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Mazyrski [523]
3 years ago
13

A person holds a portable fire extinguisher that ejects 1.0kg of water per second horizontally at a speed of 6.0 m/s. What newto

ns must the person exert on the extinguisher in order to prevent it from accelerating?
(A) 0N
(B) 6N
(C) I0N
(D) 18N
(E) 36N
Physics
1 answer:
Leviafan [203]3 years ago
7 0

Answer:

6 N

Explanation:

\dfrac{m}{t} = Mass flow rate = 1 kg/s

v = Final velocity = 6 m/s

u = Initial velocity = 0 m/s

Force is obtained when we divide change in momentum by time

F=\dfrac{m(v-u)}{t}\\\Rightarrow F=\dfrac{m}{t}(v-u)\\\Rightarrow F=1(6-0)\\\Rightarrow F=6\ N

The force the person exert on the extinguisher in order to prevent it from accelerating is 6 N

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3 years ago
A roller coaster car is traveling at a constant 3 m/s when it reaches a downward slope. On the slope, the car accelerates at a c
kirill [66]

Answer:

Velocity of the car at the bottom of the slope: approximately 20.3\; \rm m \cdot s^{-2}.

It would take approximately 3.9\; \rm s for the car to travel from the top of the slope to the bottom.

Explanation:

The time of the travel needs to be found. Hence, make use of the SUVAT equation that does not include time.

  • Let v denote the final velocity of the car.
  • Let u denote the initial velocity of the car.
  • Let a denote the acceleration of the car.
  • Let x denote the distance that this car travelled.

v^2 - u^2 = 2\, a\cdot x.

Given:

  • u = 3\; \rm m \cdot s^{-1}.
  • a = 4.5\; \rm m \cdot s^{-2}.
  • x = 45\; \rm m.

Rearrange the equation v^2 - u^2 = 2\, a\cdot x and solve for v:

\begin{aligned}v &= \sqrt{2\, a \cdot x + u^2} \\ &= \sqrt{2 \times 4.5\; \rm m \cdot s^{-2} \times 45\; \rm m + \left(3\; \rm m \cdot s^{-1}\right)^{2}} \\ &\approx 20.3\; \rm m \cdot s^{-1}\end{aligned}.

Calculate the time required for reaching this speed from u = 3\; \rm m \cdot s^{-1} at a = 4.5\; \rm m \cdot s^{-2}:

\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{20.3\; \rm m \cdot s^{-1} - 3\; \rm m \cdot s^{-1}}{4.5\; \rm m \cdot s^{-2}} \approx 3.9\; \rm m \cdot s^{-1}\end{aligned}.

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3 years ago
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