PHYSICS COLLEGE

A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the magnitude of the tension in the string is F. The boy then speeds up the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions every second. What happens to the tension in the sting?

Answers

Answer 1

Answer:

Answer

given,

Tension of string is F

velocity is increased and the radius is not changed.

the string makes two complete revolutions every second

consider the centrifugal force acting on the stone

= $(mv^2)/(r)$

now centrifugal force is balanced by tension

T = $(mv^2)/(r)$

From the above expression we can clearly see that tension is directly proportional to velocity and inversely proportional to radius.

When radius is not changing velocity is increasing means tension will also increase in the string.

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A train accelerates at -1.5 m/s2 for 10 seconds. If the train had an initialspeed of 32 m/s, what is its new speed?
A. 17 m/s
B. 15 m/s
C. 47 m/s
D. 32 m/s

Answers

Answer:

17 m/s

Explanation:

Using formula a = (v-u) /t

acceleration a = -1.5 m/s2

final velocity v = unknown

initial velocity u = 32 m/s

time t = 10s

-1.5 = (v-32)/10

-15 = v - 32

-15 + 32 = v

v = 17 m/s

Assuming the same initial conditions as described in FNT 2.2.1-1, use the energy-interaction model in two different ways (parts (a) and (b) below) to determine the speed of the ball when it is 4 meters above the floor headed down: a) Construct a particular model of the entire physical process, with the initial time when the ball leaves Christine’s hand, and the final time when the ball is 4 meters above the floor headed down.
b) Divide the overall process into two physical processes by constructing two energy-system diagrams and applying energy conservation for each, one diagram for the interval corresponding to the ball traveling from Christine’s hand to the maximum height, and then one diagram corresponding to the interval for the ball traveling from the maximum height to 4 meters above the floor headed down.
c) Did you get different answers (in parts (a) and (b)) for the speed of the ball when it is 4 meters above the floor headed down?

Answers

Answer:

(a). Vf = 7.14 m/s

(b). Vf = 7.14 m/s

(c). same answer

Explanation:

for question (a), we would be applying conservation of energy principle.

but the initial height is h = 1.5 m

and the initial upward velocity of the ball is Vi = 10 m/s

Therefore

(a). using conservation law

Ef = Ei

where Ef = 1/2mVf² + mghf ........................(1)

also Ei = 1/2mVi² + mghi ........................(2)

equating both we have

1/2mVf² + mghf = 1/2mVi² + mghi

eliminating same terms gives,

Vf = √(Vi² + 2g (hi -hf))

Vf = √(10² + -2*9.8*2.5) = 7.14 m/s

Vf = 7.14 m/s

(b). Same process as done in previous;

Ef = Ei

but here the Ef = mghf ...........(3)

and Ei = 1/2mVi² + mghi ...........(4)

solving for the final height (hf) we relate both equation 3 and 4 to give

mghf = 1/2mVi² + mghi ..............(5)

canceling out same terms

hf = hi + Vi²/2g

hf = 1.5 + 10²/2*9.8 = 6.60204m ............(6)

recalling conservation energy,

Ef = Ei

1/2mVf² + mghf = mghi

inputting values of hf and hi we have

Vf = √(2g(hi -hf)) = 7.14 m/s

Vf = 7.14 m/s

(c). From answer in option a and c, we can see there were no changes in the answers.

A laser (electromagnetic wave) has the maximum electric field strength of 1.0x1011 V/m. What is the force the laser applies on a mirror (totally reflective) of 5.0 mm2 area? A. 2.76 x105N B. 1.21 x106N C. 1.94 x106N D.4.43 x105 N E. 7.82 x104N

Answers

Answer:

The correct option is D

Explanation:

From the question we are told that

The maximum electric field strength is $E = 1.0 *10^(11) \ V/m$

The area is $A = 5.0 \ mm^2 = 5.0 *10^(-6) \ m^2$

Generally the force the laser applies is mathematically represented as

$F = \epsilon_o * E ^2 * A$

Here $\epsilon_o = 8.85*10^(-12) C/(V \cdot m)$

$F = 8.85*10^(-12) * (1.0 *10^(11)) ^2 * 5.00*10^(-6 )$

=> $F = 4.43 *10^(5) \ N$

A battery lighting a bulb is an example of _ energy converting to _ energy.

Answers

the power source is electrical ( whether the light is plugged in or has a battery)

the light bulb converts the electricity to light and heat.

in a fluorescent bulb, it is different, but the electricity is again converted to light, very little heat, though.

Electric energy converting into light and heat energy.

So, for the first blank electric, and the second blank the better answer is light.

A 30 kg child on a 2 m long swing is released from rest when the swing supports make an angle of 34 ◦ with the vertical. The acceleration of gravity is 9.8 m/s 2 . If the speed of the child at the lowest position is 2.31547 m/s, what is the mechanical energy dissipated by the various resistive

Answers

Answer:

Energy dissipated = 13.453 Joules

Explanation:

In order to solve this problem, we first compute the gravitational potential energy the child has, and then find the kinetic energy at the lowest position.

The gravitational potential energy (relative to lowest position) is found as follows:

G.P.E = mass * gravity * height

Where, Height = 2 - 2 * Cos(34°)

Height = 0.3193 meters

G.P.E = 30 * 9.8 * 0.3193

G.P.E = 93.874 J

Kinetic energy:

K.E = 0.5 * mass * velocity^2

K.E = 0.5 * 30 * 2.31547^2

K.E = 80.421 J

Energy dissipated = G.P.E - K.E

Energy dissipated = 93.874 - 80.421

Energy dissipated = 13.453 J

A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then the speed of efflux of non viscous water through the opening will be

Answers

Answer:

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

Explanation:

Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ( $P_(atm)$ ), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:

$P_(1) + \rho\cdot (v_(1)^(2))/(2) + \rho\cdot g \cdot z_(1) = P_(2) + \rho\cdot (v_(2)^(2))/(2) + \rho\cdot g \cdot z_(2)$

Where:

$P_(1)$ , $P_(2)$ - Water total pressures inside the tank and at ground level, measured in pascals.

$\rho$ - Water density, measured in kilograms per cubic meter.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_(1)$ , $v_(2)$ - Water speeds inside the tank and at the ground level, measured in meters per second.

$z_(1)$ , $z_(2)$ - Heights of the tank and ground level, measured in meters.

Given that $P_(1) = P_(2) = P_(atm)$ , $\rho = 1000\,(kg)/(m^(3))$ , $g = 9.807\,(m)/(s^(2))$ , $v_(1) = 0\,(m)/(s)$ , $z_(1) = 6.9\,m$ and $z_(2) = 4.9\,m$ , the expression is reduced to this: