The center of ?

A box sits at the edge of a spinning disc. The radius of the disc is 0.5 m, and it is initially spinning at 5 revolutions per se

Furkat [3]

Answer:

a) α = 0.375 rad/s²

b) at = 0.1875 m/s²

c) ac =79 m/s²

d) θ = 52 rad

Explanation:

The uniformly accelerated circular movemeis a circular path movement in which the angular acceleration is constant.

Tangential acceleration is calculated as follows:

at = α*R Formula (1)

Centripetal acceleration is calculated as follows:

ac =ω² *R Formula (2)

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t Formula (3)

θ= ω₀*t + (1/2)*α*t² Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

R : radius of the circular path (m)

at: tangential acceleration, (m/s²)

ac: centripetal acceleration, (m/s²)

Data:

R= 0.5 m : radius of the disk

t₀=0 , ω₀ = 5 rev/s

1 revolution = 2π rad

ω₀ = 5*(2π)rad/s =10π rad/s = 31.42 rad/s

ωf = 2*(2π)rad/s =4π rad/s = 12.57 rad/s

t = 8 s

(a) angular acceleration of the box

We replace data in the formula (3)

ωf= ω₀ + α*t

2 = 5 + α*(8)

2 -5 = α*(8)

-3 = (8)α

α=3 /8

α = 0.375 rad/s²

(b) Tangential acceleration of the box

We replace data in the formula (1)z

at =(α)*R

at = (0.375)*(0.5)

at = 0.1875 m/s²

c) Centripetal acceleration of the box at t = 8 s

We replace data in the formula (2)

ac =ω² *R

ac =(12.57)² *(0.5)

ac = 79 m/s²

d) Radians that the box has rotated over after t = 8 s

We replace data in the formula (4)

θ = ω₀*t + (1/2)*α*t²

θ = (5)*(8)+ (1/2)*( 0.375)*(8)²

θ = 52 rad

9 0

3 years ago

A car brakes from 25 m/s to 16 m/s in 2.0s. what is its acceleration?<br>

Mars2501 [29]

Acceleration = (16-25) / 2 = - 4.5 m/s^2

3 0

3 years ago

During an ultrasound, sound waves are sent by a transducer through muscle tissue at a speed of 1300 m/s. Some of the sound waves

alina1380 [7]

$2.3 \times 10^{-5} \text { seconds }$ is the time taken by the transducer to detect the reflected waves from the metal fragment after they were first emitted

Option C

<u>Explanation:</u>

Given data:

speed, v = 1300 m/s

distance, d = 3.0 cm = $3.00 \times 10^{-2} \mathrm{m}$

We need to calculate the time taken by the transducer to detect the reflected waves from the metal fragment after they were first emitted.

As we know, the velocity is the ratio of distance and the time travelled by an object. The equation form is given by,

$\text {velocity, } v=\frac{\text {distance }(d)}{\text {time}(t)}$

By applying the given values to the above equation, we get

$1300=\frac{3.00 \times 10^{-2}}{t}$

$t=\frac{3.00 \times 10^{-2}}{1300}=0.002307 \times 10^{-2}=2.3 \times 10^{-5} \text { seconds }$

7 0

4 years ago

The cart now moves toward the right with an acceleration toward the right of 2.50 m/s2. What does spring scale Fz read? Show you

katrin2010 [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The spring scale $F_2$ reads $F_2 = 2.4225 \ N$

Explanation:

From the question we are told that

The first force is $F_1 = 10.5 \ N$

The acceleration by which the cart moves to the right is $a = 2.50 \ m/s^2$

The mass of the cart is m = 3.231 kg

Generally the net force on the cart is

$F_{net} = F_1 - F_2$

This net force is mathematically represented as

$F_{net} = m * a$

So

$m* a = 10 - F_2$

$F_2 = 10.5 - 2.5 (3.231)$

$F_2 = 2.4225 \ N$

3 0

3 years ago

Yoooooooooooooooooourrrrrrrrrrrrrrrrrr

lora16 [44]

Answer:

Hey

Explanation:

5 0

3 years ago