All categories

2 years ago

How are joules measured ?

Physics

2 answers:

saw5 [17]2 years ago

4 0

Answer:You can calculate it by finding the amount of work exerted.

Explanation:

olasank [31]2 years ago

4 0

Answer:

Explanation:

amount of work done

You might be interested in

Which is the phase change in which a substance changes from a gas to liquid?

Aleksandr [31]

Answer:

Condensation

Explanation:

The phase change in which a substance changes from a gas to liquid is Condensation.

There are 3 natural states of matter;

Solids, Liquids, and Gases.

Matter under certain conditions can change from one state to another. When Solids are heated they change into Liquids through the process of Heating. Upon cooling, the Liquids convert to Solid through freezing. When a Liquid such as water is heated, it converts into the Gaseous state through evaporation. Upon cooling, the Gas converts back into the Liquid state through Condensation.

3 0

2 years ago

A load of bricks is being lifted by a crane at a steady velocity of 5.0m/s when one bricks falls off 6.0 m above the ground

Ganezh [65]

LOOK OUT BELOW ! ! !

3 0

2 years ago

A tow rope pulls a 1450 kg truck, giving it an acceleration 1.25 m/s^2. What force does the rope exert?

bagirrra123 [75]

force=mass × acceleration

mass=force ÷ acceleration

acceleration=force ÷ mass

4 0

2 years ago

Suppose a capacitor is fully charged by a battery and then disconnected from the battery. The positive plate has a charge +q and

dimaraw [331]

Answer:-q

Explanation:

Given

Capacitor is charged to a battery and capacitor acquired a charge of q i.e.

+q on Positive Plate and -q on negative Plate.

If the plate area is doubled and the plate separation is reduced to half its initial separation then capacitor becomes four times of initial value because capacitor is given by

$c=\epsilon_0 \cdot \frac{A}{d}$

where A=area of capacitor plate

d=Separation between plates

This change in capacitance changes the Potential such that new charge on the negative plate will remain same -q

8 0

2 years ago

A 6.72 particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive direction,

Romashka [77]

Answer:

$v_{y} = -104 m/s$

Explanation:

Using:

Force = electric field * charge

$F=e*q$

Force = magnitude of charge * velocity * magnetic field * sin tither

$F_{x2}= |q|*v*B*sin \alpha$

Force on particle due to electric field:

$F_{x1}= E*q = (1270N/C)*(-6.72*10^{-6} ) = -8.53*10^{-3}$

Force on particle due to magnetic field:

$F_{x2}= |q|*v*B*sin \alpha = (6.72*10^{-6} )*(1.15)*(sin90)*v = (7.728*10^{-6})*(v)$

$F_{x2}$ is in the positive x direction as $F_{x1}$ is in the negative x direction while net force is in the positive x direction.

Magnetic field is in the positive Z direction, net force is in the positive x direction.

According to right hand rule, Force acting on particle is perpendicular to the direction of magnetic field and velocity of particle. This would mean the force is along the y-axis. As this is a negatively charged particle, the direction of the velocity of the particle is reversed. Therefore velocity of particle, v, has to be in the negative y direction.

Now,

$F_{xnet}- F_{x1 } = F_{x2 }$

$(6.13*10^{-3}) - (8.53*10^{-3} ) = (7.728*10^{-6})*(v)$

$v = (F_{xnet} - F_{x1}) / (F_{x2} )$

$=((6.13*10^{-3} ) - (8.53*10^{-3})) / (7.728*10^{-6})$

$= (- 104.25) m/s$

$v_{y} = -104 m/s$

8 0

2 years ago