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soldi70 [24.7K]
3 years ago
8

How do i calculate this?

Physics
1 answer:
Lesechka [4]3 years ago
8 0

CAR 1

Momentum = Mass/Velocity

M = 2100/20

M = 105 m/s^2

CAR 2

Momentum = Mass/Velocity

M = 2100/30

M = 70 m/s^2

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Answer:

Maximum speed of the rod, v = 10.34 m/s

Explanation:

It is given that,

Voltage of the battery, V = 2.7 V

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Due to the motion of the rails inside the magnetic field, an emf will induced in it which is given by :

\epsilon=Blv

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v=\dfrac{\epsilon}{Bl}

v=\dfrac{V}{Bl}

v=\dfrac{2.7}{0.9\times 0.29}

v = 10.34 m/s

So, the maximum speed attained by the rod after the switch is closed is 10.34 m/s. Hence, this is the required solution.

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3 years ago
If you hold your arm outstretched with palm upward, the force to keep your arm from falling comes from your deltoid muscle. assu
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Your question seems to be incorrect. Please check below:

What force must the deltoid muscle provide to keep the arm in this position? By what factor does this force exceed the weight of the arm?<span>If you hold your arm outstretched with palm upward, as in (Figure 1) , the force to keep your arm from falling comes from your deltoid muscle. Assume that the arm has mass 4 kg and the distances and angles shown in (Figure 1) . 

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A robot that can move a hand left and right and rotate the hand 360º is said to have
Luda [366]

You've described two (2) axes of motion.
The third one would have been up-and-down.
 
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3 years ago
The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22
alexgriva [62]
1) The law of motion of the projectile is
h(t) = 2+22.5 t-4.9 t^2
To find the velocity, we should compute the derivative of h(t):
v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t
So now we can calculate the speed at t=2 s and t=4 s:
v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s
v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
v(t)=0
So we have
22.5-9.8 t=0
And solving this we find
t=2.30 s

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
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This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
t=4.68 s
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36  m/s
with negative sign, because it is directed downward.
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myrzilka [38]

Answer:

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Explanation:

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3 years ago
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