You do 25 J of work to move a 4N object 5 meters. Find your efficiency and write down the formula.

A ball is thrown at an angle of 38° to the horizontal. What happens to the

finlep [7]

Answer:

Vy = V0 sin 38 where Vy is the initial vertical velocity

The ball will accelerate downwards (until it lands)

Note the signs involved if Vy is positive then g must be negative

The acceleration is constant until the ball lands

t (upwards) = (0 - Vy) / -g = Vy / g final velocity = 0

t(downwards = (-Vy - 0) / -g = Vy / g final velocity = -Vy

time upwards = time downwards (conservation laws)

8 0

3 years ago

About light and energy

Sophie [7]

E = MC^2. Albert Einstein's proven formula. When mass travels at the square of speed of light, the mass gets converted into energy

7 0

3 years ago

How should the distance, D, between two point charges, q1 and q2, be changed to double their electric potential energy

Licemer1 [7]

The electrostatic potential energy, U, of one point charge q at position d in the presence of an electric field E is defined as the negative of the work W done by the electrostatic force to bring it from the reference position d to that position

$u \: = \: \frac{kq1q2}{d}$

Thus, to double the electric potential energy U we need to reduce the distance of separation by half (1/2) because they are inversely proportion

6 0

3 years ago

The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole huma

gayaneshka [121]

Explanation:

Below is an attachment containing the solution.

6 0

3 years ago

May you help me answer this

Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is $2.2 m/s^2$

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

1)

There are three Kepler's law of planetary motion:

1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, $T^2 \propto r^3$ , where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

$M g sin \theta - T = Ma$ (1)

where

$Mg sin \theta$ is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

$g=9.8 m/s^2$ (acceleration of gravity)

$\theta=35^{\circ}$

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

$T-mg sin \theta = ma$ (2)

where

m = 3.5 kg (mass of the block)

$\theta=35^{\circ}$

From (2) we get

$T=mg sin \theta + ma$

Substituting into (1),

$M g sin \theta - mg sin \theta - ma = Ma$

Solving for a,

$a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2$

2b)

The tension in the string can be calculated using the equation

$T=mg sin \theta + ma$

where

m = 3.5 kg (mass of lighter block)

$g=9.8 m/s^2$

$\theta=35^{\circ}$

$a=2.2 m/s^2$ (acceleration found in part 2)

Substituting,

$T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N$

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

$U=mgh$