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kolezko [41]
2 years ago
14

How should the student use the data collected and the known quantities from the experiment to determine the initial total mechan

ical energy of the block-ramp-Earth system for all trials in the experiment
Physics
1 answer:
fiasKO [112]2 years ago
7 0

It should be noted that the student should use K = 1/2mv² with the initial speed of the block for one trial.

<h3>Method of using the data collected.</h3>

From the complete information, the student wants to use the data collected and the known quantities from the experiment to determine the initial total mechanical energy of the block-ramp-Earth system for all trials in the experiment.

In this case, it's important to use K = 1/2mv² with the block's initial speed for one trial due to the fact that the initial speed is the same in all the trials.

Learn more about experiments on:

brainly.com/question/17274244

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Using the balanced chemical equation below, calculate how many moles of ammonia would be produced when 9.1 moles of hydrogen gas
strojnjashka [21]

6.07 moles

Explanation:

Given parameters:

Number of moles of reacting hydrogen gas = 9.1 moles

Unknown:

Number of moles of ammonia produced = ?

Solution:

  Balanced chemical equation:

           N₂   +     3H₂    →      2NH₃

We should work from the known to the unknown specie. We known the number of moles of hydrogen gas reacting. We simply relate this to that of the ammonia.

 From the reaction:

   3 mole of hydrogen gas produced 2 mole of ammonia

     9.1 moles of the hydrogen gas will produce:  \frac{2 x 9.1}{3}

                                                                            = 6.07 moles

Learn more:

Number of moles brainly.com/question/1841136

#learnwithbrainly

3 0
3 years ago
To model this process, assume two charged spherical conductors are connected by a long conducting wire and a 1.20-mC charge is p
il63 [147K]

Answer:

Part a: The electric potential of each sphere is 1.35x10⁸V

Part b: The electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

Explanation:

As the complete question is not given, the similar question is attached herewith. The values are used as indicated in the given question

Let r_1 = 6 cm=0.06 m

r2 = 2 cm = 0.02 m

Q = 1.2 mC

Let q1 and q2 are the charges on each sphere.

q1 + q2 = 1.2 mC -------(1)

In the equilibrium, V1 = V2

k*q1/r1 = k*q2/r2

q1/0.06 = q2/0.02

q1/q2 = 0.06/0.02

q1/q2 = 3 ---------(2)

On solving equation 1 and 2

we get

q1 = 0.9 mC

q2 = 0.3 mC

So

V1 = k*q1/r1 = (9*10^9*0.9*10^-3)/0.06 = 1.35*10^8 Volts

V2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02 = 1.35*10^8 Volts

So the electric potential of each sphere is 1.35x10⁸V

Part b

Now the electric potential is given as

E1 = k*q1/r1^2 = 9*10^9*0.9*10^-3/0.06^2 = 2.25*10^9 N/C

E2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02^2 = 6.75*10^9 N/C

So the electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

7 0
3 years ago
Two 2.0kg bodies A and B collide The velocities before the collision are U1=15i+30j and U2=-10j+5.0j After the collision V1=-5.0
Svetllana [295]

Use the law of conservation of momentum. Since the momentum is a linear measure, we can treat each of the dimension separately:

i-direction:

m_1u_{1i}+m_2u_{2i}=m_1v_{1i}+m_2v_{2i}\\v_{2i} = \frac{m_1u_{1i}+m_2u_{2i}-m_1v_{1i}}{m_2}=\frac{(2\cdot 15-2\cdot10+2\cdot5)kg\frac{m}{s}}{2kg}=10\frac{m}{s}

j-direction:

m_1u_{1j}+m_2u_{2j}=m_1v_{1j}+m_2v_{2j}\\v_{2j} = \frac{m_1u_{1j}+m_2u_{2j}-m_1v_{1j}}{m_2}=\frac{(2\cdot 30+2\cdot5-2\cdot20)kg\frac{m}{s}}{2kg}=15\frac{m}{s}

Answer: Final velocity is: (10i + 15j) m/s

Change in the kinetic energy:

\Delta E_k = E_{ku}-E_{kv} = \frac{1}{2}m(u_1^2+u_2^2-v_1^2-v_2^2)=\\=\frac{1}{2}2kg(1125+125-425-325)\frac{m^2}{s^2}=500J

Answer: The system lost 500J worth of kinetic energy in the collision

4 0
3 years ago
A bullet of mass 6.20 10-3 kg, moving at 1320 m/s impacts a tree stump and penetrates 11.00 cm into the wood before coming to re
sladkih [1.3K]

Answer:

  F = -49.1   10³ N

Explanation:

Let's use the kinematics to find the acceleration the acceleration of the bullet that they tell us is constant

   v_{f}² = v₀² + 2 a x

Since the bullet is at rest, the final speed is zero

    x = 11.00 cm (1 m / 100 cm) = 0.110 m

    0 = v₀² + 2 a x

   a = -v₀² / 2 x

   a = -1320²/(2 0.110)

   a = -7.92 10⁶ m / s²

With Newton's second law we find the force

   F = m a

   F = 6.20 10⁻³ (-7.92 10⁶)

   F = -49.1   10³ N

The sign means that it is the force that the tree exerts to stop the   bullet

8 0
3 years ago
CAN SOMEBODY PLEASE HELP ME! i need help and i wanna pass
umka21 [38]

Answer:it would be C

Explanation:

8 0
3 years ago
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