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IRINA_888 [86]
3 years ago
14

The diagram shows forces acting on a boat.

Physics
1 answer:
Greeley [361]3 years ago
7 0
A

I hope this helps!!:)
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What year will y'all graduate
goblinko [34]

Answer:

I will graduate on the year 2022

6 0
2 years ago
Read 2 more answers
An electric motor does 900j of work for 8 hours. Calculate the power used​
nataly862011 [7]

Answer:

0.031 W

Explanation:

The power used is equal to the rate of work done:

P=\frac{W}{t}

where

P is the power

W is the work done

t is the time taken to do the work W

In this problem, we have:

W = 900 J is the work done by the motor

t = 8 h is the time taken

We have to convert the time into SI units; keeping in mind that

1 hour = 3600 s

We have

t=8\cdot 3600 =28,800 s

And therefore, the power used is

W=\frac{900}{28800}=0.031 W

6 0
3 years ago
You've recently read about a chemical laser that generates a 20.0-cm-diameter, 26.0 MW laser beam. One day, after physics class,
aksik [14]

Answer :

(a). The speed of the block is 0.395 m/s.

(b). No

Explanation :

Given that,

Diameter = 20.0 cm

Power = 26.0 MW

Mass = 110 kg

diameter = 20.0 cm

Distance = 100 m

We need to calculate the pressure due to laser

Using formula of pressure

P_{r}=\dfrac{I}{c}

P_{r}=\dfrac{P}{Ac}Put the value into the formula[tex]P_{r}=\dfrac{26.0\times10^{6}}{\pi\times(10\times10^{-2})^2\times3\times10^{8}}

P_{r}=2.75\ N/m^2

We need to calculate the force

Using formula of force

F=P\times A

F=P\times \pi r^2

Put the value into the formula

F=2.75\times\pi (0.01)^2

F=0.086\ N

We need to calculate the acceleration

Using formula of force

F=ma

Put the value into the formula

0.086=110\times a

a=\dfrac{0.086}{110}

a=0.000781\ m/s^2

a=7.81\times10^{-4}\ m/s^2

(a). We need to calculate speed of the block

Using equation of motion

v^2=u^2+2ad

Put the value into the formula

v=\sqrt{2\times7.81\times10^{-4}\times100}

v=0.395\ m/s

(b). No because the velocity is very less.

Hence, (a). The speed of the block is 0.395 m/s.

(b). No

8 0
2 years ago
Does passing a magnet through a coil of wire break off it’s electric current
hichkok12 [17]
A magnetic field is actually generated by a moving current (or moving electric charge specifically). The magnetic field generated by a moving current can be found by using the right hand rule, point your right thumb in the direction of current flow, then the wrap of your fingers will tell you what direction the magnetic field is. In the case of current traveling up a wire, the magnetic field generated will encircle the wire. Similarly electromagnets work by having a wire coil, and causing current to spin in a circle, generating a magnetic field perpendicular to the current flow (again right hand rule).

So if you were to take a permenant magnet and cut a hole in it then string a straight wire through it... my guess is nothing too interesting would happen. The two different magnetic fields might ineteract in a peculiar way, but nothing too fascinating, perhaps if you give me more context as to what you might think would happen or what made you come up with this question I could help.

Source: Bachelor's degree in Physics.
7 0
2 years ago
A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

4 0
3 years ago
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