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Alona [7]
3 years ago
13

Friction always opposes the _____​

Physics
2 answers:
OLga [1]3 years ago
4 0

Answer:

<em>Friction </em><em>always </em><em>opposes</em><em> </em><em>the</em><em> </em>motion

<h3>I HOPE ITS RIGHT IF NOT THEN SORRY</h3>

HAVE A GREAT DAY :)

Elden [556K]3 years ago
3 0

Answer:

<h2><u>frictional force always opposes the applied force.</u></h2>

Explanation:

<h2>Frictional force acts between two surfaces in contact. It stops the </h2><h2>object from moving as the irregularities on both the surface lock into each other, thus obstructing any forward motion. Hence, frictional force always opposes the applied force.</h2><h2>_____________________(◕ᴗ◕✿)</h2>

<h2><u>PLEASE</u><u> </u><u>MARK</u><u> ME</u><u> BRAINLIEST</u><u> AND</u><u> FOLLOW</u><u> ME</u><u> LOTS</u><u> OF</u><u> LOVE</u><u> FROM</u><u> MY</u><u> HEART</u><u> AND</u><u> SOUL</u><u> DARLING</u><u> TEJASWINI</u><u> SINHA</u><u> HERE</u><u> </u><u>❤️</u></h2>
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Tcecarenko [31]
I think its B because there is no chemical change when a phase change occurs
6 0
3 years ago
Let’s say you have a cart of some mass and when pushed with 10N of force, the cart accelerates at 5.0 m/s/s. If you were to push
Korolek [52]

Answer:

a. The acceleration would increase.

Explanation:

because we know that

F=ma

m= mass and a= acceleration

so Mass is same for the cart in any situation that's why only acceleration could increase.

7 0
3 years ago
Read 2 more answers
Is a decrease in velocity considered an acceleration
Dovator [93]
Yes an change in velocity either a decrease or increase is an acceleration.
5 0
3 years ago
Find the net force FnetFnetF_net acting on the sled. Express your answer in terms of some or all of the variables mmm, sss, v1v1
kkurt [141]

Complete Question

The kinetic energy K of an object of mass m moving at a speed v is defined as . It seems reasonable to say that the speed of an object--and, therefore, its kinetic energy--can be changed by performing work on the object. In this problem, we will explore the mathematical relationship between the work done on an object and the change in the kinetic energy of that object.

Let us now consider the situation quantitatively. Let the mass of the sled be m and the magnitude of the net force acting on the sled be    The sled starts from rest.

Consider an interval of time during which the sled covers a distance s and the speed of the sled increases from v_1 to v_2. We will use this information to find the relationship between the work done by the net force (otherwise known as the net work) and the change in the kinetic energy of the sled.

Find the net force acting on the sled.

Express your answer in terms of some or all of the variables m,s,v_1, and v_2.

Answer:

The expression is  F_{net}   = \frac{1}{2s}  * m *  (v_2^2 -  v_1^2)

Explanation:

From the question we are told that

   The net force is F_{net}

    The  distance is  s

     The first velocity is  v_1

     The second velocity is  v_2

     The mass is  m

     

Generally the work energy theorem is mathematically represented as

       W =  F_{net} *  s

Also from the law energy conservation workdone is mathematically represented as

      W = \Delta  K

Here  \Delta K is the change in kinetic energy and this is mathematically represented as

      \Delta K  =  \frac{1}{2}  *  m  * \Delta v^2

So

         W  =  \frac{1}{2}  * m *  \Delta v^2

Here  

      \Delta v^2  =  v^2_2 - v^2_1

Hence

         W  =  \frac{1}{2}  * m *  (v_2^2 -  v_1^2)

So

       F_{net} *  s  = \frac{1}{2}  * m *  (v_2^2 -  v_1^2)

=>    F_{net}   = \frac{1}{2s}  * m *  (v_2^2 -  v_1^2)

4 0
3 years ago
You want to photograph a circular diffraction pattern whose central maximum has a diameter of 1.0 cm . You have a helium-neon la
Nataly [62]

Answer:

71 cm

Explanation:

The formula to apply here is θ = 1.22λ/D,

diameter of central maximum =1.0 cm

wavelength,   λ= 633 nm

Diameter of pinhole = 0.11 mm

where θ = arc length/distance = s/R

s/R = 1.22λ/D

(.005)/R = (1.22)(633 X 10^-9)/(1.1 X 10^-4)

R = .71 m   which is 71 cm

6 0
3 years ago
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