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ioda
3 years ago
15

What is the change in velocity of a 22-kg object that experiences a force of 15 N for

Physics
1 answer:
vagabundo [1.1K]3 years ago
3 0

Answer:

Force = mass × acceleration

Acceleration:

{ \tt{15 = (22 \times a)}} \\ { \tt{a =  \frac{15}{22}  \:  {ms}^{ - 2} }}

From first Newton's equation of motion:

{ \bf{v = u + at}}

Change = v - u:

{ \tt{v - u = (a \times t)}} \\ { \tt{v - u = ( \frac{15}{22} \times 1.2) }} \\ { \tt{v - u = 0.82 \:  {ms}^{ - 2} }}

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Angelina_Jolie [31]
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2 years ago
A charged particle is placed in an external magnetic field and it is moving in a circular path of radius 26m.The magnetic force
ElenaW [278]

Answer:

208 Joules

Explanation:

The radius of the circular path the charge moves, r = 26 m

The magnetic force acting on the charge particle, F = 16 N

Centripetal force, F_c = m·v²/r

Kinetic energy, K.E. = (1/2)·m·v²

Where;

m = The mass of the charged particle

v = The velocity of the charged particle

r = The radius of the path of the charged particle

Whereby the magnetic force acting on the charge particle = The centripetal force, we have;

F = F_c = m·v²/r = 16 N

(1/2) × r × F_c = (1/2) × r × m·v²/r = (1/2)·m·v² = K.E.

∴ (1/2) × r × F_c = (1/2) × 26 m × 16 N =  = (1/2)·m·v² = K.E.

∴ 208 Joules = K.E.

The kinetic energy of an particle moving in the circular path, K.E. = 208 Joules.

4 0
3 years ago
What causes differences in air pressure around the Earth?
Olegator [25]
Wind is primarily driven by differences in air pressure. These variations in air pressure are due to temperature differences caused by variations in solar energy received at the surface of the Earth.
7 0
3 years ago
A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magn
elena-s [515]

Answer:

The change in potential energy is  \Delta  PE =  -  3.8*10^{-16} \ J

Explanation:

From the question we are told that

     The  magnitude of the uniform electric field  is  E =  950 \ N/C

      The  distance traveled by the electron is  x =  2.50 \ m

Generally the force on this electron is  mathematically represented as

     F =  qE

Where F is the force and  q is the charge on the electron which is  a constant value of  q =  1.60*10^{-19} \ C

    Thus  

      F  =  950  * 1.60 **10^{-19}

      F  = 1.52 *10^{-16} \ N

Generally the work energy theorem can be mathematically represented as

          W =  \Delta  KE

Where W is the workdone on the electron by the  Electric field and  \Delta  KE  is the change in kinetic energy

Also  workdone on the electron can also  be represented as

        W =  F* x  *cos(  \theta )

Where  \theta  =  0 ^o considering that the movement of the electron is along the x-axis  

        So

             \Delta  KE  =  F  * x  cos  (0)

substituting values

         \Delta  KE  =  1.52 *10^{-16}  * 2.50   cos  (0)

          \Delta  KE   =  3.8*10^{-16} J

Now From the law of energy conservation

       \Delta PE  =  -  \Delta  KE

Where \Delta  PE is the change  in  potential energy  

Thus  

        \Delta  PE =  -  3.8*10^{-16} \ J

               

7 0
3 years ago
A horse is pulling 53.0 kg plow forward while the ground exerts a 275 N force, and the plow accelerates at 0.222 m/s^2. what is
JulsSmile [24]
Answer:
Magnitude of force the ground exerts on the plow = 263.234 N

Explanation:
Magnitude of force the ground exerts on the plow = Fground - Fplow
We are given that:
Fgound = 275 N
We will now calculate Fplow as follows:
Fplow = mass of horse * acceleration of plow
Fplow = 53 * 0.222
Fplow = 11.766 N

Now, substitute in the above equation to get magnitude of force the ground exerts on the plow as follows:
Magnitude of force the ground exerts on the plow = Fground - Fplow
Magnitude of force the ground exerts on the plow = 275 - 11.766
Magnitude of force the ground exerts on the plow = 263.234 N

Hope this helps :)
3 0
3 years ago
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