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3 years ago

A car in front of the school goes from rest to 27 m/s in 3.0 seconds. What is its acceleration (assuming

Physics

1 answer:

WITCHER [35]3 years ago

4 0

The answer is 13.5 because 27÷3.0=13.5

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kaheart [24]

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3 years ago

Read 2 more answers

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

7 0

2 years ago

In designing circular rides for amusement parks, mechanical engineers must consider how small variations in certain parameters c

Bess [88]

Answer:

Part a)

$dF = -\frac{mv^2}{r^2} dr$

Part b)

$dF = \frac{2mvdv}{r}$

Part c)

$dT = - \frac{2\pi r}{v^2} dv$

Explanation:

Part a)

As we know that force on the passenger while moving in circle is given as

$F = \frac{mv^2}{r}$

now variation in force is given as

$dF = -\frac{mv^2}{r^2} dr$

here speed is constant

Part b)

Now if the variation in force is required such that r is constant then we will have

$F = \frac{mv^2}{r}$

so we have

$dF = \frac{2mvdv}{r}$

Part c)

As we know that time period of the circular motion is given as

$T = \frac{2\pi r}{v}$

so here if radius is constant then variation in time period is given as

$dT = - \frac{2\pi r}{v^2} dv$

8 0

2 years ago

A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.25 kg to a friend standing in front of him

juin [17]

Answer:

The velocity of the student has after throwing the book is 0.0345 m/s.

Explanation:

Given that,

Mass of book =1.25 kg

Combined mass = 112 kg

Velocity of book = 3.61 m/s

Angle = 31°

We need to calculate the magnitude of the velocity of the student has after throwing the book

Using conservation of momentum along horizontal direction

$m_{b}v_{b}\cos\theta= m_{c}v_{c}$

$v_{s}=\dfrac{m_{b}v_{b}\cos\theta}{m_{c}}$

Put the value into the formula

$v_{c}=\dfrac{1.25\times3.61\times\cos31}{112}$

$v_{c}=0.0345\ m/s$

Hence, The velocity of the student has after throwing the book is 0.0345 m/s.

3 0

2 years ago

A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra

kkurt [141]

Answer:

$V_{ft}= 317 cm/s$

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

$P_i = P_f$

Where:

$P_i=M_cV_{ic} + M_tV_{it}$

$P_f = M_cV_{fc} + M_tV_{ft}$

Now:

$M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}$

Where $M_c$ is the mass of the car, $V_{ic}$ is the initial velocity of the car, $M_t$ is the mass of train, $V_{fc}$ is the final velocity of the car and $V_{ft}$ is the final velocity of the train.