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3 years ago

When are the displacement and acceleration equal to zero for the motion of a mass on a spring?

Physics

1 answer:

aliina [53]3 years ago

5 0

Answer:

Option B is the right answer.

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HELPP!! what is not true about mitosis??

Andreyy89

Answer:

I think its A

Explanation:

7 0

3 years ago

Read 2 more answers

An object dropped on Planet P falls 144 m in 6 seconds. What is the gravitational acceleration of Planet P ? Gravitational accel

Tju [1.3M]

Answer:

The gravitational acceleration of the planet is, g = 8 m/s²

Explanation:

Given data,

The distance the object falls, s = 144 m

The time taken by the object is, t = 6 s

Using the III equations of motion

S = ut + ½ gt²

∴ g = 2S/t²

Substituting the given values,

g = 2 x 144 /6²

= 8 m/s²

Hence, the gravitational acceleration of the planet is, g = 8 m/s²

7 0

3 years ago

Calculate the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lig

Nana76 [90]

Answer:

Work done = 35467.278 J

Explanation:

Given:

Height of the cone = 4m

radius (r) of the cone = 1.2m

Density of the cone = 600kg/m³

Acceleration due to gravity, g = 9.8 m/s²

Now,

The total mass of the cone (m) = Density of the cone × volume of the cone

Volume of the cone = $\frac{1}{3}\pi r^2 h$

thus,

volume of the cone = $\frac{1}{3}\pi 1.2^2\times 4$ = 6.03 m³

therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg

The center of mass for the cone lies at the $\frac{1}{4}$ times the total height

thus,

center of mass lies at, h' = $\frac{1}{4}\times4=1m$

Now, the work gone (W) against gravity is given as:

W = mgh'

W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J

4 0

3 years ago

A train travels southwest from point A to point B through the Arizona desert at 55 mi/h. How far will the train travel six-and-a

TiliK225 [7]

Explanation:

Distance = speed × time

d = (55 mi/hr) (6.5 hr)

d = 357.5 mi

4 0

3 years ago

Light is incident along the normal to face AB of a glass prism of refractive index 1.54. Find αmax, the largest value the angle

marusya05 [52]

To solve this problem it is necessary to use the concepts related to Snell's law.

Snell's law establishes that reflection is subject to

$n_1sin\theta_1 = n_2sin\theta_2$

Where,

$\theta =$ Angle between the normal surface at the point of contact

n = Indices of refraction for corresponding media

The total internal reflection would then be given by

$n_1 sin\theta_1 = n_2sin\theta_2$

$(1.54) sin\theta_1 = (1.33)sin(90)$

$sin\theta_1 = \frac{1.33}{1.54}$

$\theta = sin^{-1}(\frac{1.33}{1.54})$

$\theta = 59.72\°$

Therefore the $\alpha_{max}$ would be equal to

$\alpha = 90\°-\theta$

$\alpha = 90-59.72$

$\alpha = 30.27\°$

Therefore the largest value of the angle α is 30.27°

3 0

3 years ago