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Anastaziya [24]
2 years ago
11

a field hockey ball is launched from the ground at an angle to the horizontal. what are the ball's horizontal and vertical accel

erations at its maximum height?
Physics
1 answer:
cestrela7 [59]2 years ago
5 0
In the absence of air resistance ...

-- The ball has no horizontal acceleration at any point in its flight.

-- The ball's vertical acceleration is 9.8 meters per second-squared downward
at every point its flight, from the moment it leaves the toe of the kicker until it hits
the ground.
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Two male moose charge at each other with the same speed and meet on a icy patch of tundra. As they collide, their antlers lock t
Ksenya-84 [330]
Change in velocity of larger moose: (1/3)v - v = -(2/3)v 
<span>change in velocity of small moose: (1/3)v - (-v) = (4/3)v </span>
<span>- (change in velocity of larger moose)/(change in velocity of smaller moose) = 2

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.



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4 0
2 years ago
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As a descending elevator approaches the correct floor, it slows to a stop with a constant acceleration of magnitude 1m/s*2. The
Nookie1986 [14]

Answer:

C

Explanation:

4 0
2 years ago
Which of the following quantities provide enough information to calculate the tension in a string of mass per unit length μ that
Bad White [126]

Answer:

A. the wave speed v and Wavelength

Explanation:

Given that

Mass density per unit length=μ

Frequency = f

The velocity V given as

\mu=\dfrac{T}{V^2}\ kg/m

V=\sqrt{\dfrac{T}{\mu}}

T=Tension

V=Velocity

V= f λ

λ=Wavelength

Therefore to find the tension ,only wavelength and speed is required.

The answer is A.

8 0
2 years ago
Two stunt drivers drive directly toward each other. At time t=0 the two cars are a distance D apart, car 1 is at rest, and car 2
lesantik [10]

Answer: Hello there!

We know this:

The distance between the cars at t= 0 is D.

car 2 has an initial velocity of v0 and no acceleration.

car 1 has no initial velocity and a acceleration of ax that starts at  t = 0

then we could obtain the acceleration of the car 1 by integrating the acceleration over the time; this is v(t) = ax*t where there is not a constant of integration because the car 1 has no initial velocity.

Because the cars are moving against each other, we want to se at what time t they meet, this is equivalent to see:  

position of car 1 + position of car 2 = D

and in this way we could ignore constants of integration :D

for the position of each car we integrate again:  

P1(t) = (1/2)ax*t^2 and P2(t) = v0t

v0t + (1/2)ax*t^2 = D

v0t + (1/2)ax*t^2  - D = 0

now we can solve it for t using the Bhaskara equation.

t = \frac{-v0 +\sqrt{v0^{2} + 4*(1/2)ax*D } }{2(1/2)ax} =\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax}

that we cant solve witout knowing the values for v0, D and ax. But you could replace them in that equation and obtain the time, where you must remember that you need to choose the positive solution (because this quadratic equation has two solutions).

Now we want to know the velocity of car 1 just before the impact, this can be calculated by valuating the time in the as the time that we just found in the velocity equation for the car 1, this is:

v(\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax}) = ax*\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax} = {-v0 +\sqrt{v0^{2} + 2ax*D }

where again, you need to replace the values of v0, D and ax.

7 0
2 years ago
A 75.0 kg sailor climbs a 28.3 m rope ladder at and angle of 45.0degrees with the mast. how much work did he do?
sladkih [1.3K]
W = m · g · h
h = 28.3 m · sin 45° = 28.3 m · 0.707 = 20 m
g = 9.8 m/s²
W = 75 kg · 9.8 m/s² · 20 m
Answer:
W = 14,700 J = 14.7 kJ 

8 0
2 years ago
Read 2 more answers
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