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Feliz [49]
3 years ago
9

A box of bananas weighing 40.0 n rests on a horizontal surface. the coefficient of static friction between the box and the surfa

ce is 0.40, and the coefficient of kinetic friction is 0.20. (a if no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on the box? (b what is the magnitude of the friction force if a monkey applies a horizontal force of 6.0 n to the box and the box is initially at rest? (c what minimum horizontal force must the monkey apply to start the box in motion? (d what minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started? (e if the monkey applies a horizontal force of 18.0 n, what is the magnitude of the friction force and what is the box's acceleration?
Physics
1 answer:
77julia77 [94]3 years ago
3 0

Answer:

https://www.physicsforums.com/threads/monkey-friction.182280/

Explanation: look here

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Who actually asked Abraham to sacrifice his son?
pochemuha

I think god did ??? I searched it up okay

7 0
3 years ago
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Listed following is a set of statements describing individual stars or characteristics of stars. Match these to the appropriate
Oksana_A [137]

Answer:

Red giant or super giant → very cool but very luminous

                                      → found in the upper right of the H-R diagram.

Main sequence →The majority of stars in our galaxy

                        → Sun, for example

                        → a very hot and very luminous star

White dwarfs → very hot but very dim

                     → not much larger in radius than earth  

Explanation:

Giant:

When the stars run out of their fuel that is hydrogen for the nuclear fusion reactions then they convert into Giant stars.That's why they are very cool. Giant stars have the larger radius and luminosity then the main sequence stars.

Main Sequence:

Stars are called main sequence stars when their core temperature reaches up to 10 million kelvin and their start the nuclear fusion reactions of hydrogen into helium in the core of the star. That is why they are very hot and luminous. For example sun is known as to be in the stage of main sequence as the nuclear fusion reactions are happening in its core.

White dwarfs:

When the stars run out of their fuel then they shed the outer layer planetary nebula, the remaining core part that left behind is called as white dwarf. It's the most dense part as the most of the mass is concentrated in this part.

6 0
3 years ago
when two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of
Ahat [919]

<h2>When two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. The mass of P is half that of Q. The ratio of the specific heat capacity of P to Q​</h2>

Explanation:

Specific heat capacity

It is defined as amount of heat required to raise the temperature of a substance by one degree celsius .

It is given as :

Heat absorbed = mass of substance x specific heat capacity x rise in temperature

or ,

Q= m x c x t

In above question , it is given :

For Q

mass of Q = m

Temperature changed =T₂/2

Heat supplied = x

Q= mc t

or

X=m x C₁ X T₁

or, X =m x C₁ x T₂/2

or, C₁=X x 2 /m x T₂                                 (equation 1 )

For another quantity : P

mass of P =m/2

Temperature= T₂

Heat supplied is same that is : X

so, X= m/2 x C₂ x T₂                            

or, C₂=2X/m. T₂                                   (equation 2 )

Now taking ratio of C₂ to c₁, We have

C₂/C₁= 2X /m.T₂  /2X  /m.T₂

so, C₂/C₁= 1/1

so, the ratio is 1: 1

8 0
4 years ago
A damped mass/spring system takes 14.0 s for its amplitude of the oscillator to decrease by a factor of 9. By what factor does t
fiasKO [112]

Answer:

The correct answer is "0.246".

Explanation:

Given that the amplitude is decreased by a factor of 9, then

A \rightarrow (A-\frac{A}{9} )

A \rightarrow \frac{8A}{9}

As we know,

Energy will be:

⇒  E_{1}=\frac{1}{2}KA^2

and,

⇒  E_{2}=\frac{1}{2}K(\frac{8A}{9} )^2

          =\frac{64KA^2}{162}

⇒  \Delta E=E_1-E_2

On putting the estimated values, we get

           =\frac{1}{2}KA^2-\frac{64KA^2}{162}

⇒  \frac{\Delta E}{E}=\frac{\frac{20}{162}KA^2}{\frac{1}{2}KA^2}

          =\frac{40}{162}

          =0.246

3 0
3 years ago
A charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with p
alukav5142 [94]

A) d. 10T

When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

F=qvB

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.

This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

qvB = \frac{mv^2}{r}

which can be rewritten as

v=\frac{qB}{mr}

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

\frac{2\pi r}{T}=\frac{qB}{mr}

So, we get:

T=\frac{2\pi m r^2}{qB}

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.

B) a. f/10

The frequency of revolution of a particle in uniform circular motion is

f=\frac{1}{T}

where

f is the frequency

T is the period

We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:

T' = 10 T

Then its frequency of revolution will be:

f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}

6 0
3 years ago
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