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3 years ago

When Earth and the Moon are separated by a

2 answers:

6 0

Using the Universal Gratitation Law, we have:

$F= \frac{MmG}{d^2} \\ MmG=2*10^{20}*(3.84*10^8)^2 \\ MmG=29.4912*10^36$

Again applying the formula in the new situation, comes:

$F= \frac{MmG}{d^2} \\ F= \frac{29.4912*10^36}{(1.92*10^8)^2} \\ \boxed {F=8*10^{20}}$

Number 4

If you notice any mistake in my english, please let me know, because i am not native.

faltersainse [42]3 years ago

6 0

The strength of the gravitational forces between two masses is
inversely proportional to the square of the distance between them.

So if you change the distance to

   (1.92 x 10⁸) / (3.84 x 10⁸) = 1/2

of what it is now, then you would change the force to

   1 / (1/2)² = 4

of what it is now.

   (4) x (2 x 10²⁰) = 8.0 x 10²⁰ newtons .

You might be interested in

Part complete How long must a simple pendulum be if it is to make exactly one swing per five seconds?

shtirl [24]

Answer:

$L=6.21m$

Explanation:

For the simple pendulum problem we need to remember that:

$\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0$ ,

where $\theta$ is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:

$\omega^{2}=\frac{g}{L}$ ,

where $\omega$ is the angular frequency.

There is also an equation that relates the oscillation period and the angular frequeny:

$\omega=\frac{2\pi}{T}$ ,

where T is the oscillation period. Now, we can easily solve for L:

$(\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m$

3 0

3 years ago

An object is placed 5.00 cm beyond the focal point of a convex lens whose focal length is 10.0 cm. If the object height is 3.0 c

Aleks04 [339]

Answer:

The height of the image is, h' = 6.0 cm

The image is erect.

Explanation:

Given data,

The object distance, u = -5 cm

The focal length of convex lens, f = 10 cm

The object height, h = 3 cm

The lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{10}=\frac{1}{v}-\frac{1}{-5}$

$\frac{1}{v}=\frac{1}{10}-\frac{1}{5}$

v = -10 cm

The magnification factor of lens

$m=\frac{-10}{-5}$

m = 2

$m=\frac{h'}{h}$

$h'=h\times m$

$h'=3\times 2$

h' = 6 cm

The height of the image is, h' = 6 cm

The image is erect.

4 0

3 years ago

Part A What will be the equilibrium temperature when a 227 g block of copper at 283 °C is placed in a 155 g aluminum calorimeter

stellarik [79]

Answer:

T = 20.84°C

Explanation:

From the law of conservation of energy:

Heat Lost by Copper Block = Heat Gained by Aluminum Calorimeter + Heat Gained by Water

$m_cC_c\Delta T_c = m_wC_w\Delta T_w + m_aC_a\Delta T_a$

where,

$m_c$ = mass of copper = 227 g

$m_w$ = mass of water = 844 g

$m_a$ = mass of aluminum = 155 g

$C_c$ = specific heat capacity of calorimeter = 385 J/kg.°C

$C_w$ = specific heat capacity of water = 4200 J/kg.°C

$C_a$ = specific heat capacity of aluminum = 890 J/kg.°C

$\Delta T_c$ = change in temperature of copper = 283°C - T

$\Delta T_w$ = change in temperature of water = T - 14.6°C

$\Delta T_a$ = change in temperature of aluminum = T - 14.6°C

T = equilibrium temperature = ?

Therefore,

$(227\ g)(385\ J/kg.^oC)(283^oC-T)=(844\ g)(4200\ J/kg.^oC)(T-14.6^oC)+(155\ g)(890\ J/kg.^oC)(T-14.6^oC)\\\\24732785\ J - (87395\ J/^oC) T = (3544800\ J/^oC) T - 51754080\ J+ (137950\ J/^oC) T-2014070\ J\\\\24732785\ J +51754080\ J+2014070\ J = (3544800\ J/^oC) T+(137950\ J/^oC+(87395\ J/^oC) T\\\\78560935\ J = (3770145\ J/^oC) T\\\\T = \frac{78560935\ J}{3770145\ J/^oC}$

8 0

2 years ago

If the out come of a 1.0-kg ball moving at 3.0 m/s is 4.5 kinetic energy what would the kinetic energy be if the mass and speed

Flauer [41]

Answer: KE = 36 J

Explanation: Solution:

KE = 1/2 mv²

= 1/2 ( 2 kg)(6 m/s)²

= 36 J

7 0

2 years ago

The windshield of a speeding car hits a hovering insect. Consider the time interval from just before the car hits the insect to

HACTEHA [7]

Answer:

A. False

B True

C. False

D.False

E. True

F. False

G. False

H. False

I. True

Explanation:

A. False: The system being analyzed consists of the bug and the car. These are the two bodies involved in the collision.

B. True: The system being analyzed consists of the bug and the car

C. False: The magnitudes of the change in velocity are different from the car and the bug. The velocity of the bug changes from 0 to the velocity of the car, while there is no noticeable change in the velocity of the car

D.False: There is barely any change in the momentum of the car since the mass of the bug is very small.

E. True: Since the mass of the bug is small, and was initially at rest, the magnitude of the change in monentum will be large because the new velocity will be that of the car.

F. False: The system being analyzed consists of the bug and the car. Those are the two bodies involved in the collision

G. False: The car barely changes in velocity since the mass of the bug is small.

H. False: The car barely changes in momentum because the collision does not affect its speed so much. on the other hand the momentum change of the bug is large since its mass is small.

I. True: The bug which was initially at rest will begin moving with the velovity of the speeding car, while the car barely changes in its velocity

5 0

2 years ago