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A table exerts a 4.0 Newton force on a book which lies at rest on its top. The force exerted by the book on the table is

SOVA2 [1]

I believe the correct answer from the choices listed above is the third option. <span>The force exerted by the book on the table is equal to the force exerted by the table which is 4.0 N. The book does not move so it must be that the forces are balanced. Hope this answers the question.</span>

4 0

3 years ago

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What change will always result in an increase in gravitational force between two objects

tekilochka [14]

The gravitational force between two object depends on their masses and on their distance.

Since the formula is

$F = G\frac{m_1m_2}{d^2}$

If the masses grow, the force also grows. But I'm assuming the two objects are fixed, so you can't enlarge their mass.

So, the only option remaining is to lower their distance: since it sits at the denominator, a smaller value of d results in a bigger value for F.

So, if you reduce the distance between two objects, the gravitational force between them will always result in an increase

6 0

3 years ago

The persistence of vision for normal eye is

attashe74 [19]

Answer:

The answer is 1/16

Explanation:

1. Persistence of vision refers to the optical illusion that occurs when visual perception of an object does not cease for some time after the rays of light proceeding from it have ceased to enter the eye. 2. The persistence of vision for normal eye is 1/16 if a second.

6 0

3 years ago

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A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction f

Nat2105 [25]

Answer:

22000 N

Explanation:

Convert velocity to SI units:

98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s

Draw a free body diagram. There are three forces acting on the car. Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.

I'm going to assume the friction force is pointed down the bank. If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

∑F = ma

N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

W tan θ + F (sin θ tan θ + cos θ) = m v² / r

W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

4 0

2 years ago

A flat, wide cloud floats horizontally a few kilometers above the surface of Earth. Its lower surface carries a uniform surface

valentinak56 [21]

Answer:

$\frac{kQ}{r^2} r^$

Explanation:

Electric field strength= Force/unit charge

E= (kQq/r²)/q ₓ r

where r is the unit vector in the direction of unit charge

E= $\frac{kQ}{r^2} r^$

4 0

2 years ago