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NNADVOKAT [17]
3 years ago
14

If a car starts from rest and attains a velocity of 10m/square in 5 secondscalculate the accerleration​

Physics
2 answers:
stepan [7]3 years ago
8 0

Explanation:

Acceleration = Velocity / Time

Acceleration = 10 / 5

Acceleration = 2m/s ²

bogdanovich [222]3 years ago
7 0

Answer:

acceleration may be 34 beciase if calcuation maybr

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If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. Wh
In-s [12.5K]

Answer:

 v = -v₀ / 2

Explanation:

For this exercise let's use kinematics relations.

Let's use the initial conditions to find the acceleration of the electron

            v² = v₀² - 2a y

when the initial velocity is vo it reaches just the negative plate so v = 0

           a = v₀² / 2y

now they tell us that the initial velocity is half

          v’² = v₀’² - 2 a y’

          v₀ ’= v₀ / 2

at the point where turn v = 0              

          0 = v₀² /4  - 2 a y '

          v₀² /4 = 2 (v₀² / 2y)  y’

          y = 4 y'

          y ’= y / 4

We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.

         v² = v₀² -2a y’

         v² = 0 - 2 (v₀² / 2y) y / 4

         v² = -v₀² / 4

         v = -v₀ / 2

We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.

7 0
3 years ago
wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the
lys-0071 [83]

Answer:

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

BP = 2 m

A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

E_{A}=\frac{KQ_{A}}{AP^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{6^{2}}=1000 N/C (rightwards)

The electric field due to charge QB at P is EB leftwards

E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

5 0
3 years ago
Change in momentum is equal to force times the amount of time the force is applied? True or False?
harina [27]
The answer to this question is false
7 0
3 years ago
Increasing the number of loops in a electromagnet or solenoid will cause it to be stronger
Zielflug [23.3K]
True

The electromagnet will become stronger if we add more coils because there are more field lines in a loop then there is in a straight piece of wire. In a solenoid there are a lot of loops and they are concentrated in the middle, as more loops are added the field lines get larger, therefore making the electromagnet stronger.
5 0
3 years ago
An electron is moving at 7.4x105 m/s perpendicular to a magnetic field. It experiences a force of 2.0x10–13 N. What is the magne
alisha [4.7K]

Answer:

1.69 T

Explanation:

Applying,

F = BvqsinФ.................. Equation 1

Where F = Force, B = magnetic field, v = velocity, q = charge on an electron, Ф = angle between the electron and the field.

make B the subject of the equation,

B = F/(vqsinФ)............. Equation 2

From the question,

Given: F = 2.0×10⁻¹³ N, v = 7.4×10⁵ m/s, Ф = 90°

Constant: q = 1.60×10⁻¹⁹ C

Substitute into equation 2

B =  2.0×10⁻¹³/(7.4×10⁵×1.60×10⁻¹⁹×sin90°)

B = 0.169×10

B = 1.69 T

4 0
3 years ago
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