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3 years ago

8

What is a load force

1 answer:

vladimir2022 [97]3 years ago

8 0

Answer:

A push or pull exerted on an object

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An inventor develops a stationary cycling device by which an individual, while pedaling, can convert all of the energy expended

mr Goodwill [35]

Answer:

$Q=94185\ J$

Explanation:

Given:

mass of water, $m=0.3\ kg$

initial temperature of water, $T_i=20^{\circ}C$

final temperature of water, $T_f=95^{\circ}C$

specific heat of water, $c=4186\ J.kg^{-1}.K^{-1}$

<u>Now the amount of heat energy required:</u>

$Q=m.c.\Delta T$

$Q=0.3\times 4186\times (95-20)$

$Q=94185\ J$

Since all of the mechanical energy is being converted into heat, therefore the same amount of mechanical energy is required.

4 0

3 years ago

A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find

FrozenT [24]

Answer:

The time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

Explanation:

Given :

Current $I = 0.106$ A

Area of plate $A = 36 \times 10^{-4}$ $m^{2}$

Plate separation $d = 4 \times 10^{-3}$ m

(A)

First find the capacitance of capacitor,

$C = \frac{\epsilon _{o} A }{d}$

Where $\epsilon _{o} = 8.85 \times 10^{-12}$

$C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4} }{4 \times 10^{-3} }$

$C = 7.9 \times 10^{-12}$ F

But $C = \frac{Q}{V}$

Where $Q = It$

$C = \frac{It}{V}$

$V = \frac{It}{C}$

Now differentiate above equation wrt. time,

$\frac{dV}{dt} = \frac{I}{C}$

$= \frac{0.106}{7.9 \times 10^{-12} }$

$= 1.34 \times 10^{10}$ $\frac{V}{s}$

Therefore, the time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

8 0

3 years ago

2. Which of the following wavelength properties would require a stopwatch to measure?

katovenus [111]

Answer:

<u><em>A. wavelength</em></u>

Explanation:

The others are about sound and how high it is. That has nothing to do with time.

3 0

3 years ago

Read 2 more answers

A piston-cylinder device initially contains 0.08 m3 of nitrogen gas at 150 kPa and 200°C. The nitrogen is now expanded to a pres

Lemur [1.5K]

Answer:

$V_2 = 0.125 m^3$

Work done = = 5 kJ

Explanation:

Given data:

volume of nitrogen $v_1 = 0.08 m^3$

$P_1 = 150 kPa$

$T_1 = 200 degree celcius = 473 Kelvin$

$P_2 = 80 kPa$

Polytropic exponent n = 1.4

$\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}$

putting all value

$\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}$

$\frac{T_2} = 395.23 K = 122.08 DEGREE \ CELCIUS$

polytropic process is given as

$P_1 V_1^n = P_2 V_2^n$

$150\times 0.08^{1.4} = 80 \times V_2^{1.4}$

$V_2 = 0.125 m^3$

work done $= \frac{P_1 V_1 -P_2 V_2}{n-1}$

$= \frac{150 \times 0.8 - 80 \times 0.125}{1.4-1}$

= 5 kJ

4 0

3 years ago

What happens to the frequency of a wave of you increase the speed of the wave ?

exis [7]

Nothing happens. The frequency is determined at the source,
and it doesn't change along the way.

3 0

3 years ago