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3 years ago

10

Carlos runs with velocity \vec{v}v →= (5.6 m/s, 29o north of east) for 10 minutes. How far to the north of his starting position

does Carlos end up? (Give your answer to the nearest meter)

1 answer:

romanna [79]3 years ago

7 0

Answer:

1629 metres

Explanation:

Given that Carlos runs with velocity v = (5.6 m/s, 29o north of east) for 10 minutes.

To the north, the velocity will be:

V = 5.6 × sin 29

V = 2.715 m/s

Convert the time to second

Time = 10 × 60 = 600

Using the formula below

Velocity = distance/ time

Substitute all the parameters into the formula

2.715 = distance/600

Distance = 2.715 × 600

distance = 1629 m

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3 years ago

A donkey starts from rest and accelerates for 6.7 s at a rate of 1.6 m/s².

vitfil [10]

The maximum speed of the donkey is 10.72m/s

The question is based on the principle of motion in one dimension and hence formulas of motion in one dimension can be applied.

It is given that donkey attains an acceleration of 1.6 m/s^2

The time taken to accelerate to given speed is 6.7 seconds

We use the formula v=u + at to find the fastest speed

v is the final or maximum speed

u is the initial speed which in this case is 0 as the donkey is at rest

a is the acceleration of the donkey

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1 year ago

A 20 cm-radius ball is uniformly charged to 71 nC.

artcher [175]

Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

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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