How can a cyclist minimize friction as he or she rides?

The radius of Earth is about 6450 km. A 7070 N spacecraft travels away from Earth. What is the weight of the spacecraft at a hei

Triss [41]

Answer:

(a) 1767.43 N

(b) 182.45 N

Explanation:

Radius of earth, R = 6450 km

Weight of person, W = 7070 N

mass of person, m = W / g = 7070 / 9.8 = 721.4 kg

(a) h = 6450 km

The value of acceleration due to gravity on height is given by

$g' = g\left ( \frac{R}{R+h} \right )^2$

$g' = g\left ( \frac{6450}{6450+6450} \right )^2$

g' = g / 4 = 9.8 / 4 = 2.45 m/s^2

The weight of the person at such height is

W' = m x g' = 721.4 x 2.45

W' = 1767.43 N

(b) h = 33700 km

The value of acceleration due to gravity on height is given by

$g' = g\left ( \frac{R}{R+h} \right )^2$

$g' = g\left ( \frac{6450}{6450+33700} \right )^2$

g' = g x 0.0258 = 9.8 x 0.0258 = 0.253 m/s^2

The weight of the person at such height is

W' = m x g'

W' = 721.4 x 0.253

W' = 182.45 N

3 0

3 years ago

Electromagnetic induction

vredina [299]

Answer:

Electromagnetic Induction or Induction is a process in which a conductor is put in a particular position and magnetic field keeps varying or magnetic field is stationary and a conductor is moving. This produces a Voltage or EMF (Electromotive Force) across the electrical conductor.

3 0

2 years ago

Read 2 more answers

A skier is pulled by a towrope up a frictionless ski slope that makes an angle of 12 degrees with the horizontal. The rope moves

MArishka [77]

Answer:

Explanation:

Given,

Work done by the rope 900 m/s.
Angle of inclination of the slope = $\theta\ =\ 12^o$
Initial speed of the skier = v = 1.0 m/s
Length of the inclined surface = d = 8.0 m

part (a)

The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity

$\therefore W_r\ =\ W_g\ =\ 900\ J$

In both cases the height attained by the skier is equal. and the work done by gravity does not depend upon the speed of the skier.

part (b)

Initial speed of the skier = v = 1.0 m/s.

Rate of the work done by the rope is power of the rope.

$Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 1.0}{8.0}\\\Rightarrow P\ =\ 112.5\ Watt$

Part (c)

Initial speed of the skier = v = 2.0 m/s.

Rate of the work done by the rope is power of the rope.

$Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 2.0}{8.0}\\\Rightarrow P\ =\ 225\ Watt$

4 0

2 years ago

Convert 2 days into second

eduard

Answer:

172,000 second 

I'M NOT SURE THAT THIS IS RIGHT OR WRONG IF IT'S WRONG THEN SORRY 

3 0

2 years ago

Read 2 more answers

Using 0.500 g of nichrome, you are asked to fabricate a wire with uniform cross-section. The resistance of the wire is 0.673 Ω.

mojhsa [17]

Explanation:

Given that,

Mass of Nichrome, m = 0.5 g

The resistance of the wire, R = 0.673 ohms

Resistivity of the nichrome wire, $\rho=10^{-6}\ \Omega -m$

Density, $d=8.31\times 10^3\ kg/m^3$

(A) The length of the wire is given by using the definition of resistance as :

Volume,

$V=A\times l\\\\A=\dfrac{V}{l}\\\\Since, V=\dfrac{m}{d}\\\\V=\dfrac{m}{d}\\\\V=\dfrac{0.5\times 10^{-3}}{8.31\times 10^3}\\\\V=6.01\times 10^{-8}\ m^3$

Area,

$A=\dfrac{V}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{l}$ ....(1)

$R=\rho \dfrac{l}{A}\\\\l=\dfrac{RA}{\rho}\\\\l=\dfrac{0.673\times 6.01\times 10^{-8}}{l\times 10^{-6}}\\\\l=0.201\ m$

(b) Equation (1) becomes :

$A=\dfrac{6.01\times 10^{-8}}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{0.201}\\\\\pi r^2=3\times 10^{-7}\\\\r=\sqrt{\dfrac{3\times 10^{-7}}{\pi}} \\\\r=3.09\times 10^{-4}\ m$

Hence, this is the required solution.

5 0

3 years ago