All categories

2 years ago

When a cup is placed on a table, which force prevents the cup from falling to the ground?

Physics

2 answers:

pshichka [43]2 years ago

5 0

I believe it is normal force.

disa [49]2 years ago

3 0

Answer: Normal force prevents the cup from falling to the ground.

Explanation :

When a cup is placed on a table, the forces acting on the cup are the gravitational force and the normal force. The gravitational force acts in the downward direction while the normal force acts in an upward direction.

In the figure, N is the normal force and mg is the weight of an object acting in downward direction.

It is clear that the gravitational force is balanced by the normal force.

So, the correct option is (b) " normal force ".

You might be interested in

Temperature and pressure can change the what of a solute

igomit [66]

Temperature and pressure can change the solubility of a solute.

4 0

2 years ago

If a car accelerates from a speed of 10 m/s at an acceleration of 2 m/s2 for 3s.

Flauer [41]

Answer:

$\boxed{\sf Final \ speed \ of \ the \ car = 16 \ m/s}$

Given:

Initial speed (u) = 10 m/s

Acceleration (a) = 2 m/s²

Time taken (t) = 3s

To Find:

Final speed (v) of the car

Explanation:

From equation of motion we have:

$\boxed{ \bold{v = u + at}}$

By substituting value of u, a & t in the equation we get:

$\sf \implies v = 10 + 2\times 3 \\ \\ \sf \implies v = 10 + 6 \\ \\ \sf \implies v = 16 \: m {s}^{ - 1}$

$\therefore$

Final speed (v) of the car = 16 m/s

3 0

3 years ago

How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the

kirza4 [7]

Answer:

Approximately $6.2\; {\rm rpm}$ , assuming that the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ .

Explanation:

Let $\omega$ denote the required angular velocity of this Ferris wheel. Let $m$ denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

Weight of the passenger (downwards), $m\, g$ , and possibly
Normal force $F_\text{normal}$ that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is $0$ - that is, $F_\text{normal} = 0$ .

The net force on this passenger is $(m\, g - F_\text{normal})$ . Hence, when $F_\text{normal} = 0$ , the net force on this passenger would be equal to $m\, g$ .

Passengers on this Ferris wheel are in a centripetal motion of angular velocity $\omega$ around a circle of radius $r$ . Thus, the centripetal acceleration of these passengers would be $a = \omega^{2}\, r$ . The net force on a passenger of mass $m$ would be $m\, a = m\, \omega^{2}\, r$ .

Notice that $m\, \omega^{2} \, r = (\text{Net Force}) = m\, g$ . Solve this equation for $\omega$ , the angular speed of this Ferris wheel. Since $g = 9.81\; {\rm m\cdot s^{-2}}$ and $r = 23\; {\rm m}$ :

$\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}$ .

$\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}$ .

The question is asking for the angular velocity of this Ferris wheel in the unit ${\rm rpm}$ , where $1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s})$ . Apply unit conversion:

$\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}$ .