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2 years ago

Which of the following correctly describes the graph?

Physics

2 answers:

kobusy [5.1K]2 years ago

6 0

There nothing following and no graph.

So you could say that the list of choices describes the graph perfectly, or you could say that no answer is possible cause we got nuttin ta work wit.

tatyana61 [14]2 years ago

3 0

Answer:

Leave a picture or we can't help

You might be interested in

If you pull a resistant puppy with its leash in a horizontal direction, it takes 80 N to get it going. You can then keep it movi

netineya [11]

Answer:

The coefficient of static friction between the puppy and the floor is 0.7273.

Explanation:

The horizontal force applied to move the puppy from a steady state has to be greater than the force of static friction, after it is moving the force needs to be equal to be greater than the force of dynamic friction in order to maintain its movement. The force of static friction is given by:

$F_s = \mu_s*N$

Where $F_s$ is the static friction force, $\mu_s$ is the coefficient of static friction and $N$ is the normal force. Since there's no angle on the flor the normal force is equal to the weight of the puppy, therefore, $N = 110\text{ N}$ , to make the puppy moving we need to use a force of 80 N, therefore, $F_s = 80 \text{ N}$ , so we can solve for the coefficient as shown below:

$80 = \mu_s*110\\\mu_s = \frac{80}{110} = 0.7273\\$

The coefficient of static friction between the puppy and the floor is 0.7273.

5 0

2 years ago

In the Atwood machine shown below, m1 = 2.00 kg and m2 = 6.05 kg. The masses of the pulley and string are negligible by comparis

Rus_ich [418]

M1 descending
−m1g + T = m1a

m2 ascending
m2g − T = m2a

this gives :
(m2 − m1)g = (m1 + m2)a

a = (m2 − m1)g/m1 + m2
= (5.60 − 2)/(2 + 5.60) x 9.81
= = 4.65m/s^2

5 0

2 years ago

An amusement park ride raises people high into the air, suspends them for a moment, and then drops them at a rate of free-fall a

blsea [12.9K]

Answer: apparent weighlessness.

Explanation:

1) Balance of forces on a person falling:

i) To answer this question we will deal with the assumption of non-drag force (abscence of air).

ii) When a person is dropped, and there is not air resistance, the only force acting on the person's body is the Earth's gravitational attraction (downward), which is the responsible for the gravitational acceleration (around 9.8 m/s²).

iii) Under that sceneraio, there is not normal force acting on the person (the normal force is the force that the floor or a chair exerts on a body to balance the gravitational force when the body is on it).

2) This is, the person does not feel a pressure upward, which is he/she does not feel the weight: freefalling is a situation of apparent weigthlessness.

3) True weightlessness is when the object is in a place where there exists not grativational acceleration: for example a point between two planes where the grativational forces are equal in magnitude but opposing in direction and so they cancel each other.

Therefore, you conclude that, assuming no air resistance, a person in this ride experiencing apparent weightlessness.

3 0

2 years ago

Read 2 more answers

A single slit, which is 0.050 mm wide, is illuminated by light of 550 nm wavelength. What is the angular separation between the

likoan [24]

Answer:

The separation between the first two minima on either side is 0.63 degrees.

Explanation:

A diffraction experiment consists on passing monochromatic light trough a small single slit, at some distance a light diffraction pattern is projected on a screen. The diffraction pattern consists on intercalated dark and bright fringes that are symmetric respect the center of the screen, the angular positions of the dark fringes θn can be find using the equation:

$a\sin \theta_n=n\lambda$

with a the width of the slit, n the number of the minimum and λ the wavelength of the incident light. We should find the position of the n=1 and n=2 minima above the central maximum because symmetry the angular positions of n=-1 and n=-2 that are the angular position of the minima below the central maximum, then:

for the first minimum

$a\sin \theta_1=(1)\lambda$

solving for θ1:

$\theta_1=\arcsin (\frac{\lambda}{a})=\arcsin (\frac{550\times10^{-9}}{0.05\times10^{-3}})$

$\theta_1=0.63 degrees$

for the second minimum:

$a\sin \theta_2=(2)\lambda$

$\theta_2=\arcsin (\frac{2\lambda}{a})=\arcsin (\frac{2*550\times10^{-9}}{0.05\times10^{-3}})$

$\theta_2=1.26 degrees$

So, the angular separation between them is the rest:

$\Delta \theta =1.26-0.63$

$\Delta \theta=0.63$

4 0

2 years ago

You are moving at a speed 2/3 c toward Randy when shines a light toward you. At what speed do you see the light approaching you

lord [1]

Answer:

The speed of light will be c=3x10^8m/s

Explanation:

This is the same as the speed of light because your speed does not affecttje speed of light so you will see the light approaching you at the same speed of light c

5 0

2 years ago