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3 years ago

8

Find the speed of a bicyclist who took an hour and a half to travel 10km

1 answer:

Natalija [7]3 years ago

7 0

distance = 10km

time = 1 1/2 = 3/2 hours

speed = distance/time = 10 / 3/2 = 20/3 = 6.67 Km/h

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You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius

Gennadij [26K]

Answer:

t = 0.0735 m

Explanation:

Angular acceleration of the flywheel is given as

$\alpha = 3 rad/s^2$

now after t = 8 s the speed of the flywheel is given as

$\omega = \alpha t$

$\omega = 3 \times 8$

$\omega = 24 rad/s$

now rotational kinetic energy of the wheel is given as

$K = \frac{1}{2}I\omega^2$

$K = \frac{1}{2}(\frac{1}{2}mR^2)(24^2)$

$800 = \frac{1}{4}m(0.23)^2(24^2)$

$m = 105 kg$

now we have

$m = \rho (\pi R^2) t$

$105 = 8600(\pi \times 0.23^2) t$

$t = 0.0735 m$

4 0

3 years ago

A set is any collection of objects. true or false.

Semenov [28]

It can be, but set also means many other things.

3 0

4 years ago

A cyclotron designed to accelerate protons has a magnetic field of magnitude 0.15 T over a region of radius 7.4 m. The charge on

klio [65]

Explanation:

It is given that,

Magnetic field, B = 0.15 T

Charge on a proton, $q=1.60218\times 10^{-19}\ C$

Mass of a proton, $m=1.67262 \times 10^{-27}\ kg$

The cyclotron frequency is given by :

$f=\dfrac{qB}{2\pi m}$

$f=\dfrac{1.60218\times 10^{-19}\ C\times 0.15\ T}{2\pi \times 1.67262 \times 10^{-27}\ kg}$

f = 2286785.40 Hz

or

$\omega=14368296.44\ rad/s$

$\omega=1.43\times 10^7 rad/s$

Hence, this is the required solution.

8 0

3 years ago

A boat travels at 30 mph for a huge, solid cliff that is about 3,000 meters away. When the horn on the boat makes a toot, you ca

AleksandrR [38]

Answer:The frequency of the echo is slightly decreased

Explanation:

Given

speed of boat $=30\ mph$

cliff is $3000\ m$ away

when boat is still , suppose t is the time taken by the echo to reach observer on the boat

But as soon as boat starts moving the distance between cliff and boat decreasing and time for echo to reach observer also decreases

and we know $time \propto \frac{1}{frequency}$

therefore frequency of the echo slightly decreased.

5 0

3 years ago

N LC circuit has an oscillation frequency of 105 Hz. If C = 0.1 F , then L must be about:

Umnica [9.8K]

Answer:

L = 22.97 H

Explanation:

Given that,

Capacitance, $C=0.1\ \mu F=0.1\times 10^{-6}\ F$

Oscillation frequency, f = 0.5 Hz

The frequency of an AC circuit is given by :

$f=\dfrac{1}{2\pi \sqrt{LC} }$

Where

L is impedance

$f^2=\dfrac{1}{4\pi ^2LC}\\\\L=\dfrac{1}{4\pi ^2 f^2 C}\\\\\text{Putting all the values}\\\\L=\dfrac{1}{4\pi^2 \times (105)^2\times 0.1\times 10^{-6}}\\\\L=22.97\ H$

So, the impedance of LC circuit 22.97 H.

7 0

3 years ago