Find the mass of a golfball that accelerates at 18m/s2 when 8.3 N force is applied to it

if a train starts from rest and attains a velocity of 100m/s in 25 seconds. calculate the acceleration produced by the train.

stira [4]

-4 km/s2

Explanation:
0-100/25
-100/25
-4

5 0

2 years ago

A satellite is spinning at 6.0 rev/s. The satellite consists of a main body in the shape of a sphere of radius 2.0 m and mass 10

bearhunter [10]

Answer:

605447.7066 kgm²/s

Explanation:

$m_1$ = Mass of sphere = 10000 kg

$m_2$ = Mass of rod = 10 kg

r = Radius of sphere = 2 m

l = Length of antenna = 3 m

Angular speed

$\omega=6\times 2\pi\\\Rightarrow \omega=37.69911\ rad/s$

Angular momentum is given by

$L=I\omega$

Moment of inertia of the satellite is

$I_s=\frac{2}{5}m_1r^2$

Moment of antenna of the satellite is

$I_a=\frac{1}{3}m_2l^2$

The angular momentum of the system is

$L=I_s\omega+I_a\omega\\\Rightarrow L=\left(\frac{2}{5}m_1r^2+2\times \frac{1}{3}m_2l^2\right)\omega\\\Rightarrow L=\left(\frac{2}{5}10000\times 2^2+2\times \frac{1}{3}\times 10\times 3^2\right)\times 37.69911\\\Rightarrow L=605447.7066\ kgm^2/s$

The angular momentum of the satellite is 605447.7066 kgm²/s

5 0

3 years ago

What object circles around a planet

agasfer [191]

A object that circles around a planet is called a satellite.<span />

4 0

3 years ago

Read 2 more answers

Which formula can be used to calculate the horizontal displacement not of a horizontally launched projectile

svetlana [45]

<span>If you are looking to get an object up the highest, shoot it straight up. If you want to go for a specific horizontal displacement, use the range equation. R = v2sin(twice the launch angle)/ g. g is the gravitaional constant, 9.8 meters per second. Use degrees for the angle. v is the launch velocity. R is the horizontal displacement. This formula only works if your start altitude and end altitude are the same, i.e. you must shoot over a level field. it depends on the gravitational force of attraction of earth and air resistance. if we are neglecting air resistance, the max.horizontal distance is according to this formulae, V0/2 * sin (2theta) where V0 is the initial velocity theta is the angle with x axis and the projection. There are a number of ways that you could find a horizontally displaced object. You could for example just look.</span>

8 0

3 years ago

Two small, identical metal balls with charges 7.0 µC and 14.0 µC are held in place 1.9 m apart. In an experiment, they are conne

Nostrana [21]

The new charge on 7.0 µC ball is 10.5 µC and the new charge on 14.0 µC ball is 10.5 µC.

The change in the electrostatic force after the experiment is 0.031 N.

The given parameters:

Charge on first metal ball, q1 = 7.0 µC
Charge on second metal ball, q2 = 14 µC
Distance between the charges, r = 1.9 m

<h3>Charge on each ball after experiment</h3>

After the experiment the charges will be at equilibrium, and the charge on each metal ball will be equal.

$Q_t = q_1 + q_2\\\\Q_t = 7 \mu C + 14 \mu C\\\\Q_t = 21 \ \mu C$

$Q_1 = Q_2 = \frac{Q_t}{2} = \frac{21 \mu C}{2} = 10.5 \ \mu C$

The new charge on 7.0 µC ball = 10.5 µC

The new charge on 14.0 µC ball = 10.5 µC

<h3>Change in electrostatic force</h3>

$F_1 = \frac{kQ_1Q_2}{r^2} \\\\F_1 = \frac{9\times 10^9}{1.9^2} (7 \times 10^{-6} \times 14 \times 10^{-6}) \ \\\\ F_1 = 0.244 \ N\\\\for \ new \ charges;\\\\F_2 = \frac{kQ_1Q_2}{r^2} \\\\F_2 = \frac{9\times 10^9}{1.9^2} (10.5 \times 10^{-6} \times 10.5 \times 10^{-6}) \ \\\\ F_2 = 0.275 \ N\\\\\Delta F = F_2 - F_2\\\\\Delta F = 0.275 \ N \ - \ 0.244 \ N\\\\\Delta F = 0.031 \ N$

Lear more about electrostatic force here: brainly.com/question/17692887

8 0

2 years ago