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SOVA2 [1]
2 years ago
14

Scouts at a camp shake the rope bridge they have just crossed and observe the wave crests to be 8.00 m apart. If they shake it t

he bridge twice per second, what is the propagation speed of the waves?
Physics
1 answer:
kaheart [24]2 years ago
8 0

Answer:

Explanation:

distance between two crests = 8 m

The distance between the two crests is called wavelength.

So, wavelength, λ = 8 m

frequency = 2 Hz

Let v be the velocity of wave.

v = f x λ

v = 2 x 8

v = 16 m/s

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A car with mass m traveling at speed v has kinetic energy k. what is the kinetic energy of a second car that has the same mass m
vampirchik [111]
Kinetic energy, KE, is modeled by the formula KE =  \frac{1}{2}mv^2, where m is the mass in kg and v is the velocity in m/s.

In this scenario, mass and one-half are constant but the velocity changes. 

You can see that by squaring twice the velocity, that is equal to four times the original KE. Therefore, the answer is 4k.
7 0
3 years ago
A long, straight metal rod has a radius of 5.75 cm and a charge per unit length of 33.3 nC/m. Find the electric field at the fol
PIT_PIT [208]

Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a)  at 4.75 cm

b)  at 20.5 cm

c) at 125.0 cm

Given that:

the charge (q) = 33.3 nC/m

= 33.3 × 10⁻⁹ c/m

radius of rod = 5.75 cm

a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.

Then, the electric field will be zero.

b) The electric field formula E = \dfrac{kq }{d}

E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}

E = 1461.95 N/C

c) The electric field E is calculated as:

E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}

E = 239.76 N/C

7 0
3 years ago
How much power does it take to lift a 30.0 n box 10.0 m high in 5.00 s, if you must apply a 62n force to lift the box?
Illusion [34]
Power is defined as the rate at which the body is doing work:
P=\frac{W}{t}
Work is defined as displacement done by the force times that displacement:
W=F\cdot h
We know that we need 62N to move the box, so when we apply this force along the path of 10m we have done:
W=62N\cdot10m=620J
of work.
Now we just divide that by 5s to get how much power is required:
P=\frac{620J}{5s}=124W
5 0
3 years ago
The air in a kitchen has a mass of 60.0kg and a specific heat of 1505J/(kg°C).
BARSIC [14]
Your answer is 632,100J which is Choice D
8 0
3 years ago
Read 2 more answers
** URGENT** The voltage across the primary winding is 350,000 V, and the voltage across the secondary winding is 17,500 V. If th
Vlad [161]

As we know that in transformers we have

\frac{V_s}{V_p} = \frac{N_s}{N_p}

here we know that

V_s = 17,500 Volts

V_p = 350,000 Volts

N_s = 600 coils

now from above equation we will have

\frac{17500}{350000} = \frac{600}{N_p}

N_p = 600\times \frac{350000}{17500}

N_p = 12000 coils

6 0
3 years ago
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