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3 years ago

8

What is the effect of reducing friction on a machine?

2 answers:

spayn [35]3 years ago

8 0

Sometimes it increases the efficiency of the machine.
Reducing the friction between the tires and the ground on a car will decrease the efficiency -- you will get more wheel slip. Or reducing the friction between the steel wheels and the rail on a train will reduce the efficiency...<span />

il63 [147K]3 years ago

7 0

It increases efficiency. Hope this helps!!!

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Explain in detail what information can be learn about atoms of different elements by examining the periodic table . For the Maxi

raketka [301]

Answer:

We know the information about atomic size, energy, electronic configuration etc. of atom from the periodic table.

Explanation:

Periodic table is the arrangement of elements that are arranged according to their properties and electronic configuration.
In periodic table, on furthest right side of the periodic table, noble gases like He, Ne, Ar etc are arranged.
The atomic number of element increases while moving from left towards right in the periodic table.
The metallic character of element decreases as we proceed the table towards right.
They readily accept electron to fill the valence shell hence becoming more metallic in character.

5 0

3 years ago

A student finds an unlabeled liquid container in his lab. He notices that the container has two liquids. Since the two liquids h

raketka [301]

Answer:

$\rho = 1848.03 kg m^{-3}$

Explanation:

given data:

density of water \rho = 1 gm/cm^3 = 1000 kg/m^3

height of water = 20 cm =0.2 m

Pressure p = 1.01300*10^5 Pa

pressure at bottom

$P = P_{fluid} + P_{h_2 o}$

$P = P_{fluid} + \rho g h$

$P_{fluid} = P - \rho g h$

= 1.01300*10^5 - 1000*0.2*9.8

= 99340 Pa

$p_{fluid} = P_{atm} + \rho g h_{fluid}$ h_[fluid} = 0.307m

$99340 = 104900 + \rho *9.8*0.307$

$\rho = 1848.03 kg m^{-3}$

5 0

4 years ago

What is the momentum of an object with a mass of 60 kg and velocity of 10 m/s?<br><br> heeelp

Pavlova-9 [17]

Answer:

Linear momentum= Mass*Velocity

P=mv

P=60*10

P=600Kg m/s

3 0

3 years ago

- How long does it take a packet of length 1,000 bytes to propagate over a link of distance 2,500 km, propagation speed 2.5 *10^

Sunny_sXe [5.5K]

Answer:

It will take 0.01 s or 10 ms

Solution:

As per the question:

Length of the packet, L = 1,000 bytes = $1000\times 8 = 8000 bits$

Distance, d = 2500 km = $2.5\times 10^{6}\ m$

Speed of propagation, s = $2.5\times 10^{8}\ m/s$

Transmission rate, R = 2 Mbps

Now,

Propagation time, t can be calculated as:

$t = \frac{d}{s} = \frac{2.5\times 10^{6}}{2.5\times 10^{8}} = 0.01\ s$

t = 10 ms

In general, propagation time, t is given by:

$t = \frac{link\ distance}{Propagation\ speed}$

No, this delay is independent of the length of the packet.
No, this delay is independent of the rate of transmission.

3 0

4 years ago

Does calm water have a regular reflection or a diffuse reflection

Shtirlitz [24]

A regular relflection

8 0

3 years ago

Read 2 more answers