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MariettaO [177]
3 years ago
9

A spring is compressed so that it has 7.2 J of elastic potential energy. A 0.3 kg ball is placed on top of the spring. When the

spring is released, how high will the ball go ?
Physics
1 answer:
VLD [36.1K]3 years ago
5 0

The height of the ball is 2.45 m.

To calculate the height of the ball, we use the formula below.

<h3>Formula</h3>
  • E = mgh................ Equation 1

Where:

  • E = Elastic energy of the spring
  • m = mass of the ball
  • h = height of the ball
  • g = acceleration due to gravity.

make h the subject of the equation

  • h = E/mg...............Equation 2

From the question,

Given:

  • E = 7.2 J
  • m = 0.3 kg
  • g = 9.8 m/s²

Substitute these  values into equation 2

  • h = 7.2/(0.3×9.8)
  • h = 7.2/2.94
  • h = 2.45 m

Hence, The height of the ball is 2.45 m.

Learn more about height here: brainly.com/question/1739912


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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
Speed of 26.7 m/s in 3.06 s. How far had the car traveled by the time the final speed was achieved?
Alik [6]

Answer:

Below

Explanation:

You can use this equation to find the distance :

     distance = velocity x time

     distance = (26.7)(3.06)

                    = 81.702 m

Rounding to 3 sig figs

     = 81.7 m

Hope this helps

6 0
3 years ago
An object is at rest in front of a compressed spring. It travels over a surface that exerts a kinetic frictional force on it and
PolarNik [594]

Answer:

the object will travel 0.66 meters before to stop.

Explanation:

Using the energy conservation theorem:

E_i+K_i+W_f=K_f+U_f

The work done by the friction force is given by:

W_f=F_f*d\\W_f=\µ*m*g*d\\W_f=0.35*4*9.81*d\\W_f=13.7d[J]

so:

\frac{1}{2}1800*(10*10^{-2})+0-13.7d=0+0\\d=0.66m

3 0
3 years ago
A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the
lana [24]
(a) Let's convert the final speed of the car in m/s:
v_f = 61 km/h = 16.9 m/s
The kinetic energy of the car at t=19 s is
K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
P= \frac{W}{\Delta t}
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
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(c) The instantaneous power is given by
P_i = Fv_f
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So we need to find the acceleration first:
a= \frac{v_f-v_i}{\Delta t}=  \frac{16.9 m/s}{19 s}=0.89 m/s^2
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
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5 0
3 years ago
A wattage rating of a lightbulb is the power it consumers when it is connected across a 120 V potential difference. How does the
navik [9.2K]

Answer:

The answer is C.

120 V with 60 W light bulb is 240 ohms.

120 V with 100 W light bulb is 144 ohms.

The 100 W bulb has less resistance :)

7 0
3 years ago
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