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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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I need help, please answer
Burka [1]

This being a perfect collision means no energy is lost during the collision. Because this question asks for speed and not velocity, the speed will be the same because the final energy is the same. The speed after the collision would therefore be 1.27 m/s.

7 0
4 years ago
A jet airliner moving initially at 693 mph (with respect to the ground) to the east moves into a region where the wind is blowin
Nesterboy [21]

Answer:

The new speed of the aircraft with respect to the ground is 1414.3 mph.

Explanation:

Given that,

Angle = 37°

Velocity of jet airliner = 693 mph

Velocity of wind = 798 mph

We know that,

The new velocity of the aircraft with respect to the ground

v=v_{a}+v_{w}

We need to calculate the new speed of the aircraft with respect to the ground

Using formula for velocity

v=\sqrt{(v_{a})^2+(v_{w})^2+2v_{a}v_{w}\cos\theta}

Put the value into the formula

v=\sqrt{(693)^2+(798)^2+2\times693\times798\cos37}

v=1414.3\ mph

Hence, The new speed of the aircraft with respect to the ground is 1414.3 mph.

7 0
3 years ago
12.0-kg block is pushed across a rough horizontal surface by a force that is angled 30.0◦below thehorizontal. The magnitude of t
drek231 [11]

Answer: Fr = 26.53 N

Explanation: The constant force exerted on the block by the surface is the frictional force.

This frictional force is as a result of interaction between the body and the surface.

According to newton's second law of motion,

F - Fr = ma

F=applied force

Fr = magnitude of frictional force

m = mass of object = 12kg

a = acceleration of object = 3.2m/s²

The applied force (F= 75 N) is inclined at an angle of 30° to the horizontal thus making it have 2 components of forces given below

Fx = 75 * cos 30 = 64.95 N (horizontal component)

Fy = 75 * sin 30 = 37.5 N ( vertical motion)

The body moves across the surface, hence the horizontal component of force is responsible for motion.

F = 64.95 N

By substituting the parameters, we have that

64.96 - Fr = 12 * 3.2

64.96 - Fr = 38.4

Fr = 64.96 - 38.4

Fr = 26.53 N

7 0
3 years ago
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450 kg is your answer middle school physic


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