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Dafna1 [17]
4 years ago
10

Can anyone please help me with answering this question? : An accelerating (speeding up) body has a Net Force acting in the direc

tion of motion. True or False? Why? ​
Physics
1 answer:
Katarina [22]4 years ago
7 0
That's true.

Netwon's second law states that the resultant of the forces F acting on a body is equal to the product between its mass m and its acceleration a:
\sum F = ma
This means that if the net force acting on an object is different from zero (term on the left), than the acceleration of the object (term on the right) must be different from zero as well, and therefore the body is accelerating.

In particular, both F and a in the equation are vectors: this means that if the acceleration is positive, F and a have the same direction. In this problem, the acceleration is positive (because the object is speeding up), therefore the force and the acceleration have same direction.
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Why do electrical devices have resistance​
Kamila [148]
As electrons move through the conductor, some collide with atoms, other electrons, or impurities in the metal.
8 0
3 years ago
Explain the different methods that can be used to model the motion of an object.
Simora [160]

Answer:

You can describe the motion of an object by its position, speed, direction, and acceleration. An object is moving if its position relative to a fixed point is changing. idk if this helps.

Explanation:  

5 0
3 years ago
Read 2 more answers
An electron (q=-1.602×10-19C) is placed .03m away from spherical object with a net charge of -7.2 C.
vovangra [49]

Answer:

Explanation:

electric field at the location of electron

= 9 x 10⁹ x 7.2 / .03²

= 72 x 10¹² N/C

force on electron = electric field x charge on electron

= 72 x 10¹² x 1.6 x 10⁻¹⁹

= 115.2 x 10⁻⁷ N .

C )

work done = charge on electron x potential difference at two points

potential at .03 m

= 9 x 10⁹ x 7.2 / .03

= 2.16 x 10¹² V

potential at .001 m

= 9 x 10⁹ x 7.2 / .001

= 64.8 x 10¹² V

potential difference = (64.8 - 2.16 )x 10¹² V

= 62.64 x 10¹² V  .

work done = 62.64 x 10¹² x 1.6 x 10⁻¹⁹

= 100.224 x 10⁻⁷ J .

D )

There will be no change in the magnitude of force on positron except that the direction of force will be reversed . In case of electron , there will be repulsion and in case of positron , there will be attraction .

Work done in case of electron will be positive and work done in case of positron will be negative .

electric field due to charge will be same in both the cases .

8 0
3 years ago
What is work done aganist gravity?​
Rudik [331]

Answer:

Work done against gravity in lifting an object becomes potential energy of the object-Earth system. The change in gravitational potential energy, ΔPEg, is ΔPEg = mgh, with h being the increase in height and g the acceleration due to gravity.

Explanation:

You're Welcome.

8 0
3 years ago
Read 2 more answers
Can someone solve this problem and explain to me how you got it​
evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒F\alpha\frac{q1*q2}{r^{2}}

∴F=k\frac{q1*q2}{r^{2}}

where F is the force of attraction or repulsion

k is Coulumb's constant=9*10^{9}Nm^{2}C^{-2}

q1 and q2 are the magnitude of the charges

r is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

F=9*10^{9}*\frac{0.041*0.029}{12^{2}}

F=74312.5N

Question6:

Given: q1=3*10^{-18}C, r=5 m, F=4*10^{-11}N

therefore by Coulumb's law,

4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}

⇒q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C

4 0
4 years ago
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