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3 years ago

Which constants and directions always apply when using gravity?

Physics

2 answers:

Lostsunrise [7]3 years ago

6 0

Answer:

Explanation:

Romashka-Z-Leto [24]3 years ago

3 0

Answer:

Explanation::

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Nitella [24]

1 A

2 C

3 B

4 D

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3 years ago

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Heat engines convert which type of energy into mechanical work

KonstantinChe [14]

Answer: Heat engines convert chemical energy to mechanical energy

Explanation:

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3 years ago

An object is hung from a spring balance attached to the ceiling of an elevator cab. The balance reads 85 N when the elevator is

Ilia_Sergeevich [38]

Answer:

Explanation:

Balance force ( W ) = 85 N

a) moving upward with constant velocity (V ) = 7.9 m/s.

The reading will not change since it is moving with constant velocity

W = 85 N

b) Moving upward with V = 7.9 m/s and decelerating at the rate of (a ) = 1.1 m /s²

The net force ( F ) = W - ma

where m = W/g = 85/9.81 = 8.6646 kg

F = 85 - 8.6646 * 1.1 = 75.4689 N

4 0

3 years ago

Scientists want to place a 3400.0 kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of 2480.0

BaLLatris [955]

Answer:

Part a)

$r = 6.96 \times 10^6 m$

Part b)

$F_g = 3.004 \times 10^3 N$

Part c)

$a = 0.88 m/s^2$

Part d)

$v = \frac{2480}{2} = 1240 m/s$

Explanation:

Part a)

As we know that the orbital speed of the satellite is given as

$v = 2480 m/s$

now we will have

$v = \sqrt{\frac{GM_{mars}}{r}}$

now we have

$M_{mars} = 6.4191 \times 10^{23} kg$

$R_{mars} = 3.397 \times 10^6 m$

now we have

$2480 = \sqrt{\frac{(6.67 \times 10^{-11})(6.4191 \times 10^{23})}{r}}$

$r = 6.96 \times 10^6 m$

Part b)

Here force between mars and satellite is the gravitational attraction force which is given as

$F_g = \frac{GM_{mars} m}{r^2}$

$F_g = \frac{(6.67\times 10^{-11})(6.4191 \times 10^{23})(3400)}{(6.96\times 10^6)^2}$

$F_g = 3.004 \times 10^3 N$

Part c)

Acceleration of satellite is the ratio of gravitational force and mass of the satellite

$a = \frac{F_g}{m}$

$a = \frac{3004}{3400}$

$a = 0.88 m/s^2$

Part d)

As we know by III law of kepler

$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$

here we know that T2 = 8 T1

$(\frac{1}{8})^2= \frac{r_1^3}{r_2^3}$

$\frac{r_1}{r_2} = (\frac{1}{2})^2$

so we have

$r_2 = 4r_1$

as we know that speed is given as

$v = \sqrt{\frac{GM}{r}}$

so we can say since radius is orbit becomes 4 times so the orbital speed must be half

$v = \frac{2480}{2} = 1240 m/s$

7 0

4 years ago

A (hypothetical) large slingshot is stretched 4.00 m to launch a 440 g projectile with speed sufficient to escape from Earth (11

RSB [31]

Answer:

(A) 3,449,600 N/m

(B) 62,720 poeple

Explanation:

extension of the sling shot (e) = 4 m

mass of projectile (m) = 440 g = 0.44 kg

speed of projectile (v) = 11.2 km/s = 11,200 m/s

average force a person can exert = 220 N

spring constant (k) = ?

(A) When all the elastic potential energy is converted to kinetic energy

$o.5ke^{2} = 0.5 mv^{2}$

rearranging the above equation

spring constant (K) = $\frac{0.5mv^{2} }{0.5e^{2}}$

spring constant (K) = $\frac{0.5x0.44x11,200^{2} }{0.5x4^{2}}$

spring constant (K) = $\frac{27,596,800}{8}$

spring constant (K) = 3,449,600 N/m

(B) force required to stretch the slingshot (F) = ke = 3,449,600 x 4 = 13,798,400 N

number of people required = required force / average force per person = 13,798,400 / 220 =62,720 poeple

5 0

3 years ago

Which constants and directions always apply when using gravity?​

Which constants and directions always apply when using gravity?