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2 years ago

13

What is the weight of the body under the influence of a force of 10N reaches a speed of 1.5 m / s in 3s?

1 answer:

MariettaO [177]2 years ago

8 0

Answer:

Explanation:

F=ma

F=10N

A=v-u upon t

=1.5-0 upon 3

=0.5m per second squared

Then do force divided by. Accelaration and multiply by gravity.

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A force of 20N changes the position of a body. If mass of the body is 2kg, find the acceleration produced in the body.2. A ball

shepuryov [24]

Explanation:

<em>Hello</em><em> </em><em>there</em><em>!</em><em>!</em><em>!</em>

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3 years ago

The shifting of the observed wavelength of light due to the motion of the source toward or away from the observer is called the

alexdok [17]

Answer:

doppler effect

Explanation:

When the relative motion of two bodies results in the wavelength becoming shorter this means that the bodies are getting closer. This is known as blue shift.

When the relative motion of two bodies results in the wavelength becoming longer this means that the bodies are getting farther. This is known as red shift.

Collectively this phenomenon is known as the Doppler effect.

7 0

3 years ago

A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li

Kitty [74]

Answer:

$E_{net} = 6.44 \times 10^5 N/C$

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

$E = \frac{2k \lambda}{r}$

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

$E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}$

now plug in all data

$\lambda_1 = 4.68 \muC/m$

$\lambda_2 = 2.48 \mu C/m$

$r = 0.200 m$

now we have

$E_{net} = \frac{2k}{r}(4.68 + 2.48)$

$E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})$

$E_{net} = 6.44 \times 10^5 N/C$

8 0

2 years ago

A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur

Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

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3 years ago

An empty parallel plate capacitor is connected between the terminals of a 18.8-V battery and charges up. The capacitor is then d

Basile [38]

Answer:

$p.d' = 37.6 V$

Explanation:

From the question we are told that:

Potential difference $p.d=18.8V$

New Capacitor $C_1=C_2/2$

Generally the equation for Capacitor capacitance is mathematically given by

$C=\frac{eA}{d}$

Generally the equation for New p.d' is mathematically given by

$C_2V=C_1*p.d'$

$p.d' = 2V$

$p.d'= 2 * 18.8$

$p.d' = 37.6 V$

7 0

3 years ago