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gogolik [260]
3 years ago
9

You throw a baseball with a mass of 0.5 kg. The ball leaves your hand with a speed of 35 m/s. Calculate the kinetic energy. (SHO

W ALL WORK)
Physics
1 answer:
mixas84 [53]3 years ago
6 0

Answer:

The kinetic energy of the baseball is 306.25 joules.

Explanation:

SInce the baseball can be considered a particle, that is, that effects from geometry can be neglected, the kinetic energy (K), in joules, is entirely translational, whose formula is:

K = \frac{1}{2}\cdot m\cdot v^{2} (1)

Where:

m - Mass, in kilograms.

v - Speed, in meters per second.

If we know that m = 0.5\,kg and v = 35\,\frac{m}{s}, then the kinetic energy of the baseball thrown by the player is:

K = \frac{1}{2}\cdot m \cdot v^{2}

K = 306.25\,J

The kinetic energy of the baseball is 306.25 joules.

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In a Broadway performance, an 84.5-kg actor swings from a R = 4.30-m-long cable that is horizontal when he starts. At the bottom
Arada [10]

Answer:

1.57772 m

Explanation:

M = Mass of actor = 84.5 kg

m = Mass of costar = 55 kg

v = Velocity of costar

V  = Velocity of actor

h_i = Intial height of actor = 4.3 m

g = Acceleration due to gravity = 9.81 m/s²

As the energy of the system is conserved

\frac{1}{2}MV^2=Mgh_i\\\Rightarrow V=\sqrt{2gh_i}\\\Rightarrow V=\sqrt{2\times 9.81\times 4.3}\\\Rightarrow V=9.18509\ m/s

As the linear momentum is conserved

MV=(m+M)v\\\Rightarrow v=\frac{MV}{m+M}\\\Rightarrow V=\frac{84.5\times 9.18509}{84.5+55}\\\Rightarrow v=5.56372\ m/s

Applying conservation of energy again

\frac{1}{2}(m+M)v^2=(m+M)gh_f\\\Rightarrow h_f=\frac{v^2}{2g}\\\Rightarrow h_f=\frac{5.56372^2}{2\times 9.81}\\\Rightarrow h_f=1.57772\ m

The maximum height they reach is 1.57772 m

3 0
3 years ago
What happens to a circuit's resistance (R), voltage (V), and current (1) when
hammer [34]

Answer:

D.

R increases

V is constant

I decreases

Explanation:

The resistance of a wire is given by the following formula:

R = \frac{(Resistivity)(L)}{A}

It is clear from this formula that resistance is directly proportional to the length of wire. So, when length of wire is increased, <u>the resistance of circuit increases</u>.

The <u>voltage in the circuit will be constant</u> as the voltage source remains same and it is not changed.

Now, we can use Ohm Law:

V = IR

at constant V:

I ∝ 1/R

it means that current is inversely proportional to resistance. Hence, the increase of resistance causes <u>the current in circuit to decrease.</u>

Therefore, the correct option will be:

<u>D.</u>

<u>R increases </u>

<u>V is constant </u>

<u>I decreases</u>

6 0
3 years ago
You are playing right field for the baseball team. Your team is up by one run in the botton of the last inning of the game when
VARVARA [1.3K]

Answer:

The correct response is "No". The further explanation is given below.

Explanation:

The given values are:

Angle

\theta = 30^{\circ}

Distance

d = 20 m

Speed

s = 8.0 m/s

Now,

⇒  y(t) = \frac{1}{2}\times a\times t^2 + v0\times t\times (sin \theta)

    0 = -4.9t^2 + v0\times t\times (sin 30)

    x = v0\times (cos \theta)\times t

    65 = v0\times (cos 30)\times t

    v0\times t = \frac{65}{Cos30}

    0 = -4.9t^2 + 65\times \frac{sin30}{cos30}

    t = 2.767 \ sec

So,

⇒  d = r\times t

    20 = 8\times t

      t=2.5 \ sec

Therefore, 2.5 < 2.767, so it won't get there.

6 0
3 years ago
A point charge of +3.0 X 10-7 coulomb is placed 2.0 X 10-2 meter from a second point charge of +4.0 X 10-7 coulomb. What is the
Mars2501 [29]

Answer:

D. 2.7 N

Explanation:

Applying

F = kq'q/r²................ Equation 1

Where F = force, k = coulomb's constant, q' = first charge, q = second charge, r = distance between the charge

From the question,

Given: q' = +3.0×10⁻⁷ C, q = +4.0×10⁻⁷C, r = 2.0×10⁻² m

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values into equation 1

F = ( +3.0×10⁻⁷)(+4.0×10⁻⁷)(8.98×10⁹)/(2.0×10⁻²)²

F = 26.94×10⁻¹ N

F = 2.694 N

F ≈ 2.7 N

7 0
3 years ago
10-10x10+10 ..............​
cestrela7 [59]

Answer:

10-10x10+10=10-100+10=20-100=-80

4 0
3 years ago
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