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Dmitry_Shevchenko [17]
3 years ago
5

Un protón se acelera por una diferencia de potencial de 2.6k V y se dirige a una región donde existe un campo magnético uniforme

de 4.5T. ¿Cuál es el valor?
Physics
1 answer:
Bond [772]3 years ago
7 0

Answer:

5.08*10^{-13}N

Explanation:

With this information we can calculate the velocity of the proton by taking into account the kinetic energy of the proton:

qV=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(1.6*10^{-19}C)(2.6*10^3V)}{1.67*10^{-27}kg}}=705835.3m/s

The magnitude of the magnetic force will be:

F=qvB=(1.6*10^{-19}C)(705835.3m/s)(4.5T)=5.08*10^{-13}N

hope this helps!

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The speed of the spacecraft in this orbit
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3 years ago
Someone please help me!
kirill [66]

I think it’s the first one

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A toy rocket is fired vertically into the air from the ground at an initial velocity of 80 feet per second. Find the time it wil
miskamm [114]

Answer: 16.3 seconds

Explanation: Given that the

Initial velocity U = 80 ft/s

Let's first calculate the maximum height reached by using third equation of motion.

V^2 = U^2 - 2gH

Where V = final velocity and H = maximum height.

Since the toy is moving against the gravity, g will be negative.

At maximum height, V = 0

0 = 80^2 - 2 × 9.81 × H

6400 = 19.62H

H = 6400/19.62

H = 326.2

Let's us second equation of motion to find time.

H = Ut - 1/2gt^2

Let assume that the ball is dropped from the maximum height. Then,

U = 0. The equation will be reduced to

H = 1/2gt^2

326.2 = 1/2 × 9.81 × t^2

326.2 = 4.905t^2

t^2 = 326.2/4.905

t = sqrt( 66.5 )

t = 8.15 seconds

The time it will take for the rocket to return to ground level will be 2t.

That is, 2 × 8.15 = 16.3 seconds

8 0
3 years ago
For which question could a testable hypothesis be developed?
Mandarinka [93]
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8 0
3 years ago
Two long, straight wires are separated by a distance of 9.15 cm . One wire carries a current of 2.79 A , the other carries a cur
Dafna1 [17]

Answer:

The force is the same

Explanation:

The force per meter exerted between two wires carrying a current is given by the formula

\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}

where

\mu_0 is the vacuum permeability

I_1 is the current in the 1st wire

I_2 is the current in the 2nd wire

r is the separation between the wires

In this problem

I_1=2.79 A\\I_2=4.36 A\\r = 9.15 cm = 0.0915 m

Substituting, we find the force per unit length on the two wires:

\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(2.79)(4.36)}{2\pi (0.0915)}=2.66\cdot 10^{-5}N

However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.

The same conclusion comes out  from Newton's third law of motion, which states that when an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A (action-reaction). If we apply the law to this situation, we see that the force exerted by wire 1 on wire 2 is the same as the force exerted by wire 2 on wire 1 (however the direction is opposite).

3 0
3 years ago
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