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kakasveta [241]

3 years ago

8

Dane is holding an 8 kilogram box 2 metres above the ground. How much energy is in the box's gravitational potential energy stor

e? Assume Dane is on Earth, where g = 10 N/kg.

1 answer:

kari74 [83]3 years ago

7 0

Answer:160j

Explanation:

PE=mgh

m=8kg

g=10N/kg

h=2m

PE=8*10*2

PE=160j

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A voltage amplifier with an input resistance of 40k ohms, an output resistance of 100 ohms, and a gain of 300 V/V is connected b

AfilCa [17]

Answer:

89.45 v/v

Explanation:

Let's take the data:

First draw the amplifier circuit.

After the circuit, the voltage division rule can be used to compute the parameters:

The input section is computed like this: $\frac{R_{in} }{(R_{sig} + R_{in} } (v_{sig})$

The output section is computed like this $R_{L}/ (R_out} + R_{in} )$

The product A $V_{in} V_{out}$ gives

A $V_{in} V_{out}$ = A× $\frac{R_{in} }{(R_{sig} + R_{in} } (v_{sig})$ × $R_{L}/ (R_out} + R_{in} )$

Computing gives output voltage = 89.45 v/v

5 0

3 years ago

How to do it? Urgent

mixer [17]

a)

F= ma

a=v/t

F=5*(35/5)

F=35N

b)

a=F/m

a=(35-2)/5

a=33/5

a=6.6N

8 0

3 years ago

An artist working on a piece of metal in his forging studio plunges the hot metal into oil in order to harden it. The metal piec

lutik1710 [3]

Answer:

The temperature of the metal is $T_m = 376.8 ^o C$

Explanation:

From the question we are told that

The mass of the metal is $M = 60 \ kg$

The specific heat of the metal is $c_p = 0.1027 kcal/(kg \cdot ^oC)$

The mass of the oil is $M_o = 810 \ kg$

The temperature of the oil is $T_o = 35^oC$

The specific heat of oil is $c_o = 0.7167 kcal/(kg \cdot ^oC )$

The equilibrium temperature is $T_e = 39 ^oC$

According to the law of energy conservation

Heat lost by metal = heat gained by the oil

So

The quantity of heat lost by the metal is mathematically represented as

$Q = - Mc_p \Delta T$

=> $Q = -Mc_p (T_m - T_c)$

Where $T_ m$ the temperature of metal before immersion

The negative sign show heat lost

The quantity of gained t by the metal is mathematically represented as

$Q = M_o c_o \Delta T$

=> $Q = M_o c_o (T_c - T_o)$

So

$Mc_p (T_m - T_c) = M_o c_o (T_c - T_o)$

substituting values

$- 60 * 0.1027 (T_m - 39) = 810 * 0.7167 * (39 - 35)$

=> $T_m = 376.8 ^o C$

6 0

3 years ago

Which color of light is diffracted at a greater angle from a diffraction grating, red or yellow light?

topjm [15]

The amount of diffraction depends on the wavelength of light, with shorter wavelengths being diffracted at a greater angle than longer ones (in effect, blue and violet<span> light are diffracted at a larger angle than is red light).

I hope my answer has come to your help. God bless and have a nice day ahead!
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3 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago