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2 years ago

As your skateboard coasts uphill, your speed changes from 3 m/s to 1 m/s in

Physics

1 answer:

vlada-n [284]2 years ago

7 0

Answer:

$a=-0.33\ m/s^2$

Explanation:

<u>Accelerated Motion</u>

The acceleration of a moving body is defined as the relation of change of speed (or velocity in vector form) with the time taken. The formula is given by

$\displaystyle a=\frac{\Delta v}{t}$

Or, equivalently

$\displaystyle a=\frac{v_f-v_o}{t}$

Where vf and vo are the final and initial speeds respectively. The problem gives us these values: v0 = 3 m/s, vf = 1 m/s, t = 3 seconds. Computing a

$\displaystyle a=\frac{1-3}{3}=-0.33\ m/s^2$

The negative sing of a indicates there is deceleration or decreasing speed

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Calculate the change in length of a Pyrex glass dish (Coefficient of linear expansion for Pyrex is 3 * 10-6 / oC) that is 0.25 m

DIA [1.3K]

9* $10^{-5}$ m

Explanation:

Step 1:

We are given the initial length of the Pyrex glass dish at a particular temperature and need to calculate the change in the length when the temperature changes by 120° C. The coefficient of linear expansion of Pyrex is provided.

Step 2:

Change in length = Coefficient of linear expansion * Change in temperature * Initial length

Step 3:

Coefficient of linear expansion = 3* $10^{-6}$ /°C

Change in temperature = 120°C = 120 K

Initial length = 0.25 m

Step 4:

Change in length = 3* $10^{-6}$ * 120 * 0.25 = 9* $10^{-5}$ m

8 0

3 years ago

On the Moon the acceleration due to gravity is about one sixth that on Earth. If a golfer on the Moon imparted the same initial

swat32

Explanation:

We know that that the range of the ball on the earth

$R_{earth}=\frac{v_o^2sin2\theta}{g_{earth}}$

therefore, range of the ball on moon

$R_{moon}=\frac{v_o^2sin2\theta}{g_{moon}}$

$R_{moon}=\frac{v_o^2sin2\theta}{g_{earth}/12}$

therefore,

$R_{moon}=6R_{earth}$

Therefore, the range of ball will be 6 times on the moon than that on earth

6 0

2 years ago

A circular cross section, d = 25 mm, experiences a torque load, T = 25 N·m, and a shear force, V = 85 kN. Calculate the shear st

Maru [420]

Answer:

The correct answer is 231 Mpa i.e option a.

Explanation:

using the equation of torsion we Have

$\frac{T}{I_{p}}=\frac{\tau }{r}\\\\\therefore \tau =\frac{T}{I_{p}}\times r$

where,

$\tau$ = shear stress at a distance 'r' from the center

T = is the applied torque

$I_{p}$ = polar moment of inertia of the section

r = radial distance from the center

Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals

$\tau _{max}=\frac{4}{3}\times \frac{V}{A}$

Applying values we get

$\tau _{max}=\frac{4}{3}\times \frac{85\times 10^{3}}{0.25\times \pi \times (25\times 10^{-3})^{2}}\\\\\therefore \tau _{max}=230.88Mpa\approx 231Mpa$