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Stells [14]
2 years ago
5

When a baseball is hit, it travels around 65 mps (meters per second). the mass of the baseball is 0.145 kg. what is the kinetic

energy of the baseball?
Physics
1 answer:
nika2105 [10]2 years ago
4 0

Answer:

306J

Explanation:

The formula for kinetic energy is given as ½mv².

Using this, we get our answer by substituting the values into the formula.

½ × 0.145kg × 65m/s

= 306.31J

can be rounded off to 306J

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I am struggling on this physics question. Brainly is my last hope. Could somebody please provide an answer to this question, wit
Aleks [24]

1) 29.4 N

The force of gravity between two objects is given by:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M and m are the masses of the two objects

r is the separation between the centres of mass of the two objects

In this problem, we have

M=5.97\cdot 10^{24} kg (mass of the Earth)

m=3.0 kg (mass of the box)

r=R=6.37\cdot 10^6 m (Earth's radius, which is also the distance between the centres of mass of the two objects, since the box is located at Earth's surface)

Substituting into the equation, we find F:

F=\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24})(3.0)}{(6.37\cdot 10^6)^2}=29.4 N

2) g=9.8 m/s^2

Let's now calculate the ratio F/m. We have:

F = 29.4 N

m = 3.0 kg

Subsituting, we find

\frac{F}{m}=\frac{29.4}{3.0}=9.8 N/kg = 9.8 m/s^2

This is called acceleration of gravity, and it is the acceleration at which every object falls near the Earth's surface. It is indicated with the symbol g.

We can prove that this is the acceleration of the object: in fact, according to Newton's second law,

F=ma

where a is the acceleration of the object. Re-arranging,

a=\frac{F}{m}

which is exactly equal to the quantity we have calculated above.

5 0
3 years ago
A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien
n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

\phi=NBA

Put the value into the formula

\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2

\phi=7.85\times10^{-7}\ Tm^2

We need to calculate the induced emf

Using formula of induced emf

\epsilon=\dfrac{d\phi}{dt}

Put the value into the formula

\epsilon=\dfrac{7.85\times10^{-7}}{dt}

Put the value of emf from ohm's law

\epsilon =IR

IR=\dfrac{7.85\times10^{-7}}{dt}

Idt=\dfrac{7.85\times10^{-7}}{R}

Idt=\dfrac{7.85\times10^{-7}}{0.50}

Idt=0.00000157=1.57\times10^{-6}\ C

We know that,

Idt=dq

dq=1.57\times10^{-6}\ C

We need to calculate the voltage across the capacitor

Using formula of charge

dq=C dV

dV=\dfrac{dq}{C}

Put the value into the formula

dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}

dV=1.57\ V

Hence, The voltage across the capacitor is 1.57 V.

5 0
3 years ago
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