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mixas84 [53]
3 years ago
12

Calculate the speed for a car that went a distance of 125 kilometers in 2 hours time.

Physics
1 answer:
Murljashka [212]3 years ago
7 0
S=125km
t=2h
v=s/t=125/2=62,5km/h
or 62,5/3,6=17,36m/s
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We want to know what is the power supplied by the power cell if the current I=0.5 A and the voltage V=0.43 V. The equation for power P is P= I*V, so:

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So the correct answer is that the power cell is supplying the motor with P=0.215 W of power. 
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3 years ago
What is the magnitude of the electric field strength between them, if the potential 7.95 cm from the zero volt plate (and 2.05 c
Dmitriy789 [7]

Answer:

ΔVab = Ed

ΔVab = Va-Vb = Va-V0 = Va

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3 years ago
Find the time it takes for each object to accelerate from 0m/s to 40 m/s when pushed with 100N of force
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40s

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F=mv-mu

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4 0
3 years ago
The relative highness or lowness of a sound is called ______. Multiple Choice pitch timbre dynamics octave
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8 0
2 years ago
One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has
Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE)  = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = \sqrt{6gL}

Hence, v0 = 6.86 m/s

4 0
3 years ago
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