All categories

2 years ago

A 0.450 kg soccer ball has a kinetic energy of 119 J.

Physics

1 answer:

Anastaziya [24]2 years ago

8 0

Answer:

V is approximately = 23m/s

Explanation:

Kinetic energy = ½ mv²

Where m= mass = 0.450kg

V= velocity =?

K. E = 119J

Therefore

K. E = ½ mv²

Input values given

119= ½ × 0.450 × v²

Multiply both sides by 2

119 ×2 = 2 × 1/2 × 0.450 × v²

238= 0.450v²

Divide both sides by 0.450

238/0.450 = 0.450v²/0.450

v² = 528.89

Square root both sides

Sq rt v² = sq rt 528.89

V = 22.998m/s

V is approximately = 23m/s

I hope this was helpful, please rate as brainliest

You might be interested in

Calculate the acceleration of the body of mass 3kg on which a force of 42N has been applied for 5s

____ [38]

F=ma

42/3 = a
a = 14m/s^2

5 0

2 years ago

Calculate the distance the marble travels during the first 3.0 seconds. [Show all work, including the equation and substitution

Alenkinab [10]

D = V0t + 0.5at^2

Where d is the distance

V0 is the initial velocity

A is the acceleration

T is time

From the graph a = 4/3 m/s2

D = 0(3) + 0.5( 4/3 m/s2) ( 3 s)^2

D = 6 m

3 0

2 years ago

A 0.300kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.800m/s when it makes a head-o

e-lub [12.9K]

Answer:

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

0.010935 J

0.0858675 J

Explanation:

$m_1$ = Mass of first glider = 0.3 kg

$m_2$ = Mass of second glider = 0.15 kg

$u_1$ = Initial Velocity of first glider = 0.8 m/s

$u_2$ = Initial Velocity of second glider = 0 m/s

$v_1$ = Final Velocity of first glider

$v_2$ = Final Velocity of second glider

As momentum and Energy is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

${\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}$

From the two equations we get

$v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.3-0.15}{0.3+0.15}\times 0.8+\frac{2\times 0.15}{0.3+0.15}\times 0\\\Rightarrow v_1=0.27\ m/s$

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

$v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.3}{0.3+0.15}\times 0.8+\frac{0.3-0.15}{0.3+0.15}\times 0\\\Rightarrow v_2=1.067\ m/s$

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

Kinetic energy is given by

$K=\frac{1}{2}m_1v_1^2\\\Rightarrow K=\frac{1}{2}0.3\times 0.27^2\\\Rightarrow K=0.010935\ J$

Final kinetic energy of first glider is 0.010935 J

$K=\frac{1}{2}m_2v_2^2\\\Rightarrow K=\frac{1}{2}0.15\times 1.07^2\\\Rightarrow K=0.0858675\ J$

Final kinetic energy of second glider is 0.0858675 J

6 0

2 years ago

Assuming the starting height is 0.0 m, calculate the potential energy of the cart after it has been elevated to a height of 0.5

Bogdan [553]

The potential energy is most often referred to as the "energy at rest" and is dependent on the elevation of an object. This can be calculated through the equation,

E = mgh

where E is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. In this item, we are not given with the mass of the cart so we assume it to be m. The force is therefore,

E = m(9.8 m/s²)(0.5 m) = 4.9m

Hence, the potential energy is equal to 4.9m.

8 0

2 years ago

The graph below shows the sunspot number observed between 1750 and 2000. The graph shows sunspot number on the y axis and years

OlgaM077 [116]

1850 to 1900 because the slope would be 105. It says what is the greatest fall, so the upward slope of 120 wouldn't count.

3 0

2 years ago

Read 2 more answers