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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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ikadub [295]

Answer:

f_{fr}=1590.85 N

Explanation:

Here the total force at the horizontal components will be equal to the centripetal force on the car. So we will have:

f_{fr}cos(25)+Nsin(25)=m\frac{v^{2}}{r} (1)

  • f(fr) is the friction force
  • N is the normal force

Now, the sum of forces at the vertical direction is equal to 0.

Ncos(25)-mg-f_{fr}sin(25)=0 (2)          

Let's combine (1) and (2) to find f(fr)

f_{fr}=\frac{(mv^{2}/r)-mgtan(25)}{cos(25)+tan(25)sin(25)}

f_{fr}=\frac{(600*30^{2}/120)-600*9.81*tan(25)}{cos(25)+tan(25)sin(25)}  

f_{fr}=1590.85 N

I hope it helps you!

5 0
3 years ago
A train moving at 5 m/sec passed a track gang and then accelerated uniformly a rate of 1.2 m/sec/sec for 5 min. How far did the
Amanda [17]

Answer:

S = V0 t + 1/2 a t^2

S = 5 m/s * 300 s + 1/2 * 1.2 m/s * (300 s^2)

S = 1500 m + .6 * 90000 m = 55,500 m

Check:     V0 = 5 m/s

                V2 = V0 + a t  = 5 + 1.2 * 300 = 365 m/s

Vav = (V1 + V2) / 2 = (5 + 365) / 2 = 185 m/s     (note uniform motion)

S = 185 * 300 = 55,500 m

We calculated V2 above at 365 m/s  the speed after 300 sec

3 0
3 years ago
An 800 kHz radio signal is detected at a point 3.2 km distant from a transmitter tower. The electric field amplitude of the sign
hammer [34]

Answer:

1.07 nT

Explanation:

We know that E/B = c where E = electric field amplitude = 320 mV/m = 0.32 V/m, B = magnetic field amplitude and c = speed of light = 3 × 10⁸ m/s.

So, B = E/c

Substituting E and c into B, we have

B = E/c

= 0.32 V/m ÷ 3 × 10⁸ m/s

= 0.1067 × 10⁻⁸ T

= 1.067 × 10⁻⁹ T

= 1.067 nT

≅ 1.07 nT

3 0
3 years ago
Give an example of a reputable website
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6 0
3 years ago
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An old light bulb draws only 54.3 W, rather than its original 60.0 W, due to evaporative thinning of its filament. By what facto
Lemur [1.5K]

Answer:

The factor of the diameter is 0.95.

Explanation:

Given that,

Power of old light bulb = 54.3 W

Power = 60 W

We know that,

The resistance is inversely proportional to the diameter.

R\propto\dfrac{1}{D}

The power is inversely proportional to the resistance.

P\propto\dfrac{1}{R}

P\propto D^2

We need to calculate the factor of the diameter of the filament reduced

Using relation of power and diameter

\dfrac{P_{i}}{P_{f}}=\dfrac{D_{i}^2}{D_{f}^2}

Put the value into the formula

\dfrac{D_{i}^2}{D_{f}^2}=\dfrac{54.3}{60}

\dfrac{D_{i}}{D_{f}}=0.95

D_{i}=0.95 D_{f}

Hence, The factor of the diameter is 0.95.

7 0
3 years ago
Read 2 more answers
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