Can someone answer these question for me?

If you went to a planet that had the twice the radius as Earth, but the same mass, a 1 kg pineapple would have a weight of

kicyunya [14]

Use the law of universal gravitation, which says the force of gravitation between two bodies of mass m₁ and m₂ a distance r apart is

F = G m₁ m₂ / r²

where G = 6.67 x 10⁻¹¹ N m²/kg².

The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is

F = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²

F ≈ 9.81 N

Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)g, where g = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.

This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of

F = G m₁ m₂ / (2r)² = 1/4 G m₁ m₂ / r²

i.e. 1/4 of the weight on Earth, which would be about 2.45 N.

7 0

3 years ago

If you jumped out of a plane, you would begin speeding up as you fall downward. Eventually, due to wind resistance, your velocit

MrRa [10]

Answer:

Mg or your weight.

Explanation:

When your velocity is constant, the net force acting on you is 0. That means the upwards force of air resistance must fully balance the downwards force of gravity on you, which is Mg.

5 0

3 years ago

A person's center of mass is easily found by having the person lie on a reaction board. A horizontal, 2.1-m-long, 6.6 kg reactio

MrRissso [65]

Answer is x hopefully this helps your stupid head

5 0

3 years ago

By how much does the volume of an aluminum cube 4.00 cm on an edge increase when the cube is heated from 19.0°C to 67.0°C? The l

Genrish500 [490]

Answer:

The volume of an aluminum cube is 0.212 cm³.

Explanation:

Given that,

Edge of cube = 4.00 cm

Initial temperature = 19.0°C

Final temperature = 67.0°C

linear expansion coefficient $\alpha=23.0\times10^{-6}/C^{\circ}$

We need to calculate the volume expansion coefficient

Using formula of volume expansion coefficient

$\beta=3\alpha$

Put the value into the formula

$\beta=3\times23.0\times10^{-6}$

$\beta=0.000069=69\times10^{-6}/C^{\circ}$

We need to calculate the volume

$V= a^3$

$V=4^3$

$V=64\ cm^3$

The change temperature of the cube is

$\Delta T=T_{f}-T_{i}$

Put the value into the formula

$\Delta T=67-19 = 48^{\circ}C$

We need to calculate the increases volume

Using formula of increases volume

$\Delta V=V\beta\Delta T$

Put the value into the formula

$\Delta V=64\times69\times10^{-6}\times48$

$\Delta V=0.212\ cm^3$

Hence, The volume of an aluminum cube is 0.212 cm³.

5 0

3 years ago

A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If

Salsk061 [2.6K]

Complete question:

A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?

Answer:

The current in the circuit 7 ms later is 0.2499 A

Explanation:

Given;

Ideal inductor, L = 45-mH

Resistor, R = 60-Ω

Ideal voltage supply, V = 15-V

Initial current at t = 0 seconds:

I₀ = V/R

I₀ = 15/60 = 0.25 A

Time constant, is given as:

T = L/R

T = (45 x 10⁻³) / (60)

T = 7.5 x 10⁻⁴ s

Change in current with respect to time, is given as;

$I(t) = I_o(1-e^{-\frac{t}{T}})$

Current in the circuit after 7 ms later:

t = 7 ms = 7 x 10⁻³ s

$I(t) = I_o(1-e^{-\frac{t}{T}})\\\\I =0.25(1-e^{-\frac{7*10^{-3}}{7.5*10^{-4}}})\\\\I = 0.25(0.9999)\\\\I = 0.2499 \ A$

Therefore, the current in the circuit 7 ms later is 0.2499 A

6 0

3 years ago

Can someone answer these question for me?​

Can someone answer these question for me?