Earth is the third planet from the Sun. This placement most affects Earth's unique ____ and ___

A thin cylindrical shell is released from rest and rolls without slipping down an inclined ramp that makes an angle of 30° with

NNADVOKAT [17]

Answer: 1.59 sec

Explanation:

The kinetic energy contained in the rotation of the cylinder is 1/2 m v^2

The kinetic energy of translation is also 1/2 m v^2

so the total energy is m v^2

The force applied is mg sin (theta)

= m x 9.8 x 1/2

= 4.9 m

Now equate

F x d = m v^2

4.9 m x 3.1 = m v^2

v^2 = 4.9 x 3.1

v = sqrt(4.9 x 3.1) = 3.9 m/s

Acceleration.

V^2 = 2 a s

4.9 x 3.1 = 2 x a x 3.1

4.9 = 2 a

a = 2.45 m/s^2

Time

T = v/a

= 3.9/2.45

= 1.59 sec

6 0

4 years ago

A mass of 267 g is attached to a spring and set into simple harmonic motion with a period of 0.176 s. If the total energy of the

Gnoma [55]

Answer:

(a) 7.1 m /sec

(b) 339.9 N/m

(c) 19.91 cm

Explanation:

We have given mass m = 267 gram = 0.267 kg

Time period T = 0.176 sec

Total energy of the oscillating system = 6.74 J

We know that energy is given by

(a) $Ke=\frac{1}{2}mv_{max}^2$

$6.74=\frac{1}{2}\times 0.267\times v_{max}^2$

$v_{max}=7.1m/sec$

(b) Now $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{0.176}=35.681rad/sec$

We know that $\omega =\sqrt{\frac{k}{m}}$

$35.68=\sqrt{\frac{k}{0.267}}$

$k=339.9N/m$

(c) We know that energy is given by

$E=\frac{1}{2}KA^2$

$6.74=\frac{1}{2}\times 339.9\times A^2$

$A=19.91cm$

4 0

3 years ago

What is the mass of a bicycle that has 90 kg.m/s of momentum at a speed of 6 m/s?

OLEGan [10]

Answer:

m = 15 kg

Explanation:

p = m × v

90 = m × 6

6m = 90

m = 90/6

m = 15 kg

6 0

3 years ago

Can I improve the design of my simple machine? How?

blsea [12.9K]

Answer:

12345

Explanation:

yan na po answer ko hehehe

5 0

3 years ago

The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi

Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

$\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$

Where,

B= Magnetic Field

l = length

$\mu_0$ = Vacuum permeability

$\epsilon_0$ = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

$B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}$