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3 years ago

What is the torque about the center of the sun due to the gravitational force of attraction of the sun on the planet?

1 answer:

3 0

Torque, moment, or moment of force is the tendency of a force to rotate an object around an axis,fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist to an object. Mathematically, torque is defined as the cross product of the vector by which the force's application point is offset relative to the fixed suspension point (distance vector) and the force vector, which tends to produce rotational motion. So torque about the center of the sun due to the gravitational force of attraction of the sun on the planet = (Gxm1xm2 / r^2). r sin(theta) = Gxm1xm2 /r^2). r sin0° = 0

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An alpha particle can be produced in certain radioactive decays of nuclei and consists of two protons and two neutrons. The part

Varvara68 [4.7K]

Answer:

Explanation:

charge, q = 2e = 2 x 1.6 x 10^-19 C = 3.2 x 10^-19 C

mass, m = 4 u = 4 x 1.661 x 10^-27 kg = 6.644 x 10^-27 kg

Radius, r = 4.5 cm = 0.045 m

Magnetic field, B = 1.20 T

(a) Let the speed is v.

$v=\frac{Bqr}{m}$

$v=\frac{1.20\times 3.2\times 10^{-19}\times 0.045}{6.644\times 10^{-27}}$

v = 2.6 x 10^6 m/s

(b) Let T be the period of revolution

$T=\frac{2\pi r}{v}$

$T=\frac{2\times 3.14\times 0.045}{2.6\times 10^{6}}$

T = 1.09 x 10^-7 s

$K=\frac{B^{2}\times q^{2}\times r^{2}}{2m}$

$K=\frac{\left ( 1.20\times 3.2 \times 10^{-19}\times 0.045 \right )^{2}}{2\times 6.644\times 10^{-27}}$

K = 2.25 x 10^-14 J

(d) Let the potential difference is V.

K = qV

$V = \frac{K}{q}$

$V= \frac{2.25\times 10^-14}{3.2\times 10^{-19}}$

V = 70312.5 V

5 0

3 years ago

I have a picture of the question I need help with,

eduard

Given:

The initial velocity of the object, v=30 m/s

a_t=0

a_c≠0

The time period is Δt.

To find:

The right conclusion among the given choices.

Explanation:

a_t represents the tangential accleration on the object and a_c represents the centripetal acceleration on the object.

The centripetal acceleration is the acceleration that keeps the object in its circular path. The centripetal force only changes the direction of the velocity and not the magnitude.

Thus the magnitude of the velocity of the object remains the same after a time interval of Δt. But the direction of the velocity of the object will be changed and will be unknown after Δt seconds.

Final answer:

Thus the object will be traveling at 30 m/s in some unknown direction.

Therefore, the correct answer is option a.

7 0

9 months ago

Problem 2: (15 pts) A 10-m high cylindrical container is half-filled at the bottom with water of density =1000 kg/m3 while the t

Step2247 [10]

Answer:

$\Delta p = 90.7 kPa$