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nalin [4]
3 years ago
14

Bob Beamon's 1968 Olympic long jump set a world record which remains unbroken to this day. This amazing jump resulted from an in

itial velocity of 9.5 m/s at an angle of 40 degrees from the horizontal.
1. Calculate the initial horizontal velocity (V_ix) to two significant figures:

2. Calculate the initial vertical velocity (V_iy) to two significant figures:

3. Calculate the time needed to reach the highest point of the jump (t_1/2) to two significant figures:

4. Calculate the total time (t_TOT) needed to complete the jump to two significant figures:

5. Calculate the maximum height (h) reached during the jump to two significant figures:

6. Calculate the range (total horizontal distance) of his jump to two significant figures:

Please answer today! Thanks!
Physics
2 answers:
drek231 [11]3 years ago
8 0

Answer:

1.) 7.3 m/s 2.) 6.1 m/s

Explanation:

To calculate the initial horizontal velocity with just degrees and velocity alone is pretty simple. The formula is     Velocity*cos(degrees)

eg 9.5*cos(40)

2.  To calculate the initial vertical velocity with just degrees and velocity alone is pretty simple. The formula Velocity*sin(degrees)

eg 9.5*sin(40)

notsponge [240]3 years ago
6 0
9-+6 i think !!!! Or 2-+9x1
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kotegsom [21]
Angular momentum is conserved, just before the clay hits and just after; 
<span>mv(L/2) = Iw </span>

<span>I is the combined moment of inertia of the rod, (1/12)ML^2 , and the clay at the tip, m(L/2)^2 ; </span>
<span>I = [(1/12)ML^2 + m(L/2)^2] </span>

<span>Immediately after the collision the kinetic energy of rod + clay swings the rod up so the clay rises to a height "h" above its lowest point, giving it potential energy, mgh. From energy conservation in this phase of the problem; </span>
<span>(1/2)Iw^2 = mgh </span>

<span>Use the "w" found in the conservation of momentum above; and solve for "h" </span>
<span>h = mv^2L^2/8gI </span>

<span>Next, get the angle by noting it is related to "h" as; </span>
<span>h = (L/2) - (L/2)Cos() </span>

<span>So finally </span>
<span>Cos() = 1- 2h/L = 1 - mv^2L/4gI </span>

<span>m=mass of clay </span>
<span>M=mass of rod </span>
<span>L=length of rod </span>
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7 0
3 years ago
How are liquids like gasses
Marta_Voda [28]

Answer: A liquid is made up of tiny vibrating particles of matter, such as atoms, held together by intermolecular bonds. Like a gas, a liquid is able to flow and take the shape of a container. Most liquids resist compression, although others can be compressed.

Explanation:

5 0
3 years ago
A ball is dropped from a height of 20 meters. At what helght does the ball have a velocity of 10 meters/second?
Sunny_sXe [5.5K]

Answer:

5.10 meters.

Explanation:

v²=u²+2gh

or, (10)²=(0)²+2×9.8×h

or, 19.6h=100

or, h=5.10 meters

Hope, this helps you.

8 0
2 years ago
Bill and amy want to ride their bikes to school which is 14.4 kilometers away. It takes Amy 49 minutes to get to school and bill
Drupady [299]

3.4m/s

Explanation:

Given parameters:

Distance to school  = 14.4km

Time taken by Amy = 49min

Time taken by bill = 20min after Amy = 20+49 = 69min

Unknown parameters:

How much faster is Amy's average speed = ?

Solution:

Average speed is the rate of change of total distance with total time taken.

 Average speed = \frac{total distance }{total time taken}

convert units to meters and seconds

      1000m = 1km

       60s = 1min

Distance to school  = 14.4 x 1000 = 14400m

Time taken by Amy = 49 x 60 = 2940s

Time taken by Bill = 69 x 60 = 4140s

Average speed of Amy = \frac{14400}{2940}  = 4.9m/s

Average speed of Bill = \frac{4140}{2940}  = 1.4m/s

Differences in speed = 4.9 - 1.5 = 3.4m/s

Amy was 3.4m/s faster than Bill

learn more:

Average speed brainly.com/question/8893949

#learnwithBrainly

5 0
3 years ago
A 4.30 g bullet moving at 943 m/s strikes a 730 g wooden block at rest on a frictionless surface. The bullet emerges, traveling
Ymorist [56]

Answer:

(a)2.7 m/s

(b) 5.52 m/s

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Initial momentum = final momentum

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⇒ 730 ×v = (4054.9 - 2081.2) =1973.7

⇒v=2.7 m/s

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(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.

V_{COM} = \frac{m_b}{m_b+m_{bl}}v_{bi}=\frac{4.30}{4.30+730}\times 943 m/s = 5.52 m/s

Thus, the speed of the bullet-block center of mass is 5.52 m/s.

4 0
4 years ago
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