All categories

2 years ago

A 1,000 kg car is driving on a 15 m high bridge at 5 m/s. What is the kinetic energy of the car?

Physics

1 answer:

yanalaym [24]2 years ago

7 0

Answer:

$KE=12,500J$

Explanation:

The formula for kinetic energy is:

$KE = \frac{1}{2}mv^2$

We can plug in the given values into the equation:

$KE = \frac{1}{2}*1000kg*(5m/s)^2$

$KE = 500kg*25m^2/s^2$

$KE=12,500J$

You might be interested in

A car of mass 1600 kg can just be lifted what is the least force that the electronicmagnet must use to lift the car ?

Mkey [24]

The car's mass is 1600 kg.

Its weight is (mass) x (gravity).

On Earth, that's (1600 kg) x (9.8 m/s²) = 15,680 Newtons.

At the moment, that's the only force acting on the car, directed downward and provided by gravity.

If you want to lift the car, then the net force has to be directed upward, and must either exactly cancel or exceed the force of gravity.

So the minimum force required to lift the car is <em>15,680 Newtons</em>, directed vertically upward.

5 0

3 years ago

A conveyor belt is used to move sand from one place to another in a factory. The conveyor is tilted at an angle of 18° above the

brilliants [131]

Answer:

x = 2.044 m

Explanation:

given data

initial vertical component of velocity = Vy = 2sin18

initial horizontal component of velocity = Vx = 2cos18

distance from the ground yo = 5m

ground distance y = 0

from equation of motion

$y = yo+ V_y t +\frac{1}{2}gt^2$

$0 = 5 + 2sin18+ \frac{1}{2}*9.8t^2$

solving for t

t = 1.075 sec

for horizontal motion

$x = V_x t$

x = 2cos18*1.075

x = 2.044 m

8 0

3 years ago

4. The Earth is the only planet that contains storms. Tor F 5. Lightning is a form of energy made up of protons that move from c

Mumz [18]

Answer:

4. F

5. F

Explanation:

jupiter constantly has hurricanes and uranus has storms often too.

lightning can move many ways including from cloud to cloud and even from ground ton sky (seriously look it up it's cool)

6 0

3 years ago

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

2 years ago

True or false. there would be no like in earth without the sun

alisha [4.7K]

hello!

True

have a goed day

6 0

3 years ago