Which equation would you need to rearrange to calculate the specific heat of an object

A mass is attached to an ideal spring. At time t = 0 the spring is at its natural length and the mass is given an initial veloci

JulijaS [17]

Answer:

t = T/4

Explanation:

The power delivered to the mass by the spring is work done by the spring per second.

$P = \frac{dW}{dt}$

The work done by the spring is equal to the elastic potential energy stored in the spring.

$U = \frac{1}{2}kx^2$

The maximum energy stored in the spring is at the amplitude of the oscillation.

$U_{max} =\frac{1}{2}kA^2$

So the first time the mass reaches to its amplitude can be found by the following equation of motion:

$x = A\cos(\omega t + \phi)\\\phi = \pi/2 ~because ~at ~t= 0, ~ x = 0\\0 = A\cos(0 + \pi/2)\\x = A\cos(\omega t + \pi/2)$

When the mass reaches the amplitude:

$A = A\cos(\omega t + \pi/2)\\1 = \cos(\omega t + \pi/2)\\\omega t + \pi/2 = \pi$

because cos(π) = 1.

$\omega t = \pi/2$

Using ω = 2π/T,

$\omega t = \pi/2\\\frac{2\pi}{T}t = \pi/2\\t = \frac{T}{4}$

4 0

4 years ago

An electrical motor spins at a constant 2695.0 rpm. If the armature radius is 7.165 cm, what is the acceleration of the edge of

djyliett [7]

Answer:

Acceleration will be $5706.77rad/sec^2$

So option (D) will be correct answer

Explanation:

We have given angular speed of the electrical motor $\omega =2695rpm$

We have to change this angular speed in rad/sec for further calculation

So $\omega =2695rpm=2695\times \frac{2\pi }{60}=282.2197rad/sec$

Armature radius is given r = 7.165 cm = 0.07165 m

We have to find the acceleration of edge of motor

Acceleration is given by $a=\omega ^2r=282.2197^2\times 0.07165=5706.77rad/sec^2$

So acceleration will be $5706.77rad/sec^2$

So option (D) will be correct answer

5 0

3 years ago

mass of the planet is 12 times that of earth and its radius is thrice that of earth , then find the escape velocity on that plan

Over [174]

Answer:

The escape velocity on the planet is approximately 178.976 km/s

Explanation:

The escape velocity for Earth is therefore given as follows

The formula for escape velocity, $v_e$ , for the planet is $v_e = \sqrt{\dfrac{2 \cdot G \cdot m}{r} }$

Where;

$v_e$ = The escape velocity on the planet

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m = The mass of the planet = 12 × The mass of Earth, $M_E$

r = The radius of the planet = 3 × The radius of Earth, $R_E$

The escape velocity for Earth, $v_e_E$ , is therefore given as follows;

$v_e_E = \sqrt{\dfrac{2 \cdot G \cdot M_E}{R_E} }$

$\therefore v_e = \sqrt{\dfrac{2 \times G \times 12 \times M}{3 \times R} } = \sqrt{\dfrac{2 \times G \times 4 \times M}{R} } = 16 \times \sqrt{\dfrac{2 \times G \times M}{R} } = 16 \times v_e_E$

$v_e$ = 16 × $v_e_E$

Given that the escape velocity for Earth, $v_e_E$ ≈ 11,186 m/s, we have;

The escape velocity on the planet = $v_e$ ≈ 16 × 11,186 ≈ 178976 m/s ≈ 178.976 km/s.

3 0

3 years ago

Suppose you must design an experiment to test how a paper airplane's shape affects the distance it will fly. Your plane must fly

Rina8888 [55]

Answer:

Explanation:

Your plane must fly halfway across your classroom to get the best grade. The teacher reminds you to use both imagination and logical reasoning in your design. Write a procedure for an experiment that tests how a paper airplane's shape affects the distance it will fly. Then, describe how you might use both ...

5 0

3 years ago

Read 2 more answers

A 397-N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.

AleksAgata [21]

Answer:

h = 14.4 m

Explanation:

The height can be calculated by energy conservation:

$K_{r} + K_{t} - W = E_{p}$