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kotykmax [81]
3 years ago
13

An object is thrown straight down with an initial speed of 4 m/s from a window which is 8 m above the ground. Calculate: a) The

time it takes the object to reach the ground b) Its velocity when it reaches the ground.
Physics
1 answer:
frosja888 [35]3 years ago
8 0

Answer:

0.41s

8.01m/s

Explanation:

Using the formula; v = u + at

Where;

u = initial velocity (m/s)

v = final velocity (m/s)

t = time (s)

a = acceleration (m/s²)

According to the provided information, u = 4m/s, s = 8m,

V = u + at

4 = 0 + 9.8t

4 = 9.8t

t = 0.41s

b) v = u + at

v = 4 + 9.8(0.41)

v = 4 + 4.018

v = 8.018m/s

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You drive a beat-up pickup truck along a straight road for 8.4 km at 70 km/h, at which point the truck runs out of gasoline and
andreev551 [17]

Answer:

v_avg = 37 km/h

Explanation:

To find the average velocity in the complete trajectory you use the following formula:

v_{avg}=\frac{v_1+v_2}{2}   ( 1 )

v1: velocity in the first part of the trajectory = 70 km/h

v2: velocity in the second part of the trajectory = ?

You calculate v2 by using the following equation for a motion with constant velocity:

v_2=\frac{2.0km}{30min}*\frac{60min}{1h}=4\frac{km}{h}

you replace the values of v1 and v2 in (1) and you obtain:

v_{avg}=\frac{70km/h+4km/h}{2}=37\frac{km}{h}

hence, the average velocity is 37 km/h

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You find a pressure gauge in the lab and the spec sheet says that the maximum pressure it can read a maximum of 2 inches of wate
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Answer:

Maximum pressure in pascal is 497.84 pa

Explanation:

Pressure is given by P=\rho g h

where \rho is density of water = 1000kg/m^3

height of the water h is 2 inches = 2\times 25.4=50.8 mm=50.8\times 10^{-3} m

Maximum pressure in pascal is

P=1000\times 9.8\times 50.8\times 10^{-3}\\p=497.84 Pa

8 0
3 years ago
A plank rests on top of the axles of two identical wheels. Each wheel's outer radius is 0.25 meters and each wheel's axle has ra
kaheart [24]

Answer:

<u><em>The plank moves 0.2m from it's original position</em></u>

Explanation:

we can do this question from the constraints that ,

  • the wheel and the axle have the same angular speed or velocity
  • the speed of the plank is equal to the speed of the axle at the topmost point .

thus ,

<em>since the wheel is pure rolling or not slipping,</em>

<em>⇒v=wr</em>

where

<em>v - speed of the wheel</em>

<em>w - angular speed of the wheel</em>

<em>r - radius of the wheel</em>

<em>since the wheel traverses 1 m let's say in time 't' ,</em>

<em>v_{w}=\frac{distance}{time} =\frac{1}{t}</em>

∴

⇒w=\frac{v_{w}}{r} = \frac{1}{t*0.25}

the speed at the topmost point of the axle is :

⇒v_{a}=w*r\\v_{a}=\frac{1}{t*0.25} *0.05\\v_{a}=\frac{1}{5t}

this is the speed of the plank too.

thus the distance covered by plank in time 't' is ,

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8 0
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