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3 years ago

15. The

Physics

1 answer:

ratelena [41]3 years ago

6 0

That TIME is called the "period" of the wave.

(It's not one of the choices.)

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which planet should punch travel to if his goal is to weigh in at 118 lb? refer to the table of planetary masses and radii given

Harrizon [31]

The planet that Punch should travel to in order to weigh 118 lb is Pentune.

<h3 /><h3 /><h3>The given parameters:</h3>

Weight of Punch on Earth = 236 lb
Desired weight = 118 lb

The mass of Punch will be constant in every planet;

$W = mg\\\\m = \frac{W}{g}\\\\m = \frac{236}{g}$

The acceleration due to gravity of each planet with respect to Earth is calculated by using the following relationship;

$F = mg = \frac{GmM}{R^2} \\\\g = \frac{GM}{R^2}$

where;

M is the mass of Earth = 5.972 x 10²⁴ kg
R is the Radius of Earth = 6,371 km

For Planet Tehar;

$g_T =\frac{G \times 2.1M}{(0.8R)^2} \\\\g_T = 3.28(\frac{GM}{R^2} )\\\\g_T = 3.28 g$

For planet Loput:

$g_L =\frac{G \times 5.6M}{(1.7R)^2} \\\\g_L = 1.94(\frac{GM}{R^2} )\\\\g_L = 1.94g$

For planet Cremury:

$g_C =\frac{G \times 0.36M}{(0.3R)^2} \\\\g_C = 4(\frac{GM}{R^2} )\\\\g_C = 4 g$

For Planet Suven:

$g_s =\frac{G \times 12M}{(2.8R)^2} \\\\g_s = 1.53(\frac{GM}{R^2} )\\\\g_s = 1.53 g$

For Planet Pentune;

$g_P =\frac{G \times 8.3 }{(4.1R)^2} \\\\g_P = 0.5(\frac{GM}{R^2} )\\\\g_P = 0.5 g$

For Planet Rams;

$g_R =\frac{G \times 9.3M}{(4R)^2} \\\\g_R = 0.58(\frac{GM}{R^2} )\\\\g_R = 0.58 g$

The weight Punch on Each Planet at a constant mass is calculated as follows;

$W = mg\\\\W_T = mg_T\\\\W_T = \frac{236}{g} \times 3.28g = 774.08 \ lb\\\\W_L = \frac{236}{g} \times 1.94g =457.84 \ lb\\\\ W_C = \frac{236}{g}\times 4g = 944 \ lb \\\\ W_S = \frac{236}{g} \times 1.53g = 361.08 \ lb\\\\W_P = \frac{236}{g} \times 0.5 g = 118 \ lb\\\\W_R = \frac{236}{g} \times 0.58 g = 136.88 \ lb$

Thus, the planet that Punch should travel to in order to weigh 118 lb is Pentune.

<u>The </u><u>complete question</u><u> is below</u>:

Which planet should Punch travel to if his goal is to weigh in at 118 lb? Refer to the table of planetary masses and radii given to determine your answer.

Punch Taut is a down-on-his-luck heavyweight boxer. One day, he steps on the bathroom scale and "weighs in" at 236 lb. Unhappy with his recent bouts, Punch decides to go to a different planet where he would weigh in at 118 lb so that he can compete with the bantamweights who are not allowed to exceed 118 lb. His plan is to travel to Xobing, a newly discovered star with a planetary system. Here is a table listing the planets in that system (<em>find the image attached</em>).

<em>In the table, the mass and the radius of each planet are given in terms of the corresponding properties of the earth. For instance, Tehar has a mass equal to 2.1 earth masses and a radius equal to 0.80 earth radii.</em>

Learn more about effect of gravity on weight here: brainly.com/question/3908593

5 0

2 years ago

Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.300 m and carries a current

Marrrta [24]

Answer:

Explanation:

Since the wires attract each other , the direction of current will be same in both the wires .

Let I be current in wire which is along x - axis

force of attraction per unit length between the two current carrying wire is given by

$\frac{\mu_0}{4\pi}$ x $\frac{2 I_1\times I_2}{d}$

where I₁ and I₂ are currents in the wires and d is distance between the two

Putting the given values

285 x 10⁻⁶ = 10⁻⁷ x $\frac{2\times25.5\times I_2}{.3}$

I₂ = 16.76 A

Current in the wire along x axis is 16.76 A

To find point where magnetic field is zero due the these wires

The point will lie between the two wires as current is in the same direction.

Let at y = y , the neutral point lies

k 2 x $\frac{16.76}{y}$ = k 2 x $\frac{25.5}{.3-y}$

25.5y = 16.76 x .3 - 16.76y

42.26 y = 5.028

y = .119

= .12 m

3 0

3 years ago

What forces act upon the moon as it orbits the earth

ValentinkaMS [17]

the axis acts against and it would be a contact force

7 0

3 years ago

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What has been something positive for you that has been a result of social distancing? (Answer thoughtfully and throughly.)

Sunny_sXe [5.5K]

Answer: I have more time to think about the wonderful things in life that we don't always appreciate.

Explanation:

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3 years ago

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A 10-kg cart moving at 5 m/s collides with a 5-kg cart at rest and causes it to move 10 m/s. Which principle explains the result

Virty [35]

Hello!

A 10-kg cart moving at 5 m/s collides with a 5-kg cart at rest and causes it to move 10 m/s. Which principle explains the result? A) law of differential mass B) law of conservation of momentum C) law of unequal forces D) law of accelerated collision

We have the following data¹:

ΔP (momentum before impact) = ?

mA (mass) = 10 kg

vA (velocity) = 5 m/s

mB (mass) = 5 kg

vB (velocity) = 0 m/s

Solving:

ΔP = mA*vA + mB*vB

ΔP = 10 kg*5 m/s + 5 kg*0 m/s

ΔP = 50 kg*m/s + 0 kg*m/s

Δp = 50 kg*m/s ← (momentum before impact)

We have the following data²:

ΔP (momentum after impact) = ?

mA (mass) = 10 kg

vA (velocity) = 0 m/s

mB (mass) = 5 kg

vB (velocity) = 10 m/s

Solving:

Δp = mA*vA + mB*vB

Δp = 10 kg*0 m/s + 5 kg*10 m/s

Δp = 0 kg*m/s + 50 kg*m/s

Δp = 50 kg*m/s ← (momentum after impact)

*** Then, which principle explains the result ?

Law of conservation of momentum, <u>since the total momentum of body A and B before impact is equal to the total momentum of body A and B after impact.</u>

Note: Bodies of different masses and velocities may have the same kinetic energy, if proportionality between the units is maintained it can occur that they have the same kinetic energy.

Answer:

B) law of conservation of momentum

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5 0

3 years ago