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Delvig [45]
3 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]3 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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3 years ago
A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1
aivan3 [116]

Answer:

Part a)

F_n = 306 N

Part b)

v = 12.1 m/s

So this speed is independent of the mass of the rider

Explanation:

Part a)

By force equation on the rider at the position of the hump we can say

mg - F_n = ma_c

now we will have

mg - F_n = \frac{mv^2}{R}

F_n = mg - \frac{mv^2}{R}

now we have

F_n = 100(9.81) - \frac{100(9^2)}{12}

F_n = 981 - 675

F_n = 306 N

Part b)

At the top of the loop if the minimum speed is required so that it remains in contact so we will have

F_n + mg = ma_c

F_n = 0 at minimum speed

mg = \frac{mv^2}{R}

v = \sqrt{Rg}

v = \sqrt{15 \times 9.81}

v = 12.1 m/s

So this speed is independent of the mass of the rider

5 0
3 years ago
A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the
BartSMP [9]

Answer:

r=1.14m

Explanation:

\theta is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

F_m=F_c\\qvB=F_c(1)

F_c is the centripetal force and is defined as:

F_c=m\frac{v^2}{r}

Here v is the proton's speed and r is the radius of the circular motion. Replacing this in (1) and solving for r:

qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}

Recall that 1 J is equal to 6.242*10^{12}MeV, so:

4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J

We can calculate v from the kinetic energy of the proton:

K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}

Finally, we calculate the radius of the proton path:

r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m

8 0
3 years ago
State two applications of electrostatic
kaheart [24]
Atmospheric electricity and storms, electrostatic control filters, and industrial electrostatic seperation as well as spark discharge. these are just a few. hope it helps. 
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3 years ago
1. The process of producing energy by utilizing heat trapped inside the earth's
babymother [125]

Answer:

B

Explanation:

The process is called Geo-Thermal energy because it's an inexhaustible source of energy

6 0
3 years ago
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