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2 years ago

Can waves move or transport matter? Why or why not?

Physics

1 answer:

OLga [1]2 years ago

6 0

They can.
Surfers use waves to move themselves around.

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A driver of a car enters a new 110 km/h speed zone on the highway. The driver begins to accelerate immediately and reaches 110 k

levacccp [35]

Answer:

30Km/h

Explanation:

acceleration is the change of speed in a given time so when we substract the accelerations we can know how much the car goes per an hour

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2 years ago

Imagine imagining an imagination.

ozzi

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3 years ago

A man of 60kg moves in a lift of constant velocity 5m/s .What is the reactive force acting on the man by the elevator?

Flauer [41]

Answer:

588 N

Explanation:

Since the 60 kg is moving at a constant velocity there is no acceleration. In order for the system to be balanced, both the normal force and the force of gravity must be equal. In this case the man has a mass of 60 kg. So to find the force you multiply mass by gravitys constant (9.81). And you end up with an answer of 588.6 but I rounded to 588.

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2 years ago

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What are various names of tropical storms<br> based upon the ocean they occur in?

SpyIntel [72]

Answer:

hurricanes,Typhoons and cyclones

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2 years ago

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

1 year ago