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3 years ago

Question 3

Physics

1 answer:

pickupchik [31]3 years ago

4 0

When a police officer is trying to decide if a driver is speeding, what is his point of reference. The speed limit

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Which describes the movement of a fluid during convection?

VikaD [51]

The movement of a fluid during convection is a circular/oval motion since the fluid at the top sinks and the fluid at the bottom rises.

Hope this helps :)

3 0

4 years ago

A pool is to be filled with 60 m3 water from a garden hose of 2.5 cm diameter flowing water at 2 m/s. Find the mass flow rate of

stealth61 [152]

Answer:

Time= 6.12*10^4s

mass flow rate m=0.98kg/s

Explanation:

Given

Volume= 60m^3

diamter= 2.5cm= 0.025m

radius= 0.0125m

area A= πr^2

area A= 3.142*0.0125^2

area A= 4.9*10^-4m^2

the velocity of the flow 2m/s

volume flow rate

V=vA

V=2* 4.9*10^-4

V=9.82*10^-4 m^3/s

Time taken to fill the pool

time= volume/volume flow rate

time= 60/9.82*10^-4

time= 6.12*10^4s

Mass flow rate

m= density *volume flow rate

Assuming the density of water to be 997kg/m^3

m= 997*9.82*10^-4

m=0.98kg/s

6 0

3 years ago

when a 0.622kg basketbll hits the floor its velocit changes from 4.23m/s down to 3.85m/s up. if the averge force was 72.9N how m

scoray [572]

Answer:

Time, t = 3.2 ms

Explanation:

It is given that,

Mass of basketball, m = 0.622 kg

Initial velocity, u = 4.23 m/s

Final velocity, v = 3.85 m/s

Average force acting on the ball, F = 72.9 N

We need to find the time of contact of the ball with the floor. Let t is the time of contact. So,

$F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{0.622\times (3.85-4.23)}{72.9}\\\\t=0.0032\ s\\\\\text{or}\\\\t=3.2\ ms$

So, the ball is in contact with the floor for 3.2 ms.

8 0

4 years ago

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30˚ angle. The block’s initial speed is 10 m/s. What vert

WARRIOR [948]

Answer:

The wood block reaches a height of 4.249 meters above its starting point.

Explanation:

The block represents a non-conservative system, since friction between wood block and the ramp is dissipating energy. The final height that block can reach is determined by Principle of Energy Conservation and Work-Energy Theorem. Let suppose that initial height has a value of zero and please notice that maximum height reached by the block is when its speed is zero.

$\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g\cdot h + \mu_{k}\cdot m\cdot g\cdot s \cdot \sin \theta$

$\frac{1}{2}\cdot v^{2} = g\cdot h + \mu_{k}\cdot g\cdot \left(\frac{h}{\sin \theta} \right)\cdot \sin \theta$

$\frac{1}{2}\cdot v^{2} = g\cdot h +\mu_{k}\cdot g\cdot h$

$\frac{1}{2}\cdot v^{2} = (1 +\mu_{k})\cdot g\cdot h$

$h = \frac{v^{2}}{2\cdot (1 + \mu_{k})\cdot g}$ (1)

Where:

$h$ - Maximum height of the wood block, in meters.

$v$ - Initial speed of the block, in meters per second.

$\mu_{k}$ - Kinetic coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$m$ - Mass, in kilograms.

$s$ - Distance travelled by the wood block along the wooden ramp, in meters.

$\theta$ - Inclination of the wooden ramp, in sexagesimal degrees.

If we know that $v = 10\,\frac{m}{s}$ , $\mu_{k} = 0.20$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the height reached by the block above its starting point is: