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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

You might be interested in
wo lacrosse players collide in midair. Jeremy has a mass of 120 kg and is moving at a speed of 3 m/s. Hans has a mass of 140 kg
Julli [10]

2.71 m/s fast Hans is moving after the collision.

<u>Explanation</u>:

Given that,

Mass of Jeremy is 120 kg (M_J)

Speed of Jeremy is 3 m/s (V_J)

Speed of Jeremy after collision is (V_{JA}) -2.5 m/s

Mass of Hans is 140 kg (M_H)

Speed of Hans is -2 m/s (V_H)

Speed of Hans after collision is (V_{HA})

Linear momentum is defined as “mass time’s speed of the vehicle”. Linear momentum before the collision of Jeremy and Hans is  

= =\mathrm{M}_{1} \times \mathrm{V}_{\mathrm{J}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{H}}

Substitute the given values,

= 120 × 3 + 140 × (-2)

= 360 + (-280)

= 80 kg m/s

Linear momentum after the collision of Jeremy and Hans is  

= =\mathrm{M}_{\mathrm{J}} \times \mathrm{V}_{\mathrm{JA}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{HA}}

= 120 × (-2.5) + 140 × V_{HA}

= -300 + 140 × V_{HA}

We know that conservation of liner momentum,

Linear momentum before the collision = Linear momentum after the collision

80 = -300 + 140 × V_{HA}

80 + 300 = 140 × V_{HA}

380 = 140 × V_{HA}

380/140= V_{HA}

V_{HA} = 2.71 m/s

2.71 m/s fast Hans is moving after the collision.

4 0
3 years ago
What does the term calorie refer to
OLga [1]
Hey there,
<span>It means- the amount of energy required to raise the temperature of a liter of water one degree centigrade at sea level
</span>
Hope this helps :))

~Top
6 0
4 years ago
Dean Potter is known for slacklinging across a 915 metter deep ravine in Yosemite National Park with no safety gear. If his mass
Natasha2012 [34]

Explanation:

answer 689910

M x G x H

mass = 77

G= 9.8

Meter = 915 this is the height

77 x 9.8= 754.6

754.6 X 915 = 689910 J

8 0
3 years ago
Calculate the ratio of the drag force on a jet flying at 1190 km/h at an altitude of 7.5 km to the drag force on a prop-driven t
Bond [772]

Answer:

\frac{D_{jet}}{D_{prop}}=2.865

Explanation:

Given data

Speed of jet Vjet=1190 km/h

Speed of prop driven Vprop=595 km/h

Height of jet 7.5 km

Height of prop driven transport 3.8 km

Density of Air at height 10 km p7.8=0.53 kg/m³

Density of air at height 3.8 km p3.8=0.74 kg/m³

The drag force is given by:

D=\frac{1}{2}CpAv^2\\

The ratio between the drag force on the jet to the drag force  on prop-driven transport is then given by:

\frac{D_{jet}}{D_{prop}}=\frac{(1/2)Cp_{7.5}Av_{jet}^2}{1/2)Cp_{3.8}Av_{prop}^2} \\\frac{D_{jet}}{D_{prop}}=\frac{p_{7.5}v_{jet}^2}{p_{3.8}v_{prop}}\\\frac{D_{jet}}{D_{prop}}=\frac{(0.53)(1190)^2}{(0.74)(595)^2}\\   \frac{D_{jet}}{D_{prop}}=2.865

4 0
3 years ago
Rotational dynamics physics question please help.
soldier1979 [14.2K]
Looking at the force-time graph, wouldn't the force be integral fdt between 0 and 10s, a sort of "smoothed out pulse" ? And it looks like a familiar bell shaped curve. doesn't that produce a turning effect/torque and isn't there something about a circular analogue to F=ma in newtonian linear mechs ???
Looks like a difficult question for 5 points
8 0
3 years ago
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