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ad-work [718]
4 years ago
11

A ski lift carries people along a 220 meter

Physics
1 answer:
Amanda [17]4 years ago
6 0
Yes a ski lift carries people among a 220 meter
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An object falls from rest at a rate of 10 m/s / s . How fast was it traveling after 5 seconds of fall?
Vesna [10]

Answer:

v = u + at \\ v = 10 + 9.81 \times 5 \\ v = 59.05m {s}^{ - 1}

7 0
3 years ago
What is the net force required to accelerate a 884 kg car at 2 m/sec2?
EastWind [94]

Answer:

1768 N

Explanation:

We can solve the problem by using Newton's second law:

F=ma

where

F is the net force acting on an object

m is the mass of the object

a is its acceleration

In this problem, we have a car of mass

m = 884 kg

And its acceleration is

a=2 m/s^2

Substituting into the equation, we find the net force on the car:

F=(884)(2)=1768 N

3 0
3 years ago
Please help !!!!!!!!!!
Natalka [10]

Answer:

Lighthouse 1 during the day will be warmer, lighthouse 2 during the night will be warmer.

Explanation:

As the paragraph stated land absorbs heat and heats up faster than water. So during the day the lighthouse farthest away from the water will be hotter. But then the converse is true also land losses heat faster than water at night. So the water retains the heat from the day better making the lighthouse by the water warmer at night.

8 0
3 years ago
A 1-meter-long wire consists of an inner copper core with a radius of 1.0 mm and an outer aluminum sheathe, which is 1.0 mm thic
Mazyrski [523]

Answer:

The total resistance of the wire is = 1.917\times10^{-3}

Explanation:

Since the wires will both be in contact with the voltage source at the same time and the current flows along in their length-wise direction, the two wires will be considered to be in parallel.

Hence, for resistances in parallel, the total resistance, R_{Total}

\frac{1}{R_{Total}}  =\frac{1}{R_{cu}  }+\frac{1}{R_{al}}

Parameters given:

Length of wire = 1 m

Cross sectional area of copper A_{cu}= \pi r^{2}= \pi \times (1\times 10^{-3}  )^{2} =3.142\times10^{-6} m^{2}

Cross sectional area of aluminium wire  

A_{al}= \pi( R^{2}-r^{2})\\\\ = \pi \times [ (2\times 10^{-3}  )^{2}-(1\times 10^{-3}  )^{2}] =9.42\times10^{-6} m^{2}\\

Resistivity of copper \rho _{cu}=1.7\times 10^{-8}  \Omega .m

Resistivity of Aluminium \rho _{al}=2.8\times 10^{-8}  \Omega .m

Resistance of copper R_{cu}= \frac{\rho_{cu} \times l}{A_{cu} }  =\frac{1.7\times 10^{-8} \times 1}{3.142\times10^{-6} } =5.41\times 10^{-3}\Omega

Resistance of aluminium R_{al}= \frac{\rho_{al} \times l}{A_{al} }  =\frac{2.8\times 10^{-8} \times 1}{9.42\times10^{-6} } =2.97\times 10^{-3}\Omega

The total resistance of the wire can be obtained as follows;

\frac{1}{R_{Total}}  =\frac{1}{5.41\times10^{-3}  }+\frac{1}{2.97\times10^{-3}}=521.52\frac{1}{\Omega}

R_{Total}= 1.917\times 10^{-3}\Omega

∴ The total resistance of the wire = 1.917\times 10^{-3}\Omega

4 0
3 years ago
Read 2 more answers
What is another word for cycles per second??
Hatshy [7]
Frequensey or hertz, I looked this up on the internet!
5 0
3 years ago
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