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3 years ago

Which of these is not a type of force?

Physics

2 answers:

coldgirl [10]3 years ago

6 0

Inertia
the awnswer is inertia b

Greeley [361]3 years ago

6 0

The answer is,

<u>B. inertia</u> (:

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What part of the hammer acts as the fulcrum when the hammer is used to remove a nail

Natali5045456 [20]

The end of it with the dent

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3 years ago

How many atoms of carbon would two molecules of glucose (C6H12O6) have?

Sauron [17]

Well according to the molecular formula of glucose, one molecule would have 6 carbon atoms, and thus 2 molecules of glucose would have 12 carbon atoms.

The correct response would be B. 12.

4 0

2 years ago

Read 2 more answers

Two forces of magnitude 8N and 4N act at right angle to each other. The angle between the resultant and the 8N force is?

Marta_Voda [28]

Answer:

Draw the vector triangle (head to tail)

Let 8 be adjacent and 4 the opposite side

tan theta = 4 / 8 = .5

theta = 26.6 deg

4 0

1 year ago

A ball is thrown at an angle of 38° to the horizontal. What happens to the

finlep [7]

Answer:

Vy = V0 sin 38 where Vy is the initial vertical velocity

The ball will accelerate downwards (until it lands)

Note the signs involved if Vy is positive then g must be negative

The acceleration is constant until the ball lands

t (upwards) = (0 - Vy) / -g = Vy / g final velocity = 0

t(downwards = (-Vy - 0) / -g = Vy / g final velocity = -Vy

time upwards = time downwards (conservation laws)

8 0

2 years ago

Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce

lutik1710 [3]

Answer:

speed of electrons = 3.25 × $10^{7}$ m/s

acceleration in term g is 3.9 × $10^{17}$ g.

radius of circular orbit is 2.76 × $10^{-4}$ m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = $\sqrt{\frac{2qV}{m}}$

v = $\sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}$

v = 3.25 × $10^{7}$ m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = $\frac{Bqv}{m}$

a = $\frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}$

a = 3.82 × $10^{18}$ m/s²

and acceleration in term g

a = $\frac{3.82\times 10^{18}}{9.81}$

a = 3.9 × $10^{17}$ g

acceleration in term g is 3.9 × $10^{17}$ g.

and

electron moving in circular orbit has centripetal force

F = $\frac{mv^2}{r}$

Bqv = $\frac{mv^2}{r}$

r = $\frac{mv}{Bq}$

r = $\frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}$

r = 2.76 × $10^{-4}$ m

radius of circular orbit is 2.76 × $10^{-4}$ m

8 0

2 years ago