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LiRa [457]
3 years ago
5

Harry and Hagrid needed to get money out of Gringotts bank. The bank cart was accelerating at a rate of 4 m/s2 and it had a mass

of 8,000 kg. What is the net force acting on the cart?
USE THE FORMULA: F=m∙a Show all your work!
Physics
2 answers:
spayn [35]3 years ago
7 0

Your teacher must be a fan of Harry Porter!


The cart had a mass of 8,000 kg and it was accelerating at a rate of 4 m/s2.

Using the formula Force = mass * acceleration

Net force on the cart = 8000*4

= 32000 Newton

cricket20 [7]3 years ago
5 0

F=m∙a

m, mass = 8,000 kg

a, accelerating rate = 4 m/s2

F=8000*4

=32000 kg m/s2

=32000 N

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If a net force is greater than 0 Newtons, what is the force?
goldenfox [79]

Answer:disequilibrium

Explanation:

When the net force is not zero it is called disequilibrium

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Its simple use formuila ,
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4 0
3 years ago
Light is incident along the normal to face AB of a glass prism of refractive index 1.54. Find αmax, the largest value the angle
marusya05 [52]

To solve this problem it is necessary to use the concepts related to Snell's law.

Snell's law establishes that reflection is subject to

n_1sin\theta_1 = n_2sin\theta_2

Where,

\theta = Angle between the normal surface at the point of contact

n = Indices of refraction for corresponding media

The total internal reflection would then be given by

n_1 sin\theta_1 = n_2sin\theta_2

(1.54) sin\theta_1 = (1.33)sin(90)

sin\theta_1 = \frac{1.33}{1.54}

\theta = sin^{-1}(\frac{1.33}{1.54})

\theta = 59.72\°

Therefore the \alpha_{max} would be equal to

\alpha = 90\°-\theta

\alpha = 90-59.72

\alpha = 30.27\°

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3 0
3 years ago
Blaine steps onto a ski slope with an angle of 25°. There is a coefficient of kinetic friction of 0.15 between him and the groun
Ymorist [56]

θ = angle of the incline surface from the horizontal surface = 25⁰

μ = Coefficient of friction = 0.15

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3 0
3 years ago
For an object in uniform circular motion, what can you say about the directions of the velocity, acceleration, and net force vec
tigry1 [53]

Answer: The velocity vector is perpendicular to the acceleration vector; the acceleration vector is parallel to the net force vector.

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In a uniform circular motion, the meaning of "uniform"is the same as for an uniform straight motion, i.e., the module of the velocity vector (its speed) is constant.

Now, if the object were not describing a circular trajectory, it should move at constant speed, in a straight line, provided no external forces acted upon it.

If there is an external force acting on it, making it to follow a circular trajectory, this force doesn't change the instantaneous value of the velocity, but it changes his direction instead.

While the direction is changing, it always keep tangential to the trajectory, due to if at any moment the force disappears, the body must continue in a straight line at constant speed, following a line tangent to the circle.

It can be showed, that the acceleration vector, defined as the change in velocity over time, always aims towards the center of the circle, and is perpendicular to the velocity vector.

As the only net force acting on the object (assuming a horizontal trajectory), is the one that causes the acceleration, the acceleration vector has the same direction as the net force.

7 0
3 years ago
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