Question 8 of 25

NISA [10]

Player A needs the least amount of energy. The ball is light weight and she is closest to the goal so the momentum need to kick the ball will be the least and the distance is has to travel is the shortest. But player C needs the most amount of energy. The ball is heavy so it will take the most momentum to move the ball and over such a long distance. Hope this help idrk.

5 0

3 years ago

A soccer ball is traveling at a velocity of 50 m/s. The kinetic energy of the ball is 500 J. What is the mass of the soccer ball

Savatey [412]

A soccer ball is traveling at a velocity of 50 m/s. The kinetic energy of the ball is 500 J.The mass of the soccer ball is 0.4 kgs. This answer is derived from the formula K=1/2 MV^2.So velocity and kinetic energy are given from that mass of the ball is calculated.By substituting the values 500=1/2*M*50*50 which gives M=0.4 Kgs.<span>
</span>

4 0

3 years ago

Read 2 more answers

A motorcyclist heading east through a small town accelerate at constant 4.0meter per seconds square after he leaves the limits.

SVETLANKA909090 [29]

A) The position at t = 2.0 sec is 43.0 m east

B) The position is 55 m east

Explanation:

A)

In order to solve the problem, we take the east direction as positive direction.

We know that:

- at t = 0, the motorcyclist is at a position of $x_0 = 5.0 m$

- at t = 0, the initial velocity of the motorcyclist is $v_0 = 15.0 m$ east

- The acceleration of the motorcyclist is constant and it is $a=4.0 m/s^2$

Since the motion is a uniformly accelerated motion, the position of the motorcylist is given by the expression

$x(t)=x_0 + v_0t + \frac{1}{2}at^2$

where t is the time.

Substituting t = 2.0 s, we find the position:

$x(2.0)=(5.0)+(15)(2.0)+\frac{1}{2}(4.0)(2.0)^2=43 m$

B)

The velocity of the motoryclist can be found by calculating the derivative of the position. Therefore, it is:

$v(t)=x'(t)=v_0 + at$

where:

$v_0=15.0 m/s$ is the initial velocity

$a=4.0 m/s^2$ is the acceleration

We want to find the time t at which the velocity is

v = 25 m/s

Solving the equation for t,

$t=\frac{v-v_0}{a}=\frac{25-15}{4}=2.5 s$

And therefore, the position at t = 2.5 s is:

$x(2.5s)=5.0+(15.0)(2.5)+\frac{1}{2}(4)(2.5)^2=55 m$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

3 0

3 years ago

Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This

vova2212 [387]

Answer:

$1.554\times 10^{32}\ \text{kg}$

Explanation:

M = Mass of each star

T = Time period = 15.5 days

v = Orbital velocity = 230 km/s

G = Gravitational constant = $6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2$

Radius of orbit is given by

$R=\dfrac{vT}{2\pi}$

We have the relation

$\dfrac{Mv^2}{R}=\dfrac{GM^2}{(2R)^2}\\\Rightarrow M=\dfrac{4Rv^2}{G}\\\Rightarrow M=\dfrac{4\dfrac{vT}{2\pi}v^2}{G}\\\Rightarrow M=\dfrac{2v^3T}{\pi G}\\\Rightarrow M=\dfrac{2\times 230000^3\times 15.5\times 24\times 60\times 60}{\pi\times 6.674\times 10^{-11}}\\\Rightarrow M=1.554\times 10^{32}\ \text{kg}$

The mass of each star is $1.554\times 10^{32}\ \text{kg}$

6 0

3 years ago

A mass hanging from a spring oscillates with a period of 0.35 s. Suppose the mass and spring are swung in a horizontal circle, w

Annette [7]

Answer:

66 rpm

Explanation:

The period of oscillation is given by

$T=2\pi \sqrt{\frac {m}{k}}$

$\frac {k}{m}=\frac {4\pi^{2}}{T^{2}}$ where T is time period of oscillation which is given as 0.35 s, k s spring constant and m is the mass of the object attached to the spring.