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2 years ago

If a car starts at rest and accelerates at 3.4 m/s^2 over a distance of 200 meters, how fast will it be travelling at the end?

Physics

1 answer:

CaHeK987 [17]2 years ago

8 0

Answer:

36.9 m/s

Explanation:

From;

v^2 = u^2 + 2as

Where;

v = final velocity =?

a = acceleration = 3.4 m/s^2

u = initial velocity = 0 m/s

s = distance covered = 200 meters

v^2 = 0^2 + 2 * 3.4 * 200

v^2 = 1360

v = √1360

v = 36.9 m/s

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d = 2m
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t= d / v
= 2 / 100
= 0.02sec

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3 years ago

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A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const

grin007 [14]

Answer:

Explanation:

we can consider an element of radius r < a and thickness dr. and Area of this element is

$dA=2\pi r dr$

since current density is given

$J=kr$

then , current through this element will be,

$di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr$

integrating on both sides between the appropriate limits,

$\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr
\\\\
I=\frac{2\pi\,ka^3}{3} -------------------------------(1)$

Magnetic field can be found by using Ampere's law

$\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}$

for points inside the wire ( r<a)

now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

$\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

$B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=
\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}
\\\\B_{in}=\frac{\mu_0kl^2}{3}
$

now using equation 1, putting the value of k,

$B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}
\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}
$

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

applying the Ampere's law

$\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

$B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr)
\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}
\\\\B_{out}=\frac{\mu_0ka^3}{3r}
$

again using,equaiton 1,

$B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}
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3 years ago

How fast would you say the earth rotates near the equator?

PtichkaEL [24]

460 meters per second, or about 1,000miles per hour.

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3 years ago

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netineya [11]

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2 years ago

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A simple pendulum has length of 820mm. Calculate the frequency (g = 9.8 ms -2)

Vadim26 [7]

Answer:

$\huge\boxed{\sf f=0.55 \ Hz}$

Explanation:

Given Data:

Length = l = 820 mm = 0.82 m

Acceleration due to gravity = g = 9.8 ms⁻²

Required:

Frequency = f = ?

Formula:

$\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} }$

Solution:

$\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} } \\\\Put\ the\ givens\\\\f=\frac{1}{2 \pi} \sqrt{\frac{9.8}{0.82} }\\\\ f = 0.159 \times \sqrt{11.95} \\\\f=0.159 \times 3.457\\\\f=0.55 \ Hz\\\\\rule[225]{225}{2}$

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