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sesenic [268]
3 years ago
6

"As you sit in a fishing boat, you notice that 12 waves pass the boat every 45 s. If the distance from one crest to the next is

9.2 m , what is the speed of these waves?"
Physics
1 answer:
Alchen [17]3 years ago
4 0

Answer:

2.4564 m/s

Explanation:

Number of waves in 1 second

\dfrac{12}{45}=0.267\ Hz

This is the frequency 0.267 Hz

The distance mentioned in the question is the

\lambda = Wavelength = 9.2 m

The speed of wave is

v=f\lambda

\Rightarrow v=0.267\times 9.2

\Rightarrow v=2.4564\ m/s

The speed of the wave is 2.4564 m/s

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A driver in a car traveling at a speed of 26.8 m/s sees a deer 100 m away on the road. Calculate the minimum constant accelerati
ZanzabumX [31]
-3.5912m/s^2Use vf^2 = vi^2 + 2ax in your calculations, knowing vf, vi, and x. 
6 0
3 years ago
A 50 g copper calorimeter contains 250 g of water at 20 C. How much steam be condensed into the water to make the final temperat
Nostrana [21]

Answer:

Approximately 13\; \rm g of steam at 100\; \rm ^\circ C (assuming that the boiling point of water in this experiment is 100\; \rm ^\circ C\!.)

Explanation:

Latent heat of condensation/evaporation of water: 2260\; \rm J \cdot g^{-1}.

Both mass values in this question are given in grams. Hence, convert the specific heat values from this question to \rm J \cdot g^{-1}.

Specific heat of water: 4.2\; \rm J \cdot g^{-1}\cdot \rm K^{-1}.

Specific heat of copper: 0.39\; \rm J \cdot g^{-1}\cdot K^{-1}.

The temperature of this calorimeter and the 250\; \rm g of water that it initially contains increased from 20\; \rm ^\circ C to 50\; \rm ^\circ C. Calculate the amount of energy that would be absorbed:

\begin{aligned}& Q(\text{copper}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 0.39\; \rm J \cdot g^{-1}\cdot K^{-1} \times 50\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 585\; \rm J  \end{aligned}.

\begin{aligned}& Q(\text{cool water}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; \rm J \cdot g^{-1}\cdot K^{-1} \times 250\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 31500\; \rm J  \end{aligned}.

Hence, it would take an extra 585\; \rm J + 31500\; \rm J = 32085\; \rm J of energy to increase the temperature of the calorimeter and the 250\; \rm g of water that it initially contains from 20\; \rm ^\circ C to 50\; \rm ^\circ C.

Assume that it would take x grams of steam at 100\; \rm ^\circ C ensure that the equilibrium temperature of the system is 50\; \rm ^\circ C.

In other words, x\; \rm g of steam at 100\; \rm ^\circ C would need to release 32085\; \rm J as it condenses (releases latent heat) and cools down to 50\; \rm ^\circ C.

Latent heat of condensation from x\; \rm g of steam: 2260\; {\rm J \cdot g^{-1}} \times (x\; {\rm g}) = (2260\, x)\; \rm J.

Energy released when that x\; {\rm g} of water from the steam cools down from 100\; \rm ^\circ C to 50\; \rm ^\circ C:

\begin{aligned}Q = \;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; {\rm J \cdot g^{-1}\cdot K^{-1}} \times (x\; \rm g) \times (100\;{\rm ^\circ C} - 50\;{\rm ^\circ C}) \\ =\; & (210\, x)\; \rm J  \end{aligned}.

These two parts of energy should add up to 32085\; \rm J. That would be exactly what it would take to raise the temperature of the calorimeter and the water that it initially contains from 20\; \rm ^\circ C to 50\; \rm ^\circ C.

(2260\, x)\; {\rm J} + (210\, x)\; {\rm J} = 32085\; \rm J.

Solve for x:

x \approx 13.

Hence, it would take approximately 13\; \rm g of steam at 100\; \rm ^\circ C for the equilibrium temperature of the system to be 50\; \rm ^\circ C.

4 0
2 years ago
What is the mass in kg of a leopard in a tree if the tree branch is 36 m up and the leopard's gravitational potential energy is
ElenaW [278]

Answer:

83.3kg

Explanation:

GPE = m × g × h

GPE = mass of leopard × 10 × 36m

29988J = 360 × mass

mass = 83.3kg

5 0
3 years ago
Explain why you can "stack" water slightly above the rim of a glass if you pour the water in very carefully?
jeka57 [31]
I believe Isaac Newtons Law " Gravity ".
8 0
2 years ago
Find the average force exerted by the bat on the ball if the two are in contact for 0.00129 s. Answer in units of N.
Semmy [17]

Answer:

Explanation:

Given

time of contact between bat and ball is t=0.00129 s

suppose u is the incoming velocity and v is the final velocity after collision

m=mass\ of\ ball

Impulse exerted is given by change in momentum of the particle.

Initial momentum P_i=m\times u

Final momentum P_f=m\times v

Change in momentum \Delta P=P_i-P_f

Impulse    J=F_{avg}\cdot t=\Delta P

J=F_{avg}\cdot t=m(u-v)

F_{avg}=\frac{m(v-u)}{t}

F_{avg}=775.2\times m(v-u)\ N                

5 0
3 years ago
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