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MA_775_DIABLO [31]
2 years ago
10

Titus drives his jetski a distance of 1000 meters in 7.045 seconds. How fast was he moving in meters per second? How fast was he

moving in miles per hour?
Physics
1 answer:
Ksenya-84 [330]2 years ago
5 0

Answer:

a) v = 141.9 m/s

b) v = 317.4 miles/h

Explanation:

a) How fast was he moving in meters per second?

v = \frac{d}{t} = \frac{1000 m}{7.045 s} = 141.9 m/s

Hence, the jet ski is moving at 141.9 meters per second.

b) How fast was he moving in miles per hour?

v = \frac{d}{t} = \frac{1000 m}{7.045 s} = 141.9 m/s*\frac{3600 s}{1 h}*\frac{1 mile}{1609.34 m} = 317.4 miles/h      

Therefore, the jet ski is moving at 317.4 miles per hour.

I hope it helps you!

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How far from a -7.80 μC point charge must a 2.40 μC point charge be placed in order for the electric potential energy of the pai
Naddik [55]

Answer:

Distance between two point charges, r = 0.336 meters

Explanation:

Given that,

Charge 1, q_1=-7.8\ \mu C=-7.8\times 10^{-6}\ C

Charge 2, q_2=2.4\ \mu C=2.4\times 10^{-6}\ C

Electric potential energy, U = -0.5 J

The electric potential energy at a point r is given by :

U=k\dfrac{q_1q_2}{r}

r=k\dfrac{q_1q_2}{U}

r=9\times 10^9\times \dfrac{-7.8\times 10^{-6}\times 2.4\times 10^{-6}}{-0.5}

r = 0.336 meters

So, the distance between two point charges is 0.336 meters. Hence, this is the required solution.

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A toy car accelerates at a constant rate from rest to a speed of 4 m/s in a time of 0.55 s. What was the magnitude of the accele
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  • Final velocity=v=4m/s
  • Time=t=0.5s

\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \sf\longmapsto Acceleration=\dfrac{4-0}{0.5}

\\ \sf\longmapsto Acceleration=\dfrac{4}{0.5}

\\ \sf\longmapsto Acceleration=8m/s^2

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2. Kevin works as a janitor, and he is pushing a fully-
dybincka [34]

The time taken for him to move the bin 6.5 m is 2.30 s.

The given parameters;

  • <em>weight of the load, w = 557 N</em>
  • <em>force applied , F = 410 N</em>
  • <em>angle of force, =  15°</em>
  • <em>coefficient of kinetic friction  = 0.46</em>
  • <em>distance moved, d = 6.5 m</em>

The net horizontal force on the recycling bin is calculated as follows;

Fcos\theta - F_k = ma

where;

  • <em>m is the mass of the recycling bin</em>
  • <em />F_k<em> is the frictional force </em>

W = mg

557 = 9.8m\\\\m = \frac{557}{9.8} \\\\m = 56.84 \ kg

The net horizontal force on the recycling bin is calculated as;

Fcos \theta - F_k = ma\\\\Fcos\theta - \mu_kF_n  = ma\\\\410\times cos(15) \ - \ 0.46(557) = 56.84 a\\\\139.8 = 56.84a\\\\a = \frac{139.8}{56.84} \\\\a = 2.46 \ m/s^2

The time taken for him to move the bin 6.5 m is calculated as follows;

s = v_0t + \frac{1}{2} at^2\\\\6.5 = 0 + \frac{1}{2}  \times 2.46\times t^2\\\\6.5 = 1.23 t^2\\\\t^2 = \frac{6.5 }{1.23} \\\\t^2 = 5.285\\\\t = \sqrt{5.285} \\\\t = 2.30 \ s

Thus, the time taken for him to move the bin 6.5 m is 2.30 s.

Learn more here:brainly.com/question/21684583

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