A student rolls these four items on the same surface and at the same speed. Which item will have the greatest kinetic energy?

What is the net force acting on the buggy. ?The net force is pointing to the ?

Papessa [141]

Answer: 390, right

explanation: The net force is just the sum of all of these forces acting on an object. ... This equation is the sum of n forces acting on an object. The magnitude of the net force acting on an object is equal to the mass of the object multiplied by the acceleration of the object, as shown in this formula.

6 0

3 years ago

Would u rather/content creator or a rap artist

BartSMP [9]

a content creator because if i was a rapper i probably wouldn't make good songs lol

7 0

3 years ago

A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial sp

Dimas [21]

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m = 0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2m $U^{2}$

K.E = 1/2 x 0.17 x $6^{2}$

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/ $s^{2}$

To calculate the final velocity, let us use third equation of motion.

$V^{2}$ = $U^{2}$ + 2as

$V^{2}$ = $6^{2}$ + 2 x 0.3 x 61

$V^{2}$ = 36 + 36

$V^{2}$ = 72

V = $\sqrt{72}$

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

Learn more about dynamics here: brainly.com/question/402617

5 0

2 years ago

A parallel-plate capacitor with circular plates of radius R is being charged by a battery, which provides a constant current. At

NikAS [45]

To solve this problem it is necessary to apply the concepts related to the magnetic field.

According to the information, the magnetic field INSIDE the plates is,

$B=\frac{1}{2} \mu \epsilon_0 r$

Where,

$\mu =$ Permeability constant

$\epsilon_0 =$ Electromotive force

r = Radius

From this deduction we can verify that the distance is proportional to the field

$B \propto r$

Then the distance relationship would be given by

$\frac{r}{R} = \frac{B}{B_{max}}$

$r =\frac{B}{B_{max}} R$

$r = \frac{0.5B_{max}}{B_{max}}R$

$r = 0.5R$

On the outside, however, it is defined by

$B = \frac{\mu_0 i_d}{2\pi r}$

Here the magnetic field is inversely proportional to the distance, that is

$B \not\propto r$

Then,

$\frac{r}{R} = \frac{B_{max}{B}}$

$r = \frac{B_{max}{B}}R$

$r = \frac{B_{max}{0.5B_{max}}}R$

$r = 2R$

7 0

3 years ago

8. How did the measured angular magnification of the telescope compare with the theoretical prediction?

Genrish500 [490]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The theoretical angular magnification lies within the angular magnification range

Explanation:

From the question we are told that

The focal length of B is $f_{objective } = 43.0 \ cm$

The focal length of A is $f_{eye} = 10.4 \ cm$

The theoretical angular magnification is mathematically represented as

$m = \frac{f_{objective }}{f_{eye}} = \frac{43.0}{10.4}$

$m = \frac{f_{objective }}{f_{eye}} = 4.175$

Form the question the measured angular magnification ranges from 4 -5

So from the value calculated and the value given we can deduce that the theoretical angular magnification lies within the angular magnification range

3 0

3 years ago