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3 years ago

You are on a frozen pond, and the ice starts to crack. if you lie down on the ice and begin to crawl, this will

1 answer:

6 0

If you lie down on the ice and begin to crawl, this will spread out your weight more evenly and make it less likely for the ice to suffer a critical failure, in which you might fall below the surface of the ice and into the water. You have better weight distribution by being on more than two points - only your two feet. By crawling you disperse your weight to possibly up to eight points or more. This includes hands, elbows, knees, feet, chest. The dispersing of the weight takes significant pressure off of just one or two points, because as you are walking you are also most likely lifting a foot. By crawling, you remain in constant contact with your weight spread more evenly.

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You and your partner want to investigate how the mass of various sports balls will impact their force as they accelerate down a

luda_lava [24]

The mass of the ball is inversely proportional to acceleration of a ball.

<h3>What is friction?</h3>

This question is incomplete but I will try to help you the much I can. Friction is the force that opposes motion. Friction depends on the nature of the surfaces in contact.

When the mass of the ball is large, the acceleration of the ball decreases since mass is inversely proportional to acceleration of a body.

Learn more about acceleration: brainly.com/question/2437624

6 0

2 years ago

What average power must be supplied to the rope to generate sinusoidal waves that have amplitude 0.200 m and wavelength 0.600 m

pychu [463]

Complete question:

A taut rope has a mass of 0.123 kg and a length of 3.54 m. What average power must be supplied to the rope to generate sinusoidal waves that have amplitude 0.200 m and wavelength 0.600 m if the waves are to travel at 28.0 m/s ?

Answer:

The average power supplied to the rope to generate sinusoidal waves is 1676.159 watts.

Explanation:

Velocity = Frequency X wavelength

V = Fλ ⇒ F = V/λ

F = 28/0.6 = 46.67 Hz

Angular frequency (ω) = 2πF = 2π (46.67) = 93.34π rad/s

Average power supplied to the rope will be calculated as follows

$P_{avg} =\frac{1}{2} \mu \omega^2 A^2 V$

where;

ω is the angular frequency

A is the amplitude

V is the velocity

μ is mass per unit length = 0.123/3.54 = 0.0348 kg/m

$P_{avg} =\frac{1}{2} ( 0.0348)(93.34 \pi )^2 (0.2)^2 (28)$ = 1676.159 watts

The average power supplied to the rope to generate sinusoidal waves is 1676.159 watts.

6 0

2 years ago

A point from which the position of other objects can be described is called what?

Alisiya [41]

Answer:

Reference Point

Explanation:

3 0

3 years ago

Ah electron is a negatively charged particle that has a charge of magnitude, e - 1.60 x 10-19 C. Which one of the following stat

Advocard [28]

Answer:

The correct statement is "The electric field is directed toward the electron and has a magnitude of $\rm \dfrac{ke}{r^2}$ ".

Explanation:

According to Coulomb's law, the magnitude of the electric field due to a static point charge q at a point r distance away from it is given by

$\rm E = \dfrac{k|q|}{r^2}.$

k is the Coulmob's constant.

The direction of the electric field along the line joining the charge and the point where electric field is to be found and it is directed from positive charge to negative charge.

Conventionally, we assume a positive test charge placed at the point where electric field is to be found, the test charge has very small charge such that its charge does not affect the electric field due to the given charge.

The charge on the electron = -e.

The electric field due to an electron is given by

$\rm E = \dfrac{k|-e|}{r^2}=\dfrac{ke}{r^2}.$

The direction of this electric field is from positive test charge, placed at the point where electric field is to be found, towards the electron along the line joining the two.

Thus, the correct statement is "The electric field is directed toward the electron and has a magnitude of $\rm \dfrac{ke}{r^2}$ ".

5 0

2 years ago

¡¡¡AYUDA CON ESTOS EJERCICIOS DE FÍSICA!!!

asambeis [7]

Answer:

(a) 8 V, (b) 144000 V, (c) 2 x 10^(-8) C

Explanation:

(a) charge, q = 5 μC , Work, W = 40 x 10-^(-6) J

The electric potential is given by

W = q V

$40\times10^{-6}=5 \times10^{-6}\times V\\\\V = 8 V$

(b)

charge, q = 8 x 10^(-6) C, distance, r = 50 cm = 0.5 m

Let the potential is V.

$V =\frac{k q}{r}\\\\V =\frac{9\times 10^{9}\times 8\times 10^{-6}}{0.5}\\\\V =144000 V$

(c)

Work, W = 8 x 10^(-5) J, Potential difference, V = 4000 V

Let the charge is q.

W= q V

$8\times10^{-5}= q\times 4000\\\\q =2\times 10^{-8} C$

3 0

2 years ago