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qwelly [4]
3 years ago
10

What force does a trampoline have to apply to a gymnast to accelerate her straight up at ? Note that the answer is independent o

f the velocity of the gymnastâshe can be moving either up or down, or be stationary.
Physics
1 answer:
Andrew [12]3 years ago
5 0

Answer: Force applied by trampoline = 778.5 N

<em>Note: The question is incomplete.</em>

<em>The complete question is : What force does a trampoline have to apply to a 45.0 kg gymnast to accelerate her straight up at 7.50 m/s^2? note that the answer is independent of the velocity of the gymnast. She can be moving either up or down or be stationary. </em>

Explanation:

The total required the trampoline by the trampoline = net force accelerating the gymnast upwards + force of gravity on her.

= (m * a) + (m * g)

= m ( a + g)

= 45 kg ( 7.50 *  9.80) m/s²

Force applied by trampoline = 778.5 N

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If a 126 gram marble rolls down a hill at an acceleration of 3.0m/s2, what would the force be on the marble?​
adelina 88 [10]

Answer:

Explanation:

Net force would be the one causing the acceleration or

F = ma = 0.126(3.0) = 0.378 N

4 0
3 years ago
How far will a rubber ball fall in 10 seconds?
Inessa [10]

If it's not moving at all at the beginning of the 10 seconds, then it falls 490 meters straight down in 10 seconds.

(Note: This is true of all objects on Earth . . . rubber balls, feathers, grains of sand, school buses, battle ships . . . everything.  As long as air doesn't hold them back.  Anything falling from rest falls 490 meters in the first 10 seconds.)

4 0
3 years ago
A proton of mass is released from rest just above the lower plate and reaches the top plate with speed . An electron of mass is
xenn [34]

Answer:

  v = √ 2e (V₂-V₁) / m

Explanation:

For this exercise we can use the conservation of the energy of the electron

At the highest point. Resting on the top plate

         Em₀ = U = -e V₁

At the lowest point. Just before touching the bottom plate

        Emf = K + U = ½ m v² - e V₂

Energy is conserved

         Em₀ = Emf

          -eV₁ = ½ m v² - e V₂

           v = √ 2e (V₂-V₁) / m

Where e is the charge of the electron, V₂-V₁ is the potential difference applied to the capacitor and m is the mass of the electron

3 0
4 years ago
Estimate the volume of each ball. Use the formula
Mila [183]

The given question is incomplete. The complete question is:

Estimate the volume of each ball. Use the formula v=\frac{4\pi\times r^3}{3} where v is the volume and r is the radius. record the volume in table A of your student guide. The radius of the tennis ball is 2.1 cm and the radius of thr golf ball is 2.0 cm. What is the estimated volume of the table tennis ball in cm^3&#10; What is the estimated volume of the golf ball in

Answer: Volume of the tennis ball is 38.8cm^3 and  Volume of the golf ball is 33.5cm^3

Explanation:

We have to find the Volume of tennis ball and golf ball by using the formula v=\frac{4\pi\times r^3}{3}

Radius of the tennis ball = 2.1 cm

Radius  of the golf ball =2.0 cm.

Putting the value of radius in the formula , we get:

Volume of the tennis ball =  \frac{4\times 3.14\times (2.1cm)^3}{3}&#10;=38.8cm^3

Volume of the golf ball =   \frac{4\times 3.14\times (2.0cm)^3}{3}&#10;=33.5cm^3

Volume of the tennis ball is 38.8cm^3 and  Volume of the golf ball is 33.5cm^3

6 0
3 years ago
Read 2 more answers
A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa
ivann1987 [24]

Answer:

The moment of inertia of the wheel is 0.593 kg-m².

Explanation:

Given that,

Force = 82.0 N

Radius r = 0.150 m

Angular speed = 12.8 rev/s

Time = 3.88 s

We need to calculate the torque

Using formula of torque

\tau=F\times r

\tau=82.0\times0.150

\tau=12.3\ N-m

Now, The angular acceleration

\dfrac{d\omega}{dt}=\dfrac{12.8\times2\pi}{3.88}

\dfrac{d\omega}{dt}=20.73\ rad/s^2

We need to calculate the moment of inertia

Using relation between torque and moment of inertia

\tau=I\times\dfrac{d\omega}{dt}

I=\dfrac{I}{\dfrac{d\omega}{dt}}

I=\dfrac{12.3}{20.73}

I= 0.593\ kg-m^2

Hence, The moment of inertia of the wheel is 0.593 kg-m².

5 0
3 years ago
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