PHYSICS HIGH SCHOOL

Two boxes of masses m and 3m are stacked. The surface between the more massive box and the horizontal surface is smooth and the surface between the boxes is rough. If the less massive box does not slide on the more massive box, what is the static friction force on the less massive box?

Answers

Answer 1

Answer:

Answer:

The friction coefficient is $(F)/(4mg)$ .

Explanation:

Given that,

Mass of first box = m

Mass of second box = 3m

Let the static friction force on the less massive box is $F_(1)$ .

Since the small box is not slide, they have the same acceleration.

We need to calculate the acceleration

Using balance equation

$F-F_(1)=3ma$ ....(I)

$F_(1)=ma$ ....(II)

Adding both the equation

$F=4 ma$

$a=(F)/(4m)$

Put the value of a in equation (II)

$F_(1)=m*(F)/(4m)$

$F_(1)=(F)/(4)$

We need to calculate the friction coefficient

Using formula of frictional force

$F=\mu mg$

$\mu=(F)/(4mg)$

Hence, The friction coefficient is $(F)/(4mg)$ .

Answer 2

Answer: If the less massive box does not slide on the more massive box, the static friction force on the less massive box is (D) F/4.
F = ma
F=(3m+m)a
a = F/4m

Here are the following choices:
A) F
B) F/2
C) F/3
D) F/4

Related Questions

Why is it important for scientists to share information from their investigation?
Who established the 1st college course in radiography for dental students?
Chanel has some cotton candy that came in a cloudy shape. She wants to make it more dense. Which describes the candy before and after Chanel manipulated it?
Most of the Earth's volcanoes occurA. near the center of tectonic plates.B. where there are no tectonic plates.C. along tectonic plate boundaries.D. before tectonic plates can form.
What is the difference between magnetic field and magnetic flux

Select all that apply.A longitudinal wave is characterized by _____.
peaks compressions dips rarefactions

Answers

A longitudinal wave ischaracterized by rarefactions. A longitudinal wave is a wave motion wherein theparticles in the wave medium are displaced parallel to the advancing wave. When motionis detected from the source, the particle next to it vibrates from its rest positionand a progressive change in phase vibration is observed at each particle withinthat wave. The result is that the energy is transported from one region to theother. These combined motions result in the movement of alternating regions of rarefactionin the direction of transport of energy.

It's characterized by both compression and rarefaction.

(the compression in front of the waves induces a rarefaction behind the wave)

The mass of the sun is 1.99×1030kg and its distance to the Earth is 1.50×1011m. What is the gravitational force of the sun on the earth?

Answers

In order to calculate the gravitational force of the two bodies we use the formula which is expressed as:

F = GMm/R²

where G = 6.67 x 10^-11 in SI unit, M and m are the mass of the two bodies and R is the distance between them.

F = 6.67 x 10^-11 (1.99×10^30) (6×10^24) / (1.50×10^11)²
F = 3.53×10^22N

The gravitational force of attraction acting on the earth due to sun is $\boxed{3.539*{{10}^(22)}\,{\text{N}}}$ or $\boxed{3.54*{{10}^(22)}\,{\text{N}}}$ .

Further Explanation:

The Newton’s law of Gravitation states that the gravitational force of attraction acting between two bodies is directly proportional to the product of the mass of the two bodies and inversely proportional to the square of the distance between the two bodies.

Given:

The mass of the sun is $1.99*{10^(30)}\,{\text{kg}}$ .

The distance between the Earth and the sun is $1.50*{10^(11)}\,{\text{m}}$ .

Concept:

The gravitational force of attraction between the two bodies can be represented mathematically as:

$F=\frac{{G{M_1}{M_2}}}{{{r^2}}}$

Here, $F$ is the gravitational force of attraction, $G$ is the gravitational constant, ${M_1}$ is the mass of the first body, ${M_2}$ is the mass of the second body and $r$ is the distance between them.

Since the two bodies are Sun and the Earth so the masses of the two are considered.

The mass of the Earth is $6* {10^(24)}\,{\text{kg}}$ and the value of Gravitational constant is $6.67*{10^( - 11)}\,{{{\text{N}}\cdot{{\text{m}}^{\text{2}}}}\mathord{\left/{\vphantom{{{\text{N}}\cdot{{\text{m}}^{\text{2}}}}{{\text{k}}{{\text{g}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace} {{\text{k}}{{\text{g}}^{\text{2}}}}}$ .

Substitute $6.67*{10^( - 11)}\,{{{\text{N}}\cdot{{\text{m}}^{\text{2}}}}\mathord{\left/{\vphantom{{{\text{N}}\cdot{{\text{m}}^{\text{2}}}}{{\text{k}}{{\text{g}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace} {{\text{k}}{{\text{g}}^{\text{2}}}}}$ for $G$ , $1.99*{10^(30)}\,{\text{kg}}$ for ${M_1}$ , $6* {10^(24)}\,{\text{kg}}$ for ${M_2}$ and $1.50*{10^(11)}\,{\text{m}}$ for $r$ in above equation.

$\begin{aligned}F&=\frac{{\left({6.67*{{10}^(-11)}\,}\right)*\left({1.99*{{10}^(30)}\,{\text{kg}}}\right)*\left({6*{{10}^(24)}\,{\text{kg}}}\right)}}{{{{\left({1.50*{{10}^(11)}\,{\text{m}}}\right)}^2}}}\n&=\frac{{7.964*{{10}^(44)}}}{{2.25*{{10}^(22)}}}\n&=3.539*{10^(22)}\,{\text{N}}\n\end{aligned}$

Therefore, the gravitational force of attraction acting on the earth due to sun is $\boxed{3.539*{{10}^(22)}\,{\text{N}}}$ or $\boxed{3.54* {{10}^(22)}\,{\text{N}}}$ .

Learn More:

1. Calculate the total force on the earth due to Venus, Jupiter and Saturn brainly.com/question/2887352

2. A rocket being thrust upward as the force of the fuel being burned pushes downward brainly.com/question/11411375

3. A 50-kg meteorite moving at 1000 m/s strikes earth. Assume the velocity is along the line brainly.com/question/6536722

Answer details:

Grade: High school

Subject: Physics

Chapter: Newton’s law of Gravitation

Keywords:

Gravitational force, attraction, newton’s law, earth, sun, mass of the sun, distance, 1.99*10^30 kg, 1.50*10^11 m, 6.67x10^-11, gravitational constant.

Suggest one reason why the experiment might not have given a correct value forthe specific heat capacity of the metal.

Answers

Answer:

It is because some of the heat is lost in the surroundings.

Explanation:

To calculate the specific heat of the metal, the following steps are performed.

Take a piece of metal whose mass and the initial temperature is known.

Now take a beaker filled with some measured amount of water and the temperature of water is known.

Drop the heated metal piece in the water so that the equilibrium is reached.

According to the principle of calorimetery,

Heat lost by the metal = heat gained by the water

here, some of the heat is lost in the surroundings so we don't get the correct value of the specific heta of the metal.

A trombone has a variable length. When a musician blows into the mouthpiece and causes air in the tube of the horn to vibrate, the waves set up by the vibrations reflect back and forth in the horn to create standing waves. As the length of horn is made shorter, what happens to the frequency?A) The frequency will increase or decrease depending on how hard the horn player blows.
B) The frequency will increase
C) The frequency will decrease
D) The frequency will increase or decrease depending on the diameter of the horn.
E) The frequency will remain same

Answers

Answer:

B) The frequency will increase

Explanation:

The length of the instrument is given by

$L=(\lambda)/(2)\n\Rightarrow \lambda=2L$

Wavelength is given by

$\lambda=(v)/(f)$

$(v)/(f)=2L\ or\ L\propto (1)/(f)$

It can be seen that the length is inversely proportional to the frequency.

Here, the length of the trombone is decreasing so the frequency will increase.

An elevator is initially moving upward at a speed of 12.60 m/s. The elevator experiences a constant downward acceleration of magnitude 3.60 m/s2 for 2.97 s.

Answers

I'm assuming you're asking for the elevator's final velocity.

Consider the following equation:

Vf = Vi + aΔt

Vf is the final velocity.

Vi is the initial velocity.

a is the acceleration.

Δt is the amount of elapsed time.

Given values:

Vi = 12.60m/s

a = -3.60m/s²

Δt = 2.97s

Substitute the terms in the equation with the given values and solve for Vf:

Vf = 12.60 - 3.60×2.97

Vf = 1.91m/s

What causes the earths crust to move ?

Answers

Upwelling currents in the molten material beneath the crust.

The rocks that make up the crust are light, compared with the metal-rich material beneath. The crust floats on top like an iceberg. Slow-moving currents underneath propel the continents around the surface

The movement of magma underneath earth's crust.

Other Questions

What velocity will move an object from 50 m to −50 m in 20 s?

iffany is told that shape A is bigger than Shape B and that Shape B is bigger than shape C. She is then asked about the difference between shape A and Shape C. If Tiffany is in the concrete operational stage, her answer will most likely be: a) "A is bigger than B, but I don't know about C." b) "A is bigger than B, and C is bigger than B, so A is bigger than C." c) "I need a ruler to measure them." d) "I can't tell the difference."

what are some sources of error that could explain differences between your prediction and what you measured experimentally for sound speed

the three small spheres shown in (figure 1) carry charges q1 = 3.50 nc , q2 = -7.50 nc , and q3 = 2.05 nc .