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Margaret [11]
3 years ago
6

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 19.0 m/s when the

hand is 2.50 m above the ground.
Required:
How long is the ball in the air before it hits the ground?
Physics
1 answer:
german3 years ago
5 0

Answer: 4 s

Explanation:

Given

The ball leaves the hand of student with a speed of u=19\ m/s

When the hand is h=2.5\ m above the ground

Using the equation of motion we can write

h=ut+\dfrac{1}{2}at^2

Substitute the values

\Rightarrow 2.5=-19t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-19t-2.5=0\\\\\Rightarrow t=\dfrac{19\pm \sqrt{(-19)^2-4\times 4.9\times (-2.5)}}{2\times 19}\\\Rightarrow t=4.0049\quad [\text{Neglecting the negative value of }t]

Thus, the ball will take 4 s to hit the ground.

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An alpha particle can be produced in certain radioactive decays of nuclei and consists of two protons and two neutrons. The part
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Answer:

Explanation:

charge, q = 2e = 2 x 1.6 x 10^-19 C = 3.2 x 10^-19 C

mass, m = 4 u = 4 x 1.661 x 10^-27 kg = 6.644 x 10^-27 kg

Radius, r = 4.5 cm = 0.045 m

Magnetic field, B = 1.20 T

(a) Let the speed is v.

v=\frac{Bqr}{m}

v=\frac{1.20\times 3.2\times 10^{-19}\times 0.045}{6.644\times 10^{-27}}

v = 2.6 x 10^6 m/s

(b) Let T be the period of revolution

T=\frac{2\pi r}{v}

T=\frac{2\times 3.14\times 0.045}{2.6\times 10^{6}}

T = 1.09 x 10^-7 s

(c) The formula for the kinetic energy is

K=\frac{B^{2}\times q^{2}\times r^{2}}{2m}

K=\frac{\left ( 1.20\times 3.2 \times 10^{-19}\times 0.045 \right )^{2}}{2\times 6.644\times 10^{-27}}

K = 2.25 x 10^-14 J

(d) Let the potential difference is V.

K = qV

V = \frac{K}{q}

V= \frac{2.25\times 10^-14}{3.2\times 10^{-19}}

V = 70312.5 V

5 0
3 years ago
Which situation is work not being done? A) A bookcase is slid across carpeting. B) A stack of books is carried at waist level ac
Nastasia [14]

Answer:

B) A stack of books is carried at waist level across a room

Explanation:

Work is defined as:

W=Fd cos \theta

where

F is the force applied

d is the displacement of the object

\theta is the angle between the direction of the force and of the displacement

From the formula, we see that the work done is zero when the force and the displacement are perpendicular to each other. Let's now analyze each situation:

A) A bookcase is slid across carpeting. --> work is done, because the force that pushes the bookcase is in the same direction of the displacement

B) A stack of books is carried at waist level across a room. --> no work is done, because the force to carry the book is vertical, while the displacement of the books is horizontal

C) A chair is lifted vertically with respect to the floor. --> work is done, because the force that lifts the chair is vertical, and the displacement is vertical as well

D) A table is dropped onto the ground. --> work is done, because the force of gravity (that makes the table falling down) is vertical and the displacement of the table is also vertical.

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3 years ago
A one-piece cylinder has a core section protruding from the larger drum and is free to rotate around its central axis. A rope wr
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Answer:

Magnitude the net torque about its axis of rotation is 2.41 Nm

Solution:

As per the question:

The radius of the wrapped rope around the drum, r = 1.33 m

Force applied to the right side of the drum, F = 4.35 N

The radius of the rope wrapped around the core, r' = 0.51 m

Force on the cylinder in the downward direction, F' = 6.62 N

Now, the magnitude of the net torque is given by:

\tau_{net} = \tau + \tau'

where

\tau = Torque due to Force, F

\tau' = Torque due to Force, F'

tau = F\times r

tau' = F'\times r'

Now,

\tau_{net} = - F\times r + F'\times r'

\tau_{net} = - 4.35\times 1.33 + 6.62\times 0.51 = - 2.41\ Nm

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

|\tau_{net}| = 2.41\ Nm

 

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