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nalin [4]
3 years ago
11

23. If a jogger runs 100 meters west and then turns around and runs 30 meters east. What

Physics
2 answers:
marshall27 [118]3 years ago
7 0

Answer:

A

Explanation:

if he goes to the west, the east is opposite so 100-30

In-s [12.5K]3 years ago
4 0
70 meters because west is the opposite of East’s so just 100-30
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State pascal law and write three instrument which base unit?​
OLEGan [10]

Answer:

Pascal's law (also Pascal's principle[1][2][3] or the principle of transmission of fluid-pressure) is a principle in fluid mechanics given by Blaise Pascal that states that a pressure change at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere.[4] The law was established by French mathematician Blaise Pascal in 1653 and published in 1663.[5][6]

7 0
3 years ago
What type of ion is formed by a non-metal atom?
Verizon [17]
Non metals produce something called protons they are ions 
7 0
3 years ago
A weight lifter lifts a dumbbell a certain height in 2.0 s, while a competitor does the same workin 1.0 s. Compared to the power
Aloiza [94]

Answer:

a. one-half as great

Explanation:

The power developed by the first lifter is one-half as great as that of the second person.

  Power is defined as the rate at which work is done;

          Power  = \frac{workdone}{time}

Since the two lifters do the same work at different time, let us estimate their power;

       P₁ = \frac{workdone}{2}                     P₂ = \frac{workdone }{1}

   We see that for P₁, power is half of the work done whereas in P₂ power is the same as the work done.

Therefore,

           The power of the first weight lifter is one-half the second lifter.

4 0
3 years ago
Consider a balloon of mass 0.030kg being inflated with a gas of density 0.54kg/m. What will be the volume of the balloon when it
Nonamiya [84]

the weight of the balloon is .030 * 10 = 0.3 N

the weight of the gas of volume v is 0.54*10 N

The lifting force of a volume of v m³ of displaced air is 1.29v N

so, we need

1.29*10*v = 0.3 + 0.54*10*v

or

1.29v = 0.03+0.54v

7 0
1 year ago
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:

a)

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).

b)

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

c)

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

d)

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

e)

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
3 years ago
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