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2 years ago

Problem: Asteroid Field.

Physics

2 answers:

Vesna [10]2 years ago

6 0

Answer:

An asteroid is a minor planet of the inner Solar System. Historically, these terms have been applied to any astronomical object orbiting the Sun that did not resolve into a disc in a telescope and was not observed to have characteristics of an active comet such as a tail.

olchik [2.2K]2 years ago

6 0

Answer:

a place in our solar system where small bodies – mostly rocky and some metallic – orbit the sun. Sometimes scientists call these little worlds minor planets. One (Ceres) is even categorized as a dwarf planet. These objects move mostly between the orbits of our solar system’s 4th planet, Mars, and 5th planet, Jupiter. Astronomers once thought they were leftovers of a rocky planet that Jupiter’s gravity tore apart long ago. Now they think differently.

Explanation:

is the small thing that separated from the star and was near the planet Earth

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Which of the following happens during cell division? Question 7 options: It becomes more difficult for the cell to get rid of wa

kolbaska11 [484]

Answer;

B. Each daughter cell receives its own copy of the parent cell's DNA.

Explanation;

Cell division is the process in which a parent cell divides, giving rise to two or more daughter cells. It is an essential biological process in many organisms.

There are two types of cell division. Mitosis is used for growth and repair and produces diploid cells identical to each other and the parent cell. Meiosis is used for sexual reproduction and produces haploid cells different to each other and the parent cell.

6 0

3 years ago

A particle moving on a circle has a velocity of 5 m/s and a normal acceleration of 10 m/s^2. What is the radius of the circle?

dybincka [34]

Answer:

Radius of the circle will be 2.5 m

Explanation:

We have given velocity of particle moving in the circle v = 5 m/sec

Acceleration of particle in the circle $a=10m/sec^2$

We have to find the radius of the circle

We know that acceleration is given by $a=\frac{v^2}{r}$

So $10=\frac{5^2}{r}$

$r=\frac{25}{10}=2.5m$

So radius of the circle will be 2.5 m

3 0

3 years ago

Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a

lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

5 0

3 years ago

What are the top three osha cited ladder violations

bonufazy [111]

Answer:

Explanation:

1. FALL PROTECTION-GENERAL REQUIREMENTS (29 CFR 1926.501) 6,010 VIOLATIONS

2. HAZARD COMMUNICATION (29 CFR 1910.1200). 3,671

3. SCAFFOLDING (29 CFR 1926.451). 2,813

6 0

3 years ago

Hemoglobin (Hb) is the O2-carrying protein in our blood. Unlike myoglobin, it has four sites allowing it to bind up to four O2 m

bogdanovich [222]

Answer:

Yes, the energy is not simply the sum of the individual binding energies at each site, it is the product of energy at each binding site of hemoglobin.

Explanation:

Myoglobin and hemoglobin are two different cells. Myoglobin binds only one oxygen while the hemoglobin has the ability to binds four oxygen atoms at its four sides. Myoglobin present in muscle tissue only while hemoglobin is present in the whole body. Oxyhemoglobin is formed when oxygen binds with hemoglobin cell. This oxygen is take to all cells and energy is released due to the breakdown of glucose molecules with this oxygen.

4 0

3 years ago

Problem: Asteroid Field.​

Problem: Asteroid Field.