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Elden [556K]
2 years ago
10

An 80 N bowling ball has an apparent weight of 60 N when completely submerged in water. What is the average density of the bowli

ng ball
Physics
1 answer:
Allisa [31]2 years ago
3 0

Answer:

V * ρB = WB        volume of ball * density of ball = weight of ball

V * ρW = Ww        volume of ball * density of water = buoyant force

ρB / ρW = WB / Ww = 80 / 20 = 4    water provides 20 N of buoyant force

ρB = 4 ρW = 4 gm/cm^3

ρW = 1000 kg / m^3 = 1 gm / cm^3

(1 gm/cm^3) = .001 kg / .000001 m^3 = 1000 kg/^3

ρW = 1 gm / cm^3

multiply by 9.8 or 980 to get weight densities

In this case

4 * 1000 kg/m^3 * 9.8 m/s^2 = 39200 N /m^3      weight density of ball

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Work done is when a force is exerted to cause a displacement in a certain object. 
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work done = force applied * displacement of the object 
when the force applied is not in the same direction as that of the displacement of the object then the effect of the force is not its whole value. The force is then applied at an angle to that of the displacement of the object, then the resultant force is the force exerted* cos of the angle between force and displacement, in this instance the angle is 40 °.
the new equation is then;
work done = force cos 40° * displacement
after substitution,
work = 25 N * 0.76 * 50 m
         = 957.55 J
round it off
         = 9.6 *10² J
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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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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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