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3 years ago

What other forces affect the soccer ball after the student's foot stops touching it?

Physics

2 answers:

hodyreva [135]3 years ago

8 0

Gravity, acceleration, velocity

svet-max [94.6K]3 years ago

4 0

Gravity I hope this helps you I'm not very good at this kind of stuff

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At a certain instant, the earth, the moon, and a station- ary 1250-kg spacecraft lie at the vertices of an equilateral triangle

Murrr4er [49]

Answer:

a). F = 3.376 N, θ = 59.18°

b). W = 1.3x $10^{9}$ J

Explanation:

We know

Gravitational constant, G = 6.673 x $10^{-11}$ N- $m^{2}$ / $kg^{-2}$

Mass of the earth, M = 5.97 x $10^{24}$ kg

mass of the moon, m = 7.35 x $10^{22}$ kg

Mass of the satellite, $m_{s}$ = 1250 kg

Distance between the objects, r = 3.84 x $10^{5}$ km

= 3.84 x $10^{8}$ m

Now

The force on the satellite due to moon

$F_{m}= \frac{G\times m\times m_{s}}{r^{2}}$

$F_{m}= \frac{6.673\times 10^{-11}\times 7.35\times 10^{22}\times 1250}{(3.84\times 10^{8})^{2}}$

$F_{m}$ = 0.0415 N ( in the positive x direction )

The force on the space craft due to the earth

$F_{m}= \frac{G\times M\times m_{s}}{r^{2}}$

$F_{m}= \frac{6.673\times 10^{-11}\times 5.97\times 10^{24}\times 1250}{(3.84\times 10^{8})^{2}}$

$F_{m}$ = 3.377 N ( at 60° to x axis )

Now component of force of earth along x axis

$F_{e_{x}} = F_{e}\times cos 60$

= 3.377 x 0.5

= 1.6885 N

Now component of force of earth along y axis

$F_{e_{y}} = F_{e}\times sin 60$

= 3.377 x 0.86

= 2.90 N

∴ Net force on the space craft due to earth and moon along x axis

$F_{x}$ = $F_{e}$ cos 60+ $F_{m}$

= 1.3885+0.0415

= 1.73 N

Net force on the space craft due to earth and moon along y axis

$F_{x}$ = $F_{e_{y}}$

= 2.90 N

Therefore, total force F = $\sqrt{(F_{x}^{2})+(F_{y}^{2})}$

F = $\sqrt{(1.73^{2})+(2.90^{2})}$

F = 3.376 N

∴ Magnitude of the net gravitational force on the space craft is 3.376 N

Direction of net force on the space craft is given by

$\Theta = \arctan \left (\frac{F_{y}}{F_{x}}\right )$

$\Theta = \arctan \left (\frac{2.90}{1.73}\right )$

$\Theta = 59.18$ °

Therefore this direction is 59.18° from the line joining earth and the space craft.

b).

∴ Gravitational potential energy of the space craft is given by

$E = \frac{G.M.m_{s}}{r}+\frac{G.m.m_{s}}{r}$

$E = \frac{G\times m_{s}\left ( M+m \right )}{r}$

$E = \frac{6.673\times 10^{-11}\times 1250\left ( 5.97\times 10^{24}+7.35\times 10^{22} \right )}{3.84\times 10^{8}}$

E = 1312769385 J

E = 1.3 x $10^{9}$ J

Therefore minimum work done is 1.3x $10^{9}$ J

8 0

3 years ago

The established value for the speed of light in a vacuum is 299,792,458 m/s. What is the order-of-magnitude of this number?

iogann1982 [59]

The order of magnitude is 10⁸ .

6 0

3 years ago

According to the law of conservation of energy, which statement must be true?

Nitella [24]

Answer:

B. A system cannot take in additional matter.

Explanation: The total amount of energy in the universe remains constant, it can merely change from one form to another.

Hope it helps you:)))

have a good day

5 0

2 years ago

Young's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a s

jok3333 [9.3K]

Answer:

0.88 mm

Explanation:

Given

Wavelength of the atoms, λ = 502 nm = 502*10^-9 m

Radius of the screen away from the double slit, r = 1 m

We know that Y(20) = 11.4*10^-3 m

d = (20 * R * λ) / Y(20)

d = (20 * 1 * 502*10^-9)/11.4*10^-3

d = 1*10^-5 / 11.4*10^-3

d = 0.88 mm

Therefore, the distance of separation between the two slits is 0.88 mm

7 0

4 years ago

What is the weight on Earth of a girl with a mass of 32 kg (F = ma)?

GenaCL600 [577]

Answer:

F = 313.6 N

Explanation:

weight is the force acting on an object due to gravity.

so,

f = ma (m = mass, a = acceleration)
f = 32 × 9.8
f = 313.6 Newton

( note : i have taken acceleration due to gravity as 9.8, it can be taken as an approx of 10, so if you have to take 10, the weight of girl will be 320 Newton )

i hope it helped....

4 0

3 years ago