What must an object have in order to exert a gravitational field?

How does changes in distance affect the gravitational pull between two objects? Describe and give one example.

maxonik [38]

The formula is

F_grav = G * m1 * m2 / r^2

G m1 and m2 are going to stay the same once chosen no matter what the distance is. The only thing that will change is the distance.

As the distance increases, the Gravitational Force will decrease. It will decrease by quite a bit.

As the distance decreases, the gravitational force will Increase.

The relationship is inverse. The moon travelling around the earth is one example. The earth travelling around the sun is another.

8 0

3 years ago

A common flashlight bulb is rated at 0.32 A and 4.3 V (the values of the current and voltage under operating conditions). If the

sleet_krkn [62]

Answer:

1176.01 °C

Explanation:

Using Ohm's law,

V = IR................. Equation 1

Where V = Voltage, I = current, R = Resistance when the bulb is on

make R the subject of the equation

R = V/I.................. Equation 2

R = 4.3/0.32

R = 13.4375 Ω

Using

R = R'(1+αΔθ)............................. Equation 3

Where R' = Resistance of the bulb at 20°, α = Temperature coefficient of resistivity, Δθ = change in temperature

make Δθ the subject of the equation

Δθ = (R-R')/αR'.................. Equation 4

Given: R = 13.4375 Ω, R' = 1.6 Ω, α = 6.4×10⁻³ K⁻¹

Substitute into equation 4

Δθ = (13.4375-1.6)/(1.6×0.0064)

Δθ = 11.8375/0.01024

Δθ = 1156.01 °C

But,

Δθ = T₂-T₁

T₂ = T₁+Δθ

Where T₂ and T₁ = Final and initial temperature respectively.

T₂ = 20+1156.01

T₂ = 1176.01 °C

5 0

2 years ago

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What provides the centripetal force needed to keep Earth in orbit?

kkurt [141]

The centripetal force needed to keep earth in orbit is gravity.

3 0

3 years ago

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Pls help answer embed <br>

Savatey [412]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, $v_t=60\ m/s$

Area of cross section, $A=0.33\ m^2$

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

Weight of the object = drag force

R = W

or

$\dfrac{1}{2}\rho CAv_t^2=mg$

Where

$\rho$ is the density of air = 1.225 kg/m³

C is drag coefficient

So,

$C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01$

So, the drag coefficient is 1.01.

5 0

3 years ago

a Car with a mass of 2000 kg is moving around a circular curve at a uniform velocity of 25 m/s the curve has a radius of 80 m wh

Viefleur [7K]

Well, first of all, the car is not moving at a uniform velocity, because,
on a curved path, its direction is constantly changing. Its speed may
be constant, but its velocity isn't.

The centripetal force on a mass 'm' that keeps it on a circle with radius 'r' is

   F = (mass) · (speed)² / (radius).

For this particular car, the force is

(2,000 kg) · (25 m/s)² / (80 m)

   = (2,000 kg) · (625 m²/s²) / (80 m)

   = (2,000 · 625 / 80) (kg · m / s²)

   = 15,625 newtons .

5 0

3 years ago