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puteri [66]
2 years ago
9

Consider a basketball player spinning a ball on the tip of a finger. If a player performs 1.91 J1.91 J of work to set the ball s

pinning from rest, at what angular speed ????ω will the ball rotate? Model a basketball as a thin-walled hollow sphere. For a men's basketball, the ball has a circumference of 0.749 m0.749 m and a mass of 0.624 kg0.624 kg .
Physics
1 answer:
Black_prince [1.1K]2 years ago
7 0

Answer:

ω = 4.07 rad/s

Explanation:

By conservation of the energy:

W = ΔK

1.91J = I/2*\omega^2

where I = 2/3*m*R^2=0.23kg.m^2

Solving for ω:

\omega = \sqrt{W*2/I} =4.07rad/s

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When a test charge q0 = 2 nC is placed at the origin, it experiences a force of 8 times 10-4 N in the positive y direction. What
ser-zykov [4K]

Answer:

Electric field, E=4\times 10^5\ N/C

Explanation:

It is given that,

Magnitude of charge, q_o=2\ nC=2\times 10^{-9}\ C

Force experienced, F=8\times 10^{-4}\ N

We need to find the electric field at the origin. It is given by :

F=q_o\times E

E=\dfrac{F}{q_o}

E=\dfrac{8\times 10^{-4}}{2\times 10^{-9}}

E=4\times 10^5\ N/C

So, the electric field at the origin is 4\times 10^5\ N/C. Hence, this is the required solution.

3 0
2 years ago
Oceanographers use submerged sonar systems, towed by a cable from a ship, to map the ocean floor. In addition to their downward
KATRIN_1 [288]

Answer:

Tension in the cable is T = 16653.32 N

Explanation:

Give data:

Cross section Area A = 1.3 m^2

Drag coefficient CD = 1.2

Velocity V = 4.3 m/s

Angle made by cable with horizontal  =30 degree

Density \rho \ of\  water= 1000 kg/m3

 Drag force FD is given as

F_{D} = \fracP{1}{2} \rho v^{2} C_{D} A

        = 0.5\times 1000\times 4.32\times  1.2\times 1.3

Drag force = 14422.2 N acting opposite to the motion

As cable made angle  of 30 degree with horizontal  thus horizontal component is take into action to calculate drag force

TCos30 = F_D

T = \frac{F_D}{cos30}

T =\frac{ 14422.2}{cos 30}

T = 16653.32 N

7 0
2 years ago
Calculate the average charge on arginine when ph=9.20. (hint : find the average charge for each ionizable group and sum these to
weqwewe [10]
Arginine is a basic aminoacid, because it has two amino groups and one acid group. At a low pH, every ionizable group is protoned. At a little higher pH, the acid group looses its proton. A little higher pH, one amino group looses its proton. At a very high pH, all ionizable groups are not protoned.

Pkas

<span> <span><span> <span> pka1 = 1.82 </span> <span> pka2 = 8.99 </span> <span> pka3 = 12.48 </span> </span> </span></span> So 9.20 is higher tan the second pKa and lower than the third pka. This means the acid has already lost its proton, and one of the aminos too, but the second amino hasn’t. When an acid is not protoned, it has a negative charge. When an amino is not protoned, it’s neutral. When an amino is protoned, it has a positive charge. So this amnino acid has one positive charge (one of the aminos) and one negative charge (the acid), what makes it neutral.
4 0
2 years ago
A baseball of mass m = 0.49 kg is dropped from a height h1 = 2.25 m. It bounces from the concrete below and returns to a final h
Brilliant_brown [7]

Answer:

Explanation:

Impulse = change in momentum

mv - mu , v and u are final and initial velocity during impact at surface

For downward motion of baseball

v² = u² + 2gh₁

= 2 x 9.8 x 2.25

v = 6.64 m / s

It becomes initial velocity during impact .

For body going upwards

v² = u² - 2gh₂

u² = 2 x 9.8 x 1.38

u = 5.2 m / s

This becomes final velocity after impact

change in momentum

m ( final velocity - initial velocity )

.49 ( 5.2 - 6.64 )

= .7056 N.s.

Impulse by floor in upward direction

= .7056 N.s

6 0
2 years ago
A truck collides with a car on horizontal ground. At one moment during the collision, the magnitude of the acceleration of the t
Mice21 [21]

Answer:

The magnitude of the acceleration of the car is 35.53 m/s²

Explanation:

Given;

acceleration of the truck, a_t = 12.7 m/s²

mass of the truck, m_t = 2490 kg

mass of the car, m_c = 890 kg

let the acceleration of the car at the moment they collided = a_c

Apply Newton's third law of motion;

Magnitude of force exerted by the truck = Magnitude of force exerted by the car.

The force exerted by the car occurs in the opposite direction.

F_c = -F_t\\\\m_ca_c = -m_t a_t\\\\a_c =- \frac{m_ta_t}{m_c} \\\\a_c = -\frac{2490 \times 12.7}{890} \\\\a_c = - 35.53 \ m/s^2

Therefore, the magnitude of the acceleration of the car is 35.53 m/s²

3 0
2 years ago
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