What type of friction is using chalk in the summer to draw on the ground in Copley square?

The action of property being taken directly from a person or in that person's presence must be an element in which type of crime

grin007 [14]

B, larceny because that's theft of personal property.

5 0

3 years ago

Change the following data into scientific notation.<br><br> 5 000 000 km

RoseWind [281]

To find out scientific notation, you want to make sure that number is less than 10. So do 5.000000, you don't rally need the zeros but I just want to make my point. So use 10^x meaning ten the whatever power adds zeros like 5.000000x10^6 meaning it is increasing it by six zeros moving it out of the decimals and letting become 5,000,000.

8 0

3 years ago

HELPPP WILL MARK BRAINLIST What type of image is formed by a lens if m= 2.7?

Gelneren [198K]

Answer:

the answer is C

Explanation:

So, the formed image will be real and inverted.

6 0

3 years ago

Help please 20 points ? ASAP

KengaRu [80]

Answer:

21: Mass

22: Unbalanced Force

23: Friction

24: Balanced

Explanation:

3 0

3 years ago

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:

$\Delta U = mg \Delta h$