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Rasek [7]
3 years ago
12

A softball is thrown from the origin of an x-y coordinate system with an initial speed of 18 m/s at an angle of 35∘ above the ho

rizontal. Part A Find the x positions of the softball at the times t = 0.50 s, 1.0 s, 1.5 s, and 2.0 s.
Physics
1 answer:
podryga [215]3 years ago
6 0
<span>What we need to first do is split the ball's velocity into vertical and horizontal components. To do that multiply by the sin or cos depending upon if you're looking for the horizontal or vertical component. If you're uncertain as to which is which, look at the angle in relationship to 45 degrees. If the angle is less than 45 degrees, the larger value will be the horizontal speed, if the angle is greater than 45 degrees, the larger value will be the vertical speed. So let's calculate the velocities sin(35)*18 m/s = 0.573576436 * 18 m/s = 10.32437585 m/s cos(35)*18 m/s = 0.819152044 * 18 m/s = 14.7447368 m/s Since our angle is less than 45 degrees, the higher velocity is our horizontal velocity which is 14.7447368 m/s. To get the x positions for each moment in time, simply multiply the time by the horizontal speed. So 0.50 s * 14.7447368 m/s = 7.372368399 m 1.00 s * 14.7447368 m/s = 14.7447368 m 1.50 s * 14.7447368 m/s = 22.1171052 m 2.00 s * 14.7447368 m/s = 29.48947359 m Rounding the results to 1 decimal place gives 0.50 s = 7.4 m 1.00 s = 14.7 m 1.50 s = 22.1 m 2.00 s = 29.5 m</span>
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Acceleration = change of velocity divided by time interval = Δv/Δt.
MariettaO [177]

Answer:

a=2.378 m/s^2

Explanation:

a=Δv/Δt------eq(1)

Δv=Vf-Vi=120 km/h-0 km/h=120 km/h

or Δv=33.3 m/sec

or time=t=14s

putting values in eq(1)

a=33.3/14

a=2.378 m/s^2

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3 years ago
When the activation energy of an exothermic reaction decreases at a given temperature, the reaction rate increases because the _
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3 years ago
Four children pull on the same stuffed toy at the same time yet there is no net force on the toy. How is this possible.
Stels [109]

Answer:

Net force is Zero.

Explanation:

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3 0
2 years ago
Read 2 more answers
What force is needed to accelerate a 0.5kg football at a rate of 40m/s
GREYUIT [131]
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6 0
3 years ago
A car starts from rest and accelerates uniformly for a five seconds along a straight road. If speed obtained by the car is 72 km
Step2247 [10]

Answer:

50 meters

Explanation:

Let's start by converting to m/s. There are 3600 seconds in an hour and 1000 meters in a kilometer, meaning that 72km/h is 20m/s.

v_f=v_o+at

Since the car starts at rest, you can write the following equation:

20=0+a(5) \\\\a=20\div 5=4 m/s^2

Now that you have the acceleration, you can do this:

d=v_o+\dfrac{1}{2}at^2

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d=\dfrac{1}{2}(4)(5)^2=2 \cdot 25=50m

Hope this helps!

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3 years ago
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