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Ostrovityanka [42]
3 years ago
5

A standard baseball has a mass of 144.3 g. Determine the weight, in newtons, of a baseball.

Physics
1 answer:
alex41 [277]3 years ago
7 0
Fg = ma
Fg = (0.1143 kg) (9.81 N/kg)
Fg = 1.12 N
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Why do you think the temperature does not change much during a phase change?
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It depends on where the temperature is dropping, in which body so to speak. Generally, the temperature adapts to the two bodies, for example if a hot piece of metal meets a cold one, the two will continue until they are at an equal temperature, an intermediate temperature.

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What best describes the dropping height of a ball that bounced back up to a height of 45 centimeters?
murzikaleks [220]

Answer:C:Less than 45 centimeters, as the ball transforms some of its potential energy into thermal energy and sound energy

Less than 45 centimeters, as the ball transforms some of its potential energy into thermal energy and sound energy.

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Every time the ball bounces, its height will be less than its previous height.

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3 years ago
6.6 kg block initially at rest is pulled to theright along a horizontal, frictionless surfaceby a constant, horizontal force of
neonofarm [45]

Answer:

v = 3.04 m/s

Explanation:

given,

mass of the block, M = 6.6 Kg

horizontal force, F = 12.2 N

distance, L = 2.5 m

initial speed  = 0 m/s

speed of the block,v = ?

we now

Work done is equal to change in Kinetic energy.

Work done = Force x displacement

W = Δ K E

Δ K E = Force x displacement

\dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2= F .s

\dfrac{1}{2}\times 6.6 \times v^2 - 0= 12.2\times 2.5

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3 years ago
An astronaut lands on a new, recently discovered planet in a different star system. The astronaut measures the acceleration due
Nezavi [6.7K]

Answer:

The radius of the new planet is ~2.04 * 10⁶ m, or 2,041,752 m.

Explanation:

We can use Newton's Law of Universal Gravitation:

  • \displaystyle F_g=G\frac{Mm}{r^2}

Let's look at Newton's 2nd Law:

  • F=ma

We can set these equations equal to each other:

  • \displaystyle G\frac{Mm}{r^2} =ma

The mass of the second mass (astronaut) cancels out. We are left with:

  • \displaystyle G\frac{M}{r^2} =a

We are solving for the radius of the new planet, so we can rearrange the equation:

  • \displaystyle r=\sqrt{\frac{GM}{a} }

Substitute in our known values given in the problem (<u><em>G = 6.67 * 10⁻¹¹ </em></u><em> ; </em><u><em>M = 7.5 * 10²³</em></u><em> ; </em><u><em>a = 12</em></u>).

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The radius of the new planet is ~2.04 * 10⁶ m.

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