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Ray Of Light [21]

3 years ago

11

A rotating fan completes 1200 revolutions every minute. consider the tip of a blade, at a radius of 0.19 m. through what distanc

e does the tip move in one revolution?

1 answer:

ser-zykov [4K]3 years ago

5 0

The tip of the fan moves through the outer side of the circle.

So it moves a distance of perimeter of circle in one revolution.

Perimeter of circle = 2 $\pi$ r, where r is the radius of circle.

In this case radius of circular motion = 0.19 meter

So perimeter of circle = 2 $\pi$ *0.19 = 0.38 $\pi$ = 1.194 m

So distance does the tip move in one revolution = 1.194 meter

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Two students are sitting 1.50 m apart. One student has a mass of 70.0 kg and

Galina-37 [17]

Answer:

1.08x10⁻⁷

Explanation:

F=(GM₁M₂)/r²

=((6.67x10⁻¹¹)(70)(52))/(1.5²)

=2.42788x10⁻⁷/2.25

=1.07905778x10⁻⁷

≈1.08x10⁻⁷

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3 years ago

147.5 ml of ethyl alcohol (coefficient of volume expansion = 1120 x 10-6K-1) at a temperature of 273.1 K is measured into a 150.

marishachu [46]

Answer:

288.2 K

Explanation:

$\beta$ = Volumetric expansion coefficient = $1120\times 10^{-6}\ /K$

$T_i$ = Initial temperature = 273.1 K

$T_f$ = Final temperature

$v_0$ = Original volume = 150 mL

Change in volume is given by

$\Delta v=\beta v_0(T_f-T_i)\\\Rightarrow T_f=\frac{\Delta v}{\beta v_0}+T_i\\\Rightarrow T_f=\frac{150-147.5}{1120\times 10^{-6}\times 147.5}+273.1\\\Rightarrow T_f=288.2\ K$

The temperature of the ethyl alcohol should be 288.2 K to reach 150 mL

8 0

3 years ago

What is the effect of using split rings in a simple d.c generator

Aliun [14]

Answer:

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5 0

3 years ago

A. How many calories are needed to raise the temperature of 1 gram of water by 1 °C?

sweet-ann [11.9K]

A) 1 cal

B) 80 cal

C) 540 cal

Explanation:

A)

The amount of heat energy needed to raise the temperature of a certain mass of a substance is given by

$Q=mC\Delta T$

where

m is the mass of the substance

C is the specific heat capacity

$\Delta T$ is the change in temperature

In this problem:

m = 1 g is the mass of water

$C=1 cal/g^{\circ}C$ is the specific heat capacity of water

$\Delta T=1^{\circ}C$ is the change in temperature

So, the heat needed is

$Q=(1)(1)(1)=1 cal$

B)

For a solid substance at its melting point, the amount of heat needed to melt completely the substance is given by

$Q=m\lambda_f$

where

m is the mass of the substance

$\lambda_f$ is the specific latent heat of fusion of the substance

In this problem:

- The ice is already at melting point, 0 °C

- Mass of the ice: $m=1g$

- Specific latent heat of fusion of ice: $\lambda_f=80 cal/g$

So, the heat needed is

$Q=(1)(80)=80 cal$

C)

For a liquid substance at its boiling point, the amount of heat needed to boil completely the substance is given by

$Q=m\lambda_v$

where

m is the mass of the substance

$\lambda_v$ is the specific latent heat of vaporization of the substance

In this problem:

- The water is already at boiling point, 100 °C

- Mass of the water: $m=1g$

- Specific latent heat of vaporization of water: $\lambda_v=540 cal/g$

So, the heat needed is

$Q=(1)(540)=540 cal$

5 0

3 years ago