Setting up a home network using wireless connections is creating a ____

What force causes a 1-kg mass to accelerate at a rate of 1 meter per second each second?

Zigmanuir [339]

Answer:

$F=1N$

Explanation:

Conceptual analysis

To solve this problem we apply Newton's second law:

The acceleration of an object is proportional to the force F acting on it and inversely proportional to its mass m.

a = F / m

Where,

F = m * a Formula (1)

F: Force in Newtons (N)

m: mass in kg

a: acceleration in m/(s^2)

a = v / t Formula (2)

v: speed in m/s

t: time in seconds (s)

Known information

We know the following data:

m = 1kg

v = 1 m/s

t = 1s

Development of the problem:

In the Formula (2): $a = \frac{\frac{1m}{s}}{1s} = 1 \frac{m}{s^2}$

In the Formula (1): $F=1kg* 1 \frac{m}{s^2}=1N$

8 0

3 years ago

Effciency of a lever is never 100% or more. why?Give reason

Troyanec [42]

Answer:

Ideally, the work output of a lever should match the work input. However, because of resistance, the output power is nearly always be less than the input power. As a result, the efficiency would go below $100\%$ .

Explanation:

In an ideal lever, the size of the input and output are inversely proportional to the distances between these two forces and the fulcrum. Let $D_\text{in}$ and $D_\text{out}$ denote these two distances, and let $F_\text{in}$ and $F_\text{out}$ denote the input and the output forces. If the lever is indeed idea, then:

$F_\text{in} \cdot D_\text{in} = F_\text{out} \cdot D_\text{out}$ .

Rearrange to obtain:

$\displaystyle F_\text{in} = F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}}$

Class two levers are levers where the perpendicular distance between the fulcrum and the input is greater than that between the fulcrum and the output. For this ideal lever, that means $D_\text{in} > D_\text{out}$ , such that $F_\text{in} < F_\text{out}$ .

Despite $F_\text{in} < F_\text{out}$ , the amount of work required will stay the same. Let $s_\text{out}$ denote the required linear displacement for the output force. At a distance of $D_\text{out}$ from the fulcrum, the angular displacement of the output force would be $\displaystyle \frac{s_\text{out}}{D_\text{out}}$ . Let $s_\text{in}$ denote the corresponding linear displacement required for the input force. Similarly, the angular displacement of the input force would be $\displaystyle \frac{s_\text{in}}{D_\text{in}}$ . Because both the input and the output are on the same lever, their angular displacement should be the same:

$\displaystyle \frac{s_\text{in}}{D_\text{in}} =\frac{s_\text{out}}{D_\text{out}}$ .

Rearrange to obtain:

$\displaystyle s_\text{in}=s_\text{out} \cdot \frac{D_\text{in}}{D_\text{out}}$ .

While increasing $D_\text{in}$ reduce the size of the input force $F_\text{in}$ , doing so would also increase the linear distance of the input force $s_\text{in}$ . In other words, $F_\text{in}$ will have to move across a longer linear distance in order to move $F_\text{out}$ by the same $s_\text{out}$ .

The amount of work required depends on both the size of the force and the distance traveled. Let $W_\text{in}$ and $W_\text{out}$ denote the input and output work. For this ideal lever:

$\begin{aligned}W_\text{in} &= F_\text{in} \cdot s_\text{in} \\ &= \left(F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}}\right) \cdot \left(s_\text{out} \cdot \frac{D_\text{in}}{D_\text{out}}\right) \\ &= F_\text{out} \cdot s_\text{out} = W_\text{out}\end{aligned}$ .

In other words, the work input of the ideal lever is equal to the work output.

The efficiency of a machine can be measured as the percentage of work input that is converted to useful output. For this ideal lever, that ratio would be $100\%$ - not anything higher than that.

On the other hand, non-ideal levers take in more work than they give out. The reason is that because of resistance, $F_\text{in}$ would be larger than ideal:

$\displaystyle F_\text{in} = F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}} + F(\text{resistance})$ .

As a result, in real (i.e., non-ideal) levers, the work input will exceed the useful work output. The efficiency will go below $100\%$ ,

4 0

3 years ago

Look at the diagram showing resistance and flow of electrons.

poizon [28]

The answer would be:

D.

X: Low potential energy

Y: High Potential energy

Z: Flow of electrons

Z is clearly the flow of electrons, as shown by the arrow demonstrating the direction of the flow. So you can easily cross out choices B and C. Now, you can see that Y has more energy stored and X has a lot less, so you can conclude that Y has high potential energy while X has low potential energy.

8 0

3 years ago

Read 2 more answers

Which term describes the amount of charge that passes a point in a circuit<br> each second?

Flura [38]

Answer:

volts

Explanation:

8 0

3 years ago

Read 2 more answers

Ethane and ethanol are isomers.