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3 years ago

Qwertyuiop hhffhhhhf fiuefhasuhs afbauyhsbvzskhbdv

Physics

2 answers:

Crank3 years ago

8 0

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mars1129 [50]3 years ago

4 0

Answer:

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Ray Of Light [21]

Acceleration = Force \ mass

0,375N/0,60kg=0.6ms-2

4 0

2 years ago

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

One advantage to using this form of circuit is that A) voltage available to each light remains constant regardless of the number

7nadin3 [17]

I believe the answer is A Voltage available to each light remains constant regardless of the number of lights that are on.

3 0

3 years ago

Read 2 more answers

If A is the set of even integers less than 10, and B is the set of multiples of 3 less than 20, what is the union of sets A and

tatyana61 [14]

Answer:

(D)

Explanation:

A(2,3,4,6,8)

B(6,9,12,15,18)

(AuB)=(2,3,4,6,8,9,12,15,18)

5 0

2 years ago

A water balloon is thrown at an angle of 30° with a velocity of 40 m/s. How long do you have to move out of the way if the ballo

boyakko [2]

Answer:

t=0.58s

Explanation:

We can consider only the horizontal component since the horizontal component of the velocity is constant ( $v_x=vcos\theta$ ), and we want to know how much time it takes for the balloon to travel a horizontal distance $d_x$ at that speed. The definition of (horizontal) velocity is $v_x=\frac{d_x}{t}$ , so we have:

$t=\frac{d_x}{v_x}=\frac{d_x}{vcos\theta}=\frac{(20m)}{(40m/s)cos(30^{\circ})}=0.58s$

6 0

3 years ago