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Viefleur [7K]
2 years ago
10

What is called the process of finding exact quantity of a substance?​

Physics
1 answer:
Tema [17]2 years ago
7 0

Answer:

Measurement is called the process of finding exact quantity of a substances.

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A bicycle wheel rotates at a constant 25 rev/min. What is true about its angular acceleration?
Nostrana [21]

Answer:

The angular acceleration is zero

Explanation:

When an object is in rotational motion, it has a certain angular velocity, which is the rate of displacement of its angular position.

This angular velocity can change or remain constant - this is given by the angular acceleration, which is:

\alpha =\frac{\Delta \omega}{\Delta t}

where

\Delta \omega is the change in angular velocity

\Delta t is the time elapsed

Therefore, the angular acceleration is the rate of change of angular velocity.

In this problem, the bicycle rotates at a constant angular velocity of

\omega=25 rev/min

This means that the change in angular velocity is zero:

\Delta \omega=0

And so, that the angular acceleration is zero:

\alpha=0

8 0
3 years ago
What is a element that tends to be shiny, easy shaped, and a good conductor of heat?
ozzi
The answer to your question is Metal
8 0
3 years ago
The water is reflecting light, Is this specular or diffuse reflection? explain your answer​
FinnZ [79.3K]
Yes the answer is true
8 0
3 years ago
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
3 years ago
If you want a charged object to become neutral,then you have to ————-? A.Touch it with a positively charged object B.Touch it wi
JulsSmile [24]
A) Touch it with a positively charged object
6 0
3 years ago
Read 2 more answers
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