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Darya [45]
2 years ago
14

a stone is thrown vertically upwards with an initial velocity of 20 metre per second find the maximum height it reaches and the

time taken by to reach the height( g =10m/s)​
Physics
1 answer:
ziro4ka [17]2 years ago
4 0

Answer:

y = 20 [m]

Explanation:

To solve this type of problem we must use the following equation of kinematics.

v_{f}^{2} =v_{o}^{2}-2*g*y

where:

Vf = final velocity = 0

Vo = initial velocity = 20 [m/s]

g = gravity acceleration = 10 [m/s²]

y = elevation [m]

Now the key to solving this problem is to know that when the stone reaches the maximum height its speed is zero. Therefore the final velocity term is zero.

0 = (20)^{2}-2*10*y\\20*y=400\\y=20 [m]

Another important fact is that the sign of gravitational acceleration was taken as negative, since it acts in the opposite direction to the movement of the stone.

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a pennyfarthing is a style of a bicycle with a very large front wheel and a small real wheel, the cyclist who sit high above and
s344n2d4d5 [400]

Answer:

In one rotation, the large wheel turns 4m.

Explanation:

The given values are:

Input distance,

= 0.64 m

Mechanical advantage,

= 0.16

As we know,

⇒ Out. \ Distance = \frac{Inp. \ distance}{Mechanical \ advantage}

On putting the values, we get

⇒                         =\frac{0.64}{0.16}

⇒                         =4 \ m

4 0
3 years ago
A tennis ball and a solid steel ball of the same diameter are dropped at the same time. Ignoring the air resistance effects, whi
Paha777 [63]
Yes I have the answer
6 0
3 years ago
Calculate the pressure of water in a will if the deep of the water is 10 m​
Bas_tet [7]

Answer:

98,000 pa

Explanation:

The formula for water pressure is as follows:

pressure = pgh

Where <em>p </em>is the density of water (in kg/m3), <em>g </em>is the gravitational field strength, and <em>h </em>is the height of the water.

The density of water is 1000kg/m3, the gravitational field strength is 9.8, and the height is 10. Substituting in these values:

pressure = 1000 \times 9.8 \times 10

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7 0
1 year ago
A sample of 4.50 g of methane occupies 12.7 dm3 at 310 K. (a) Calculate the work done when the gas expands isothermally against
valkas [14]

Answer:

(A) Work done will be 87.992 KJ

(B) Work done will be 167.4 KJ            

Explanation:

We have given mass of methane m = 4.5 gram = 0.0045 kg

Volume occupies V_1=12.7dm^3=12.7liters

And volume is increased by 3.3dm^3 so V_2=12.7+3.3=16liters

Temperature T = 310 K

Pressure is given as 200 Torr = 26664.5 Pa

(a) At constant pressure work done is given by

W=P(V_2-V_1)=26664.5\times (16-12.7)=87992.85J=87.992kj

(b) At reversible process work done is given by W=nRTln\frac{V_2}{V_1}

We have given mass = 4.5 gram

Molar mass of methane = 16

So number of moles n=\frac{mass\ in\ gram}{mol;ar\ mass}=\frac{4.5}{16}=0.28125

So work done W=0.28125\times 8.314\times 310ln\frac{16}{12.7}=167.4J

7 0
2 years ago
3. Your friend says your body is made up of more than 99.9999% empty space. What do you think?
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I would agree with the statement. it's not just the body, but everything that we see is almost 99.9999% empty space

8 0
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