Ask question

Ask question

All categories

3 years ago

13

A child riding a bicycle at 5 meters per seconds accelerates at 2.0 meters per second for 4.0 seconds. What is the child's speed

at the end of this 4.0 second interval

1 answer:

slega [8]3 years ago

5 0

Explanation:

Using Kinematics,

We have v = u + at.

=> v = (5m/s) + (2.0m/s²)(4.0s)

=> v = 13m/s.

Hence the child's speed is now 13m/s.

You might be interested in

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$

where

m = 950 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

$\Delta h$ is the change in height, which is

$\Delta h = L sin 30^{\circ}$

where L = 1250 m is the total distance covered.

Substituting, we find the energy transferred:

$\Delta U = mg L sin \theta = (950)(9.8)(1250)(sin 30^{\circ})=5.82\cdot 10^6 J$

8 0

3 years ago

What is the magnitude of an electric field that balances the weight of a plastic sphere of mass 2.1 g that has been charged to -

Liula [17]

Answer:

Electric field, $E=6.86\times 10^6\ N/C$

Explanation:

It is given that,

Mass of sphere, m = 2.1 g = 0.0021 kg

Charge, $q=-3\ nC=-3\times 10^{-9}\ C$

We need to find the magnitude of electric field that balances the weight of a plastic spheres. So,

$ma=qE$

a = g

$E=\dfrac{mg}{q}$

$E=\dfrac{0.0021\ kg\times 9.8\ m/s^2}{-3\times 10^{-9}\ C}$

$E=6860000\ N/C$

or

$E=6.86\times 10^6\ N/C$

Hence, the magnitude of electric field that balances its weight is $6.86\times 10^6\ N/C$ . Hence, this is the required solution.

4 0

3 years ago

the speed of travel of the moon around the earth, using the formula for the speed of a moving object in a circular path

Svetach [21]

Answer: 1018.26 m/s

Explanation:

Approaching the orbit of the Moon around the Earth to a circular orbit (or circular path), we can use the equation of the speed of an object with uniform circular motion:

$V=\sqrt{G\frac{M}{r}}$

Where:

$V$ is the speed of travel of the Moon around the Earth

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24} kg$ is the mass of the Earth

$r=384400(10)^{3} m$ is the distance from the center of the Earth to the center of the Moon

Solving:

$V=\sqrt{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{5.972(10)^{24} kg}{384400(10)^{3} m}}$

$V=1018.26 m/s$ This is the speed of travel of the Moon around the Earth

5 0

3 years ago

1. What function does the skull serve for the skeleton?

Free_Kalibri [48]

1.c
2.b
3.a
4.c
5.b
I hop this helps

8 0

3 years ago

Our eyes can see light with an angular resolution of about 1’—equivalent to about a third of a millimeter at arm’s length. Suppo

olya-2409 [2.1K]

Answer:

We would not be able to make our way around the earth's surface, to read, to sculpt or to create technology because we cannot see!

Explanation:

The minimum angular separation that can be distinguished by an eye gives the angular resolution of the eye.

Given that the Angular resolution with infrared radiation is = $1.0^0$ equal to $60'$

This resolution is very much greater than that of the eye $(1')$

The angular resolution that our eyes can see is about $\frac{1}{3}mm$ at arms length

Angular resolution of infrared = $\frac{1}{3} * 60 = 20mm$ at arms length

We therefore cannot read, sculpt or create technology because we cannot see.

7 0

3 years ago