All categories

3 years ago

Whiteout friction you could not, write, drive or fly and airplane. Why not?

1 answer:

7 0

friction is the resistance that one surface or object encounters when moving over another. Due to gravity pulling everything down things need to friction in order to move

i hope this helps :/

You might be interested in

Help me please, ;) I could use it

erastova [34]

Answer:

The solution(s) are in order with respect to the attachments

$2.613\:\cdot10^5$ Joules ; 5. Adding the same amount of heat to two different objects will produce the same increase in temperature ; 2. Same speed in both ; 2. A

Explanation:

Diagram 1 ( Liquid Nitrogen ) : So as you can see, we want our units in Joules here, and can therefore multiply the mass of gaseous nitrogen and the latent heat of liquid nitrogen, to cancel the units kg, and receive our solution - in terms of Joules. Let's do it.

q ( energy removed ) = mass of nitrogen $*$ latent heat of liquid nitrogen,

q = 1.3 kg $*$ 2.01 $*$ 10⁵ J / kg = $1.3\:\cdot \:2.01\:\cdot \:\:10^5$ = $10^5\cdot \:2.613$ = $100000\cdot \:2.613$ = $261300$ Joules = $261.3$ kiloJoules = 2.613 $*$ 10⁵Joules is the energy that must be removed

Diagram 2 : The same amount of heat does not necessarily mean the same increase in temperature for two different objects. The increase in temperature depends on the specific heat capacity of the substance. Therefore your solution is 5 ) Adding the same amount of heat to two different objects will produce the same increase in temperature.

Diagram 3 : The temperatures in both glasses are the same, and hence the molecules have the same average speed. Therefore your solution is 2 ) Same speed in both.

Diagram 4 : Glass A has more water molecules, and hence has more thermal energy. Your solution is 2 ) A.

7 0

3 years ago

Read 2 more answers

What is the direction of the sum of these two vectors?

gayaneshka [121]

Answer:

what is the direction of the sum of these two vectors?

7 0

2 years ago

Read 2 more answers

What is the magnification of an object that is 4.15 m in front of a camera that has an image position of 5.0 cm?

Stolb23 [73]

-0.012

Done !!!!!!!!

4 0

3 years ago

What is the answer for this question (4x³)²

Delvig [45]

16x to the power of 6

6 0

3 years ago

Read 2 more answers

Usain Bolt's world-record 100 m sprint on August 16, 2009, has been analyzed in detail. At the start of the race, the 94.0 kg Bo

ZanzabumX [31]

a) 893 N

b) 8.5 m/s

c) 3816 W

d) 69780 J

e) 8030 W

Explanation:

The net force acting on Bolt during the acceleration phase can be written using Newton's second law of motion:

$F_{net}=ma$

where

m is Bolt's mass

a is the acceleration

In the first 0.890 s of motion, we have

m = 94.0 kg (Bolt's mass)

$a=9.50 m/s^2$ (acceleration)

So, the net force is

$F_{net}=(94.0)(9.50)=893 N$

And according to Newton's third law of motion, this force is equivalent to the force exerted by Bolt on the ground (because they form an action-reaction pair).

Since Bolt's motion is a uniformly accelerated motion, we can find his final speed by using the following suvat equation:

$v=u+at$

where

v is the final speed

u is the initial speed

a is the acceleration

t is the time

In the first phase of Bolt's race we have:

u = 0 m/s (he starts from rest)

$a=9.50 m/s^2$ (acceleration)

t = 0.890 s (duration of the first phase)

Solving for v,

$v=0+(9.50)(0.890)=8.5 m/s$

First of all, we can calculate the work done by Bolt to accelerate to a speed of

v = 8.5 m/s

According to the work-energy theorem, the work done is equal to the change in kinetic energy, so

$W=K_f - K_i = \frac{1}{2}mv^2-0$

where

m = 94.0 kg is Bolt's mass

v = 8.5 m/s is Bolt's final speed after the first phase

$K_i = 0 J$ is the initial kinetic energy

So the work done is

$W=\frac{1}{2}(94.0)(8.5)^2=3396 J$

The power expended is given by

$P=\frac{W}{t}$

where

t = 0.890 s is the time elapsed

Substituting,

$P=\frac{3396}{0.890}=3816 W$

First of all, we need to find what is the average force exerted by Bolt during the remaining 8.69 s of motion.

In the first 0.890 s, the force exerted was

$F_1=893 N$

We know that the average force for the whole race is

$F_{avg}=820 N$

Which can be rewritten as

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}$

And solving for $F_2$ , we find the average force exerted by Bolt on the ground during the second phase:

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}\\F_2=\frac{(0.890+8.69)F_{avg}-0.890F_1}{8.69}=812.5 N$

The net force exerted by Bolt during the second phase can be written as

$F_{net}=F_2-D$ (1)

where D is the air drag.

The net force can also be rewritten as

$F_{net}=ma$

where

$a=\frac{v-u}{t}$ is the acceleration in the second phase, with

u = 8.5 m/s is the initial speed

v = 12.4 m/s is the final speed

t = 8.69 t is the time elapsed

Substituting,

$a=\frac{12.4-8.5}{8.69}=0.45 m/s^2$

So we can now find the average drag force from (1):

$D=F_2-F_{net}=F_2-ma=812.5 - (94.0)(0.45)=770.2 N$

So the increase in Bolt's internal energy is just equal to the work done by the drag force, so:

$\Delta E=W=Ds$

where

d is Bolt's displacement in the second part, which can be found by using suvat equation:

$s=\frac{v^2-u^2}{2a}=\frac{12.4^2-8.5^2}{2(0.45)}=90.6 m$

And so,

$\Delta E=Ds=(770.2)(90.6)=69780 J$

The power that Bolt must expend just to voercome the drag force is given by

$P=\frac{\Delta E}{t}$

where

$\Delta E$ is the increase in internal energy due to the air drag

t is the time elapsed

Here we have:

$\Delta E=69780 J$

t = 8.69 s is the time elapsed

Substituting,

$P=\frac{69780}{8.69}=8030 W$

And we see that it is about twice larger than the power calculated in part c.

3 0

2 years ago