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SCORPION-xisa [38]
3 years ago
8

Wondering if you have enough rope to rappel to the ground, you drop a rock off the top, and hear the sound of it hitting the bot

tom 4.2 seconds later. Find the height of the cliff ignoring the time that the sound takes to travel back to you from the bottom.
Physics
2 answers:
Lena [83]3 years ago
5 0

Answer:

86.5 m

Explanation:

miskamm [114]3 years ago
4 0

Answer:

86.5 m

Explanation:

s =  ut +  \frac{1}{2} at {}^{2}  \\ s = (0)(4.2) +  \frac{1}{2} (9.81)(4.2) {}^{2} \\ s = 86.5m

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152,155 J

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If displacement covered by a particle is zero then distance cover by it<br>​
sleet_krkn [62]

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When displacement is zero, the particle may be at rest, therefore, distance travelled = 0.

Again, when displacement is zero, the final position matches with the initial position after some time, but the distance travelled will not be zero.

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Read 2 more answers
A T-shirt cannon can shoot a 0.085 kg T-shirt at nearly 30 m/s. The T-shirt cannon has a mass of 33 kg. If the initial net momen
IgorLugansk [536]

Answer:

Approximately 0.077\; {\rm m\cdot s^{-1}} (assuming that external forces on the cannon are negligible.)

Explanation:

If an object of mass m is moving at a velocity of v, the momentum p of that object would be p = m\, v.

Momentum of the t-shirt:

\begin{aligned} p(\text{t-shirt}) &= m(\text{t-shirt}) \, v(\text{t-shirt}) \\ &= 0.085\; {\rm kg} \times 30\; {\rm m \cdot s^{-1}} \\ &= 2.55 \; {\rm kg \cdot m \cdot s^{-1}} \end{aligned}.

If there is no external force (gravity, friction, etc.) on this cannon, the total momentum of this system should be conserved. In other words, if p(\text{cannon}) denote the momentum of this cannon:

p(\text{t-shirt}) + p(\text{cannon}) = 0.

p(\text{cannon}) = -p(\text{t-shirt}) = -2.55\; {\rm kg \cdot m \cdot s^{-1}}.

Rewrite p = m\, v to obtain v = (p / m). Since the mass of this cannon is m(\text{cannon}) = 33\; {\rm kg}, the velocity of this cannon would be:

\begin{aligned} v(\text{cannon}) &= \frac{p(\text{cannon})}{m(\text{cannon})} \\ &= \frac{-2.55\; {\rm kg \cdot m \cdot s^{-1}}}{33\; {\rm kg}} \\ &\approx 0.077\; {\rm m \cdot s^{-1}}\end{aligned}.

8 0
2 years ago
A spherical bowling ball with mass m = 4.1 kg and radius R = 0.117 m is thrown down the lane with an initial speed of v = 8.9 m/
weqwewe [10]

Answer:Given mass = 4.1kg

Radius = 0.0117m

Velocity V = 8.4m/s

Coefficient of friction = 0.25

Explanation: Below is an attached solution to the problem stated above.

1. The angular acceleration is equal to 524rad/s^2

2. The linear acceleration is equal to 2.45m/s^2

3.the time it takes the ball to begin rolling = 0.98s

4. The distance the ball slides before it begins to roll = 7.05m

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