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kakasveta [241]
3 years ago
15

What does Newton’s law of universal gravitation say about the distance between objects?

Physics
1 answer:
SashulF [63]3 years ago
3 0
Newton law of universal gravitation states that a particle attracts every other particle in the universe using a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
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A circular 10-turn coil with a radius of 5.0 cm carries a current of 5 A. It lies in the xy plane in a uniform magnetic field =
Tju [1.3M]

Answer:

U = – 0.12J

Explanation:

Given N = 10 turns, I = 5A, r = 5×10-²m

B^ = 0.05 T iˆ+ 0.3 T kˆ

Magnitude of the magnetic field vector B = √(0.05²+0.3²) = 0.304T

Area = πr² = π(5×10-²)² = 7.85×10-³m²

Magnetic moment μ = NIA

μ = 10×5×7.85×10-³ = 0.3925Am²

U = -μ•B = –0.3925×0.304 = –0.12J

The sign is negative because the magnetic moment is aligned with the magnetic field.

3 0
3 years ago
Which quantities define momentum?
grigory [225]

Answer:

A. How much matter an object has, plus the magnitude and direction  of its motion

Explanation:

Momentum is defined as the product of mass by velocity, in the international system of measurements (SI) momentum has the following Units [kg*m/s].

P = m*v

where:

P = momentum Lineal [kg*m/s]

m = mass [kg]

v = velocity [m/s]

Therefore the answer is A) How much matter an object has, plus the magnitude and direction  of its motion

7 0
3 years ago
Which force causes send dunes? A. Rivers B. Ice C. Wind D. Gravity
lutik1710 [3]
Wind I believe cause it carries the sand to different places
5 0
3 years ago
Read 2 more answers
A wheel 33 cm in diameter accelerates uniformly from 240 rpm to 360 rpm in 6.5 s. how far will a point on the edge of the wheel
melisa1 [442]

First we find the angular acceleration \alpha. The angular velocities are \frac{240}{60} =4 \;rps and \frac{360}{60} =6 \;rps The angular velocities and \alpha are related as

\omega=\omega_0+\alpha t\\ \alpha =\frac{\omega - \omega_0}{t} \\ \alpha =\frac{6-4}{(6.5)} \\ \alpha =0.3077 \;rps^2\\ \alpha =(0.3077 )2\pi =1.933 \;rad/s^2

Angle turned in 6.5 seconds is

2 \alpha\theta =   \omega^2-\omega_0^2\\ \theta =  \frac{ \omega^2-\omega_0^2}{2 \alpha}\\  \theta =  \frac{ 6^2-4^2}{2 (1.933)}\\  \theta = 5.173 \; rad

The distance traveled by a point on the edge of the wheel is r \theta = \frac{33}{2}(5.173)= 85.35 \;cm

7 0
3 years ago
A cylindrical piece of wood of height H, radius r, and uniform density rhoc is bobbing up and down in still water with its axis
Keith_Richards [23]

Answer:

r\sqrt{\frac{g\pi\rho_w}{m}}

Explanation:

As the buoyant force is proportional to the length of which the wood is submerged in water, we can model the buoyant force as a spring force and the bobbing wood is a simple harmonic motion described as A cos (ωt + φ) where

\omega = \sqrt{\frac{k}{m}}

where k (N/m) is the "spring" buoyant constant and m is the mass of wood

The buoyant force is basically the weight of water displaced by the submerged wood, which is the gravity acting on the cylindrical volume

F_b = W_w = m_wg = V_w\rho_w g

Since the cylindrical has a form of AL where A is the base area and L is the length submerged

F_b = AL\rho_w g = (\pi r^2\rho_w g) L

As L can be treated as the spring "stretched/compressed" length, the rest is k:

F_b = kL = (\pi r^2\rho_w g) L

k = (\pi r^2\rho_w g)

Therefore

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{\pi r^2\rho_w g}{m}} = r\sqrt{\frac{g\pi\rho_w}{m}}

8 0
3 years ago
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