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Juli2301 [7.4K]

3 years ago

13

Clois what is the weight of a body in the earth, if its weig is 5Nin moon?

1 answer:

Debora [2.8K]3 years ago

8 0

Explanation:

because the moon has less mass than earth, the force due to gravity at the lunar surface is only about 1/6 that on earthso,the weight of a body on earth is 6×5N =30N

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A police officer is using her radar to check speeds. She is moving south at 50 mph. You are moving

Luda [366]

Answer:

the relative speed should be 40mph

Explanation:

8 0

3 years ago

A car starts to move from rest and covers a distance of 360m in one minute. Calculate the acceleration of the car.

Romashka-Z-Leto [24]

<h2>The acceleration of car is 0.2 ms⁻²</h2>

Explanation:

When the car moves , the distance covered is calculated by the relation

S = u t + $\frac{1}{2}$ a t²

In this question u = 0 , because car was at rest initially

Thus S = $\frac{1}{2}$ a t²

here S is displacement and a is the acceleration of car

Therefore 360 = $\frac{1}{2}$ a ( 60 )²

Because time taken is one minute or 60 seconds

Therefore a = $\frac{360x2}{3600}$

or a = 0.2 m s⁻²

4 0

3 years ago

Which statement is the best example of pseudoscience?

Nady [450]

<span>pseudoscience includes beliefs theories, or practices that have been or are considered scientific, but have no basis in scientific fact.
This could mean the were disproved scientifically, can't be tested or lack evidence to support them.
Examples:Channeling- involves communicating with a spirit through a person

Astrology- beliefs that humans are affected by the position of celestial bodies
</span>

5 0

3 years ago

A circular loop of wire has radius of 9.50 cmcm. A sinusoidal electromagnetic plane wave traveling in air passes through the loo

m_a_m_a [10]

Answer:

The induced emf is $\epsilon = 0.1041 \ V$

Explanation:

From the question we are told that

The radius of the circular loop is $r = 9.50 \ cm = 0.095 \ m$

The intensity of the wave is $I = 0.0215 \ W/m^2$

The wavelength is $\lambda = 6.90\ m$

Generally the intensity is mathematically represented as

$I = \frac{ c * B^2 }{ 2 * \mu_o }$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4 \pi *10^{-7} N/A^2$

B is the magnetic field which can be mathematically represented from the equation as

$B = \sqrt{ \frac{ 2 * \mu_o * I }{ c} }$

substituting values

$B = \sqrt{ \frac{ 2 * 4\pi *10^{-7} * 0.0215 }{ 3.0*10^{8}} }$

$B = 1.342 *10^{-8} \ T$

The area is mathematically represented as

$A = \pi r^2$

substituting values

$A = 3.142 * (0.095)^2$

$A = 0.0284$

The angular velocity is mathematically represented as

$w = 2 * \pi * \frac{c}{\lambda }$

substituting values

$w = 2 * 3.142 * \frac{3.0*10^{8}}{ 6.90 }$

$w = 2.732 *10^{8} rad \ s^{-1}$

Generally the induced emf is mathematically represented as

$\epsilon = N * B * A * w * sin (wt )$

At maximum induced emf $sin (wt) = 1$

So

$\epsilon = N * B * A * w$

substituting values

$\epsilon = 1 * 1.342 *10^{-8} * 0.0284 *2.732 *10^{8}$

$\epsilon = 0.1041 \ V$

7 0

3 years ago

Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin

Arte-miy333 [17]

Answer:

$x=\dfrac{r}{\sqrt2}$

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

$E=K\dfrac{Q}{(r^2+x^2)^{3/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$E=K{Q}{(r^2+x^2)^{-3/2}}$

$\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0$

$r^2+x^2-3x^2=0$

$x=\dfrac{r}{\sqrt2}$

At $x=\dfrac{r}{\sqrt2}$ the electric field will be maximum.

3 0

3 years ago