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3 years ago

How is your guys day going?

Physics

2 answers:

vichka [17]3 years ago

7 0

Answer:

Awesome and u made it even better when you gave me these free points

Explanation:

faltersainse [42]3 years ago

4 0

Good i’m tired how about you

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How much work do you do on a 15 N book in lifting it straight up for a distance of<br> 0.40 meters?

astra-53 [7]

Answer:

Work done, W = 6 J

Explanation:

It is given that,

Force of gravity acting on the book, weight of the book is 15 N

We need to find the work done in lifting the book straight up for a distance of 0.4 meters.

The weight of the book is acting in downward direction and the book is lifted straight up, it means angle between them is 180 degrees. Work done is given by :

$W=Fd\cos180\\\\W=15\times 0.4\times \cos180\\\\W=-6\ J$

So, the magnitude of work done in lifting the book is 6 joules.

7 0

3 years ago

What is the mass of a rock lifted 2 meters off the ground that has 196 J of potential energy?

eimsori [14]

Answer:

10kg

Explanation:

Let PE=potential energy

PE=196J

g(gravitational force)=9.8m/s^2

h(change in height)=2m

m=?

PE=m*g*(change in h)

196=m*9.8*2

m=10kg

4 0

3 years ago

3.How can you find the velocity of the object just before impact without directly measuring it?

e-lub [12.9K]

Ugudbis sinwajdiekbdudjwndjid

7 0

3 years ago

A 12 volt car battery is connected to a 3 ohm brake light. What is the current carrying energy to the lights?

zloy xaker [14]

Answer:

4 A

Explanation:

V = IR, where V=voltage, I=current, R=resistance. This is Ohm's Law. (remember that for units V = volts, Ω = ohms, A = amperes.)

V = IR

12 V = I * 3 Ω

12/3 = I

8 0

3 years ago

At what distance from Earth does the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity ex

telo118 [61]

Answer:

346 * 10⁶ m

Explanation:

The force of gravity of the earth that will cancel the the force of gravity exerted by the moon will be equal to each other

Let $F_{e}$ be the force of gravity exerted by the earth

and let $F_{m}$ be the force of gravity exerted by the moon

According to Newton's law of universal gravitation, the force of attraction between two different masses, m₁ and m₂ separated by a distance, d, is given by:

$F = \frac{Gm_{1} m_{2} }{d^{2} }$

Mass of the earth, $m_{e} = 5.97 * 10^{24} kg$

Mass of the moon, $m_{m} = 7.348 * 10^{22} kg$

Mass of the satellite, $m_{s} = ?$

$F_{e} = \frac{G*5.97 * 10^{24} M }{d^{2} }$ ...............................(1)

The earth and the moon are separated by a distance, 3.844 * 10⁸ m

$F_{m} = \frac{G*7.348 * 10^{22} M }{(3.844 * 10^{8} - d) ^{2} }$ ............................(2)

Equating equations (1) and (2)

$\frac{5.97 * 10^{24} }{d^{2} } = \frac{7.348 * 10^{24} }{(3.844* 10^{8} -d)^{2} }$

$(5.97 * 10^{24})(14.78 * 10^{16} -7.688*10^{8}d + d^{2}) = 7.348 * 10^{24} d^{2} \\88.24*10^{40} - 45.9 * 10^{32}d + 5.97 * 10^{24}d^{2} = 7.348 * 10^{24} d^{2}\\ 1.378 * 10^{24}d^{2} + 45.9 * 10^{32}d + 88.24*10^{40} = 0\\$

Factorising out $10^{24}$

$1.378d^{2} + 45.9 * 10^{8}d + 88.24*10^{16} = 0$

Solving for d in the quadratic equation above:

d = 346 * 10⁶ m

4 0

3 years ago

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