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joja [24]
3 years ago
11

You drop two balls of equal diameter from the same height at the same time. Ball 1 is made of metal and has a greater mass than

ball 2, which is made of wood. The upward force due to air resistance is the same for both balls. Is the drop time of ball 1 greater than, less than, or equal to the drop time of ball 2? Explain why
Physics
2 answers:
Lelechka [254]3 years ago
8 0

Answer:

The drop time ball 1 is less than the drop time of ball 2. A further explanation is provided below.

Explanation:

The net force acting on the ball will be:

⇒ F_{net}=mg-F_r

Here,

F = Force

m = mass

g = acceleration

Now,

According to the Newton's 2nd law of motion, we get

⇒ F_{net} = ma

To find the value of "a", we have to substitute "F_{net}=ma" in the above equation,

⇒ ma=mg-F_r

⇒    a=g-\frac{F_r}{m}

We can see that, the acceleration is greater for the greater mass of less for the lesser mass. Thus the above is the appropriate solution.

KiRa [710]3 years ago
5 0

Answer:

Both the ball takes equal time to reach to the ground.  

Explanation:

Two balls of same diameter

Let the height is h.

Mass of ball 1 is more than the mass of ball 2.

The second equation of motion is

h = u t +0.5 gt^2

Here, the buoyant force due to air is same. So, the time of fall is independent of the mass.

So, both the ball takes equal time to reach to the ground.  

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Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e
blondinia [14]

Answer:

Explanation:

Electron's kinetic energy = 2 eV

= 2 x 1.6 x 10⁻¹⁹ J

1/2 m v² = 3.2 x 10⁻¹⁹

1/2 x 9.1 x 10⁻³¹ x v² = 3.2 x 10⁻¹⁹

v² = .703 x 10¹²

v = .8385 x 10⁶ m/s

Electrons revolve in a circular orbit when forced to travel in a magnetic field whose radius can be expressed as follows

r = mv / Bq

where m , v and q are mass , velocity and charge of electron .

here given magnetic field B = 90 mT

= 90 x 10⁻³ T

Putting these values in the expression above

r =  mv / Bq

= \frac{9\times10^{-31}\times.8385\times10^6}{90\times10^{-3}\times1.6\times10^{-19}}

= .052 mm.

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3 years ago
You drop your cell phone. Prior to hitting the ground, the phone's kinetic energy will ________ and its potential energy will __
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Answer:

The kinetic energy of the phone would increase. The gravitational potential energy of the phone would decrease.

Explanation:

The kinetic energy {\rm KE} of an object is proportional to the square of the speed of that object. If air resistance is negligible, the phone would accelerate under gravitational pull and speed up. Hence, the kinetic energy of the phone would increase.

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A 70kg man runs at a pace of 4 m/s and a 50g meteor travels at 2 km/s. Which has the most kinetic energy?.
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Answer:

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Explanation:

because he has more weight and is moving faster

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Before Mt. Everest was discovered, what was the highest mountain in the world?
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3 0
3 years 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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