To verify the law of solid friction project work of physics

A cyclist accelerates from a velocity of 10 miles/hour east until reaching a velocity of 20 miles/hour east in 5 seconds. What w

Sliva [168]

Answer:

$a = 0.894\ m/s^2$

Explanation:

<u>Motion with Constant Acceleration</u>

A body moves with constant acceleration when the speed changes uniformly in time. The equation used to find the final speed vf is

$v_f=v_o+at$

Where vo is the initial speed, a is the acceleration, and t is the time.

The cyclist has an initial speed of vo=10 miles/hour and ends up at vf=20 miles/hour in t=5 seconds.

Both speeds are given in miles/hour and we must convert it to m/s:

1 mile/hour = 0.44704 m/s

10 mile/hour = 4.47 m/s

20 mile/hour = 8.94 m/s

The acceleration is calculated by solving for a:

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

$\displaystyle a=\frac{8.94-4.47}{5}$

$a = 0.894\ m/s^2$

3 0

2 years ago

A car starting from rest (i.e. initial velocity = 0.0 m/s), moves in the positive X-direction with a constant average accelerati

olga_2 [115]

Answer:

Final speed of the car, v = 24.49 m/s

Explanation:

It is given that,

Initial velocity of the car, u = 0

Acceleration, $a=3.1\ m/s^2$

Time taken, t = 7.9 s

We need to find the final velocity of the car. Let it is given by v. It can be calculated using first equation of motion as :

$v=u+at$

$v=0+3.1\times 7.9$

v = 24.49 m/s

So, the final speed of the car is 24.49 m/s. Hence, this is the required solution.

8 0

3 years ago

A plane stops from 250 mph in 25 seconds. Calculate the plane's acceleration.

Lelechka [254]

Explanation:

Given:

v₀ = 250 mph

v = 0 mph

t = 25 s

Find: a

v = at + v₀

(0 mph) = a (25 s) + (250 mph)

a = -10 mph/s

6 0

3 years ago

A basketball has a volume of 300 cm3. If Michael pumped 200 cm3 and Fandi pumped another 200 cm3 into it then the total volume o

photoshop1234 [79]

Answer:

...

Explanation:

8 0

3 years ago

The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete

oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\$

make shear stress (τ) the subject of the formula

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}$

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of 0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2$

Part (b) shear stress at a distance of 0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0}{2*12} =0$

3 0

4 years ago

To verify the law of solid friction project work of physics ​

To verify the law of solid friction project work of physics