If the diameter of the black marble is 3.0cm, and bye using the formula for volume, what is a good approximation if it’s volume? Record to the ones place

Answers

Answer 1

Answer:

Complete question is;

If the diameter of the black marble is 3.0 cm, and by using the formula for volume, what is a good approximation of its volume?

Answer:

14 cm³

Explanation:

We will assume that this black marble has the shape of a sphere from online sources.

Now, volume of a sphere is given by;

V = (4/3)πr³

We are given diameter = 3 cm

We know that radius = diameter/2

Thus; radius = 3/2 = 1.5 cm

So, volume = (4/3)π(1.5)³

Volume ≈ 14.14 cm³

A good approximation of its volume = 14 cm³

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The asteroid 234 Ida has a mass of about 4 × 1016 kg and an average radius of about 16 km. What is the acceleration due to gravity on 234 Ida? Assume that the asteroid is spherical; use G = 6.67 × 10–11 Nm2/kg2.A. 1 cm/s2
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C. 5 cm/s2
D. 6 cm/s2

Answers

The asteroid 234 Ida has a mass of about 4×1016 kg and an average radius of about 16 km. The acceleration due to gravity will be 1.04 cm/s². Hence, option A is correct.

What is the acceleration due to gravity?

The acceleration an object experiences as a result of gravitational force is known as acceleration due to gravity. M/s² is its SI unit. Its vector nature—which includes both magnitude and direction—makes it a quantity. The unit g stands for gravitational acceleration. At sea level, the standard value of g on the earth's surface is 9.8 m/s².

The formula for the acceleration due to gravity is g=GM/r².

According to the question, the given values are :

Mass, M = 4 × 1016 kg or

M = 4 × 10¹⁶.

Radius, r = 16 km or,

r = 16000 meter.

G = 6.67 × 10⁻¹¹ Nm²/kg²

g = (6.67 × 10⁻¹¹ ) (4 × 10¹⁶) / 16000²

g = 0.0104 m/s² or,

g = 1.04 cm/s².

Hence, the acceleration due to gravity will be 1.04 m/s²

To get more information about Acceleration due to gravity :

brainly.com/question/13860566

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Answer:

1 cm/s²

Explanation:

I just took the quiz

A sphere has a charge of −84.0 nC and a radius of 5.00 cm. What is the magnitude of its electric field 3.90 cm from its surface?

Answers

Answer:

$E = -9.5* 10^4~N/C$

Explanation:

Gauss' Law should be applied to find the E-field 3.9 cm from the surface of the sphere.

In order to apply Gauss' Law, an imaginary spherical shell (Gaussian surface) should be placed around the original sphere. The exact position of the shell must be 3.9 cm from the surface of the original sphere.

Gauss' Law states that

$\int {\vec{E}d\vec{a}} = (Q_(enc))/(\epsilon_0)$

Here, the integral in the left-hand side is equal to the area of the imaginary surface. After all, the reason behind choosing the imaginary surface a spherical shell is to avoid this integral. The enclosed charge in the right-hand side is equal to the charge of the sphere, -84.0 nC. The radius of the imaginary surface must be 5 + 3.9 = 8.9 cm.

So,

$E4\pi r^2 = (-84* 10^(-9))/(8.8* 10^(-12))\nE4\pi (8.9 * 10^(-2))^2 = (-84* 10^(-9))/(8.8* 10^(-12))\n\nE = -9.5* 10^4~N/C$

Explain why Planck’s introduction of quantization accounted for the properties of black-body radiation.

Answers

Explanation:

The classic model of a black body made predictions of the emission at small wavelengths in open contradiction with what was observed experimentally, this led Planck to develop a heuristic model. This assumption allowed Planck to develop a formula for the entire spectrum of radiation emitted by a black body, which matched the data.

Consider three drinking glasses. All three have the same area base, and all three are filled to the same depth with water. Glass A is cylindrical. Glass B is wider at the top than at the bottom, and so holds more water than A. Glass C is narrower at the top than at the bottom, and so holds less water than A. In which glass is the pressure on the base greatest liquid pressure at the bottom?a. Glass Ab. Glass B
c. Glass C
d. All three have equal non-zero pressure at the bottom.
e. All three have zero pressure at the bottom.

Answers

Answer:

Explanation:

Pressure of the fluid in any container is a function of density of the fluid in the container, and depth of the fluid.

The static pressure of fluid in a container with depth h is given by:

P = p * g* h

Where p : density of fluid

g: gravitational constant 9.81 m/s^2

h : depth of the fluid

Since, all the glasses filled have same Area base, same depth and same density of fluid and g is constant. The pressure at the bottom of each drinking glass is equal for all cases. As supported by the relationship given above.

Tell uses of cancave mirror and convex mirror.

Answers

Answer:

Uses of concave mirror:

Shaving mirrors.

Head mirrors.

Ophthalmoscope.

Astronomical telescopes.

Headlights.

Solar furnaces.

Uses of convex mirror:

Convex mirrors always form images that are upright, virtual, and smaller than the actual object. They are commonly used as rear and side view mirrors in cars and as security mirrors in public buildings because they allow you to see a wider view than flat or concave mirrors.

please give me full points.

You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has stopped due to an accident up ahead. You step on your brakes to slow down as quickly as possible. Assume that you to slow down to 30 mph in about 5 seconds. A) With this same average acceleration, how much longer would it take you to stop?B) What total distance would you travel from when you first apply the brakes until the car stops?

Answers

A.The time taken for the car to stop is 8.75 s

B.The distance travelled when the brakes were applied till the car stops is 136.89 m

A. Determination of the time taken for the car to stop.

We'llbegin bycalculatingthedecelerationof thecar

Initial velocity (u) = 70 mph = 0.447 × 70 = 31.29 m/s

Final velocity (v) = 30 mph = 0.447 × 30 = 13.41 m/s

Time (t) = 5 s

Deceleration (a) =?

$a \: = (v \: - u)/(t) \n \n a = (13.41 - 31.29)/(5) \n \n a \: = ( - 17.88)/(5) \n \n$

a = –3.576 m/s²

Finally,we shall determine the time taken for the car to stop.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Time (t) =?

$v \: = u \: + at \n 0 \: = 31.29 \: + \: ( - 3.576 * t) \n 0 \: = 31.29 \: - 3.576 * t \n collet \: like \: terms \n 0 - 31.29 \: = - 3.576 * t \n - 31.29 \: = - 3.576 * t \n divide \: both \: side \: by \: - 3.576 \n t \: = (- 31.29)/(- 3.576) \n$

t = 8.75 s

Thus, the time taken for the car to stop is 8.75 s

B.Determination of the total distance travelled when the brakes were applied.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Distance (s) =?

${v}^(2) = {u}^(2) + 2as \n {0}^(2) = {31.29}^(2) + (2 * - 3.576 * s) \n 0 = 979.0641 - 7.152 s \n collect \: like \: terms \n 0 - 979.0641 = - 7.152 s \n - 979.0641 = - 7.152 s \n divide \: both \: side \: by \: - 7.152 \n s = (- 979.0641)/(- 7.152) \n \n$