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vladimir1956 [14]
3 years ago
10

.A coin rolls off the edge of a table. The coin

Physics
1 answer:
geniusboy [140]3 years ago
5 0

Answer:

Apply the following formulae horizontally And get A value for time

Remember horizontal acceleration is zero

s  = ut +  \frac{1}{2}a {t}^{2}   \\ 0.8 = 1.7 \times t \\  \frac{0.8}{1.7}  = t \\ t = 0.47s

and then to find the height apply the same above equation vertically...remember vertical initial velocity is zero

s = ut +  \frac{1}{2} a {t}^{2}  \\ s =  \frac{1}{2}  \times 10 \times (0.47) ^{2}  \\ s = 1.1045m

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Evaluate ( 64800 ms ) 2 to three significant figures and express the answer in Si units. Express your answer using three signifi
Sergeu [11.5K]

Answer:

42.0×10² second²

Explanation:

Here, time is given in milisecond

(64800 ms)²

= 4199040000 ms²

The SI unit is seconds

1 second = 1000 milisecond

1\ milisecond=\frac{1}{1000}\ second

\\\Rightarrow 1\ milisecond^2=\left(\frac{1}{1000}\right)^2\ second^2

4199040000\ ms^2=4199040000\times \left(\frac{1}{1000}\right)^2\ second^2=4199.04\ second^2

42.0×10² second²

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3 years ago
Write your opinion about achievement made by during rana rule​
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Answer:

Your opinion about achievement made by during rana rule

Explanation:

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3 years ago
The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at
denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

A)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m  _____ eqn (1)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s  _____ eqn (2)

<u>FOR OXYGEN:</u>

velocity of oxygen = √(3KT)/(mass of oxygen)  

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

<u>velocity of oxygen = (1/4)(2000 m/s) = 500 m/s</u>

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

<u>K.E of Oxygen = 4000 J</u>

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

<u>v = 2149.24 m/s</u>

<u></u>

8 0
4 years ago
Which benefit outweighs the risks in the technological design of headphones?
Dmitriy789 [7]

Answer:

The Answer is A They can damage hearing.

Explanation:

I took the Test

7 0
3 years ago
A sled starts off with an initial velocity of 8 m/s and slows down to 2 m/s after 3 seconds. What was its acceleration?
melomori [17]

Answer:

-2m/s

Explanation:

3 0
3 years ago
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