All categories

3 years ago

Imagine a box sitting on a shelf. What forces are acting on the box?

1 answer:

3 0

In case of an object sitting at rest on another base, there are two equal and opposite forces – Normal force and the gravity.

Answer: Option A

Explanation:

When an object is placed at rest position on another object, there is a force exerted by the surfaces of the two contact objects. This force is denoted as Normal Force.

When an object such as a box is placed on a shelf, its surface exerts a contact force on the base of the shelf- The Normal force directed upward. Meanwhile, the gravity stays at its action and tries to pull the box towards itself.

Both of these forces however are equal and opposite and therefore, there is zero net force on the box. That's why it remains at rest, holding on Newton's third law.

You might be interested in

You fill two balloons with gas, one with hydrogen and one with carbon dioxide. You hold a match to each balloon. The hydrogen ba

miv72 [106K]

Answer:E. Hydrogen was able to participate in an exergonic reaction and carbon dioxide couldn't

Explanation:

An exergonic reaction releases energy to the environment. The combustion of hydrogen contained in the balloon is a chemical reaction. The reaction can take place because hydrogen combines with oxygen in air, that is, the gas is combustible. CO2 does not support combustion, it does not combine with oxygen in air and it is also denser than air, hence does not participate in the exergonic reaction.

4 0

3 years ago

How many significant digits are in the number 204.0920

monitta

Answer:

Seven

Explanation:

The rules for significant digits are:

Non-zero digits are always significant.
Zeros between significant digits are also significant.
Trailing zeros are significant only after a decimal point.

Here, the 2, 4, 9, and 2 are significant because they are non-zero digits.

The first two 0s are significant because they are between significant digits.

The last 0 is significant because it is a trailing zero after a decimal point.

Therefore, all seven digits are significant.

3 0

3 years ago

Read 2 more answers

A balloonist happens to drop his metal ballpoint pen from his balloon as he is taking notes on his flight. Because his pen has a

Lerok [7]

So this is dealing with the conservation of energy. So you set kinetic energy equal to potential energy, so it looks like this:

1/2mv^2=mgh. The m's cancel out, so it is 1/2v^2=gh.

To find out what the height h is, divide g on both sides, so...

h=0.5v^2/g. v=22m/s, g=9.81m/s^2, so h=(0.5)(22^2)/(9.81)=24.67m

5 0

3 years ago

A uniformly charged ball of radius a and charge –Q is at the center of a hollowmetal shell with inner radius b and outer radius

vlabodo [156]

Answer:

r < a:

$E = \frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}$

r = a:

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

a < r < b:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

r = b:

$E = \frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

b < r < c:

E = 0

r = c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

r < c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

Explanation:

Gauss' Law will be applied to each region to find the E-field.

$\int \vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}$

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

r<a:

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

$\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}$

Applying Gauss' Law:

$E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}$

The minus sign determines the direction of the field, which is towards the center.

At r = a:

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

At a < r < b:

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

$E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

At r = b:

$E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

Again, the minus sign indicates the direction of the field towards the center.

At b < r < c:

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

E = 0.

At r < c:

The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

$E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

At r = c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

3 0

3 years ago

Balok diam diatas bidang miring pada sudut kemiringan 40° balok mulai bergerak,tentukan koefisien gesek statis antara balok dan

Pachacha [2.7K]

Answer:

0.84

Explanation:

m = Massa balok

g = Percepatan gravitasi

$\theta$ = Sudut kemiringan

$\mu$ = Koefisien gesekan statik antara balok dan bidang miring

Gaya balok karena beratnya diberikan oleh

$F=mg\sin\theta$

Gaya gesekan diberikan oleh

$f=\mu mg\cos\theta$

Kondisi dimana balok mulai bergerak adalah ketika gaya balok akibat beratnya sama dengan gaya gesek pada balok.

$mg\sin\theta=\mu mg\cos\theta\\\Rightarrow \mu=\dfrac{mg\sin\theta}{mg\cos\theta}\\\Rightarrow \mu=\tan\theta\\\Rightarrow \mu=\tan40^{\circ}\\\Rightarrow \mu=0.84$

Koefisien gesekan statik antara balok dan bidang miring adalah 0.84.

7 0

3 years ago