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3 years ago

Is there ever a situation where an ant will have more momentum than an elephant? Explain why or why not?

Physics

2 answers:

NeX [460]3 years ago

7 0

Answer: The correct answer is D.

Explanation:

Momentum is defined as the product of the mass and the velocity.

To have more momentum, there should be both mass and the velocity.

If the object has more mass as compared to the object having lighter mass and the heavier mass object is standing still and lighter one is moving then the lighter object will have more momentum.

In the given problem, if the elephant is still standing but an ant is moving then an ant will have more momentum than an elephant.

Therefore, the correct answer is D.

kipiarov [429]3 years ago

6 0

An ant is having higher momentum than elephant only if elephant is standing still.

Explanation:

The word momentum is known as product of mass $m$ of object and its velocity $\overrightarrow v$ by which it moves and the momentum can be denoted by vector term $\overrightarrow p$ .

The object whose product of mass and velocity are higher will have more momentum.

If elephant stands still, means its velocity is equal to zero, then product of its mass and velocity will be zero. Hence, it will have zero momentum.

In the meantime, the mass of ant is insignificantly small but have some velocity for moving. So, the ant will have small momentum but somewhat greater than zero.

Therefore, an ant will have more momentum than elephant if the elephant stands still.

If an ant will stand still it have 0 velocity, so the momentum will be equal to 0. Hence, this option is also incorrect.

The option “The elephant has much high mass” is incorrect because having high mass but zero velocity will result in zero momentum.

Thus, an ant will have higher momentum than elephant if elephant is standing still.

Learn More:

1. Momentum brainly.com/question/11896510

2. More about momentum brainly.com/question/11947870

3. Force and motion brainly.com/question/11942991

Answer Details:

Grade: High School

Subject: Physics

Chapter: Momentum

Keywords:

Ant, more, explain, standing, mass, fast, still, change, object, velocity, force, times, impulse, acting, negligible, product, zero, elephant, momentum.

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A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with

Komok [63]

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

$\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\ g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s$

$\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM$

Therefore the speed in RPM will be 62.64 RPM.

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2 years ago

A sound is recorded at 19 decibels. What is the intensity of the sound?

sp2606 [1]

$1 \times 10^{-10.1} \mathrm{Wm}^{-2}$ is the intensity of the sound.

Answer: Option B

Explanation:

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about $1 \times 10^{-12} \mathrm{Wm}^{-2}$ ). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB). Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

$\text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)$

Where,

I = Intensity of the sound produced

$I_{0}$ = Standard Intensity of sound of 60 decibels = $1 \times 10^{-12} \mathrm{Wm}^{-2}$

So for 19 decibels, determine I as follows,

$19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9$

When log goes to other side, express in 10 to the power of that side value,

$\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}$

$I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}$

5 0

3 years ago

A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu

dalvyx [7]

Answer:

The frictional torque is $\tau = 0.2505 \ N \cdot m$

Explanation:

From the question we are told that

The mass attached to one end the string is $m_1 = 0.341 \ kg$

The mass attached to the other end of the string is $m_2 = 0.625 \ kg$

The radius of the disk is $r = 9.00 \ cm = 0.09 \ m$

At equilibrium the tension on the string due to the first mass is mathematically represented as

$T_1 = m_1 * g$

substituting values

$T_1 = 0.341 * 9.8$

$T_1 = 3.342 \ N$

At equilibrium the tension on the string due to the mass is mathematically represented as

$T_2 = m_2 * g$

$T_2 = 0.625 * 9.8$

$T_2 = 6.125 \ N$

The frictional torque that must be exerted is mathematically represented as

$\tau = (T_2 * r ) - (T_1 * r )$

substituting values

$\tau = ( 6.125 * 0.09 ) - (3.342 * 0.09 )$

$\tau = 0.2505 \ N \cdot m$

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3 years ago

A tire has a pressure of 325 kPa at 10°C.

musickatia [10]

5 times that of initial pressure i.e 1625 kpa

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2 years ago

Read 2 more answers

The heat capacity of 0.125Kg of water is measured to be 523j/k at a room temperature.Hence, calculate the heat capacity of water

Naily [24]

Answer:

A. 4148 J/K/Kg

B. 4148 J/K/L

Explanation:

A. Heat capacity per unit mass is known as the specific heat capacity, c.

C = Heat capacity/mass(kg)

C = (523 J/K) / 0.125 Kg = 4148 J/K/Kg

B. Volume of water = mass/density

Density of water = 1 Kg/L

Volume of water = 0.125 Kg/ 1Kg/L

Volume of water = 0.125 L

Heat capacity per unit volume = (523 J/K) / 0.125 L

Heat capacity per unit volume = 4148 J/K/L

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3 years ago