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anastassius [24]
2 years ago
13

7. What is the velocity of an object that has a mass of 4.5 kilograms and

Physics
1 answer:
labwork [276]2 years ago
7 0

Answer:

266.66 m/s

Explanation:

p=mv

1200=4.5v

v=266.66 m/s

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vazorg [7]

Each station can detect how far away the epicenter was. So each station basically has a circle made of possible epicenters. When you have three, you narrow it down to one, final point.

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3 years ago
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8 0
3 years ago
a ball rolls horizontally of the edge of the cliff at 4 m/s, if the ball lands at a distance of 30 m from the base of the vertic
algol13

Answer:

Approximately 281.25\; \rm m. (Assuming that the drag on this ball is negligible, and that g = 10\; \rm m \cdot s^{-2}.)

Explanation:

Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:

  • Horizontal: no acceleration, velocity is constant (at v(\text{horizontal}) is constant throughout the descent.)
  • Vertical: constant downward acceleration at g = 10\; \rm m \cdot s^{-2}, starting at 0\; \rm m \cdot s^{-1}.

The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: x(\text{horizontal}) = 30\; \rm m. Combine these two quantities to find the duration of this descent:

\begin{aligned}t &= \frac{x(\text{horizontal})}{v(\text{horizontal})} \\ &= \frac{30\; \rm m}{4\; \rm m \cdot s^{-1}} = 7.5\; \rm s\end{aligned}.

In other words, the ball in this question start at a vertical velocity of u = 0\; \rm m \cdot s^{-1}, accelerated downwards at g = 10\; \rm m \cdot s^{-2}, and reached the ground after t = 7.5\; \rm s.

Apply the SUVAT equation \displaystyle x(\text{vertical}) = -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t to find the vertical displacement of this ball.

\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}.

In other words, the ball is 281.25\; \rm m below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be 281.25\; \rm m\!.

5 0
3 years ago
On a journey of 600 km, a train was delayed 1 hour and 30 min after having covered 1 4 of the way. To arrive on time at the dest
mars1129 [50]

Answer:

The train travelled 10hours.

Explanation:

Using speed= distance/time ...eq1

Let the time taken by train to cover the journey be t.

Let the speed of train be s

Time= distance/speed ...eq2

Time t =600/s ...eq3

The train is delayed for 1 1/2 hours=3/2 hours

Train increased by 15km/hr.

Train travelled 1/4 of 600k.= 150km.

Speed increased by s + 15 to cover the remaining 450km

t = 150/s + 400/(s + 15) + 3/2 ..eq4

Equating eq3 and 4

600/s= 150/s + 460/(s + 15) + 3/2

450/s = 3/2 + 450/(s + 15)

3s^2 + 45s -13500=0

Solving the quadratic equation

S= -45 +-sqrt(45^2-4(3×13500)/2×3

S= 60 , -75

Hence speed of train is 60km/hr

Using eq 2 distance/speed=time

600/60=10

t= 10hours

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Which of the following is transferred to an object when work is done? A. motion B. energy C. force D. friction
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