All categories

3 years ago

An object resistance to any change in its motion is tye _ of the object

Physics

2 answers:

Degger [83]3 years ago

8 0

Answer:

Inertia

Explanation:

Anna [14]3 years ago

7 0

Answer:

inertia

Explanation:

You might be interested in

Does the response of the light bulb depend on how fast you move the bar magnet? if so, how?

Illusion [34]

Yes , the response of the light bulb depend on how fast you move the bar magnet

Flux is the presence of a force field in a specified physical medium, or the flow of energy through a surface

Lenz's law states that the induced electromotive force with different polarities induces a current whose magnetic field opposes the change in magnetic flux through the loop in order to ensure that the original flux is maintained through the loop when current flows in it.

Yes, waving a magnet around does create an electromagnetic wave which does affect the light bulb .

Due to motion of the bar, there will be a constant change in flux and due to Lenz's Law a current within the coil will be induced . This induced current can be used to power the light bulb.

As we know that the greater the speed, the greater the magnitude of the current, and the current is zero when there is no motion.

There will be change in brightness as the bar moves with faster speed.

To learn more about Lenz's Law

brainly.com/question/10048453

#SPJ4

7 0

1 year ago

Light of wavelength 550 nm comes into a thin slit and produces a diffraction pattern on a board 8.0 m away. The first minimum da

FrozenT [24]

Answer:

Width of the slit will be equal to 1.47 mm

Explanation:

We have given wavelength of the light $\lambda =550nm=550\times 10^{-9}m$

Distance D = 8 m

Distance between first minimum dark fringe and the central maximum is 2 mm

So $x=3\times 10^{-3}m$

We have to find the width of the slit

For the first order wavelength is equal to $\lambda =\frac{x}{D}\times a$ , here a width of slit

So $a=\frac{\lambda D}{x}=\frac{550\times 10^{-9}\times 8}{3\times 10^{-3}}=1466.666\times 10^{-6}=1.47mm$

So width of the slit will be equal to 1.47 mm

5 0

3 years ago

What might be collected to monitor the amount of excess nitrogen in the local water system

Debora [2.8K]

Answer:

If excess nitrogen is found in the crop fields, the drainage water can introduce it into ... have aquifers that can supply a lot of freshwater very near the land surface.

4 0

3 years ago

The answer please? I have exam tomo

makkiz [27]

Please be determined and being hardworking person do not rely on the other people to make your problems solved

Explanation:

Ok?

5 0

2 years ago

PLEASE HELP ME 45 POINTS

sergij07 [2.7K]

Answer:

a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

c) The tension force in the cord is 28.02 newtons.

d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is 3.551 meters per second.

Explanation:

a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

b) Both masses are connected to each other by the same cord, the direction of acceleration will be dominated by the mass of greater mass (mass A) and both masses have the same magnitude of acceleration. By the 2nd Newton's Law, we create the following equation of equilibrium:

Mass A

$\Sigma F = T - m_{A}\cdot g = -m_{A}\cdot a$ (1)

Mass B

$\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a$ (2)

Where:

$T$ - Tension force in the cord, measured in newtons.

$m_{A}$ , $m_{B}$ - Masses of blocks A and B, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration of the each block, measured in meters per square second.

By subtracting (2) by (1), we get an expression for the acceleration of each mass:

$m_{B}\cdot a +m_{A}\cdot a = T-m_{B}\cdot g -T + m_{A}\cdot g$

$(m_{B}+m_{A})\cdot a = (m_{A}-m_{B})\cdot g$

$a = \frac{m_{A}-m_{B}}{m_{B}+m_{A}} \cdot g$

If we know that $m_{A} = 5\,kg$ , $m_{B} = 2\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the acceleration of the masses is:

$a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)$

$a = 4.203\,\frac{m}{s^{2}}$

The acceleration of the masses is 4.203 meters per square second.

c) From (2) we get the following expression for the tension force in the cord:

$T = m_{B}\cdot (a+g)$

If we know that $m_{B} = 2\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $a = 4.203\,\frac{m}{s^{2}}$ , then the tension force in the cord:

$T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)$

$T = 28.02\,N$

The tension force in the cord is 28.02 newtons.

d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:

$\Delta y = \frac{1}{2}\cdot a\cdot t^{2}$ (3)