In the pie chart below, which of the following represents the amount of Earth's surface covered by water?
A only
B only
B and C
D only

Answer: The Senate provides advice and consent on presidential nominations.

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

mass = 80 kg

acceleration = 2 m/s²

We have

force = 80 × 2 = 160

We have the final answer as

Hope this helps you

Answer:

Resolve the vectors:

F1 = 30 cos 30 i + 30 sin 30 j = 26 i + 15 j

F2 = 80 cos 135 i + 80 sin 135 j = -56.6 i + 56,6 j

F1 + F2 = (26 - 56.6) i + (15 + 56.6) j = -30.6 i + 71.6 j

The offsetting force would be

F3 = 30.6 i - 71.6 j

The magnitude of this vector is (30.6^2 + 71.6^2)^1/2 = 77.9

cos theta 3 = 30.6 / 77.9 = 66.9 deg

Check:

77.9 * sin 66.9 = 71.6  agreeing in magnitude with F3

Now we need the quadrant of the offsetting vector

sin theta 3 = -         cos theta 3 = +

This occurs in the fourth quadrant and theta 3 = 360 - 66.9 = 293.1 deg

Answer: i think its a bullfrog forgive me if im wrong i dont exactly know

Explanation:

Answer:

that is an African bullfrog

Answer:

a.true

Explanation:

because the gravitational strength of earth and moon is different

Answer:

no

Explanation:

because the heaver an object the harder it is to move s the force will not be able to move it

Answer:

G M1 m1 / R1^2 = G M2 m2 / R2^2          gravitational equation

We can try Kepler's harmonic law

P2^2 = K R2^3

P1^2 = K R1^3     for earth

(P2 / P1)^2 = (R2 / R1)^3       dividing equations

R2^3 = (P2 / P1)^2 * R1^3

R2^3 = (53 / 365)^2 * R1^3

R2^3 = .145 R1^3

R2 = .526 R1

If one uses 93E6 miles as distance of sun from earth then

R2 = 49E6 miles

Answer:

Explanation:

im sorry if this doesn't help but i think  its A

Answer:

15 N :D

Explanation:

The size of the drag force on this object falling at a constant speed is determined as 15 N.

The drag force on the object is determined by calculating the net force on the object;

F(net) = W - D

where;

ma = W - D

at constant speed, acceleration, a = 0

0 = W - D

W  = D

15 N = D

Thus, the size of the drag force on this object falling at a constant speed is determined as 15 N.

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Answer:

ytyug

Explanation:

Answer:

60 Ns

Explanation:

change in momentum = Fxt

6*10

60 Ns

A force of 6.0 N acts on a 4.0 Kg object for 10.0 s. The object's change in momentum is 60 kg-m/sec.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

change in momentum = force . time

change in momentum = 6 . 10 = 60

A force of 6.0 N acts on a 4.0 Kg object for 10.0 s. The object's change in momentum is 60 kg-m/sec.

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Answer:

it look the same just to tell you

The speed of the man and the chair after the book is thrown is 0.2 m/s.

The given parameters:

The total mass of the man and the book is calculated as follows;

m₂ = 70 kg + 9.2 kg

m₂ = 79.2 kg

The speed of the man and the chair after the book is thrown is determined by applying the principle of conservation of linear momentum;

[tex]m_1 u_1 = m_2u_2\\\\u_2 = \frac{m_1u_1}{m_2} \\\\u_2 = \frac{2.1 \times 7.2}{79.2} \\\\u_2 = 0.2 \ m/s[/tex]

Thus, the speed of the man and the chair after the book is thrown is 0.2 m/s.

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I only know the formula which is ω = 2πf

Answer:

It’s a

Explanation:

The magnitude of the acceleration of the box is 2 m/s².

The given parameters;

The resultant of the two given forces is calculated as follows;

[tex]F= \sqrt{F_1 ^2 + F_2 ^2} \\\\F = \sqrt{60^2 + 80^2} \\\\F = 100 \ N[/tex]

The magnitude of the acceleration of the box is calculated as follows;

[tex]F = ma\\\\a = \frac{F}{m} \\\\a = \frac{100}{50} \\\\a = 2 \ m/s^2[/tex]

Thus, the magnitude of the acceleration of the box is 2 m/s².

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Answer:

− 2.08

Explanation:

Answer:

apagado, así que lo más destacado es el problema, oh, está bien, esa es la respuesta, cariño.

how both affect the ozone layer and affect temperature in a grobal way

Answer:

you need kinetic energy because it is the energy that is used when In motion

The type of energy that one acquire when you eat a bowl of hot vegetable soup is chemical energy.

Chemical power the bonds of chemical molecules contain energy. Exothermic reactions are those in which chemical energy is released during the reaction, frequently in the form of heat.

Some of the heat energy needed for a reaction to continue can be stored as chemical energy in newly created bonds.

Food particles go through a chemical reaction that releases chemical energy when we eat it. The body then uses this energy to carry out its regular tasks.

As a result, when you eat a cup of vegetable soup, you also consume the chemical energy it contains.

Thus, the answer is chemical energy.

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The force diagram on the desk consists of normal reaction and weight of the pencil.

The net force on the desk in vertical direction is 1 N.

The net force on the desk in horizontal direction is 0.

The given parameters;

The force diagram of the desk is sketched as follows;

                                             ↑ N

                                             ⊕

                                              ↓ W

Where;

The net force on the desk in vertical direction = 1 N

The net force on the desk in horizontal direction = 0

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unlock your third eye and youll be able to see it

Explanation:

it will not change it's motion because it's force is less than the force applied

Answer:

add 44m/s and 22m/s then multiply it by 11

Explanation:

A race car traveling at 44m/s slows at a constant rate to a velocity of 22m/s over 11 seconds, Acceleration is -2 m/s² (deceleration), and distance is 242 meters.

To solve this problem, it is given:

Initial velocity ([tex]\(v_i\)[/tex] ) = 44 m/s

Final velocity ([tex]\(v_f[/tex] ) = 22 m/s

Time ([tex]\(t\)[/tex] ) = 11 s

We have two tasks: finding the acceleration (a) and then using it to calculate the distance (d).

1. Calculate Acceleration (a):

[tex]\[a = \frac{v_f - v_i}{t} \\\\= \frac{22 \, \text{m/s} - 44 \, \text{m/s}}{11 \, \text{s}} \\\\= -2 \, \text{m/s}^2\][/tex]

The negative sign indicates deceleration (slowing down).

2. Calculate Distance (d):

Using the kinematic equation:

[tex]\[d = v_i \cdot t + \frac{1}{2} \cdot a \cdot t^2\][/tex]

Plugging in the values:

[tex]\[d = (44 \, \text{m/s}) \cdot (11 \, \text{s}) + \frac{1}{2} \cdot (-2 \, \text{m/s}^2) \cdot (11 \, \text{s})^2\][/tex]

Calculating the first part:

[tex]\[d = 484 \, \text{m} - 121 \, \text{m} \\\\= 363 \, \text{m}\][/tex]

Calculating the second part:

[tex]\[d = 363 \, \text{m} + \frac{1}{2} \cdot (-2 \, \text{m/s}^2) \cdot (121 \, \text{s}^2) \\\\= 363 \, \text{m} - 121 \, \text{m} \\\\= 242 \, \text{m}\][/tex]

The total distance the race car moves during this time is 242 meters.

Thus, acceleration is -2 m/s² (deceleration), and distance is 242 meters.

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Answer:

like 23 or something

Explanation:

Differences between earthquakes at continental plate boundaries and earthquakes at mid-ocean ridges discussed in the answer.

When plates travel in the same direction and collide, this happens. The thinner, denser, and more flexible oceanic plate dips beneath the thicker, more rigid continental plate when a continental plate meets an oceanic plate. It's known as subduction.

Deep ocean trenches, like the one that runs down South America's west coast, are created by subduction. The continent's undercut rocks start to melt. On occasion, a chain of volcanoes forms as the molten rock rises to the surface and passes through the continent. Nearly 80% of earthquakes happen along convergent borders, when plates are being forced together.

Shallow earthquakes, often less than 30 km deep, occur in small bands adjacent to plate borders along mid-ocean ridges and transform edges. There are earthquakes in subduction zones at various depths, some of which are more than 700 km deep. Because they occur everywhere throughout the subducting slab that stretches beneath the opposing plate, earthquake bands are wider along subduction zones. Wide areas of distributed earthquakes may be associated with collision zones between continents, such as those between the Eurasian plate and the plates to the south of the African, Arabian, and Indian tectonic arcs.

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Answer:

c

Explanation: