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pychu [463]
3 years ago
5

Two ice boats (one of mass m, one of mass 2m) hold a race on a frictionless, horizontal, frozen lake. Bath ice boats start at re

st, and the wind exerts the same constant force on both iceboats. After moving a same distance s, which iceboat crosses the finish line with a quicker speed v
Physics
1 answer:
Elenna [48]3 years ago
5 0

Answer:

They both cross the finish line with the same kinetic energy

Explanation:

Same force, same displacement, so same KE at the location of the finish line. They don’t cross the line at the same time, but that was not the question!

∆KA= (mA/2)vF,A2

∆KA=∆KBso vF,B/vF,A= (MA/MB)1/2∆KB= (mB/2)vF,B2

Lighter boat goes faster so reaches finish line 1st

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You are working at a company that manufactures electrical wire. Gold is the most ductile of all metals: it can be stretched into
devlian [24]

Answer:R=1.424 M\Omega

Explanation:

Given

mass of gold m=1.6 gm

Length of wire L=2.2 km

Resistivity of gold \rho =2.44\times 10^{-8}

density of gold =19.3\times 10^3 kg/m^3

and mass=volume\times density

1.6\times 10^{-3}=volume\times 19.3\times 10^3

volume=8.29\times 10^{-8} m^3

And Resistance R=\frac{\rho L}{A}

also be written as

R=\frac{\rho L^2}{V}

where L=length

V=volume

\rho =resistivity\ of\ gold

R=\frac{2.44\times 10^{-8}\times (2200)^2}{8.29 \times 10^{-8}}

R=1.42\times 10^{6} \Omega

R=1.424 M\Omega

5 0
3 years ago
The position of a ball as a function of time is given by
Andre45 [30]

Answer:

x = -6.5 meters

Explanation:

The position of a ball as a function of time t is given by :

x=3\ m+(-5\ m/s)t..................(1)

Where t is time in seconds

We need to find the position of the ball at 1.9 s. It can be simply calculated putting t = 1.9 s in equation (1) as :

x=3+(-5)(1.9)

x = -6.5 meters

So, the position of the ball at 1.9 seconds is -6.5 meters. Hence, this is the required solution.

6 0
4 years ago
Write the dimension of a / b in the x = at + bt2. Where x is the distance and t is the time?​
Luba_88 [7]

The dimension of a/b where x is the distance and t is the time is T

Given the expression

x = at + bt²

where

x is the distance

t is the time

Based on the homogeneity principle, the expression on the left-hand side must be equal to that on the right. Hence;

x = at

a = \frac{x}{t}

Since x is the distance and distance is measured in metres, the dimension equivalent will be the length 'L'

Since t is the time and time is measured in seconds, the dimension equivalent will be the seconds 'T'

a=\frac{L}{T}

Similarly;

x  = bt²

b=\frac{x}{t^2}\\b=\frac{L}{T^2}

Next is to get a/b;

\frac{a}{b} =  \frac{L}{T} \div \frac{L}{T^2}\\\frac{a}{b} = \frac{L}{T}*\frac{T^2}{L}  \\\frac{a}{b} =\frac{T^2}{T}\\\frac{a}{b} =T

Hence the dimension of a/b is T

4 0
3 years ago
19. Use a scale ray diagram to find the image position of
luda_lava [24]
I hope that helps you!!!!!!

7 0
3 years ago
Use this formula to solve this problem: Power (P) = Work (W) ÷ time (t) Peter's body supplies a force of 500 N to run up a 5 met
sesenic [268]
As you said p=w/t.
but, w=f×s
=500×5=2500j
t=10s
p=w/t
=2500/10=250 watts
4 0
3 years ago
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