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Veronika [31]
3 years ago
8

Patrick Mahomes is throwing a football with an applied force of 230 N but is experiencing air resistance of 30 N. The ball has a

mass of 5kg. What is the acceleration of the ball?
Physics
1 answer:
Ganezh [65]3 years ago
7 0

Answer:

www.chiefs.com

Web results

Patrick Mahomes

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"a needle can be made to "float" on the surface tension of water. what causes this surface tension to form"
astra-53 [7]
The cohesion of water molecules to each other
5 0
4 years ago
CAN SOMEONE PLEASE HELP ME
Anna71 [15]

Answer:

No, it is not fair because they could overdo it and get injured and others may not have it.

3 0
3 years ago
a water line starts the service with an altitude of 1200m over the sea level, what is the velocity of the water above 1050 m ove
Ilia_Sergeevich [38]

Answer:

Velocity = 94.85m/s

Explanation:

<u>Given the following data ;</u>

Height = 1200m

Vertical distance = 1050m

To find the time, we would use the second equation of motion;

S = ut + \frac {1}{2}at^{2}

Substituting into the equation, we have;

1200 = 0(t) + \frac {1}{2}*9.8*t^{2}

1200 = 0 + 4.9*t^{2}

1200 = 4.9*t^{2}

t^{2} = \frac {1200}{4.9}

t = \sqrt{122.45}

t = 11.07 secs

To find the velocity;

Mathematically, velocity is given by the equation;

Velocity = \frac{distance}{time}

Substituting into the above equation;

Velocity = \frac{1050}{11.07}

Velocity = 94.85m/s

Therefore, the velocity of the water above 1050 m over the sea level is 94.85m/s.

7 0
3 years ago
A​ hot-air balloon is 150 ft above the ground when a motorcycle​ (traveling in a straight line on a horizontal​ road) passes dir
Katyanochek1 [597]

Answer:

 dR/dt = 10.2 ft / s

Explanation:

Let's work this problem by finding the distance between the balloon and the motorcycle and then drift for the speed change of the distance

Balloon

      y = y₀ +v_{oy} t

Motorcycle

      x = v₀ₓ t

Distance, let's use Pythagoras' theorem

      R² = x² + y²

      R² = (v₀ₓ t)² + (y₀ + v_{oy} t)²

     v₀ₓ = 88 ft / s

     v_{oy} = 8 ft / s

     y₀ = 150 ft

     R² = (8 t)² + (150 + 8 t )²

     R² = 64 t² + (150 + 8t )²

This is the expression for the distance between the two bodies, the rate of change is the derivative with respect to time (d / dt)

         2RdR / dt = 64 2 t + 2 (150 + 8t) 8

        dR / dt = [64 t + (1200 + 64t )] / R

dR/dt = (1200 +128 t)/R

Let's calculate for the time of 10 s

        dR / dt = (1200 + 128 10) / R = 2480 /R

       R = √ [64 10² + (150 + 8 10)²

       R = √ [6400 + 52900]

       R = 243.5 ft

       dR / dt = (2480) / 243.5

       dR / dt = 10.2 ft / s

8 0
4 years ago
What is the energy in joules and ev of a photon in a radio wave from an am station that has a 1500 khz broadcast frequency?
LekaFEV [45]
E=hf
E=6.626e-34[Js]•1500e3[Hz]=9.939e-28[J]
For eV divide this by electron charge
9.939e-28[J]/1.6022e-19[C/e]=
6.203e-9eV
4 0
4 years ago
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