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Nat2105 [25]
2 years ago
9

Một electron di chuyển theo đường tròn vuông góc với từ trường đều 1mT. Moment động lượng của electron đối với tâm vòng tròn là

4×〖10〗^(−25) 〖^2〗∕. Xác định
bán kính của đường tròn
tốc độ của electron
Physics
1 answer:
8090 [49]2 years ago
7 0

Answer:

can you translate in English

Explanation:

maybe if u did I wouldv help u?

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I) Give two practical<br>application of static friction<br>​
bulgar [2K]

1).  Walking / Driving

If there were no static friction between the soles of your shoes and the ground, then you could move your feet back and forth but your body would never go anywhere.

Same for using tires to move a car, a bus, a bicycle or a motorcycle.

2).  Sleeping

If there were no static friction between your jammies and the sheet, you would slide right off of the bed whenever there was the slightest breeze of air in the room.  

7 0
2 years ago
Negatively charged particle that orbits the nucleus
Ede4ka [16]

Answer: Electrons.

Explanation:

Electrons are the negatively charged particles that orbit the nucleus of an atom. Protons, on the other hand, are the positively charged particles that orbit the nucleus of an atom.

in the picture, the protons and neutrons are “in” the nucleus and you’ll also see the the electrons are orbiting all around the nucleus.

If this helped please mark me brainliest.

6 0
3 years ago
Determine the launch speed of a horizontally launched projectile that lands 26.3m from the base of a 19.3m high cliff.
atroni [7]

The launch velocity of the projectile is 13.28 m/s.

What is projectile motion?

The motion of an object thrown in the air under the force of gravity is known as projectile motion.

Since the object is launched horizontally, its initial velocity along the vertical direction is zero. From the second kinematic equation,

s=u*t+(1/2)at^2.

where s is the displacement, t is the time, u is the initial velocity and a is the acceleration. Since the height is decreasing, so it will be taken negative.

For the vertical motion, s=-19.3 m, a=-9.8 m/s^2 and u=0. Put the values in the above equation and solve it.

-19.3 = (0)*t+(1/2)*(-9.8)*t^2

19.3 = (1/2)*(9.8)*t^2

t=1.98 s

Since the velocity along the horizontal direction is constant, the displacement along the horizontal direction is given by the formula,

X=vt

where X is the horizontal displacement, v is the initial horizontal velocity and t is the time.

For the horizontal motion, X=26.3 m and t=1.98 s. Put the values in this equation and solve it.

26.3=v*(1.98)

v=13.28 m/s

The launch velocity is equal to the initial horizontal velocity, so it is equal to 13.28 m/s.

Learn more about projectile motion.

brainly.com/question/11049671

#SPJ4

4 0
2 years ago
Assume that our computer stores decimal numbers using 16 bits - 10 bits for a sign/magnitude mantissa and 6 bits for a sign/magn
madreJ [45]

Explanation:

a) 7.5= 111.1×2°= 0.1111×2^3

which can also be written as

(1/2+1/4+1/8+1/16)×8

sign of mantissa:=0

Mantissa(9 bits): 111100000

sign of exponent: 0

Exponent(5 bits): 0011

the final for this is:011110000000011

b) -20.25= -10100.01×2^0= -0.1010001×2^5

sign of mantissa: 1

Mantissa(9 bits): 101000100

sign of exponent: 0

Exponent(5 bits): 00101

the final for this is:1101000100000101

c)-1/64= -.000001×2^0= -0.1×2^{-5}

sign of mantissa: 1

Mantissa(9 bits): 100000000

sign of exponent: 0

Exponent(5 bits): 00101

the final for this is:1100000000100101

5 0
3 years ago
Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with an
Alona [7]

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

F=\dfrac{kQ^2}{r^2} ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}.....(2)

Dividing equation (1) and (2), we get :

\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}

Hence, the correct option is (d) i.e. " 4F/9"

7 0
3 years ago
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