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777dan777 [17]
4 years ago
11

A fisherman sees his boat moving up and down, owing to waves on the water. It takes 2.5 s for the boat to travel from its highes

t to lowest point, a distance of 0.62 m. The fisherman sees that the wave crests are spaced 6 m apart. How fast are the waves traveling?
Physics
1 answer:
nata0808 [166]4 years ago
7 0

Answer:

The waves are traveling in a speed of

v=1.20 m/s

Explanation:

The fisherman see a wave at 6m apart so that's the wavelength

L=6m

Now the period is the time between two successive waves so

t=2*2.5s=5s

The velocity of the waves is describe by:

v=f*L

f=\frac{1}{t}=\frac{1}{5s}=20s^{-1}

v=0.20s^{-1}*6m

v=1.20 \frac{m}{s}

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Consider a system of a cliff diver and the Earth. The gravitational potential energy of the system decreases by 25,000 J as the
salantis [7]

Answer:

568.18 N

Explanation:

From the question,

The formula for gravitational potential is given as

Ep = mgh........................ Equation 1

Where Ep = Gravitational potential, m = mass of the diver,h = Height.

But,

W = mg.................... Equation 2

Where W = weight of the diver.

Substitute equation 2 into equation 1

Ep = Wh

Make W the subject of the equation

W = Ep/h................... Equation 3

Given: Ep = 25000 J, h = 44 m

Substitute into equation 3

W = 25000/44

W = 568.18 N.

Hence the weight of the diver = 568.18 N

5 0
3 years ago
If the mass of a
mixer [17]

Answer:

The final acceleration becomes (1/3) of the initial acceleration.

Explanation:

The second law of motion gives the relationship between the net force, mass and the acceleration of an object. It is given by :

F=ma

m = mass

a = acceleration

According to given condition, if the mass of a  sliding block is tripled while a constant net force is applied. We need to find how much does the acceleration decrease.

a=\dfrac{F}{m}

Let a' is the final acceleration,

a'=\dfrac{F}{m'}

m' = 3m

a'=\dfrac{F}{3m}

a'=\dfrac{1}{3}\times \dfrac{F}{m}

a'=\dfrac{1}{3}\times a

So, the final acceleration becomes (1/3) of the initial acceleration. Hence, this is the required solution.

5 0
3 years ago
A 50-kg copper block initially at 140°C is dropped into an insulated tank that contains 90 L of water at 10°C. Determine the fin
Kaylis [27]

Answer:

16.33°C

Explanation:

Applying,

Heat lost by copper = heat gained by water

cm(t₁-t₃) = c'm'(t₃-t₂).............. Equation 1

Where c = specific heat capacity of copper, m = mass of copper, c' = specific heat capacity of water, m' = mass of water, t₁ = initial temperature of copper, t₂ = initial temperature of water, t₃ = final equilibrium temperature.

From the question,

Given: m = 50 kg, t₁ = 140°C, m' = 90 L = 90 kg, t₂ = 10°C

Constant: c = 385  J/kg°C, c' = 4200J/kg°C

Substitute these values into equation 1

50(385)(140-t₃) = 90(4200)(t₃-10)

(140-t₃) = 378000(t₃-10)/19250

(140-t₃) = 19.64(t₃-10)

140-t₃ = 19.64t₃-196.6

19.64t₃+t₃ = 196.4+140

20.64t₃ = 336,4

t₃ = 336.4/20.6

t₃ = 16.33°C

7 0
3 years ago
The principle of resistance training that suggests that muscles should be gradually required to do more than they are used to do
nadya68 [22]
The principle of resistance training that suggests that muscles should be gradually required to do more than they are used to doing is D. OVERLOAD.

The principle of overload states that a greater than normal stress or load on the body is required for training adaptation to take place.

Overload refers to the amount of load or resistance, providing a greater stress, or load, on the body than it is normally accustomed to in order to increase fitness.
5 0
4 years ago
You set a small eraser at a distance of 0.27 m from the center of a turntable you find in the attic and establish that when the
Licemer1 [7]

Answer with Explanation:

We are given that

Distance,r=0.27 m

Tangential speed=v=0.49 m/s

a.Angular speed ,\omega=\frac{v}{r}

Using the formula

\omega=\frac{0.49}{0.27}=1.8 rad/s

\omega=\frac{2\pi}{T}

T=\frac{2\pi}{\omega}

Time period,T=\frac{2\pi}{1.8}=3.49 s

b.Amplitude,A=Distance of small eraser from the center of a turnable =0.27 m

c.Maximum speed,V_{max}=0.49 m/s

d.Maximum acceleration=a=r\omega^2=0.27(1.8)^2=0.87 m/s^2

5 0
3 years ago
Read 2 more answers
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