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torisob [31]
2 years ago
6

Which terms describe the motion of most objects in our solar system?

Physics
2 answers:
olga2289 [7]2 years ago
6 0

Answer:Geography, Earth Science, Astronomy

Rotation describes the circular motion of an object around its center.

Explanation:

Shalnov [3]2 years ago
3 0

Answer:

it is Cyclic and predictable

You might be interested in
A golf ball is hit horizontally off the edge of a 30 m high cliff and lands a distance of 25 m from the edge of the cliff. What
Ratling [72]

Answer:

V₀y = 0 m/s

t = 2.47 s

V₀ₓ = 61.86 m/s

Vₓ = 61.86 m/s

Explanation:

Since, the ball is hit horizontally, there is no vertical component of velocity at initial point. So, the initial vertical velocity (V₀y) will beL

<u>V₀y = 0 m/s</u>

For the initial vertical velocity of golf ball we consider the vertical motion and apply 2nd equation of motion:

Y = V₀y*t + (0.5)gt²

where,

Y = Height = 30 m

g = 9.8 m/s²

t = time to hit the ground = ?

Therefore,

30 m = (0 m/s)(t) + (0.5)(9.8 m/s²)t²

t² = 30 m/4.9 m/s²

t = √6.122 s²

<u>t = 2.47 s</u>

For initial vertical velocity we analyze the horizontal motion of the ball. We neglect the frictional effects in horizontal motion thus the speed remains uniform. Hence,

V₀ₓ = Xt

where,

V₀ₓ = Initial vertical Velocity = ?

X = Horizontal Distance = 25 m

Therefore,

V₀ₓ = (25 m)(2.47 s)

<u>V₀ₓ = 61.86 m/s</u>

<u></u>

Due, to uniform motion in horizontal direction:

Final Vertical Velocity = Vₓ = V₀ₓ

Vₓ = 61.86 m/s

4 0
3 years ago
A basketball player does 2.43 x 105 J of work during her time in the game, and evaporates 0,1 '10 kg of water. Assuming a latent
olchik [2.2K]

The change in the player's internal energy is -491.6 kJ. The number of nutritional calories is -117.44 kCal

For this process to take place, some of the basketball player's perspiration must escape from the skin. This is because sweating relies on a physical phenomenon known as the heat of vaporization.

The heat of vaporization refers to the amount of heat required to convert 1g of a liquid into a vapor without causing the liquid's temperature to increase.

From the given information,

  • the work done on the basketball is dW = 2.43 × 10⁵ J

The amount of heat loss is represented by dQ.

where;

  • dQ = -mL

∴

Using the first law of thermodynamics:b

dU = dQ - dW

dU = -mL - dW

dU = -(0.110 kg × 2.26 × 10⁶ J/kg - 2.43 × 10⁵ J)

dU = -491.6 × 10³ J

dU = -491.6 kJ

The number of nutritional calories the player has converted to work and heat can be determined by using the relation:

\mathbf{dU = -491.6 \ kJ \times (\dfrac{1 \ cal}{ 4.186 \ J})}

dU = -117.44 kcal

Learn more about first law of thermodynamics here:

brainly.com/question/3808473?referrer=searchResults

5 0
2 years ago
To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T
Fantom [35]

Answer:

V_1= 3.4*10^7m/s

Explanation:

From the question we are told that

Nucleus diameter d=5.50-fm

a 12C nucleus

Required kinetic energy K=2.30 MeV

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

            K_2 +U_2=K_1+U_1

where

K_1 =initial kinetic energy

K_2 =final kinetic energy

U_1 =initial electric potential

U_2 =final electric potential

mathematically

   U_2 = \frac{Kq_pq_c}{r_2}

where

r_f=distance b/w charges

q_c=nucleus charge =6(1.6*10^-^1^9C)

K=constant

q_p=proton charge

Generally kinetic energy is know as

         K=\frac{1}{2}  mv^2

Therefore

         U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2}  mv_1^2 +U_1

Generally equation for radius is d/2

Mathematically solving for radius of nucleus

         R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})

         R=2.75*10^-^1^5m

Generally we can easily solving mathematically substitute into v_1

   q_p=6(1.6*10^-^1^9C)

   K_1=9.0*10^9 N-m^2/C^2

   U_1= 0

   R=2.75*10^-^1^5m

   K=2.30 MeV

   m= 1.67*10^-^2^7kg

   V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }

    V_1= 3.4*10^7m/s

Therefore the proton must be fired out with a speed of V_1= 3.4*10^7m/s

8 0
3 years ago
A father fashions a swing for his children out of a long rope that he fastens to the limb of a tall tree. As one of the children
trasher [3.6K]

Answer:

The centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

                     

Explanation:

The centripetal acceleration is given by:

a_{c} = \frac{v^{2}}{r}

Where:

v^{2}: is the tangential speed = 9.50 m/s

r: is the distance = 6.00 m

Hence, the centripetal acceleration is:

a_{c} = \frac{v^{2}}{r} = \frac{(9.50 m/s)^{2}}{6.00 m} = 15.04 m/s^{2}

Therefore, the centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

I hope it helps you!

3 0
3 years ago
Read 2 more answers
Hulk starts at -2 m, jumps to -15 m, and settles at 6 m. What is Hulk's displacement?
Lorico [155]

Answer:

Sorry don't know the answer

6 0
3 years ago
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