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3 years ago

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A ball experiences forces of 14 N [N] and 9.2 N [W]. A Free Body diagram is required. What is the acceleration of the ball if it

s mass is 225 g?

1 answer:

GrogVix [38]3 years ago

5 0

Answer:

First, we can define the North as the positive y-axis, and the East as the positive x-axis.

Here we have two forces:

F₁ = 14N to north

F₂ = 9.2 N to west.

Then the free body diagram will be the one that can be seen below:

Now we want to find the acceleration that the ball experiences if the mass is 225g

Knowing that 1k = 1000g, we can rewrite the mass as:

M = 0.225g

Here we also need to remember the second Newton's law:

F = M*a

Net force equals mass times acceleration.

First, let's find the net force:

F = F₁ + F₂ = 14N*(0, 1) + 9.2N*(-1, 0 )

F = (-9.2N, 14N)

The module of the force is then:

I F I = √( (-9.2N)^2 + (14N)^2) = 16.75 N

Then the acceleration is given by

16.75 N = 0.225kg*a

(16.75 N)/(0.225 kg) = a = 74 m/s^2

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A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the s

strojnjashka [21]

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

$m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v$

Where:

$m_{M}$ , $m_{S}$ - Masses of the small meteorite and the communication satellite, measured in kilograms.

$v_{M}$ , $v_{S}$ - Speeds of the small meteorite and the communication satellite, measured in meters per second.

$v$ - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

$v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}$

If $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ and $v_{S} = 0\,\frac{m}{s}$ , the final speed is now calculated:

$v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}$

$v = 0.1\,\frac{m}{s}$

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

$K_{S} + K_{M} - K - Q_{disp} = 0$

$Q_{disp} = K_{S}+K_{M}-K$

Where:

$K_{S}$ , $K_{M}$ - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

$K$ - Kinetic energy of the satellite-meteorite system, measured in joules.

$Q_{disp}$ - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

$Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]$

Given that $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ , $v_{S} = 0\,\frac{m}{s}$ and $v = 0.1\,\frac{m}{s}$ , the dissipated heat is:

$Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]$ $Q_{disp} = 200000\,J$

$Q_{disp} = 200\,kJ$

The energy coverted to heat is 200 kilojoules.

4 0

3 years ago

HELPPPPPP PLEASEEEEEEE-

Alex787 [66]

Answer:

One piece has a north pole only, and the other piece has à soutn pole only.

Explanation:

mark me brainliest!!

8 0

2 years ago

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A certain water wave has a wavelength of 25 m and a frequency of 4.0 Hz. What is its velocity?

babymother [125]

Wave speed = frequency * wavelength
Wave speed = 4 * 25
Wave speed = 100 m/s

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3 years ago

A tug boat pulls a ship with a constant net horizontal force of 5.00•10*3 N and causes the ship to move through a harbor. How mu

Murljashka [212]

The work done on the ship is $1.5\cdot 10^7 J$

Explanation:

The work done by a force on an object is given by:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

In this problem, we have:

$F=5.00\cdot 10^3 N$ (force acting on the ship)

d = 3.00 km = 3000 m (displacement of the ship)

$\theta=0^{\circ}$ (because the force is horizontal, and the displacement is horizontal as well)

Therefore, the work done on the ship is

$W=(5.00\cdot 10^3)(3000)(cos 0^{\circ})=1.5\cdot 10^7 J$

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3 years ago

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Julli [10]

The answer is the first one

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3 years ago

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