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alexdok [17]
4 years ago
15

What kind of simple machine is a catapult?

Physics
1 answer:
jok3333 [9.3K]4 years ago
7 0
A LEVER is one type of simple machine.
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A wire loop of radius 0.50 m lies so that an external magnetic field of magnitude 0.40 T is perpendicular to the loop. The field
vazorg [7]

The magnitude of the induced emf is given by:

ℰ = |Δφ/Δt|

ℰ = emf, Δφ = change in magnetic flux, Δt = elapsed time

The magnetic field is perpendicular to the loop, so the magnetic flux φ is given by:

φ = BA

B = magnetic field strength, A = loop area

The area of the loop A is given by:

A = πr²

r = loop radius

Make a substitution:

φ = B2πr²

Since the strength of the magnetic field is changing while the radius of the loop isn't changing, the change in magnetic flux Δφ is given by:

Δφ = ΔB2πr²

ΔB = change in magnetic field strength

Make another substitution:

ℰ = |ΔB2πr²/Δt|

Given values:

ΔB = 0.20T - 0.40T = -0.20T, r = 0.50m, Δt = 2.5s

Plug in and solve for ℰ:

ℰ = |(-0.20)(2π)(0.50)²/2.5|

ℰ = 0.13V

3 0
3 years ago
Question 2 of 5 Instructions:Select the correct answer. What do scientists use to determine the temperature of a star? thermomet
Lena [83]
Scientists use the color of a star to determine its temperature.
PreviousReset C:\format dir\..<span />
4 0
3 years ago
Read 2 more answers
What replaces a cold current that sinks to the ocean floor?
FrozenT [24]
Well simple the warm water then replaces the cold current that sinks to the ocean floor.
5 0
3 years ago
Pam, wearing a rocket pack, stands on frictionless ice. She has a mass of 49 kg. The rocket supplies a constant force for 22.0 m
ss7ja [257]

Answer:

Magnitude of the force is 4350N

Explanation:

As the woman accelerates at a distance of 22 m to go from rest to 62.5 m / s, we can use the kinematics to find the acceleration

v² = v₀² + 2 a x

v₀ = 0

a = v² / 2x

 a = 62.5²/(2 × 22)

 a = 88.78m/s²

the time you need to get this speed

     v = v₀ + a t

     t = v / a

     t = 62.5 / 88.78

     t = 0.704s

Let's caculate the magnitude of the force

F = ma

= 49 × 88.78

= 4350.22

≅ 4350N

Magnitude of the force is 4350N

     t = 1,025 s

      a = 55.43 m / s²

4 0
3 years ago
Calculate the magnitude of the force exerted by each wire on a 1.20-m length of the other.
Zielflug [23.3K]

Incomplete question.The complete question is attached below as screenshot along with figure

Answer:

F=6.00*10^{-6}N

Force is repulsive

Explanation:

Given data

Current I₁=5.00A

Current I₂=2.00A

Length L=1.20 m

Radius r=0.400m

To find

Force F

Solution

As the force is repulsive because currents are in opposite direction

From repulsive force we know that:

F=\frac{u_{o}I_{1}I_{2}L}{2\pi r}

Substitute the given values

F=\frac{u_{o}(5.00A)(2.00A)(1.20m)}{2\pi (0.400m)}\\ F=6.00*10^{-6}N

4 0
3 years ago
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