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Delvig [45]
2 years ago
12

You throw a ball upward with a speed of 14m/s. What is the acceleration of the ball after it leaves your hand? Ignore air resist

ance and provide an explanation for your answer.
Physics
1 answer:
omeli [17]2 years ago
5 0

The acceleration of the ball after leaving the hand is 9.8 m/s^2 downward

Explanation:

In order to find the acceleration of the ball during its motion, we have to study which forces are acting on it.

After the ball leaves the hand, if we neglect air resistance, there is only one force acting on the ball: the force of gravity, whose magnitude is

F=mg

where m is the mass of the ball and g is the acceleration of gravity (g=9.8 m/s^2), acting in the downward direction.

According to Newton's second law, the acceleration of the ball is given by

a=\frac{\sum F}{m}

where

\sum F is the net force acting on the ball

After the ball leaves the hand, the only force acting on it is the force of gravity, so we can substitute (mg) into the previous equation:

a=\frac{mg}{m}=g=9.8 m/s^2

This means that the acceleration of the ball remains 9.8 m/s^2 downward for the entire motion, after leaving the hand.

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this
wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

r=r_o\times A^{\frac{1}{3}}

r_o=1.25 \times 10^{-15} m = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}

r=4.6656\times 10^{-15} m=4.6656 fm

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) F=k\times \frac{q_1q_2}{a^2}

k=9\times 10^9 N m^2/C^2 = Coulombs constant

q_1,q_2 = charges kept at distance 'a' from each other

F = electrostatic force between charges

q_1=+1.602\times 10^{-19} C

q_2=+1.602\times 10^{-19} C

Force of repulsion between two protons on opposite sides of the diameter

a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m

F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}

F=2.6527 N

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

6 0
2 years ago
Behavioral Adaptations: Behavior that animals begin life with that helps them meet their needs is called ________.
Arturiano [62]
I believe the blank would simply be behaviour adaptations. Behavioural adaptations are behaviours that organisms demonstrate to help them better survive and reproduce in a habitat. Hope that helps!!
6 0
2 years ago
Read 2 more answers
If opposing forces acting on an object are equal, the net force is
makkiz [27]

Answer:

0 N

Explanation:

suppose, you push a box with 5 N, and another person pushes the box on the opposite side of the box with 5 N, the net force (resultant ) is 0 N, the box will not move if it wasn't moving

hope this helps

8 0
2 years ago
For both resonance curves and Fourier spectra, amplitude is plotted vs frequency, but these two types of plots are not the same.
andrey2020 [161]

Answer:

he peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

Explanation:

In a resonance experiment, the amplitude of the system is plotted as a function of the frequency, finding maximums for the values ​​where some natural frequency of the system coincides with the excitation frequency.

In a Fourier transform spectrum, the amplitude of the frequencies present is the signal, whereby each peak corresponds to a natural frequency of the system.

From this explanation we can see that in the first case the peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

7 0
3 years ago
A circular loop of wire with a diameter of 0.626 m is rotated in a uniform electric field to a position where the electric flux
Mila [183]

Answer:

C) 2.44 × 106 N/C

Explanation:

The electric flux through a circular loop of wire is given by

\Phi = EA cos \theta

where

E is the electric field

A is the cross-sectional area

\theta is the angle between the direction of the electric field and the normal to A

The flux is maximum when \theta=0^{\circ}, so we are in this situation and therefore cos \theta =1, so we can write

\Phi = EA

Here we have:

\Phi = 7.50\cdot 10^5 N/m^2 C is the flux

d = 0.626 m is the diameter of the coil, so the radius is

r = 0.313 m

and so the area is

A=\pi r^2 = \pi (0.313 m)^2=0.308 m^2

And so, we can find the magnitude of the electric field:

E=\frac{\Phi}{A}=\frac{7.50\cdot 10^5 Nm^2/C}{0.308 m^2}=2.44\cdot 10^6 N/C

3 0
3 years ago
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