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2 years ago

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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

2 answers:

Lelechka [254]2 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]2 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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Vector C has a magnitude of 24.6 m and points in the − y ‑ direction. Vectors A and B both have positive y ‑ components, and mak

frez [133]

Answer:

A= 61.35

B= -44.40

Explanation:

1. Using the components method we have:

$A_{x}= A cos \alpha\\B_{x}= B cos \beta\\C_{x}= 0\\\\A_{y}= A sin \alpha\\B_{y}= B sin \beta\\C_{y}= 24.6\\$

Considering that the vector sum $A+B+C=0$ , then:

$|V|=\sqrt{V_{x}^{2} +V_{y}^{2} }=0$

Then:

$V_{x} ^{2} =0; V_{x} =0\\V_{y} ^{2} =0; V_{y} =0$

It means the value of x and y component is 0.

2. Determinate the equations that describe each component:

$V_{x}= A cos \alpha -B cos \beta=0 (1)\\V_{y}= A sin \alpha +B sin \beta - C=0 (2)$

Form Eq. (1):

$A=B \frac{cos \beta}{cos \alpha} (3)$

Replacing A in Eq. (2):

$(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-C=0\\(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-=C\\\\B(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)=C\\B=C(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)^{-1} (4)$

Replacing values of C, α and β in (4):

$B= 24.6 (\frac{(cos 27.7)(sin 44.9)}{cos 44.9}+sin 27.7)^{-1} \\B= -44.4$

Replacing value of B in (3)

$A=-44.40\frac{cos 27.7}{sin 49.9} \\A= 61.35$

5 0

2 years ago

You connect a 100-resistor, a 800-mH inductor, and a 10.0-uF capacitor in series across a 60.0-Hz, 120-V (peak) source. The impe

steposvetlana [31]

Answer:

Impedance, Z = 107 ohms

Explanation:

It is given that,

Resistance, R = 100 ohms

Inductance, $L=800\ mH=800\times 10^{-3}\ H=0.8\ H$

Capacitance, $C=10\ \mu F=10\times 10^{-6}\ F=10^{-5}\ F$

Frequency, f = 60 Hz

Voltage, V = 120 V

The impedance of the circuit is given by :

$Z=\sqrt{R^2+(X_C-X_L)^2}$ ...........(1)

Where

$X_C$ is the capacitive reactance, $X_C=\dfrac{1}{2\pi fC}$

$X_C=\dfrac{1}{2\pi \times 60\times 10^{-5}}=265.65\ \Omega$

$X_L$ is the inductive reactance, $X_L={2\pi fL}$

$X_L={2\pi \times 60\times 0.8}=301.59\ \Omega$

So, equation (1) becomes :

$Z=\sqrt{(100)^2+(265.65-301.59)^2}$

Z = 106.26 ohms

or

Z = 107 ohms

So, the impedance of the circuit is 107 ohms. Hence, this is the required solution.

8 0

3 years ago

The age of the universe is around 100,000,000,000,000,000s. A top quark has a lifetime of roughly 0.000000000000000000000001s. W

Maslowich

Answer:

10⁴¹ s quark top lives have been in the history of the universe.

Explanation:

You need to determine how many quark top lives there have been in the history of the universe, that is, what is the age of the universe divided by the lifetime of a top quark. Expressed in a formula, this is:

$t\frac{Age of the universe}{Lifetime of a top quark}$

Yo know that the "Age of the universe" is 100,000,000,000,000,000 which can also be expressed as 10¹⁷ s .

You also know that the "Lifetime of a top quark" is 0.000000000000000000000001 which can also be expressed as 10⁻²⁴ s.

Then $t=\frac{10^{17} }{10^{-24} }$

Recalling that the result of dividing two powers of the same base is another power with the same base where the exponent is the subtraction of the initial exponents, it is possible to calculate this division as follows:

$t=10^{17-(-24)}$

$t=10^{17+24}$

<u><em>t=10⁴¹ s</em></u>

So <u><em>10⁴¹ s quark top lives have been in the history of the universe.</em></u>

5 0

2 years ago

Which circuit would have the most electrical power?

aliya0001 [1]

the answer is c because it has the most volts

a

6 0

1 year ago

The sound from a bolt of lightning travelled 4.08 km in 12.0 s. What was the speed

LenaWriter [7]

Answer:

83.67 m/s

Explanation:

Set up a calculation to convert units of measure to what you need.

You have km/s and you need m/s.

4.08km 1000 m 83.67m

----------- X ---------- = --------------- the km will cancel out and you are left

12.0 s 1 km s with m/s

6 0

2 years ago