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3 years ago

14

a force is applied to a block causing it to accelerate along a horizontal, frictionless surfaces. the energy gained by the block

is equal to the

1 answer:

GaryK [48]3 years ago

3 0

Energy given to the block

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A particle of mass 6.3 x 10-8kg and charge 7.1 muC is traveling due east. It enters perpendicularly a magnetic field whose magni

ASHA 777 [7]

Answer:

t=0.016s

Explanation:

we can use

$\frac{v}{r}=\omega=\frac{qB}{m}=\frac{7.1*10^{-6}C*1.7T}{6.3*10^{-8}kg}=191.58s^{-1}$

the time that the particle is in the magnetic field is one half oa period. Hence

$t=\frac{T}{2}=\frac{\pi}{\omega}=0.016s$

I hope this is useful for you

regards

8 0

3 years ago

We know that charged objects with the same sign repel each other and objects charged with opposite signs attract. What happens w

choli [55]

solution:

When an uncharged conducting object brought near to a charged insulating object there is a force on the conducting object to move the electrons within it to opposite sides of the conductor. That means there is a separation of charges in the conducting object in the presence of the charged insulating object near to it but the charge on the conducting object is neutral.

Thus, the conducting object is uncharged.

There is a force of attraction between the uncharged conducting object and the insulating object when it brought near to the insulating object.

Thus, there is a force on the conducting object.

The conductor remains uncharged and a force is exerted on it.

3 0

3 years ago

1. What net force is required to accelerate a car at a rate of 2 m/s2 if the car has a mass of 3,000 kg?

nataly862011 [7]

Answer:

Net force required to accelerate the car is 6000 N

Explanation:

Force is calculated by the equation, F = Mass × Acceleration

This is based on Newton's Second Law of Motion which states that the force acting on an object is its mass times the acceleration of the object.

Here, mass = 3000 kg and acceleration = 2 m/s²

⇒ Force = Mass × Acceleration

= 3000 × 2 = 6000 N

⇒ F = 6000 N

⇒ M = 3000 kg

⇒ a = 2 m/s²

7 0

3 years ago

An experiment on the earth's magnetic field is being carried out 1.00 m from an electric cable. What is the maximum allowable cu

Whitepunk [10]

Answer:

The allowed current in the cable is 1.15 A.

Explanation:

Given that,

Distance = 1.00 m

Suppose the magnetic field is $2.3\times10^{-5}\ T$ and if the experiment is to be accurate to 1.0 %

We need to calculate the current

Using formula of magnetic field

$B=\dfrac{\mu_{0}I}{2\pi r}$

$I=\dfrac{B\times2\pi r}{\mu_{0}}$

Put the value into the formula

$I=\dfrac{2.3\times10^{-5}\times2\times\pi\times1.00}{4\pi\times10^{-7}}$

$I=115\ A$

If the experiment is to be accurate to 1.0%

Then,

We need to calculate the allowed current in the cable

$I'=(1.0%)\times I$

$I'=\dfrac{1.0}{100}\times115$

$I'=1.15\ A$

Hence, The allowed current in the cable is 1.15 A.

4 0

3 years ago

Male Rana catesbeiana bullfrogs are known for their loud mating call. The call is emitted not by the frog's mouth but by its ear

Sidana [21]

Answer:

The amplitude of the eardrum's oscillation is 6.65×10^-13 m.

Explanation:

Given data:

The sound has a frequency of 262 Hz

The sound level is 84 dB

The air density is 1.21 kg/m^3

The speed of sound is 346 m/s

Solution:

As, Intensity of sound is given by,

I = Io×10^(s/10 db)

I = 2×π^2×ρ×v×f^2×Sm^2

Thus,

Sm = √(Io×10^(s/10 db)) / √( 2×π^2×ρ×v×f^2)

Now, put the values,

Sm = √( 10^-12 × 10^(84/10) ) / √( 2×(3.14)^2×1.21×346×(262)^2 )

Sm = √(2.51×10^-4 / 5.66×10^8)

Sm = √0.443×10^-12

Sm = 6.65×10^-13 m.

8 0

3 years ago