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1 year ago

How much work needs to be done to launch a baseball (145 g) at a speed of 108 MPH? The unit is Joule.

Physics

2 answers:

nalin [4]1 year ago

8 0

The work done in launching the baseball is 169 J.

<h3>What is work-energy theorem ?</h3>

The work-energy theorem states that, the net work done on a system is equal to the change in kinetic energy of the system.

Here,

Mass of the baseball, m = 145g

Speed of the baseball, v = 108 Mph = 48.28 m/s

According to work-energy theorem,

Work done, W = Kinetic energy

W = 1/2 mv²

W = 1/2 x 145 x 10⁻³ x 48.28²

W = 169 J

Hence,

The work done in launching the baseball is 169 J.

To learn more about work-energy theorem, click:

brainly.com/question/30759174

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sergiy2304 [10]1 year ago

6 0

Answer:

Given

m=145g=0.145kg (cause mass has unit kg)

v=108mph=48.28m/s

Find the work need

W=K (Work-Energy theorem)

K=1/2•m•v²

K=1/2•0.145•48.28²

K=168.99J

So, W=K=168.99J

Please give Brainliest

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To balance a chemical equation we use______. <br> -coefficients <br> -subscripts

Vitek1552 [10]

Answer:

Coefficients

Explanation:

Coefficients are used to balance chemical equations.
According to the law of conservation of mass, the mass of the reactants in a chemical equation should be equivalent to the mass of the products.
The conservation of mass is achieved in chemical equations is achieved through balancing chemical equations.
<em><u>Balancing chemical equations is a try and error of putting appropriate coefficients to the reactants or products to ensure that the number of atoms of each element are equal on the side of the reactants and that of products. </u></em>

3 0

3 years ago

The modernization of tennis occurred in which<br> country?

Kobotan [32]

Answer:

The International Lawn Tennis Federation, now known simply as the International Tennis Federation, the sport's governing body, was founded in 1913, composed of 13 national tennis associations. The Modern Olympics , growing out of the ancient tradition, resurfaced under the direction of the International Olympic Committee in Athens in 1896.

Explanation:

8 0

3 years ago

A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav

gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

$K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2$ (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

$v^2=v_o^2+2gy$

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}$

With this value you can find the spring constant k from the equation (1):

$mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}$

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

$W=F_e\\\\mg=kx$

you solve the last expression for x:

$x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m$

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}$

Next, you calculate x from the equation (1):

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m$

The net fire is stretched 2.72 m

5 0

3 years ago

Two objects attract each other gravitationally. If the mass of each object doubles, how does the gravitational force between the

madam [21]

Answer:

The gravitational force between them quadruples

Explanation:

According to law of gravitation, the force of attraction (F) between two masses m1 and m2 is directly proportional to the product of the masses and inversely proportional to the square of the distance(r) between them. Mathematically,

F1 = Gm1m2/r²... 1

If their masses doubles, the formula becomes;

F2 = G(2m1)(2m2)/r²

F2 = 4Gm1m2/r² ... 2

Dividing equation 2 by 1, we have;

F2/F1 = {4Gm1m2/r²}÷{Gm1m2/r²}

F2/F1 = 4Gm1m2/r²×r²/Gm1m2

F2/F1 = 4

F2 = 4F1

The gravitational force between the masses when they doubles quadruples.

6 0

4 years ago

A neutron star and a black hole are 3.34 x 1012 m from each other at a certain point in their orbit. The neutron star has a mass

m_a_m_a [10]

Answer:

F=1.65 x 10²⁶ N

Explanation:

Given that

Distance ,R= 3.34 x 10¹² m

Mass m₁= 2.78 x 10³⁰ kg

Mass ,m₂= 9.94 x 10³⁰ kg

we know that gravitational force F given as

$F=G\dfrac{m_1m_2}{R^2}$

G=Constant

G=6.67 x 10⁻¹¹ Nm²/kg²

Now by putting the values

$F=6.67\times 10^{-11}\times \dfrac{2.78\times 10^{30}\times 9.94\times 10^{30}}{(3.34\times 10^{12})^2}\ N$

F=1.65 x 10²⁶ N

Therefore the force between these two mass will be 1.65 x 10²⁶ N.

5 0

3 years ago