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Vinvika [58]
3 years ago
5

How do you calculate acceleration

Physics
2 answers:
never [62]3 years ago
8 0

Answer:

As we know that acceleration is defined as the rate of change in velocity

So in order to find the acceleration of a given system we need to find the change in velocity of the given system with respect to time.

So here the formula is given as

a = \frac{v_2 - v_1}{\Delta t}

so in order to find the acceleration we need

v_2 = final velocity

v_1 = initial velocity

\Delta t = time interval

so here we can say that acceleration calculation must need the change in velocity and the time taken to change the velocity both and the ration of this will give us the value of the acceleration

AfilCa [17]3 years ago
7 0
A=f/m
Example a=10/2
A=5
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With what minimum speed must you toss a 110 g ball straight up to just touch the 11-m-high roof of the gymnasium if you release
Sholpan [36]

Answer:

 v = 13.79 m/s

Explanation:

given,

mass of ball = 110 g

height = 11 m

ball is released from = 1.3 m

minimum speed = ?

using conservation of energy

Potential energy is conserved in the form of kinetic energy

P E =KE

m g h= \dfrac{1}{2}mv^2

v^2 = 2 g h

v=\sqrt{2 g h}

v=\sqrt{2\times 9.8 \times (11-1.3)}

v=\sqrt{2\times 9.8 \times 9.7}

v=\sqrt{190.12}

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3 years ago
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The correct choice is Buffers.

Buffers contains both acids and bases in equilibrium.when base is added to buffers, they adjust the pH value by releasing hydrogen ions.when acid is added to buffers, they adjust the pH value by consuming hydrogen ions. hence these substances resist any change in their pH value by releasing or absorbing hydrogen ions.This process keeps the pH value constant.

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3 years ago
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If two identical trees are cut down, one with a hand saw, and one with an electric saw...
nataly862011 [7]
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At what height h above the ground does the projectile have a speed of 0.5v?
maw [93]

Answer:

h=\dfrac{3v^2}{8g}

Explanation:

It is given that,

Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.

v=u+at

v=u-gt

Initial speed of the projectile is v and final speed is 0.5 v.

0.5v=v-gt

t=\dfrac{v}{2g}

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Let h is the height above the ground. Using the second equation of motion as :

h=vt-\dfrac{1}{2}gt^2

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So, the height of the projectile above the ground is \dfrac{3v^2}{8g}. Hence, this is the required solution.

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3 years ago
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