F = 98000 × 2

F = 196000 N

Therefore, the force of gravity acting on the column of water is 196000 N

4 0
3 years ago
Read 2 more answers
A 50.1 kg diver steps off a diving board and drops straight down into the water. The water provides an average net force of resi
Varvara68 [4.7K]

Answer:

The total distance is 16.9 m

Explanation:

We understand work in physics as certain force exerted through certain distance. To reach that point below the water, the work done by the diver must be equal to the work done by the water's force of resistance. Therefore, we determine both work expressions and we solve the equation for the diver distance, which is the total distance between the diving board and the stopping point underwater.

W_{d}: work done by diver\\W_{R}: work due the force of resistance\\W_{d} = W_{R}\\F_{d}\times d_{d}=F_R \times d_{R}\\d_{d}= \frac{F_R \times d_R}{F_d}= \frac{F_R \times d_R}{m_d \times g}= \frac {1598 N \times 5.2 m}{50.1 kg \times 9.81 m/s^2}\\d_{d}= 16. 91 m

7 0
3 years ago
A small plane flies 40.0 km in a direction 60° north of east and then flies 30.0 km in a direction 15° north of east. Use the an
lana66690 [7]

Answer:

d= 64.7 km

\theta = 40.9^{o}

displacement vector=r_xi + r_yj =  48.9i + 42.4j

Explanation:

total distance = 40 + 30 = 70 km

during 1st flight

r_1 x = 40*cos60

r_1 x = 20 km

r_1 y = 40*sin60

r_1 y = 34.64 km

during 2nd flight

r_2 x = 30*cos15

r_2 x = 28.9 km

r_2 y = 30*sin15

r_2 y = 7.76 km

the two component of r are:

r_x = r_1x + r_2x = 20 + 28.9 = 48.9 km

r_y = r_1y + r_2y = 34.64 + 7.76 = 42.4 km

Geographical Direction \theta = tan^{-1}\frac{r_y}{r_x} [tex]\theta = 40.9^{o}

Displacement d= \sqrt{r_x^2 + r_y^2}

                     d = \sqrt{48.9^2+42.4^2} = 64.7 km

d= 64.7 km

displacement vector=r_xi + r_yj =  48.9i + 42.4j

8 0
3 years ago
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