All categories

3 years ago

Which describes the average velocity of an ant traveling at a constant speed

Physics

1 answer:

yanalaym [24]3 years ago

3 0

Answer: A. The total displacement divided by the time and C. The slope of the ant's displacement vs. time graph.

Explanation:

Hi! The question seems incomplete, but I found the options on the internt:

A. The total displacement divided by the time.

B. The slope of the ant's acceleration vs. time graph.

C. The slope of the ant's displacement vs. time graph.

D. The average acceleration divided by the time.

Now, since we know the ant is travelling at a constant speed, its average velocity $V$ will be expressed by the following equation:

$V=\frac{d}{t}$

Where:

$d$ is the ant's total displacement

$t$ is the time it took to the ant to travel to the kitchen

Hence one of the correct options is: A. The total displacement divided by the time

On the other hand, this can be expressed by a displacement vs. time graph graph, where the slope of that line leads to the equation written above. So, the other correct option is: C. The slope of the ant's displacement vs. time graph.

You might be interested in

Problem #3: 3.6. A diode has wdo = 0.4 µm and φj = 0.85 V. (a) What reverse bias is required to triple the depletion-layer width

Alchen [17]

Answer:

a) 6.8 Volt

b) 1.21Цm

Explanation:

We are given from the question that

The zero -bias depletion layer width( $W_{do}$ ) is 0.4Цm

The built in voltage φj is 0.85V

Now to calculate the reverse voltage( $V_{R}$ ) that would be required to triple the depletion - layer width.

The depletion - layer width ( $W_{d}$ ) of the diode has the formula

$W_{d} = W_{do} \sqrt{1 + \frac{V_{R} }{Qj} }$

For three times of $W_{d}$ we have

$3W_{d} = W_{do} \sqrt{1 +\frac{V_{R} }{Qj} }$

=> $\frac{V_{R} }{Qj} = 3^{2} -1$

=> $V_{R} = 8Qj$

Substituting value of φj

We have

$V_{R} = 8(0.85V)$

= 6.8 V

The required bias voltage $V_{R}$ is 6.8 V

The solution for the b part of the question is uploaded on first image

7 0

2 years ago

How large amount of energy is produced during the fission of uranium

vova2212 [387]

Answer:

answer is a very large amount of energy is produced from a very small mass

Explanation:

nuclear energy is produced either by fusion or fission the former is fusion of lighter atoms into heavier elements while the letter is the splitting of a heavier atom into lighter atoms. both produce tremendous amount of energy fusion causes compassion of mass wild fission reduces it. and produce it. fusion does not produce radioactive particles while fission does (alpha and beta particles and neutrons)

4 0

2 years ago

A car with a mass of 2000 kg is moving around a circular curve at a uniform velocity of 25 m/s per second. The curve has a radiu

IrinaK [193]

The centripetal force is:
F = mv² / R
Where:
m: mass of the object
v: object speed
R: radius of the curve.
We have to:
m = 2000kg
v = 25 m / s
R = 80 meters.
Then the centripetal force acting on the vehicle is:
F = (2000kg * (25m / s) ²) / 80m
F = 15625 N

4 0

2 years ago

Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th

Kipish [7]

Answer:

Probability of tunneling is $10^{- 1.17\times 10^{32}}$

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

Max velocity of the tennis ball, $v_{m} = 200\ mph$ = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

$KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J$

Kinetic energy of the tennis ball, KE' = $\frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s$

Now, the distance the ball can penetrate to is given by:

$\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}$

$\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js$

Thus

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = 1.52\times 10^{-35}\ m$

Now,

We can calculate the tunneling probability as:

$P(t) = e^{\frac{- 2t}{\eta}}$

$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

$P(t) = 10^{- 1.17\times 10^{32}}$

6 0

3 years ago

A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$

Thus, the final velocity of thrower is $3.88~m/s$ and that for the catcher is $0.029~m/s$ .

8 0

2 years ago