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A 5.49 kg ball is attached to the top of a vertical pole with a 2.15 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.654.65 m/s in a horizontal circle with the string remaining taut. Calculate the angle, between 0° and 90°, that the string makes with the pole. Take ????=9.81g=9.81 m/s2. angle: °

Answers

Answer 1

Answer:

Answer: $\theta =45.73^(\circ)$

Explanation:

Given

Length of string =2.15 m

mass of ball =5.49 kg

speed of ball=4.65 m/s

Here

Tension provides centripetal acceleration

$T\cos\theta =mg$ -----1

$T\sin \theta =(mv^2)/(r)$ ------2

Divide 2 & 1

$tan\theta =(v^2)/(rg)$

$tan\theta =(4.65^2)/(2.15* 9.8)$

$tan\theta =1.026$

$\theta =45.73^(\circ)$

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Select the correct answer.Which statement is true if the refractive index of medium A is greater than that of medium B?
A.
O B.
Total internal reflection is possible when light travels from air to medium B to medium A.
Total internal reflection is possible when light travels from medium A to medium B.
Total internal reflection is possible when light travels from medium B to medium A.
D. Total internal reflection is possible when light travels from air to medium A.
E. Total internal reflection is possible when light travels from air to medium B.
C
C.

Answers

Final answer:

If the refractive index of medium A is greater than that of medium B, then total internal reflection is possible when light travels from medium B to medium A.

Explanation:

If the refractive index of medium A is greater than that of medium B, then total internal reflection is possible when light travels from medium B to medium A.

Learn more about Refraction of light

Solenoid A has total number of turns N length L and diameter D. Solenoid B has total number of turns 2N, length 2L and diameter 2D. Inductance of solenoid A is 8 times inductance of solenoid B
1/4 of inductance of solenoid B
same as inductance of solenoid B
1/8 of inductance of solenoid B
four times of inductance of solenoid B

Answers

Answer:

∴Inductance of solenoid A is $\frac18$ of inductance of solenoid B.

Explanation:

Inductance of a solenoid is

$L=N\frac\phi I$

$=N(B.A)/(I)$

$=N(\mu_0NI)/(l.I)A$

$=(\mu_0N^2A)/(l)$

$=(\mu_0N^2)/(l).\pi(\frac d2)^2$

$=\mu_0\pi(N^2d^2)/(4l)$

N= number of turns

$l$ = length of the solenoid

d= diameter of the solenoid

A=cross section area

B=magnetic induction

$\phi$ = magnetic flux

$I$ = Current

Given that, Solenoid A has total number of turns N, length L and diameter D

The inductance of solenoid A is

$=\mu_0\pi(N^2D^2)/(4L)$

Solenoid B has total number of turns 2N, length 2L and diameter 2D

The inductance of solenoid B is

$=\mu_0\pi((2N)^2(2D)^2)/(4.2L)$

$=\mu_0\pi(16 N^2D^2)/(4.2L)$

Therefore,

$\frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=(\mu_0\pi(N^2D^2)/(4L))/(\mu_0\pi(16 N^2D^2)/(4.2L))$

$\Rightarrow \frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=\frac18$

$\Rightarrow {\textrm{Inductance of A}}=\frac18* {\textrm{Inductance of B}}$

∴Inductance of solenoid A is $\frac18$ of inductance of solenoid B.

Hi there!

We can begin by calculating the inductance of a solenoid.

Recall:
$L = (\Phi _B)/(i)$

L = Inductance (H)
φ = Magnetic Flux (Wb)

i = Current (A)

We can solve for the inductance of a solenoid. We know that its magnetic field is equivalent to:
$B = \mu _0 (N)/(L)i$

And that the magnetic flux is equivalent to:
$\Phi _B = \int B \cdot dA = B \cdot A$

Thus, the magnetic flux is equivalent to:
$\Phi _B = \mu _0 (N)/(L)iA$

The area for the solenoid is the # of loops multiplied by the cross-section area, so:
$A_(total)= N * A$

$\Phi _B = \mu _0 (N^2)/(L)iA$

Using this equation, we can find how it would change if the given parameters are altered:
$\Phi_B ' = \mu_0 ((2N)^2)/(2L) i * 4A$

**The area will quadruple since a circle's area is 2-D, and you are doubling its diameter.

$\Phi'_B = (4)/(2) * 4(\mu_0 (N)/(L)iA) = 8\mu_0 (N)/(L)iA$

Thus, Solenoid B is 8 times as large as Solenoid A.

Solenoid A is 1/8 of the inductance of solenoid B.

You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has stopped due to an accident up ahead. You step on your brakes to slow down as quickly as possible. Assume that you to slow down to 30 mph in about 5 seconds. A) With this same average acceleration, how much longer would it take you to stop?B) What total distance would you travel from when you first apply the brakes until the car stops?

Answers

A.The time taken for the car to stop is 8.75 s

B.The distance travelled when the brakes were applied till the car stops is 136.89 m

A. Determination of the time taken for the car to stop.

We'llbegin bycalculatingthedecelerationof thecar

Initial velocity (u) = 70 mph = 0.447 × 70 = 31.29 m/s

Final velocity (v) = 30 mph = 0.447 × 30 = 13.41 m/s

Time (t) = 5 s

Deceleration (a) =?

$a \: = (v \: - u)/(t) \n \n a = (13.41 - 31.29)/(5) \n \n a \: = ( - 17.88)/(5) \n \n$

a = –3.576 m/s²

Finally,we shall determine the time taken for the car to stop.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Time (t) =?

$v \: = u \: + at \n 0 \: = 31.29 \: + \: ( - 3.576 * t) \n 0 \: = 31.29 \: - 3.576 * t \n collet \: like \: terms \n 0 - 31.29 \: = - 3.576 * t \n - 31.29 \: = - 3.576 * t \n divide \: both \: side \: by \: - 3.576 \n t \: = (- 31.29)/(- 3.576) \n$

t = 8.75 s

Thus, the time taken for the car to stop is 8.75 s

B.Determination of the total distance travelled when the brakes were applied.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Distance (s) =?

${v}^(2) = {u}^(2) + 2as \n {0}^(2) = {31.29}^(2) + (2 * - 3.576 * s) \n 0 = 979.0641 - 7.152 s \n collect \: like \: terms \n 0 - 979.0641 = - 7.152 s \n - 979.0641 = - 7.152 s \n divide \: both \: side \: by \: - 7.152 \n s = (- 979.0641)/(- 7.152) \n \n$

s = 136.89 m

Therefore, the total distance travelled by the car when the brakes were applied is 136.89 m

Learn more: brainly.com/question/9163788

Answer:8.75 s,

136.89 m

Explanation:

Given

Initial velocity $=70 mph\approx 31.29 m/s$

velocity after 5 s is $30 mph\approx 13.41 m/s$

Therefore acceleration during these 5 s

$a=(v-u)/(t)$

$a=(13.41-31.29)/(5)=-3.576 m/s^2$

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

$0=31.29-3.576* t$

$t=(31.29)/(3.576)=8.75 s$

(b)total distance traveled before stoppage

$v^2-u^2=2as$

$0^2-31.29^2=2* (-3.576)\cdot s$

s=136.89 m

Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate from 2.00 to 6.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him.

Answers

Answer:

69.68 N

Explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = $(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

W = $F_(total) .d$

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = $(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

$F_(total) .d =(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

$F_(total) = ((1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i))/(d)$

$F_(total=) ((1)/(2) X 62 X6^(2) -(1)/(2) X 62 X2^(2) )/(25)$

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

$F_(sprinter) = F_(total) + F_(wind) = 39.7 + 30 = 69.68 N$

Answer:

Force exerted by sprinter = 69.68 N

Explanation:

From work energy theorem, we know that, work done is equal to change in kinetic energy.

Thus,

W = ΔK = Kf - Ki = (1/2)m•(v_f)² - (1/2)m•(v_i)² - - - - eq(1)

Now,

Work done is also;

W = Force x Distance = F•d - - - (2)

From the question, we are given ;

v_f = 6 m/s

v_i = 2 m/s

d = 25m

m = 62 kg

Equating equation 1 and 2,we get;

(1/2)m•(v_f)² - (1/2)m•(v_i)² = F•d

Plugging in the relevant values to obtain ;

(1/2)(62)[(6)² - (2)²] = F x 25

31(36 - 4) = 25F

992 = 25F

F = 39.68 N

The force the sprinter exerts backward on the track will be the sum of this force and the headwind force.

Thus,

Force of sprinter = 39.68 + 30 = 69.68N

Impulse is the______of the force and time of contact

Answers

Answer:

Product

Explanation:

Impulse is defined as the average force acting on an object times the time the force acts:

Impulse = F · Δt

Calculate the linear momentum per photon,energy per photon, and the energy per mole of photons for radiation of wavelength; (a) 600 nm (red), (b) 550 nm (yellow), (c) 400 nm (violet), (d) 200 nm (ultraviolet), (e) 150 pm (X-ray), and (f ) 1.0 cm (microwave).

Answers

Answer:

The detailed explanations is attached below

Explanation:

What is applied is the De brogile equation and the equation showing a relationship between Energy, speed of light and wavelength.

The explanation is as attached below.

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