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Mariulka [41]
3 years ago
13

4. Which of the following statement is correct regarding velocity and speed of a moving body?

Physics
2 answers:
Jlenok [28]3 years ago
8 0

Answer:

d

Explanation:

Velocity is the same as speed except it's a vector quantity which means it has a direction.

lara31 [8.8K]3 years ago
7 0

Answer:

Explanation:

Hello friend!!!

The correct option is <u><em>(d) Velocity of a moving body is its speed in a given direction.</em></u>

Hope this helps

plz mark as brainliest!!!!!!!!

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Show that (a)KE=1/2mv2
evablogger [386]

Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{see \: below}}}}}

Explanation:

\underline{ \bold{ \sf{To \: prove \: that \: kinetic \: energy =  \frac{1}{2} m {v}^{2} }}}

Let us consider, a body of mass ' m ' is lying at rest ( initial velocity = 0 ) on a smooth surface. Let a constant force F displaces this body in its own direction by a displacement ' d '. Let 'v' be it's final velocity. The work done ' W ' by the force is given by :

\sf{W = FD}

⇒\sf{W = m \:  \times a \:  \times s} \:  \:  \:  \:  \:  \:  \:  \:  \: ( \: ∴ \: f \:  =  \: ma \: ; \: s \:  = d)

⇒\sf{W = m \:  \times  \frac{v - u}{t}  \times  \frac{u + v}{2}  \times t \:  \:  \:  \:  \:  \:  \:  \:  \: (∴ \: a =   \frac{v - u}{t} and \: s =  \frac{u + v}{2}  \times t}

⇒\sf{W = m \times  \frac{ {v}^{2}  -  {u}^{2} }{2} }

⇒\sf{W =  \frac{1}{2} m {v}^{2}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: (since, \: initial \: velocity(u) = 0)}

The work done becomes the kinetic energy of the body. Thus, the kinetic energy of a body of mass ' m : moving with the velocity equal to 'v ' is 1 / 2 mv²

∴ \sf{KE=  \frac{1}{2} m {v}^{2} }

\sf{ \underline{ \bold{  {proved}}}}

Hope I helped!

Best regards!!

5 0
3 years ago
Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25
mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

4 0
3 years ago
CAN SOMEONE PLZ HELP 5,6,7,8,9 plzzz show work too I’m not understanding this much appreciated
soldi70 [24.7K]

#5 is supposed to have a little graph showing for each choice. They're not there. Nothing to choose from.

6-C

7-A

8-B

9-D

7 0
4 years ago
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