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vodomira [7]
2 years ago
6

In the figure, a baseball is hit at a height h = 1.20 m and then caught at the same height. It travels alongside a wall, moving

up past the top of the wall 1.1 s after it is hit and then down past the top of the wall 4.8 s later, at distance D = 54 m farther along the wall. (a) What horizontal distance is traveled by the ball from hit to catch? What are the (b) magnitude and (c) angle (relative to the horizontal) of the ball's velocity just after being hit? (d) How high is the wall?
Physics
1 answer:
pickupchik [31]2 years ago
8 0

Answer:

D.

Explanation:

The ball travels distance D in ∆t = 2.60 s. Therefore, the x component of the initial velocity is D =v

ox

​

 ⇒ ∆t = D/∆t = 15.4 m/s Due to the symmetry of projectile motions, it takes 1.2 seconds the ball to reach the initial height after passing the top of the wall downward. Therefore, the total flight time is t

tot

​

 = 1.2 + 2.6 +1.2 = 5 s The horizontal range is R = v

ox

​

t

tot

​

 = 77.0 m  

By symmetry, the ball reaches the peak of the motion at time t

top

​

 = t

tot

​

/2 = 2.5 s At the peak, the y component of the velocity is zero. Thus we have 0 = v

oy

​

 – gt

top

​

 ⇒ v

oy

​

  = gt

top

​

= 24.5 m/s The height of the wall is equal to the height of the ball at t = 1.2 s h = y (t = 1.2 s) = v

oy

​

 +v

oy

​

t – (1/2)gt

2

  = 23.3 m

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A proton is initially at rest. After some time, a uniform electric field is turned on and the proton accelerates. The magnitude
marusya05 [52]

Answer:

a) 8.83*10⁵ m/s  b) 2.80*10⁶ m/s

Explanation:

a) Assuming no other forces acting on the proton, the acceleration on it is produced by the electric field.

By definition, the  force due to the electric field is as follows:

F = q*E = e*E (1)

where e is the elementary charge, the charge carried by only one proton, and is e = 1.6*10⁻¹⁹ C.

According to Newton's 2nd law, this force is at the same time, the product of the mass of the proton, times the acceleration a:

F = mp*a (2)

From (1) and(2), being left sides equal, right sides must be equal too:

F = e*E = mp*a

Solving for a:

a = \frac{e*E}{mp} =\frac{1.6e-19C*1.36e5N/C}{1.67e-27kg} =1.3e13 m/s2

⇒ a = 1.3*10¹³ m/s²

As we have the value of a (which is constant due to the field is uniform), the displacement x, and we know that the initial velocity is 0, in order to get the value of the speed, we can use the following kinematic equation:

vf^{2} -vo^{2} = 2*a*x

Replacing by v₀ = 0, a= 1.3*10¹³ m/s² and  x = 0.03 m, we can find vf as follows:

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⇒ vf = 8.83*10⁵ m/s

b) We can just repeat the equation from above, replacing x=0.03 m by x=0.3 m, as follows:

vf =\sqrt{2*(1.3e13 m/s2)*0.3m} = 2.80e6 m/s

⇒ vf = 2.80*10⁶ m/s

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Answer:

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Explanation:

An inflated balloon has a high pressure region on its inside. Gases always move from a region of high pressure to a region of low pressure. When a balloon is inflated its membrane stretches making it even more porous.

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