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vodomira [7]
3 years ago
9

A football player kicks a ball at a 30o angle from the ground with an initial velocity of 15 m/s. What is the final velocity of

the ball when it hits the ground?
Physics
1 answer:
navik [9.2K]3 years ago
7 0

Given that,

Angle = 30°

Initial velocity = 15 m/s

We need to calculate the time of flight

Using formula of time of flight

T=\dfrac{2u\sin\theta}{g}

Where, u = initial velocity

g = acceleration due to gravity

Put the value into the formula

T=\dfrac{2\times15\sin30}{9.8}

T=1.5\ sec

We need to calculate the final velocity of the ball

Using equation of motion

v=u+gt

v=15+9.8\times1.5

v=29.7\ m/s

Hence, The final velocity of the ball is 29.7 m/s.

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A bungee cord is 30m long and when stretched a distance x itexerts a restoring force of magnitude kx. Your father-in-law (mass95
madam [21]

Answer:

The distance the bungee cord that would be stretched 0.602 m, should be selected when pulled by a force of 380 N.

Explanation:

As from the given data

the length of the rope is given as l=30 m

the stretched length is given as l'=41m

the stretched length required is give as  y=l'-l=41-30=11m

the mass is m=95 kg

the  force is  F=380 N

the gravitational acceleration is g=9.8 m/s2

The equation of  k is given by equating the energy at the equilibrium point which is given as

U_{potential}=U_{elastic}\\mgh=\dfrac{1}{2} k y^2\\k=\dfrac{2mgh}{y^2}

Here

m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so

k=\dfrac{2mgh}{y^2}\\k=\dfrac{2\times 95\times 9.8\times 41}{11^2}\\k=630.92 N/m

Now the force is

F=kx\\ or

x=\dfrac{F}{k}\\

So here F=380 N, k=630.92 N/m

x=\dfrac{F}{k}\\x=\dfrac{380}{630.92}\\x=0.602 m

So the distance is 0.602 m

6 0
3 years ago
Joe is attempting to ramp his bicycle over a row of burning tires. If the ramp has a launch angle of 20 degrees, and Joe can ped
Luden [163]
Answer is 6 tires.

This is a projectile question.

First make sure units are consistent - express speed in m/s.

20 km/h = 20000m / 3600 s = 5.56 m/s

Assume the takeoff point of the ramp is at ground level (height, h, = 0m). We need to determine how long Joe is in the air, and use that time to calculate the horizontal distance he traveled.

Joe is traveling 5.56 m/s on a ramp angled at 20 degrees. There are vertical and horizontal components to his speed:

Vertical speed = 5.56sin20 = 1.90 m/s
Horizontal speed = 5.56cos20 = 5.22 m/s

An easy way to proceed is to calculate the time it takes for Joe’s vertical speed to reach 0m/s - this represents the time when Joe is at his maximum height and is therefore halfway through the trip. Double whatever time this is to find the total time of the trip. Remember he is decelerating due to gravity:

Time to peak:
a = Δv / Δt
-9.8 = -1.9 / Δt
Δt = 0.19s

Total trip time:
0.19 x 2 = 0.38s

Now that we have the total tome Joe is in the air, we can find the horizontal distance he traveled:

v = d / t
5.22 = d / 0.38
d = 1.98m

Now divide this total distance by the length of an individual tire to find the number of tires he will clear:

1.98 / 0.3 = 6.6 tires

Therefore he can jump 6 tires safely (he will land in the middle of the 7th tire).

Lots of steps I know but just try to think of the situation and keep track of the vertical and horizontal things!
4 0
3 years ago
1. You have a cat who has a mass of 10 kg and is
disa [49]

Answer:

1) F = 100N

2) a = 2 m/s²

3) m = 25 kg

Explanation:

1) F = ma ( F = ?, m = 10 kg, a = 10 m/s² )

  F = 10×10

  F = 100 N

2) F = ma ( F = 20N, m = 10 kg, a = ? )

   20 = 10×a

   10a = 20

   a = 20/10

   a = 2 m/s²

3)F = ma ( F = 100N, m = ?, a = 4 m/s² )

  100 = m×4

  4m = 100

  m = 100/4

  m = 25 kg

Hope that helps! Good luck!

 

7 0
3 years ago
Dr. Haviland is looking at mental rotation data but he forgot to label his chart. The data shows reaction times for 700 ms, 1400
ira [324]

Answer:

Explanation:

(C) 180 degrees, 90 degrees, 120 degrees.

6 0
3 years ago
A glacier advanced down a mountain from an elevation of 2010 m to 1780 m in 5 years. What was the glaciers rate of change in a y
Ivanshal [37]

Answer:

The solution of the given problem is provided in the following subsection.

Explanation:

According to the question,

The change in the height of the glacier will be:

= 1780-2010

= -230

Change in time,

= 5 years

Now,

The rate of change will be:

= \frac{m}{years}           ∵ (1 year = 12 months)

= \frac{230}{5}

= -46 \ m/years

or,

The rate of change will be:

= \frac{m}{months}         ∵ (5 years in months = 5×12 months)

= \frac{-230}{5\times 12}

= \frac{-230}{60}

= =3.83 \ m/months

6 0
3 years ago
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