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Darya [45]
2 years ago
7

What is the frequency of a wave that has a wavelength of 2000 meters and is traveling at 20 m/s?

Physics
1 answer:
Artist 52 [7]2 years ago
7 0

Answer:

freqiency=velocity/wavelength

ief=20/2000=0.01hz

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How many watt hours will 3-155amp hour 12 volt batteries wired in a parallel configuration produce?
Ede4ka [16]

Answer:

B 5580 W•hr

Explanation:

A Watt is a Volt times an Amp

3(12 V(155 A•hr)) = 5580 W•hr

4 0
2 years ago
PLZ SOMEONEE HELPP I’LL MARK BRANLIESTTTT
7nadin3 [17]

Answer:

I'm pretty sure it's 37.5 joules of energy

Explanation:

hope this helps!

8 0
2 years ago
We can model a lightning bolt as a very long, straight wire. If a lightning bolt carries a current of 30 kA, and you are unfortu
Liula [17]

Answer:

Magnetic field experienced = 4.5 × 10⁻⁴ T

Explanation:

The magnetic field around an infinite straight current-carrying wire at a distance r from the wire is given by

B = (μ₀I)/(2πr)

B = ?

I = 20 KA = 20000 A

r = 8.9 m

μ₀ = magnetic permeability = 1.257 × 10⁻⁶ T.m/A

B = (1.257 × 10⁻⁶ × 20000)/(2π×8.9) = 4.5 × 10⁻⁴ T

8 0
3 years ago
A crane lifts a load of 15000N to a height 40m in 50 seconds. Calculate the power of the crane​
Nataly [62]

Answer:

P = 12000 W

Explanation:

General data:

  • F = 15000 N
  • d = 40 m
  • t = 50 s
  • P = ?

Work is force times unit of distance. So in order to calculate the power we must first calculate the work.

Formula:

  • \boxed{\bold{W=\frac{F}{d}}}

Replace and solve

  • \boxed{\bold{W=\frac{15000\ N}{40\ m}}}
  • \boxed{\bold{W=600000\ J}}

once the work is found, we proceed to find the power according to the formula:

  • \boxed{\bold{P=\frac{W}{t}}}
  • \boxed{\bold{P=\frac{600000\ J}{50\ s}}}
  • \boxed{\boxed{\bold{P=12000\ W}}}

The power of the crane is <u>12000 Watts.</u>

Greetings.

6 0
2 years ago
A bobsledder pushes her sled across horizontal snow to get it going, then jumps in. After she jumps in, the sled gradually slows
anastassius [24]

Answer:

In the vertical direction the acting forces are the normal force and the weight of the bobsleder plus the sled. In the horizontal direction the acting force is the friciton force.

Explanation:

Hi there!

Please, see the attached figure for a graphic representation of the forces acting on the sled after the bobsleder jumped in.

In the vertical direction, the acting forces are the normal force (N) and the weight of the sled plus the bobsledder (W).

Since the sled is not being accelerated in the vertical direction, the sum of forces in that direction is zero:

∑Fy = W + N = 0 ⇒ W = N

The weight is calculated as follows:

W = (mb + ms) · g

Where:

mb = mass of the bobsleder.

ms = mass of the sled.

g = acceleration due to gravity.

In the horizontal direction the only acting force is the friction force (Fr). The friction force is calculated a follows:

Fr = N · μ

Where:

N = normal force.

μ = kinetic friction coefficient.

Since N = W = (mb + ms) · g

Fr = (mb + ms) · g · μ

If we want to find the acceleration of the sled after the bobsleder jumps in, we can apply Newton's second law:

∑F = m · a

Where "a" is the acceleration and "m" is the mass of the object (in this case, the mass of bobsleder plus the mass of the sled).

∑F = Fr =  (mb + ms) · g · μ =  (mb + ms) · a

(mb + ms) · g · μ =  (mb + ms) · a

Solving for "a":

g · μ = a

3 0
3 years ago
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