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3 years ago

What is the final velocity of a body if it is moving with 13 m/s in 300 seconds and its acceleration is 30 m/s 2.

Physics

1 answer:

Nutka1998 [239]3 years ago

6 0

Answer:

9013m/s

Explanation:

acceleration= v- u/ t

=> at = v-u

=> v = at + u

=> v =30*300+13

= 9013m/s

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If it were possible to remove gravity and friction, think about what would happen to a football if it were tossed into the air.

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Ignoring fluid resistance, football will <span>maintain a constant speed until other forces accelerate the football.</span>

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3 years ago

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Constants A capacitor is connected across an ac source that has voltage amplitude 59 0 V and frequency 77 0 Hz Part C What is th

Vladimir79 [104]

Answer:

$C = 1.77 \times 10^{-4} F$

Explanation:

As we know that in AC circuit we have

$V = i x_c$

here we have

V = 59 V

i = 5.05 A

so we will have

$x_c = \frac{59}{5.05}$

$x_c = 11.68 ohm$

also we know that

$x_c = \frac{1}{\omega C}$

here we will have

$11.68 = \frac{1}{(2\pi 77)C}$

$C = 1.77 \times 10^{-4} F$

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3 years ago

The police department is excited to have some new motorcycle units. One officer said that these motorcycles can go from 0 miles

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The officer is describing the motorcycle's acceleration

That is, he's describing how the motorcycle's velocity changes when compared to the time. That's the formula for calculating the average acceleration of any body.

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3 years ago

A mass of 0.5 kg hangs motionless from a vertical spring whose length is 1.10 m and whose unstretched length is 0.50 m. Next the

ser-zykov [4K]

Answer:

The maximum length during the motion is $L_{max} = 1.45m$

Explanation:

From the question we are told that

The mass is $m =0.5 kg$

The vertical spring length is $L = 1.10m$

The unstretched length is $L_{un} = 1.30m$

The initial speed is $v_i = 1.3m/s$

The new length of the spring $L_{new} = 1.30 m$

The spring constant k is mathematically represented as

$k = -\frac{F}{y}$

Where F is the force applied $= m * g = 0.5 * 9.8=4.9N$

y is the difference in weight which is $=1.10-0.50=0.6m$

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

Now substituting values accordingly

$k = \frac{4.9}{0.6}$

$= 8.17 N/m$

The elastic potential energy is given as $E_{PE} = \frac{1}{2} k D^2$

where D is this the is the displacement

Since Energy is conserved the total elastic potential energy would be

$E_T = initial \ elastic\ potential \ energy + kinetic \ energy$

$E_T = \frac{1}{2} k D_{max}^2 = \frac{1}{2} k D^2 + \frac{1}{2} mv^2$

Substituting value accordingly

$\frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2$

$4.085 * D_{max}^2 = 3.69$

$D^2_{max} = 0.9033$

$D_{max} = 0.950m$

So to obtain total length we would add the unstretched length

So we have

$L_{max} = 0.950 + 0.5 = 1.45m$

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3 years ago

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A car travel with a constant velocity of 20.5m/s for 20seconds what distance does it cover in this time ?

prohojiy [21]

Answer:

410 m

Explanation:

Given:

v₀ = 20.5 m/s

a = 0 m/s²

t = 20 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (20.5 m/s) (20 s) + ½ (0 m/s²) (20 s)²

Δx = 410 m

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3 years ago