If the slope of a line on a speed-time graph is 0 and it is not on the x-axis, what is the object doing?

A record turntable rotates through 5.0 rad in 2.8 s as it is accelerated uniformly from rest. What is the angular velocity at th

Usimov [2.4K]

Answer:

$\omega_f = 3.584\ rad/s$

Explanation:

given,

turntable rotate to, θ = 5 rad

time, t = 2.8 s

initial angular speed = 0 rad/s

final angular speed = ?

now, using equation of rotational motion

$\theta = \omega_i t + \dfrac{1}{2}\alpha t^2$

$5 = 0+ \dfrac{1}{2}\alpha\times 2.8^2$

$\alpha= \dfrac{10}{2.8^2}$

α = 1.28 rad/s²

now, calculation of angular velocity

$\omega_f = \omega_i + \alpha t$

$\omega_f =0 +1.28\times 2.8$

$\omega_f = 3.584\ rad/s$

hence, the angular velocity at the end is equal to 3.584 rad/s

4 0

2 years ago

Why doe cups break in the winter?

eimsori [14]

Answer: Glass may break at low temperatures, but this is because the contents freeze and their expansion cause the glass to crack (if the cap does not come off). ... Hot temperatures can cause the glass to break when the bottle is subject to excessive thermal variations. hope this helps can u give me brainliest

Explanation:

5 0

2 years ago

Read 2 more answers

The resistance of the coil of a

ololo11 [35]

Answer:

54 is the correct answer to this question

7 0

2 years ago

A rectangular coil of 65 turns, dimensions 0.100 m by 0.200 m, and total resistance 10.0 ? rotates with angular speed 29.5 rad/s

vovikov84 [41]

Answer:

Explanation:

N = 65

Area, A = 0.1 x 0.2 = 0.02 m^2

R = 10 ohm

ω = 29.5 rad/s

B = 1 T

(a) at t = 0

e = N x B x A x ω

e = 65 x 1 x 0.02 x 29.5

e = 38.35 V

(b) The maximum rate of change of magnetic flux is equal to the maximum value of induced emf.

Ф = 38.35 Wb/s

(c) e = NBAω Sinωt

e = 65 x 1 x 0.02 x 29.5 x Sin (29.5 x 0.05)

e = 38.174 V

(d) Maximum torque

τ = M B Sin 90

τ = N i A B

τ = N e A B / R

τ = 65 x 38.35 x 0.02 x 1 / 10

τ = 5 Nm

8 0

2 years ago

Two point charges, 3.4 μC and -2.0 μC , are placed 5.0 cm apart on the x axis. Assume that the negative charge is at the origin,

Elza [17]

The electric field is zero at x = -16.45cm

Data;

q1 = 3.4 μC
q2 = -2.0 μC
distance = 5cm

<h3>The Electric Field at point 0</h3>

As the 3μC is larger than -2.0μC and the charges are opposite sign. The electric field will be zero at the negative axis.

Let the point be at x.

For an electric field to be equal to zero;

$k(\frac{q_1}{d_1})^2 + k(\frac{q_2}{d_2})^2 = 0\\\frac{3.4}{(5-x)^2} - \frac{2}{x^2} = 0\\$

Let's solve for x using mathematical methods.

$\frac{3.4x^2 - 2(5-x)^2}{(5-x)^2x^2}= 0\\ 3.4x^2 - 2(5-x^2) = 0\\3.4x^2 - 50 - 2x^2 + 20x = 0\\1.4x^2 +20x - 50 = 0$

Solving the above quadratic equation;

$x = -16.45cm$

The electric field is zero at x = -16.45cm

Learn more on electric field at a point here;

brainly.com/question/1592046

brainly.com/question/14372859

8 0

2 years ago