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4 years ago

A roller coaster at an amusement park girls from the top of a tall hill to the bottom during this time of the roller coaster

Physics

1 answer:

Mrac [35]4 years ago

8 0

D), because the roller coaster it's falling, therefore, losing height, but gaining velocity

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Consider a cube with one corner at the origin and with sides of length 50 cm positioned along the x y z axes. There is an electr

lara [203]

Answer:

Charge in the cube = Q = (2.323 × 10⁻¹⁰) C

Explanation:

E → = < 45, 210 y , 0 > N/C

The length of the cube = 50 cm = 0.8 m

To find the charge in the cube, we need the net flux in the cube

In the z-direction, the electric field is 0, no, net flux in that direction,

In the x-direction, the electric field is constant, the same flux thay enters the face at x = 0 is the same that leaves at x= 0

50 cm, hence, no net flux in that direction too.

But the y direction, the electric field changes according to 210y from y = 0 to y = 0.50 m.

At y = 0, electric field = 210(0) = 0 N/C

At y = 0.50, electric field = 210(0.5) = 105 N/C.

Electric flux = Φ = E A = 105 (0.5 × 0.5) = 26.25 Vm or N/m²C

(the area of the xz plane that the field in this direction passes through is indeed 0.5²)

But according to Gauss' law, the net flux in a box is given by

Φ = Q/ε₀

where Q = charge in the box = ?

ε₀ = (8.85 × 10⁻¹²) C²N·m².

Q = Φε₀ = 26.25 × 8.85 × 10⁻¹²

Q = (2.323 × 10⁻¹⁰) C

Hope this Helps!!!

4 0

4 years ago

Calculate the volume of this regular solid.

Nataly [62]

Answer:

V = 42.41cm^3

Explanation:

In order to calculate the volume of the solid, you use the following formula:

$V=\frac{1}{3}\pi r^2 h$

where

r: radius of the circular base of the cone = 3 cm

h: height from the circular base to the peak of the cone = 4.5 cm

You replace the values of r and h in the formula for the volume V:

$V=\frac{1}{3}\pi(3cm)^2(4.5)=42.411cm^3\approx42.41cm^3$

hence, the volume of the solid is 42.41 cm^3

5 0

3 years ago

What is astrometry? What is astrometry? measuring the positions of stars on the sky using metric units for distance (e.g. meters

Fofino [41]

Explanation:

It is a branch of astronomy wherein measurements of the positions and movements of celestial bodies are precisely are made.

For the measurement, minute wobbling of the position of the body is detected. When the shift is detected periodically, the astronomers come to know that the star is orbited by a planet.

Also, basic data about the properties of the stars can be collected using this method.

5 0

3 years ago

A 70.0 kg ice hockey goalie, originally at rest, has a 0.170 kg hockey puck slapped at him at a velocity of 33.5 m / s . Suppose

Sergeeva-Olga [200]

Answer:

Final velocity of both goalie & puck = 0.018116 m/s

Explanation:

M1U1 + M2U2 = (M1+M2)V

70 x 0 + 0.17 x 33.5 = (70+0.17)V

V = 0.08116m/s

4 0

4 years ago

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

3 years ago