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LiRa [457]
4 years ago
6

suppose that you’re running a boy away from your dog who sits and watches you leave if you use the bicycle as a reference frame

how does your dog appear to move?
Physics
1 answer:
solong [7]4 years ago
6 0

Answer:

If you use the bicycle as a reference frame, your dog will appear to be moving backwards farther from you/your bicycle. Although the dog is sitting on one place, your bicycle is moving away from the dog

Explanation:

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What’s the net force of an object being throwing into the air
timurjin [86]

Answer: forces acting on an object being thrown into the air is gravity and possibly air resistance

Explanation:

4 0
3 years ago
If the electron just misses the upper plate as it emerges from the field, find the speed of the electron as it emerges from the
Scilla [17]

The speed of the electron as it emerges from the field is; 388.587 m/s

<h3>What is the speed of the electron?</h3>

Initial speed; v₀x = 1.1 * 10⁶ m/s

Acceleration in horizontal direction = 0 m/s²

distance; s_x = 2 cm = 0.02 m

Thus, formula to find time here is;

t = s_x/v₀x

t = 0.02/(1.1 * 10⁶)

t = 1.82 * 10⁻⁶ s

Now for the vertical distance; v,y_o = 0 m/s

Thus, the equation of motion becomes;

s_y = ¹/₂at²

0.005 = ¹/₂a(1.82 * 10⁻⁶)²

Solving for a gives;

a = 3.02 * 10¹³ m/s²

Thus the speed of the electron as it emerges from the field is;

v² = u² + 2as

v = √(0² + 2(3.02 * 10¹³ * 0.005))

v = 388.587 m/s

Read more about Electron speed at; brainly.com/question/15094100

#SPJ1

Complete Question is;

An electron is projected with an initial speed v0 = 1.1 * 10⁶ m/s into the uniform field between the parallel plates. The distance between the plates is 1 cm and the length of the plates is 2 cm. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. E = N/C

find the speed of the electron as it emerges from the field?.

6 0
1 year ago
If he completes the 200 m dash in 23 s and runs at constant speed throughout the race, what is his centripetal acceleration as h
Blababa [14]

Incomplete question.The complete one is here

A runner taking part in the 200m dash must run around the end of a track that has a circular arc with a radius curvature of 30m. The runner starts the race at a constant speed. If she completes the 200m dash in 23s and runs at constant speed throughout the race, what is her centripetal acceleration as she runs the curved portion of the track?

Answer:

a_c=2.523m/s^2

Explanation:

Given data

L_{length}=200m\\r_{radius}=30m\\t_{time}=23s

Required

Centripetal acceleration

Solution

According to the motions equation the velocity given by:

V_{velocity}=\frac{L_{length}}{t_{time}}\\V_{velocity}=\frac{200m}{23s}\\V_{velocity}=8.7m/s

The centripetal acceleration is given by:

a_{c}=\frac{V_{velocity}^2}{r_{radius}}\\ a_c=\frac{(8.7m/s)^2}{30m}\\ a_c=2.523m/s^2

4 0
4 years ago
Give a quantitative definition of being in contact.
san4es73 [151]

Two things are said to be in contact if the smallest distance between a point in one of them and a point in the other one is zero.

7 0
3 years ago
An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if
maxonik [38]
The electron is accelerated through a potential difference of \Delta V=780 V, so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
\frac{1}{2}mv^2 =  e \Delta V
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
\Delta V is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s


Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
evB=m \frac{v^2}{r}
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T
3 0
4 years ago
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