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3 years ago

Newtons third law of motion the lab questions are in the paper

2 answers:

4 0

Newton's third law states that when two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction. The third law is also known as the law of action and reaction. ... For example, a book resting on a table applies a downward force equal to its weight on the table.

Rama09 [41]3 years ago

3 0

Answer:

Sir Isaac Newton was an English scientist. He was born in 1642 and died in 1727. This was around the time of the early colonization of North America: the founding of some of the original 13 colonies, the French and Indiana wars and the Salem witch trials, but before the American Revolution.

Newton is best known for three very important principles of physics called classical mechanics. These principles describe how things move and are referred to today by his name - Newton's Laws of Motion. There are three of them, Newtons First, Second and Third Law of Motion. Today's experiment will demonstrate Newton's Third Law of Motion: for every action there is an equal and opposite re-action. SIMPLY: If you push an object, that object pushes back in the opposite direction equally hard.

Explanation:

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Un motor realiza un trabajo de 5000 j en 20 s a) ¿ cual es la potencia del motor ?B)¿en cuanto tiempo desarollan el mismo trabaj

zimovet [89]

Answer:

a. Potencia = 250 Watts

b. Tiempo = 333.33 segundos

Explanation:

Dados los siguientes datos;

Energía = 5000 Joules

Tiempo = 20 segundos.

Para encontrar el poder;

Poder = energía/tiempo

Potencia = 5000/20

Potencia = 250 Watts

B. ¿Para saber cuánto tarda una máquina en realizar el mismo trabajo;

Dado: potencia = 15 W

Tiempo = energía/poder

Tiempo = 5000/15

Tiempo = 333.33 segundos

3 0

3 years ago

A cannonball is fired at a 45.0° angle and an initial velocity of 670 m/s. Assume no air resistance. How long until the cannonba

Naddik [55]

Answer:

96.7 s

Explanation:

Time of flight in projectile can be calculated thus:

T = 2 × u × sin ϴ/ g

Where;

T = time of flight (s)

u = initial velocity (m/s)

ϴ = Angle of projectile (°)

g = acceleration due to gravity (9.8m/s²)

Based on the provided information; u = 670m/s, ϴ = 45°

Hence, using T = 2.u.sin ϴ/ g

T = 2 × 670 × sin 45° ÷ 9.8

T = 1340 × 0.7071 ÷ 9.8

T = 947.52 ÷ 9.8

T = 96.68

T = 96.7s

5 0

3 years ago

At each corner of a square of side there are point charges of magnitude Q, 2Q, 3Q, and 4Q

Bad White [126]

Answer:

$\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}$

$\displaystyle \theta =68^o$

Explanation:

Electrostatic Force

It's the force that appears between two electrical charges q1 q2 when they are placed at a certain distance d. The force can be computed by using the Coulomb's law:

$\displaystyle F=\frac{KQ_1Q_2}{d^2}$

We have an arrangement of 4 charges as shown in the image below. We need to calculate the total force exerted on the charge 2Q by the other 3 charges. The free body diagram is also shown in the second image provided. The total force on 2Q is the vectorial sum of F1, F2, and F3. All the forces are repulsive, since all the charges have the same sign. Let's compute each force as follows:

$\displaystyle |F_1|=\frac{KQ(2Q)}{l^2}=\frac{2KQ^2}{l^2}$

$\displaystyle |F_2|=\frac{K(2Q)(4Q)}{l^2}=\frac{8KQ^2}{l^2}$

The distance between 3Q and 2Q is the diagonal of the rectagle of length l:

$\displaystyle |d_3|=\sqrt{l^2+l^2}=\sqrt{2}\ l$

The force F3 is

$\displaystyle |F_3|=\frac{K(3Q)(2Q)}{(\sqrt{2l)}^2}=\frac{3KQ^2}{l^2}$

Each force must be expressed as vectors. F1 is pointed to the right direction, thus its vertical components is zero

$\displaystyle \vec{F_1}=\left \langle |F_1|,0 \right \rangle=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle$

F2 is pointed upwards and its horizontal component is zero

$\displaystyle \vec{F_2}=\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle$

F3 has two components because it forms an angle of 45° respect to the horizontal, thus

$\displaystyle \vec{F_3}=\left \langle \frac{3KQ^2}{l^2}\ cos45^o,\frac{3KQ2}{l^2} sin45^o\right \rangle$

$\displaystyle \vec{F_3}=\left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

Now we compute the total force

$\displaystyle \vec{F_t}=\vec{F_1}+\vec{F_2}+\vec{F_3}$

$\displaystyle \vec{F_t}=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle +\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle + \left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

$\displaystyle \vec{F_t}=\left \langle \left(2+\frac{3\sqrt{2}}{2}\right)\frac{KQ^2}{l^2},\left(8+\frac{3\sqrt{2}}{2}\right) \frac{KQ^2}{l^2}\right \rangle$

$\displaystyle F_t=\left \langle 4.121,10.121 \right \rangle \frac{KQ^2}{l^2}$

Now we compute the magnitude

$\boxed{\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}}$

The direction of the total force is given by

$\displaystyle tan\theta =\frac{10.121}{4.121}=2.4558$

$\boxed{\displaystyle \theta =68^o}$

6 0

3 years ago

Describe how the eye changes visible light to an image

mr_godi [17]

First, light passes through the cornea. It bends the light inward. The light then passes through the pupil. After entering the eye through the pupil, the light passes through the lens. It focuses the light to create an image on the retina. It consists of light receptor cells called rods and cones. The rods and cones send nerve impulses about the image to the brain through the optic nerve. The visual cortex of the brain receives and interprets the information and tells you what you are seeing

3 0

4 years ago

You deposit a thin film of magnesium difluoride on a glass lens (n >1.60), reducing the reflection of yellow light, at normal

JulsSmile [24]

To solve this problem it is necessary to apply the concepts related to destructive interference.

The concept refers to an overlap of two or more waves of identical or similar frequency that, when interfering, creates a new pattern of waves of lower intensity

By definition destructive interference is given by

$2Mt = (n+\frac{1}{2})\lambda$