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2 years ago

Once an object enters orbit, what keeps the object moving sideways?

Physics

2 answers:

Makovka662 [10]2 years ago

8 0

Answer:

The right answer is inertia.

Explanation:

This name is given to the tendency of an object that is moving to remain that way, or if it is resting remain that way until an external force act upon it. Basically it means an object does what they are doing unless a force changes their speed or direction.

yaroslaw [1]2 years ago

7 0

It's the natural tendency of things to keep going unless there's something trying to stop them.

It's usually called "inertia".

Don't get the idea from all of this that things stop unless there's something to keep them going. The truth is exactly the opposite: Things keep going unless there's something to make them stop.

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There are two different size spherical paintballs and the smaller one has a diameter of 5 cm and the larger one is 9 cm in diame

slavikrds [6]

Answer:

145.8 cm³ of paint

Explanation:

d₁ = Smaller diameter paintball = 5 cm

d₂ = Larger diameter paintball = 9 cm

V₂ = Volume of larger diameter paintball

Volume of smaller diameter paintball

$V_1=\frac{4}{3}\pi r_1^3\\\Rightarrow V_1=\frac{4}{3}\pi \left(\frac{d_1}{2}\right)^3\\\Rightarrow V_1=\frac{4}{24}\pi d_1^3$

Similarly

$V_2=\frac{4}{24}\pi d_2^3$

Dividing the above two equations, we get

$\frac{V_1}{V_2}=\frac{d_1^3}{d_2^3}\\\Rightarrow V_2=\frac{V_1}{\frac{d_1^3}{d_2^3}}\\\Rightarrow V_2=\frac{28}{\frac{125}{729}}\\\Rightarrow V_2=163.296\ cm^3$

∴ The larger one hold 163.296 cm³ of paint

5 0

2 years ago

In the 25-ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of int

Ugo [173]

Answer:

a. 8.33 x 10 ⁻⁶ Pa

b. 8.19 x 10 ⁻¹¹ atm

c. 1.65 x 10 ⁻¹⁰ atm

d. 2.778 x 10 ⁻¹⁴ kg / m²

Explanation:

Given:

I = 2500 W / m² , us = 3.0 x 10 ⁸ m /s

P rad = I / us

P rad = 2500 W / m² / 3.0 x 10 ⁸ m/s

P rad = 8.33 x 10 ⁻⁶ Pa

P rad = 8.33 x 10 ⁻⁶ Pa *[ 9.8 x 10 ⁻⁶ atm / 1 Pa ]

P rad = 8.19 x 10 ⁻¹¹ atm

P rad = 2 * I / us = ( 2 * 2500 w / m²) / [ 3.0 x 10 ⁸ m /s ]

P rad = 1.67 x 10 ⁻⁵ Pa

P₁ = 1.013 x 10 ⁵ Pa /atm

P rad = 1.67 x 10 ⁻⁵ Pa / 1.013 x 10 ⁵ Pa /atm = 1.65 x 10 ⁻¹⁰ atm

P rad = I / us

ΔP / Δt = I / C² = [ 2500 w / m² ] / ( 3.0 x 10 ⁸ m/s)²

ΔP / Δt = 2.778 x 10 ⁻¹⁴ kg / m²

3 0

2 years ago

An object of mass m has these three forces acting on it (there is no normal force, "no surface"). F1 = 1 N, F2 = 9 N, and F3 = 5

hammer [34]

Answer:

solved

Explanation:

a) F_net = (F2 - F3)i - F1 j

b) |Fnet| = sqrt( (F2 - F3)^2 + F1^2)

= sqrt( (9- 5)^2 + 1^2)

= 4.123 N

c) θ = tan^-1( (Fnet_y/Fnet_x)

= tan^-1( -1/(9-5) )

= -14.036°

7 0

2 years ago

In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land

sattari [20]

Answer:

a) Impulse |J|= 219.4 kgm/s

b) Force F = 2672 N

Explanation:

Given

Height of fall h = 0.50 m

Mass M = 70 kg

Period of collision t = 0.082 s

Solution

The final velocity of the person v is zero since the person will come to rest.

The initial velocity of the person can be calculated by using the "law of conservation of energy".

Initial Kinetic energy = Final potential energy

$\frac{1}{2} mu^2=mgh\\\\u = \sqrt{2gh} \\\\u = \sqrt{2 \times 9.81 \times 0.50} \\\\u = 3.13 m/s$

a) Impulse

J = final momentum - initial momentum

$J = mv -mu\\\\J = 0 - (70 \times 3.13)\\\\J = -219.2 kgm/s$

Magnitude of impulse

$|J| = 219.1 kgm/s$

b) Force

$F = \frac{J}{t} \\\\F = \frac{219.1}{0.082} \\\\F = 2672 N$

4 0

2 years ago

A roller coaster car drops a maximum vertical distance of 35.4 m. Determine the maximum speed of the car at the bottom of that d

marissa [1.9K]

Answer:

The maximum speed of the car at the bottom of that drop is 26.34 m/s.

Explanation:

Given that,

The maximum vertical distance covered by the roller coaster, h = 35.4 m

We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :

$mgh=\dfrac{1}{2}mv^2$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 35.4}$

v = 26.34 m/s

So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.

8 0

2 years ago