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3 years ago

What energy transfer will a stretched rubber band have when let go

Physics

1 answer:

GarryVolchara [31]3 years ago

3 0

Answer:

when the rubber band is realeased the potential energy is quickly converted to kinetic energy this is equal to one mass of the the rubber band multiplied by its velocity( in meters per second)

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a uniform disc and hollow right circular cone have the same formula for their moment of inertia when rotating about the central

Romashka-Z-Leto [24]

Answer:

This is as a result that about the central axis a collapsed hollow cone is equivalent to a uniform disc

Explanation:

The integration of the differential mass of the hollow right circular cone yields

$I=\int\limits dmr^2 = \int\limits^a_b {\frac{2Mxr^2}{R^2 +H^2} } \, dx = \frac{2MR^2dx}{(R^2 +H^2)^2} \frac{(R^2 +H^2)^2}{4} = \frac{1}{2}MR^2$

and for a uniform disc

I = 1/2πρtr⁴ = 1/2Mr².

6 0

4 years ago

A security guard walks at a steady pace traveling 170 m in one trip around the perimeter of a building.

WARRIOR [948]

Speed= distance/time
Speed= 170/230
Speed=0.74 m/s

6 0

3 years ago

The weight of an iron block is 8.0±0.3? and is placed on a wooden base of area, 3.5±0.2? 2. Calculate the pressure exerted by th

vesna_86 [32]

Answer:

Explanation:

Units...Nothing is accurate without UNITS. 8.0±0.3 lbs,? Newtons? 3.5±0.2 m²? ft²? cm²? km²? furlong²? fathoms²?

your answer will be 8.0±0.3 / 3.5±0.2 units / units²

8 0

3 years ago

A rock rolls down a hill. Which form of energy is this an example of? (2 points)

Bond [772]

Answer:

c.Mechanical

Explanation:

6 0

3 years ago

1. A step-up transformer increases 15.7V to 110V. What is the current in the secondary as compared to the primary? Assume 100 pe

Serjik [45]

1. $I_2 = 0.14 I_1$

Explanation:

We have:

$V_1 = 15.7 V$ voltage in the primary coil

$V_2 = 110 V$ voltage in the secondary coil

The efficiency of the transformer is 100%: this means that the power in the primary coil and in the secondary coil are equal

$P_1 = P_2\\V_1 I_1 = V_2 I_2$

where I1 and I2 are the currents in the two coils. Re-arranging the equation, we find

$\frac{I_2}{I_1}=\frac{V_1}{V_2}=\frac{15.7 V}{110 V}=0.14$

which means that the current in the secondary coil is 14% of the value of the current in the primary coil.

2. 5.7 V

We can solve the problem by using the transformer equation:

$\frac{N_p}{N_s}=\frac{V_p}{V_s}$

where:

Np = 400 is the number of turns in the primary coil

Ns = 19 is the number of turns in the secondary coil

Vp = 120 V is the voltage in the primary coil

Vs = ? is the voltage in the secondary coil

Re-arranging the formula and substituting the numbers, we find:

$V_s = V_p \frac{N_s}{N_p}=(120 V)\frac{19}{400}=5.7 V$

5 0

3 years ago