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3 years ago

What is the kinetic energy of a soccer ball which has a mass of 0.8 kg and is kicked at a velocity of 10 m/s

Physics

1 answer:

Lubov Fominskaja [6]3 years ago

5 0

Answer:

Explanation:

Kinetic energy= K.E = (1/2) × mv²

K.E = 0.5× 0.8kg× 100m²/s² = 40 N

You might be interested in

What is the electrical force between q2 and q3? Recall that k = 8. 99 × 109 N•meters squared over Coulombs squared. 1. 0 × 1011

nlexa [21]

Force on the particle is defined as the application of the force field of one particle on another particle. the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.

<h3>What is electric force?</h3>

Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.

The electric force in the second case will be the same as in the first case. Therefore the force on the particle will be the same.

$\rm F= K\frac{q_2q_3}{r^2}$

$\rm F= 9\times 10^9 \times \frac{1.6 \times 10^{-13}\times 1.6\times10^{-13}}{(0.5)^2}$

$\rm F= - 1. 1 \times 10^{11 }N$

Hence the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.

To learn more about the electric force refer to the link;

brainly.com/question/1076352

4 0

2 years ago

In order to calculate momentum we must have the object's

Vlad [161]

You need to have the Mass and velocity

5 0

3 years ago

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

DiKsa [7]

Answer:0.061

Explanation:

Given

$T_C=300 k$

Temperature of soup $T_H=340 K$

heat capacity of soup $c_v=33 J/K$

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

$\eta =\frac{dW}{Q}=1-\frac{T_C}{T}$

$dW=Q(1-\frac{T_C}{T})$

$dW=c_v(1-\frac{T_C}{T})dT$

integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

$W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT$

$W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}$

$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

$=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}$

$=\frac{300-340-300\ln (\frac{300}{340})}{300-340}$

$=\frac{-40+37.548}{-40}$

$=0.061$

4 0

3 years ago

how long does it take adrian to reach his destination if his displacement is 253 meters and he walks north with a velocity of 1.

klemol [59]

Velocity = displacement/time

1.6 = 253/t

t = 158.125

It takes them about 158 seconds

4 0

3 years ago

Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(

bezimeni [28]

Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

length of the rope = 1.33 m

mass of the rope = 3.31 g

comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

A is amplitude

now on comparing

a) Amplitude = 2.20 mm

b) frequency =

$f = \dfrac{\omega}{2\pi}$

$f = \dfrac{743}{2\pi}$

f = 118.25 Hz

c) wavelength

$k= \dfrac{2\pi}{\lambda}$

$\lambda= \dfrac{2\pi}{k}$

$\lambda= \dfrac{2\pi}{7.02}$

$\lambda= 0.895\ m$

d) speed

$v = \dfrac{\omega}{k}$

$v = \dfrac{743}{7.02}$

v = 105.84 m/s

e) direction of the motion will be in negative x-direction

f) tension

$T = \dfrac{v^2\ m}{L}$

$T = \dfrac{(105.84)^2\times 3.31 \times 10^{-3}}{1.33}$

T = 27.87 N

g) Power transmitted by the wave

$P = \dfrac{1}{2}m\ v \omega^2\ A^2$

$P = \dfrac{1}{2}\times 0.00331\times 105.84\times 743^2\ 0.0022^2$

P = 0.438 W

5 0

3 years ago