All categories

3 years ago

Calculate the kinetic energy of a motorcycle of mass 60kg travelling at a velocity of 40km/h

Physics

1 answer:

ELEN [110]3 years ago

5 0

Answer:

1848.15J

Explanation:

KE =1/2 mv^2

Mass = 60kg, velocity =40km/h =11.11m/s

Hence

KE =30 x(11.1)^2 /2 = 1848.15J

You might be interested in

PLZ HELP ME FAST A relationship between two variables is called:

Irina18 [472]

Answer:

B- Correlation

Explanation:

6 0

2 years ago

Read 2 more answers

A ball is dropped from a height of 20 meters. At what height does the ball have a velocity of 10 meters/second?

borishaifa [10]

Answer:B

Explanation:

Initial velocity, u=0m/s

Distance,s=20m

a=+g=9.8m/s*s

Using v*v=u*u+2gs

v*v=0+2*9.8*20

v*v=392

v=19.8

When s=20m, v = 19.8m/s

Therefore when v = 10m/s, s= 10*20/19.8

s =10.1m

6 0

3 years ago

A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

5 0

3 years ago

You do 120 j of work while pulling your sister back on a swing, whose chain is 5.10 m long. you start with the swing hanging ver

Goryan [66]

The work done to pull the sister back on the swing is equal to the increase in potential energy of the sister:
$W= \Delta U = mg \Delta h$ (1)

where m is the sister's mass, g is the gravitational acceleration and $\Delta h$ is the increase in altitude of the sister with respect to its initial position.

By calling $\theta$ the angle of the chain with respect to the vertical, the increase in altitude is given by
$\Delta h = L - L \cos \theta = L(1 - \cos \theta)$ (2)
where L is the length of the chain.

Putting (2) inside (1), we find
$W= m g L (1 - \cos \theta)$
from which we can find the mass of the sister:
$m = \frac{W}{g L (1 - \cos \theta)} = \frac{120 J}{(9.81 m/s^2)(5.10 m)(1- \cos 32.0^{\circ})} =15.8 kg$

5 0

2 years ago

On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the

Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

$\frac{F_E}{F_G}=\frac{qE}{mg}$ (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

$\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764$

Then, the gravitational force is 0.764 times the electric force on the particle

The acceleration of the particle is obtained by using the second Newton law:

$F_E-F_G=ma\\\\a=\frac{qE-mg}{m}$

you replace the values of all variables:

$a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}$

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

7 0

2 years ago

Calculate the kinetic energy of a motorcycle of mass 60kg travelling at a velocity of 40km/h​

Calculate the kinetic energy of a motorcycle of mass 60kg travelling at a velocity of 40km/h