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natka813 [3]
2 years ago
14

Draw a diagram showing 2 forces acting on an object in opposite directions with a total net force of 3N to the left.

Physics
2 answers:
olga55 [171]2 years ago
8 0

Answer:

3n--- Square __3n

Explanation:

i will notify u after i study it

andriy [413]2 years ago
6 0

Answer:

hope it helps u ........

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Which is likely to be more common in our Galaxy: white dwarfs or black holes? Why?
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3 years ago
our 3.80-kg physics book is placed next to you on the horizontal seat of your car. The coefficient of static friction between th
gtnhenbr [62]

Answer:

Explanation:

Maximum force of friction possible = μmg

= .65 x 3.8 x 9.8

= 24.2 N

u = 72 x 1000 / 60 x 60

= 20 m /s

v² = u² - 2as

a = 20 x 20 / (2 x 30)

= 6.67 m / s²

force acting on it

= 3.8 x 6.67

= 25.346 N

Friction force possible is less .

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It will slip off .

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A test car travels in a straight line along the x-axis. The graph in the figure shows the car’s position x as a function of time
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8 0
3 years ago
Read 2 more answers
A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th
GREYUIT [131]

Answer:

1.26\cdot 10^7 m/s

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

F=qvB

where

q is the charge

v is its velocity

B is the strength of the magnetic field

Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

qvB=m\frac{v^2}{r}

where

m is the mass of the particle

r is the radius of the orbit of the particle

The equation can be re-arranges as

v=\frac{qBr}{m}

where in this problem we have:

q=1.6\cdot 10^{-19}C is the magnitude of the charge of the electron

B=208 G=208\cdot 10^{-4}T is the strength of the magnetic field

The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,

r=3.45 mm = 3.45\cdot 10^{-3} m

m=9.11\cdot 10^{-31} kg is the mass of the electron

So, the electron's speed is

v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s

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3 years ago
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