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NeX [460]
3 years ago
13

Pweese help luvs qwq pweese look at the image?

Physics
1 answer:
hjlf3 years ago
6 0

Answer:

12 m/s

Explanation:

divide distance over time

72/6 = 12

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True or false? the potential energy of a membrane potential comes solely from the difference in electrical charge across the mem
bazaltina [42]
The statement would be False. T<span>he potential energy of a membrane potential comes solely from the difference in electrical charge across the membrane. In addition to that, membrane potential actually regulates the potential difference of nerve cells across the membrane estimated at 70 mV.</span>
7 0
3 years ago
__________energy might also be released during a chemical reaction
pentagon [3]
Kinetic energy i think
7 0
3 years ago
Using Figure 2, what is the momentum of Train Car A before the collision?
Bess [88]

Answer:

Option A. 180000 Kgm/s.

Explanation:

From the question given above, the following data were obtained:

For Train Car A:

Mass of train car A = 45000 Kg

Velocity of train car A = 4 m/s

Momentum of train car A =?

For Train Car B:

Mass of train car B = 45000 Kg

Velocity of train car B = 0 m/s

Momentum is simply defined as the product of mass and velocity. Mathematically, it can be expressed as:

Momentum = mass × velocity

With the above formula, the momentum of train car A before collision can be obtained as follow:

Mass of train car A = 45000 Kg

Velocity of train car A = 4 m/s

Momentum of train car A =?

Momentum = mass × velocity

Momentum = 45000 × 4

Momentum of train car A = 180000 Kgm/s

5 0
3 years ago
The root-mean-square speed (thermal speed) of the molecules of a gas is 200m/s at 23.0°C. At 227°C the root-mean-square speed (t
777dan777 [17]

Answer:

330 m/s  approx

Explanation:

The RMS speed of a gas is proportional to square root of its absolute temperature is

V ( RMS ) ∝ √T

\frac{V_1}{V_2} =\sqrt{\frac{T_1}{T_2} }

Here V₁ = 200 , T₁ = 23 +273 = 300K , T₂ = 227 +273 = 500 K

Putting the values

200 / V₂ = \sqrt{\frac{300}{500} }

V₂ = 330 m/s  approx

8 0
3 years ago
Interactive Solution 9.37 presents a method for modeling this problem. Multiple-Concept Example 10 offers useful background for
viva [34]

Answer:

21.67 rad/s²

208.36538 N

Explanation:

\omega_f = Final angular velocity = \dfrac{1}{6}78=13\ rad/s

\omega_i = Initial angular velocity = 78 rad/s

\alpha = Angular acceleration

\theta = Angle of rotation

t = Time taken

r = Radius = 0.13

I = Moment of inertia = 1.25 kgm²

From equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{13-78}{3}\\\Rightarrow \alpha=-21.67\ rad/s^2

The magnitude of the angular deceleration of the cylinder is 21.67 rad/s²

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=1.25\times -21.67\\\Rightarrow \tau=-27.0875

Frictional force is given by

F=\dfrac{\tau}{r}\\\Rightarrow F=\dfrac{-27.0875}{0.13}\\\Rightarrow F=-208.36538\ N

The magnitude of the force of friction applied by the brake shoe is 208.36538 N

5 0
3 years ago
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