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2 years ago

Calculate the p.e. of a 5 kg mass when it is (i) 3 m, (ii) 6 m, above the ground. (g = 10 N/kg)

Physics

1 answer:

Mashcka [7]2 years ago

4 0

Explanation:

P.E=MGH

(1) P.E=5×10(G)×3

= 150 kg m^/s^ ( ^ =sqaure)

(2) P.E=5×10×6

= 300 kg m^/s^

Hope it helps, just remember the formula u can solve any question regarding gravition of earth (P.E=Mgh)

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What happens to a liquid beyond its certain temperature

Natalka [10]

Answer: what do you mean ''beyond certain temperature'', Do you mean ask in cooling the liquid down or heating the liquid up?

Explanation:

7 0

2 years ago

A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00°C, the resistance of the carbon r

dusya [7]

Answer:

28 degree C

Explanation:

We are given that

$T_1=4.00^{\circ}$

$R_1=217.7 \Ohm$

$R_2=215.1\Ohm$

$\alpha=-5.00\times 10^{-4}C^{-1}$

We have to find the temperature on a spring day when resistance is 215.1 ohm.

We know that

$\alpha(T_2-T_1)=\frac{R_2}{R_2}-1$

Using the formula

$-5.00\times 10^{-4}(T_2-4)=\frac{215.1}{217.7}-1$

$-5\times 10^{-4}(T_2-4)=0.988-1=-0.012$

$T_2-4=\frac{0.012}{5\times 10^{-4}}=24$

$T_2=24+4=28^{\circ}C$

Hence, the temperature on a spring day 28 degree C.

7 0

2 years ago

On the surface of the earth the weight of an object is 200 lb. Determine the height of the

siniylev [52]

Answer:

The height of the object is 5007.4 miles.

Explanation:

Given that,

Weight of object = 200 lb

We need to calculate the value of $Gmm_{e}$

Using formula of gravitational force

$F=\dfrac{Gmm_{e}}{r^2}$

Put the value into the formula

$200=\dfrac{Gmm_{e}}{(3958.756)^2}$

$200\times(3958.756)^2=Gmm_{e}$

$Gmm_{e}=3.134\times10^{9}$

We need to calculate the height of the object

Using formula of gravitational force

$F=\dfrac{Gmm_{e}}{r^2}$

Put the value into the formula

$125=\dfrac{200\times(3958.756)^2}{r^2}$

$r^2=\dfrac{200\times(3958.756)^2}{125}$

$r^2=25074798.5$

$r=\sqrt{25074798.5}$

$r=5007.4\ miles$

Hence. The height of the object is 5007.4 miles.

7 0

2 years ago

A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When

natita [175]

Answer:

a) $\mu_{k} = 0.704$ , b) $R = 0.312\,m$

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s$

$\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}$

$\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}$

$\mu_{k} = 0.704$

b) The speed of the block is determined by using the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$v = x\cdot \sqrt{\frac{k}{m} }$

$v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }$

$v \approx 9.829\,\frac{m}{s}$

The radius of the circular loop is:

$\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}$

$\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}$

$R = 0.312\,m$

5 0

2 years ago

Find the resistance of a stereo: current is 7 A voltage is 49 V.

pochemuha

Answer:

R = 7 [amp]

Explanation:

To solve this problem we must use ohm's law which tells us that the voltage is equal to the product of the current by the resistance. In this way, we have the following equation.

V = I*R

where:

V = voltage = 49 [V] (units of volts)

I = current = 7 [amp] (amperes)

R = resistance [ohms]

Now clearing R.

R =V/I

R = 49/7

R = 7 [amp]

5 0

2 years ago

Calculate the p.e. of a 5 kg mass when it is (i) 3 m, (ii) 6 m, above the ground. (g = 10 N/kg)​

Calculate the p.e. of a 5 kg mass when it is (i) 3 m, (ii) 6 m, above the ground. (g = 10 N/kg)