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nalin [4]
4 years ago
6

1.) I’m bored, so I decide to play catch by myself. I throw a 1.5kg ball in the air at 25 m/s. How long do I have to wait to cat

ch it?
2.) I shoot a high velocity rifle in the air at an airplane. The plane is flying at 10000. m. The bullet leaves the rifle at 215 m/s. Can I hit the plane?????
Physics
2 answers:
masya89 [10]4 years ago
8 0

Answer:

1) 5 seconds

2) No

Explanation:

1) When you catch the ball, it returns to its initial height, so the displacement is 0.

Given:

Δy = 0 m

v₀ = 25 m/s

a = -10 m/s²

Find: t

Δy = v₀ t + ½ at²

0 m = (25 m/s) t + ½ (-10 m/s²) t²

0 = 25t − 5t²

0 = 5t (5 − t)

t = 5

2) At the maximum height, the vertical velocity is 0 m/s.

Given:

v₀ = 215 m/s

v = 0 m/s

a = -10 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (215 m/s)² + 2 (-10 m/s²) Δy

Δy ≈ 2310 m

No, the bullet cannot reach the plane.

navik [9.2K]4 years ago
4 0

Answer:

technically yes

Explanation:

with a gun depending on how fast it shoots so when you fire at something you shoot in front of it a little bit so you hit it but a plane that fast you shoot like 100 feet infront of it...

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A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizo
Alla [95]

Answer:

v_{ox}= 19.6\ m/s

Explanation:

Data provided in the question:

Height above the ground, H= 5.0m

Range of the ball, R= 20 m

Initial horizontal velocity = v_{ox}

Initial vertical velocity= v_{oy}  (Since ball was thrown horizontally only)

Acceleration acting horizontally, a_x = 0 m/s²  [ Since no acceleration acts horizontally) ]

Vertical Acceleration, a_y = 9.8 m/s² (Since only gravity acts on it)

Let 't' be the time taken to reach ground

Therefore, using equations of motion, we have

H= v_{oy}t+\frac{1}{2}a_yt^2

5= (0)t+\frac{1}{2}(9.8)t^2

t= \frac{10}{9.8}=1.02 s

Then using Equations of motion for horizontal motion,

R= v_{ox}t+\frac{1}{2}a_xt^2

20= v_{ox}(1.02)+\frac{1}{2}(0)(1.02)^2

v_{ox}= 19.6\ m/s

4 0
3 years ago
According to the Heisenberg uncertainty principle, if the uncertainty in the speed of an electron is 3.5 × 103 m/s, the uncertai
GREYUIT [131]

Explanation:

It is given that,

Uncertainty in the speed of an electron, \Delta v=3.5\times 10^3\ m/s

According to Heisenberg uncertainty principle,

\Delta x.\Delta p=\dfrac{h}{4\pi}

\Delta x is the uncertainty in the position of an electron

Since, \Delta p=m\Delta v

\Delta x=\dfrac{h}{4\pi.m \Delta v}

\Delta x=\dfrac{6.6\times 10^{-34}}{4\pi\times 9.1\times 10^{-31}\times 3.5\times 10^3}

\Delta x=1.64\times 10^{-8}\ m

So, the uncertainty in its position is 1.64\times 10^{-8}\ m. Hence, this is the required solution.

6 0
3 years ago
Please help! I'm not sure what equation or the process to do this question.
lions [1.4K]

Answer:

The momentum is 1.94 kg m/s.

Explanation:

To solve this problem we equate the potential energy of the spring with the kinetic energy of the ball.

The potential energy U of the compressed spring is given by

U = \dfrac{1}{2} kx^2,

where x is the length of compression and k is the spring constant.

And the kinetic energy of the ball is

K.E = \dfrac{1}{2}mv^2.

When the spring is released all of the potential energy of the spring goes into the kinetic energy of the ball; therefore,

\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2,

solving for v we get:

v = x \sqrt{\dfrac{k}{m} }.

And since momentum of the ball is p=mv,

p =mx \sqrt{\dfrac{k}{m} }.

Putting in numbers we get:

p =(0.5kg)(0.25m) \sqrt{\dfrac{(120N/m)}{0.5kg} }.

\boxed{p=1.94kg\: m/s}

5 0
3 years ago
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