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sergey [27]
2 years ago
14

Please help due tomorrow I will give brainliest! and points​

Physics
1 answer:
zhannawk [14.2K]2 years ago
4 0

Answer:

1.) 0.075 km/min

2.) 0 km/min

3.) -0.15 km/min

Explanation:

<u>Formula:</u>

<u />

s= \frac{d}{t}

  • s = speed
  • d = distance
  • t = time

<u />

<u>Calculating the speed:</u>

<u />

1.) \frac{1.5-0}{20-0}=\frac{1.5}{20}=0.075

2.) \frac{1.5-1.5}{40-20}=\frac{0}{20}=0

3.) \frac{0-1.5}{50-40}=\frac{-1.5}{10}=-0.15

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Static friction is the friction that exists between a stationary object and the surface on which it's resting.
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which best describes what happens with the sonar to sound pulses from a ship after they hit the ocean floor?
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Answer:

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Explanation:

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The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is th
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Answer:

1.92 x 10⁻¹²J

Explanation:

The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e

F = ma

Where;

a = \frac{v^2}{r}             [v = linear velocity, r = radius of circular path]

=> F = m\frac{v^2}{r}           ------------(i)

We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;

F = q v B sin θ       ----------(ii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = the angle between the velocity and the magnetic field.

Combining equations (i) and (ii) gives

m\frac{v^2}{r} = q v B sin θ           [θ = 90° since the proton is orbiting at the maximum orbital radius]

=> m\frac{v^2}{r} = q v B sin 90°

=> m\frac{v^2}{r} = q v B

Divide both side by v;

=> m\frac{v}{r} = qB

Make v subject of the formula

v = \frac{qBr}{m}

From the question;

B = 1.25T

m = mass of proton = 1.67 x 10⁻²⁷kg

r = 0.40m

q = charge of a proton = 1.6 x 10⁻¹⁹C

Substitute these values into equation(iii) as follows;

v = \frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}

v = 4.79 x 10⁷m/s

Now, the kinetic energy, K, is given by;

K = \frac{1}{2}mv²

m = mass of proton

v = velocity of the proton as calculated above

K = \frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )

K = 1.92 x 10⁻¹²J

The kinetic energy is 1.92 x 10⁻¹²J

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9. A 5.0 kg block on an inclined plane is acted upon by a horizontal force of 100 N shown in the figure below. The coefficient o
Helga [31]

Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).

Now the force of friction is F_s=\mu*N, where N is the normal force and from the diagram it is F_y=mg*cos(\theta).

Thus F_s=\mu*N=\mu*mg*cos(\theta).

Therefore,

F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).

Substituting the value for F_H,m,\mu, and \:\theta we get:

F_{tot}= -38.63N.

Now acceleration is simply

a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.

The negative sign indicates that the acceleration is directed up the incline.

Part B:

d=\frac{1}{2} at^2

Which can be rearranged to solve for t:

t=\sqrt{\frac{2d}{a} }

Substitute the value of d=0.50m and a=7.7m/s and we get:

t=0.36s.

which is our answer.

Notice that in using the formula to calculate time we used the positive value of a, because for this formula absolute value is needed.

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