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frez [133]
3 years ago
9

Which vector has an x-component with a length of 4?

Physics
1 answer:
matrenka [14]3 years ago
5 0

Answer:the scaler ax vertex

Explanation:

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Consider the train car described in the previous part. Another experiment is conducted in it: A net force of 20N is applied to a
nordsb [41]

Answer:

No

Explanation:

The supplied information about the object and train is incomplete. Acceleration is the rate at which the velocity of a body changes with time. Here the velocity and time is not given

7 0
3 years ago
A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr
VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

rR:\\\\E=k\frac{Q}{r^2}

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}

with this values of a you can use the following formula:

a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0
2 years ago
Un acróbata de 60.0 kg está unido a un cordón de bungee con un resorte de 10.0 m de longitud . Salta de un puente que abarca un
Ede4ka [16]

Answer:

What kind of languege is this please?

8 0
2 years ago
How does inertia affect a person who is not wearing a seatbelt during a collision?
ELEN [110]

A person who is not wearing a seatbelt during a collision will be thrown forward because it maintains forward motion

<h3>Further explanation </h3>

In Newton's law, it is stated that if the resultant force acting on an object of magnitude is zero,  it can be formulated :

\large{\boxed{\bold{\Sigma F = 0}}}

then the object tends to defend itself from its state. So for objects in a state of movement, objects tend to move forever. Likewise, for objects in a state of rest, they tend to remain forever. The tendency of objects like this is called<em> inertia </em>

The size of inertia is proportional to mass, the greater the mass of the object, the greater the inertia of the object.

In objects with mass m that move translatively, the object will maintain its linear velocity

When we are in a vehicle that moves forward, then we will still maintain a state of forwarding motion. If our vehicle stops suddenly, then we keep moving forward so we will be pushed forward. From this point, the use of a safety belt serves to hold back our movements so that there are no fatal accidents or collisions.

<h3>Learn more </h3>

Newton's law of inertia

brainly.com/question/1412777

example of Newton's First Law of inertia

brainly.com/question/1090504

law of motion

brainly.com/question/75210

Keywords: inertia, Newton's First Law

7 0
3 years ago
Read 2 more answers
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\°

Therefore the largest value of the angle α is 30.27°

3 0
3 years ago
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