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tatuchka [14]
3 years ago
14

Acceleration occurs whenever an ___________________ force acts on an object.

Physics
2 answers:
leonid [27]3 years ago
8 0
Acceleration occurs whenever the forces on an object are unbalanced.

It's the group of forces on the object that's either balanced or unbalanced.
There's no such thing as "an unbalanced force".

Elina [12.6K]3 years ago
4 0
Acceleration is known to occur whenever an unbalanced force acts on an object.
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If the volume of a container increases what<br> effect will this have on pressure? Why?
frozen [14]

As the volume of the container increases the pressure inside will decrease because the atoms have more room to move around in.

5 0
2 years ago
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A computer monitor accelerates electrons and directs them to the screen in order to create an image. If the accelerating plates
tensa zangetsu [6.8K]

Answer:

Electric field at a distance of 1.45 cm will be 172.41\times 10^4N/C

Explanation:

We have given the distance d = 1.45 cm = 0.0145 m

And the potential difference V=2.5\times 10^4volt

There is a relation between potential difference and electric field

Electric field at a distance d due to a potential difference is given by

E=\frac{V}{d}, here E is electric field, V is potential difference and d is distance

So E=\frac{V}{d}=\frac{2.5\times 10^4}{0.0145}=172.41\times 10^4N/C

8 0
3 years ago
A small glass bead has been charged to + 30.0 nC . A small metal ball bearing 2.60 cm above the bead feels a 1.80×10−2 N downwar
BlackZzzverrR [31]

Answer:

The charge on the ball bearing 4.507 × 10^-8 C

Explanation:

From Coulomb's law

F = kq1q2/r²

make q2 the subject

q2 = Fr²/kq1

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8 0
2 years ago
I really need help please just answer at least one
yuradex [85]

Answer:

9) a = 25 [m/s^2], t = 4 [s]

10) a = 0.0875 [m/s^2], t = 34.3 [s]

11) t = 32 [s]

Explanation:

To solve this problem we must use kinematics equations. In this way we have:

9)

a)

v_{f}^{2} = v_{i}^{2}-(2*a*x)\\

where:

Vf = final velocity = 0

Vi = initial velocity = 100 [m/s]

a = acceleration [m/s^2]

x = distance = 200 [m]

Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.

0 = (100)^2 - (2*a*200)

a = 25 [m/s^2]

b)

Now using the following equation:

v_{f} =v_{i} - (a*t)

0 = 100 - (25*t)

t = 4 [s]

10)

a)

To solve this problem we must use kinematics equations. In this way we have:

v_{f} ^{2} =  v_{i} ^{2} + 2*a*(x-x_{o})

Note:  The positive sign of the equation means that the car increases his speed.

5^2 = 2^2 + 2*a*(125 - 5)

25 - 4 = 2*a* (120)

a = 0.0875 [m/s^2]

b)

Now using the following equation:

v_{f}= v_{i}+a*t\\

5 = 2 + 0.0875*t

3 = 0.0875*t

t = 34.3 [s]

11)

To solve this problem we must use kinematics equations. In this way we have:

v_{f} ^{2} =  v_{i} ^{2} + 2*a*(x-x_{o})

10^2 = 2^2 + 2*a*(200 - 10)

100 - 4 = 2*a* (190)

a = 0.25 [m/s^2]

Now using the following equation:

v_{f}= v_{i}+a*t\\

10 = 2 + 0.25*t

8 = 0.25*t

t = 32 [s]

4 0
3 years ago
Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the p
ira [324]

Answer:

The force exerted on the q_1 is  F =  2.25*10^{3} \ N

Explanation:

From the question we are told that

     The area is  A =  2.34*10^{-3} \ m^2

     The magnitude of charge placed on them is  q =  7.07 * 10^{-7} C

     The charge placed between the plate is q_1 = 6.62 *10^{-5} C

   

The electric field generated around the plate  is mathematically represented as

           E =  \frac{q}{A \epsilon_o}

Substituting values

          E =  \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}

         E = 34*10^{6} \ V/m

The force exerted the charge q_1 is  mathematically represented as

        F =  q_1 * E

Substituting values  

        F =  6.62 *10^{-5} * 34*10^{6}

        F =  2.25*10^{3} \ N

8 0
3 years ago
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