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3 years ago

True or False

Physics

1 answer:

Anon25 [30]3 years ago

3 0

Explanation:

The work of the electric force does not depend on the trajectory or the path taken. Regardless of trajectory a and b we have the following

Vide picture

$W_{ab} = \frac{K Q q}{R_{a} } \frac{K Q q}{R_{b} }$

Wab: Work

K: coulomb constant (9·109 N·m2/C2)

Ra,Rb: distances

Qq: electric charges

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In order to determine the coefficients of friction between rubber and various surfaces, a student uses a rubber eraser and an in

Strike441 [17]

Answer:

μs = 0.75

μk = 0.58

Explanation:

From a force diagram:

$m*g*sin \theta - Ff=m*a$ (1)

$N-m*g*cos \theta = 0$ (2)

When it starts slipping, friction force is the maximum and acceleration is 0. Replacing these conditions on (1):

$m*g*sin \theta - \mu*m*g*cos \theta=0$ Solving for μs:

$\mu=tan \theta$

μs = tan 36.7° = 0.75

When it moves at constant speed, friction force is kinetic friction and acceleration is 0. With these conditions the coefficient is:

μk = tan 30.1° = 0.58

8 0

4 years ago

What force is required to move 7 M if the work done is 9 J

Murrr4er [49]

Answer:

1.29 N

Explanation:

The equation for force (with work and distance) is:

$Force=\frac{Work}{distance}$

We can plug in the given values into the equation:

$Force=\frac{9J}{7m}$\approx1.29 N$

5 0

2 years ago

Read 2 more answers

How are elastic and inelastic collisions different?

dimaraw [331]

Answer:

Explanation:

6 0

3 years ago

Read 2 more answers

2. A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelli

Mice21 [21]

Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 2000(12000)(9.8)+1/2(2000)(750²)

W = 235200000+562500000

W = 797700000 J

6 0

3 years ago

Jeff, a snowboarder who weighs 1000 N, is practicing on a trampoline for an upcoming competition. As Jeff bounces straight up of

NISA [10]

Answer:

Potential energy $P_E = 2000 J$

Explanation:

Given data

Weight of the person = 1000 N

Mass of the person = $\frac{1000}{9.81} = 102 kg$

Speed = 6.26 $\frac{m}{s}$

Kinetic energy is given by

$E_K = \frac{1}{2} m v^{2}$

Put the values in above formula we get

$E_K = \frac{1}{2}(102) 6.26^{2}$

$E_K = 2000 J$

When the person reach top point then at this point the velocity is zero & kinetic energy at top point is equals to the potential energy.

Therefore potential energy

$P_E = 2000 J$

5 0

3 years ago