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Fed [463]
2 years ago
3

The tape in a videotape cassette has a total length 240 m and can play for 2.1 h. As the tape starts to play, the full reel has

an outer radius of 47 mm and an inner radius of 11 mm. At some point during the play, both reels will have the same angular speed. What is this common angular speed?
Physics
1 answer:
Nesterboy [21]2 years ago
3 0

total length of videotape = 240 m

total time of play = 2.1 h = 7560 s

so linear speed of the videotape is given as

v = \frac{d}{t}

v = \frac{240}{7560} = 0.032 m/s

now we know that

v = r\omega

here both reels have same speed when radius on both sides is same

so here

r = \frac{r_o + r_i}{2}

r = \frac{47 + 11}{2} mm = 29 mm

now from above equation we will have

0.032 = 0.029\omega

\omega = 1.1 rad/s

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A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 ft/s along
babymother [125]

Answer:

Explanation:

height of pole = 15 ft

height of man = 6 ft

Let the length of shadow is y .

According to the diagram

Let at any time the distance of man is x.

The two triangles are similar

\frac{y-x}{y}=\frac{6}{15}

15 y - 15 x = 6 y

9 y = 15 x

y=\frac{5}{3}x

Differentiate with respect to time.

\frac{dy}{dt}=\frac{5}{3}\frac{dx}{dt}

As given, dx/dt = 4 ft/s

\frac{dy}{dt}=\frac{5}{3}\times 4

\frac{dy}{dt}=\frac{20}{3} ft/s

6 0
2 years ago
Platinum has a density of 21.4 g/cm3. if thieves were to steal platinum from a bank using a small truck with a maximum payload o
slega [8]
We will convert the 1dm3 in terms of cm3 as follows:

1dm^3 = (10 cm)^3
= 1000 cm^3

The mass of platinum is equal to 900 lb. 
Then we will convert the mass in terms of grams as follows:
1 lb = 453.6 g
900 = 900 x 453.6 g
= 408240 g

Then density of platinum is equal to 21.4 g/cm^3 
We will calculate the volume of platinum in mass 408240 g as follows:
Volume of platinum = mass of platinum / density of platinum
= 408240 g / 21.4 g/cm^3 
= 19076.6 cm^3  

The total volume of platinum is 19076.6 cm^3
The volume of platinum in 1 L bar is 1000cm^3
So, to calculate the number of bars we will use the formula as follows;
Number of bars = volume of platinum available / volume of platinum required in 1 L bar
= 19076.6 cm^3 / 1000 cm^3
= 19
So, the number of bars are 19.
4 0
2 years ago
Explain why is not advisable to use small values of I in performing an experiment on refraction through a glass prism?
Svetradugi [14.3K]
Explain<span> why it is </span>not advisable to use small values<span> of incident ray in </span>performing experiment<span> on the</span>refraction through a glass prism<span>.</span>
6 0
2 years ago
Read 2 more answers
An 8.0 m, 240 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground
sweet [91]

Answer:

5.7 m

Explanation:

AD = length of the ladder = L = 8 m

AB = distance of the center of mass of the ladder = (0.5) L = (0.5) 8 = 4 m

AC = distance of person on the ladder from the bottom end = x

W = weight of the ladder = 240 N

F_{g} = weight of the person = 710 N

F = force by the wall on the ladder

N = normal force by ground on the ladder = ?

Using equilibrium of force along the vertical direction

N = F_{g} + W

N = 710 + 240

N = 950 N

μ = Coefficient of static friction = 0.55

f =static frictional force on the ladder

Static frictional force is given as

f = μ N

f = (0.55) (950)

f = 522.5 N

Force equation along the horizontal direction is given as

F = f

F = 522.5 N

using equilibrium of torque about point A

F Sin50 (AD) = W Cos50 (AB) + (F_{g} Cos50 (AC))

(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)

x = 5.7 m

7 0
2 years ago
A girl swings on a playground swing in such a way that at her highest point she is 4.1 m from the ground, while at her lowest po
Umnica [9.8K]

Answer:

V1 =8.1 m/s

Explanation:

height at highest point (h2) = 4.1 m

height at lowest point (h1) = 0.8 m

acceleration due to gravity (g) = 9.8 m/s^{2}

from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore

mgh1 + 0.5mV1^{2} = mgh2 + 0.5mV2^{2}

where

  • speed at highest point = V2
  • speed at lowest point = V1
  • mass of the girl and swing = m
  • at the highest point, the  speed is minimum (V1 = 0)
  • at the lowest point the speed is maximum (V2 is the maximum speed)
  • therefore the equation becomes mgh1 + 0.5mV1^{2} = mgh2

      m(gh1 + 0.5V1^{2}) = m(gh2)

      gh1 + 0.5V1^{2} = gh2

      V1 = \sqrt{\frac{gh2 - gh1}{0.5}}

now we can substitute all required values into the equation above.

V1 = \sqrt{\frac{(9.8x4.1) - (9.8x0.8)}{0.5}}

V1 = \sqrt{\frac{32.34}{0.5}}

V1 =8.1 m/s

8 0
2 years ago
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