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Veseljchak [2.6K]
3 years ago
10

Slope or velocit ime graph represent

Physics
1 answer:
Lubov Fominskaja [6]3 years ago
7 0

Answer:

Please check the explanation.

Explanation:

The slope of the velocity-time graph illustrates the change in velocity with respect to change in time.

In other words, the acceleration of the object is defined by the slope of a velocity graph. The acceleration can be obtained by finding the slope at a particular time.

Hence, the slope of the velocity time graph represent represents acceleration.

2. Also the acceleration of any object, like a car or bike, at some discrete instant in time 't' is termed as instantaneous acceleration which can be determined if we may take the derivate of the given velocity function.

3. A vector that has a direction and magnitude of 1 is termed as a unit vector, often called a direction vector.

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artcher [175]

Answer:

Part a)

\rho = 2.12\mu C/m^3

Part b)

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q_2 = 8.88 nC

q_3 = 71 nC

Part c)

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E_2 = 7992 N/C

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Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

V = \frac{4}{3}\pi R^3

V = \frac{4}{3}\pi(0.20)^3

V = 0.0335 m^3

now the charge density of the ball is given as

\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3

Part b)

Now the charge enclosed by the surface is given as

q = \rho V

at radius of 5 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3

q = 1.11 nC

at radius of 10 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3

q = 8.88 nC

at radius of 20 cm

q = 71 nC

Part c)

As we know that electric field is given as

E = \frac{kq}{r^2}

so we have electric field at r = 5 cm

E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}

E_1 = 3996 N/C

electric field at r = 10 cm

E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}

E_2 = 7992 N/C

electric field at r = 20 cm

E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}

E_3 = 15975 N/C

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