All categories

4 years ago

How does friction affect a machine efficiency?

1 answer:

5 0

It wears down its components causing it to not work as effectively, it also creates heat which can cause(according on the machine) it to burn out. Another Immediate problem is that it can slow the machine causing it to lose efficiency.

You might be interested in

Can anybody helpp????

Oliga [24]

Answer: white shirt-reflect

top hat- reflect

elepahnt- i think absorb im quite sure

Explanation:

Reflection means that they are thrown back from a surface; absorption means that they are incorporated by a surface and transformed into heat energy. Different surfaces reflect and absorb differently. Different shirts with different colors will absorb light of different wavelength. Black shirt will absorb all colors of light while white shirt will reflect all colors of light. thats why on a hot day you wouldnt want to wear a black t shirt.

6 0

3 years ago

An electric field of 8.20 ✕ 105 V/m is desired between two parallel plates, each of area 25.0 cm2 and separated by 2.45 mm. Ther

astraxan [27]

Answer:

q = 1.815 \times 10^{-8} C

Charge on one plate is positive in nature and on the other plate it is negative in nature.

Explanation:

E = 8.20 x 10^5 V/m, A = 25 cm^2, d = 22.45 mm

According to the Gauss's theorem in electrostatics

The electric field between the two plates

$E = \frac{\sigma }{\varepsilon _{0}}$

${\sigma }= E \times {\varepsilon _{0}}$

${\sigma }= 8.20 \times 10^{5} \times {8.854 \times 10^{-12}$

${\sigma }= 7.26 \times 10^{-6} C/m^{2}$

Charge, q = surface charge density x area

$q = 7.26 \times 10^{-6} \times 25 \times 10^{-4}$

q = 1.815 \times 10^{-8} C

5 0

3 years ago

A flat roof is very susceptible to wind damage during a thunderstorm and/or tornado. If a flat roof has an area of 500 m2 and wi

Evgesh-ka [11]

Answer: The magnitude of the force exerted on the roof is 490522.5 N.

Explanation:

The given data is as follows.

Below the roof, $v_{1}$ = 0 m/s

At top of the roof, $v_{2}$ = 39 m/s

We assume that $P_{1}$ is the pressure at lower surface of the roof and $P_{2}$ be the pressure at upper surface of the roof.

Now, according to Bernoulli's theorem,

$P_{1} + 0.5 \times \rho \times v^{2}_{1} = P_{2} \times 0.5 \rho \times v^{2}_{2}$

$P_{1} - P_{2} = 0.5 \times \rho \times (v^{2}_{2} - v^{2}_{1})$

= $0.5 \times 1.29 \times [(39)^{2} - (0)^{2}]$

= $0.645 \times 1521$

= 981.045 Pa

Formula for net upward force of air exerted on the roof is as follows.

F = $(P_{1} - P_{2})A$

= $981.045 \times 500$

= 490522.5 N

Therefore, we can conclude that the magnitude of the force exerted on the roof is 490522.5 N.

5 0

3 years ago

explain how you can use a speedometer and a clock to tell how far you've traveled in a car if the cars odometer is not working

astraxan [27]

You multiply your average speed with the time for which you have been traveling

5 0

3 years ago

A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and has mass 4.00 kg, while the balls each have mass

Talja [164]

Answer:

a)1.93 kg-m^2

b) 1.45 kg-m^2

c) = 0

d) 1.15 kg-m^2

Explanation:

mass of the bar M = 4 kg

length of the bar = 2 m

mass of balls m1= m2= 0.3 kg

moment of inertia of bar $I= \frac{ML^2}{12}$

about an axis perpendicular to the bar through its center.

a) MOI of bar + 2×m×(L/2)^2

$I= \frac{ML^2}{12}$ + $2m\frac{L}{2}^2$

now putting the values of m, M and L as above and solving we get

I= 1.93 kg-m^2

b) perpendicular to the bar through one of the balls

$I=M\frac{L^2}{3} +mL^2$

$I=4\frac{2^2}{3} +0.3\times4^2$ = 1.45 kg-m^2

c) parallel to the bar through both balls

zero as the no mass distribution along the parallel to the bar through both balls.

d) parallel to the bar and 0.500 m from it.

I= $(M+2m_1)\frac{1}{2}^2$

putting values and solving we get

1.15 kg-m^2

4 0

4 years ago