Which of the filling is a fossil fuel?

Answers

Answer 1
I need a picture to see what exactly this question is asking

Related Questions

Question 4
Which of the following is unique for any given element?
O the mass of a neutron
o the number of neutrons
o the charge on the electons
O the number of protons

Answers

the number of protons

it's unique for any element because it's determined by the atomic number and no two elements have the same atomic number

Help me Please!!!!!!!

Answers

The speed is equal to the area under the line up to the point where t .=15 s.
Do find the area of the triangle and that if the rectangle and add them together.
The area of the triangle is 25 and the rectangle is also 25 so the speed is 50 m/s

A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp. How much of its original total energy (in J) survives as KE when it reaches the ground

Answers

This question is incomplete, the complete question is;

A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp.

How much of its original total energy (in J) survives as KE when it reaches the ground? m = 9.9 kg h = 4.9 m d = 5 m μ = 0.3 θ = 36.87°

Answer:

the amount of its original total energy (in J) that survives as KE when it reaches the ground will is 358.975 J

Explanation:

Given that;

m = 9.9 kg

h = 4.9 m

d = 5 m

μ = 0.3

θ = 36.87°

Now from conservation of energy, the energy is;

Et = mgh

we substitute

Et = 9.9 × 9.8 × 4.9

= 475.398 J

Also the loss of energy i

E_loss = (umg cosθ) d

we substitute

E_loss  = 0.3 × 9.9 × 9.8 × cos36.87°  × 5

= 116.423 J

so the amount of its original total energy (in J) that survives as KE when it reaches the ground will be

E = Et - E_loss

E = 475.398 J - 116.423 J

E = 358.975 J

At an amusement park, a swimmer uses a water slide to enter the main pool. You may want to review (Pages 234 - 241) . Part A If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.81 m , what is her speed at the bottom of the slide

Answers

Answer:

Her speed at the bottom of the slide is 7.42 m/s

Explanation:

From the question,

The swimmer starts at rest, that is, her initial speed, u is 0 m/s.

Since she slides without friction and descends through a vertical height, then it is a free fall motion (due to gravity).

Also, from the question,

She descends through a vertical height of 2.81 m.

To determine her speed at the bottom of the slide, that is her final speed,

From one of the equations of motion for freely falling bodies

v² = u² + 2gh

Where v is the final speed

u is the initial speed

g is acceleration due to gravity (g = 9.8 m/s²)

and h is height

From the question,

u = 0 m/s

h = 2.81 m

Putting the values into the equation

v² = u² + 2gh

v² = 0² + 2×9.8×2.81

v² = 55.076

v =√55.076

v = 7.42 m/s

Hence, her speed at the bottom of the slide is 7.42 m/s.

The power that a student generates when walking at a steady pace of vw is the same as when the student is riding a bike at vb = 3vw. The student is going to travel a distance d. The energy the student uses when walking is Ew. The energy the student uses when biking is Eb. The ratio EwEb is

Answers

Answer:

3

Explanation:

If an object is moving with a constant velocity to the right, what direction is the net force.

Group of answer choices

A.To the right

B.To the left

C.Net force is 0

D.Not enough information

Answers

Answer:

At constant velocity, his weight equals the force of friction. In other words, there is no net force. If however, he loosens his grip and decreases the friction force, he will accelerate downward.

Explanation:

Open Box. Consider a hollow box with the top missing. The sides have negligible thickness and each has length L and mass m. (a) Find the x-coordinate of the center of mass.

Answers

Answer:

x_{cm} = L / 2

Explanation:

The center of mass is defined by

         [tex]x_{cm}[/tex] = 1 / M ∑ m_{i}  x_{i}

where M is the total mass of the system

in this case the system is continuous so, for which we use the density

      ρ = dm / dx

      dm = ρ dx

substituting

        x_{cm} = 1 / M ∫ x ρ dx

        x_{cm} = ρ / M ∫ x dx

we integrate and evaluate from x = 0 to x = L

        x_{cm} = ρ / M (L² /2 -0)

       

we introduce the density which is constant

        ρ = M / L

        x_{cm} = 1 /M (M/L)  L² / 2

        x_{cm} = L / 2

A crane uses a single cable to lower a steel girder into place. The girder moves with constant speed. The cable tension does work WT and gravity does work WG. Which statement is true

Answers

Explanation:

Work done by a force is given by :

[tex]W=Fd\cos\theta[/tex]

Where

F is force, d is displacement and [tex]\theta[/tex] is the angle between F and d.

In this problem, a crane is moving in downward direction, the force gravity is in downward direction and the tension is in upward direction.

We know that if force and displacement is in same direction, work is positive while if force and displacement is in oposite direction, work is negative.

I would mean that, [tex]W_g[/tex] is positive, because gravity is parallel to the displacement and [tex]W_t[/tex] is negative, because the tension is opposite to the displacement.

At which point on the image to the right would the ball have the greatest velocity if it moved from A to G.

please help me out.

A


B


C


D


E


F


G

Answers

Answer:

Total energy = Kinetic Energy + Potential Energy = Constant

Since the potential energy is lowest at point D the kinetic energy will be greatest at point D and the velocity will be the greatest.

The Intensity level of a loud saw is 100 db at a distance of 5m. At what distance would the level be 80 db

Answers

Answer:

50 m

Explanation:

The relationship between the intensity of sound in dB and distance is given by the formula:

[tex]B_2=B_1+20log(\frac{R_1}{R_2} )\\\\Where \ B_2\ is \ the\ sound\ intensity\ at\ distance\ R_2\ and\\B_1\ is \ the\ sound\ intensity\ at\ distance\ R_1\ \\\\Given\ that: B_1=100\ dB, R_1=5\ m, B_2=80\ dB\\\\B_2=B_1+20log(\frac{R_1}{R_2} )\\\\80=100+20log(\frac{5}{R_2} )\\\\-20=20log(\frac{5}{R_2} )\\\\log(\frac{5}{R_2} )=-1\\\\\frac{5}{R_2}=10^{-1}\\\\\frac{5}{R_2}=0.1\\\\R_2=5/0.1=50\ m[/tex]

Pressure and temperature ______ with depth below Earth’s surface.

Answers

Answer:

Pressure increases as you move deeper below earth's surface.

Tempurature  increases as you move deeper below earth's surface.

Hope this helps!

Explanation:

give three factors which are responsible for the vanishing forest ​

Answers

Answer:

1. Huge wildfires

2. Deforestation

3. Reduced amount of aforestation, etc

If the body with a mass of 4kg is moved by a force of 20 N, what is the rate of its acceleration?

Answers

Answer:

The answer is 5 m/s²

Explanation:

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

[tex]acceleration = \frac{force}{mass} \\[/tex]

From the question

force = 20 N

mass = 4 kg

We have

[tex]a = \frac{20}{4} \\ [/tex]

We have the final answer as

5 m/s²

Hope this helps you

Which two types of energy does a book have as it falls to the floor

Answers

Answer:

kinetic and potential energy

Explanation:

D
5. Mariam driving at a speed of 20.0 m/s applies
brakes close to a signal and travels a distance of
200 m before coming to rest. What was her
acceleration?
A. -0.50 m/s2
B. -0.70 m/s2
C. -1.00 m/s2
D. -2.00 m/s2
6. A trollen at rest is nushed to accelerate at a

Answers

Answer:

maibi.... D

Explanation:

I think is D

N₂ + H₂
NH3
how do i balance this equation?

Answers

Answer:

N2 + 3H2 ----->  2NH3

Explanation:

Reactants side:

2 Nitrogen

5 Hydrogen

Products Side:

2 Nitrogen

5 Hydrogen

The boys are finally old enough to compete in the box car derby race at the local fair. They have been working on their cars since the conclusion of the race last year. One boy's car raced down the track and placed 2nd in his race. However, the other boy's car started well but half-way through the race a wheel came off and his car came to a complete stop. The boy was very disappointed and the other boy felt horrible for his friend. Which of the following graphs best represents the motion of boy's car that stopped?

Answers

i think it’s b because they both stumbled up a hole

(A) Electricity and Magnetism
A). Three point charges are aligned along the x axis as shown in
Fig. Find the electric field at (a) the position (2, 0) and (b) the
position (0, 2).

Answers

electricity

Explanation:

the position (2,o

A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s2 over a distance of 340 m. How fast is it going after that acceleration? (68.15 m/s)

Answers

v² - u² = 2 ax

where u = initial velocity, v = final velocity, a = acceleration, and ∆x = distance traveled.

So

v² - (15 m/s)² = 2 (6.5 m/s²) (340 m)

v² = 4645 m²/s²

v ≈ 68.15 m/s

Which famous Baroque period composer wrote 46 pieces of music while in jail?

Answers

Answer:

John Sebastian Bach

Answer:

Johann Sebastian Bach

Explanation:

please tell me if i am wrong! thank you!

If a rock is skipped into a lake at 24 m/s2, with that what force was the rock thrown if it was 1.75kg?

Answers

Answer: f= M×A

1.75kg×24= 42N

Explanation:

Because to find force you do Mass times acceleration so I did 1.75 kg times 24 would equal 42 Newtons!

This diagram shows two different forces acting on a skateboarder. The
combined mass of the skateboard and the person is 81.5 kg. Based on this
information, what is the acceleration of the skateboarder?
Air resistance = 11.40 N
mi
Applied force = 52.80
A. 0.51 m/s2 to the left
O B. 1.94 m/s2 to the right
O C. 1.94 m/s2 to the left
O D. 0.51 m/s2 to the right

Answers

Answer:

Option D. 0.51 m/s² to the right.

Explanation:

The following data were obtained from the question:

Force applied (Fₐ) = 52.8 N

Force resistance (Fᵣ) = 11.4 N

Mass = 81.5 kg

Acceleration (a) =.?

Next, we shall determine the net force (Fₙ). This can be obtained as follow:

Force applied (Fₐ) = 52.8 N

Force resistance (Fᵣ) = 11.4 N

Net force (Fₙ) =?

Fₙ = Fₐ – Fᵣ

Fₙ = 52.8 – 11.4

Fₙ = 41.4 N to the right

Finally, we shall determine the acceleration of the skateboarder as show below:

Net force (Fₙ) = 41.4 N to the right

Mass = 81.5 kg

Acceleration (a) =..?

F = ma

41.4 = 81.5 × a

Divide both side by 81.5

a = 41.4 / 81.5

a = 0.51 m/s² to the right.

Thus, the acceleration of the skateboarder is 0.51 m/s² to the right.

Answer: .51 to the right

Explanation:

a man weighing 490 n on earth weighs only 81.7 n on the moon. His mass on the moon is__kg. (Use g=9.8 m/s2

Answers

Answer:

m = 50 [kg]

Explanation:

In order to solve this problem we must be clear about the difference between weight and mass. Weight is the product of mass by the acceleration of the planet or the star. While the mass is always preserved it never changes regardless of where it is located.

So for the earth we have:

g = gravity acceleration = 9.8 [m/s^2]

m = mass [kg]

W = weigth = 490 [N]

therefore the mass will be:

m = W/g

m = 490/9.8

m = 50 [kg]

Now it is important to remember that the mass will be the same on the moon or on the earth, but the weight will be different, because the gravity acceleration of the moon is different from the gravity acceleration on earth

So the gravity on the moon is equal to:

81.7 = 50 * gm

gm = 1.634 [m/s^2]


The interaction between electrical energy and magnetism has been an important
topic in 20th century science, Which term describes this interaction?

Answers

Answer:

Maybe

Explanation:

I say maybe because it will help them still but not quite

5. A car advertisement states that a certain car can accelerate from rest to 70 m/s in 7
seconds. Find the car's average acceleration.
O-0.10 m/s^2
10 m/s^2
-10 m/s^2
O 0.10 m/s^2

Answers

Answer:

The car's average acceleration is [tex]10\ m/s^2[/tex].

Explanation:

Constant Acceleration Motion

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being vo the initial speed, a the constant acceleration, vf the final speed, and t the time, the following relation applies:

[tex]v_f=v_o+at[/tex]

If we need to find the acceleration, we solve the above equation for a:

[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]

The car accelerates from rest (vo=0) to vf=70 m/s in t=7 seconds. Substitute the values into the formula:

[tex]\displaystyle a=\frac{70-0}{7}=\frac{70}{7}=10[/tex]

[tex]a=10\ m/s^2[/tex]

The car's average acceleration is [tex]10\ m/s^2[/tex].

Note: The choices are not very clear, but the second choice seems to be the correct answer.

Two boxes of masses 3M and 5M are attached by a massless rope. They are being pulled to the right with a constant force of P = 800 N, which allows them to just overcome static friction, with a μs= 0.70 between the floor and the boxes.
a. Find M.
b. Find the Tension in the rope between the two boxes.

Answers

Answer:

  a) about 14.577 kg

  b) 300 N

Explanation:

b) In order for the acceleration to be the same for each mass, the 800 N force must be divided between the boxes in proportion to their mass. That is, the net force on the 5M mass must be 5/8 of the total force, or 500 N. Then the tension in the rope is 800 N -500 N = 300 N, which is 3/8 of 800 N.

Tension: 300 N

__

a) The total mass is 8M, and the total normal force on the floor is ...

  F = ma = (8M)(9.8 m/s^2)

The friction force is 0.7 times this, and is equal to the 800 N force pulling on the boxes.

  800 N = (8M)(9.8 m/s^2)(0.7)

  M = 800/(8·9.8·0.7) kg ≈ 14.577 kg

2. A bird flying horizontally at 10 m/s drops a branch. The bird is flying at an altitude of 20 m. Determine
the horizontal displacement it moves relative to where it was dropped.

Answers

Answer:

The horizontal displacement is 20 m.

Explanation:

Given that,

Velocity = 10 m/s

Height = 20 m

We need to calculate the time

Using equation of motion

[tex]s=ut+\dfrac{1}{2}gt^2[/tex]

Put the value into the formula

[tex]20=0+\dfrac{1}{2}\times9.8\times t^2[/tex]

[tex]t^2=\dfrac{20\times2}{9.8}[/tex]

[tex]t=\sqrt{\dfrac{20\times2}{9.8}}[/tex]

[tex]t=2.0\ sec[/tex]

We need to calculate the horizontal displacement

Using formula of horizontal displacement

[tex]\Delta x=v_{x}\times t[/tex]

Put the value into the formula

[tex]\Delta x=10\times2.0[/tex]

[tex]\Delta x=20\ m[/tex]

Hence, The horizontal displacement is 20 m.

What equation relates mechanical energy, thermal energy, and total energy when there is friction present in a system?

Answers

E total = ME + E thermal Explanation:
APEX

The coefficient of static friction between m1 and the horizontal surface is 0.50, and the coefficient of kinetic friction is 0.30. (a) If the system is released from rest, what will its acceleration be

Answers

This question is incomplete

Complete Question

m1 is 10kg, m2 is 4.0kg. The coefficient of static friction between m1 and the horizontal surface is 0.50. and the Coefficient of kinetic friction is 0.30.

a) if the system is released from rest what will be its acceleration

Answer:

0.7 m/s²

Explanation:

The coefficient of static friction between m1 and the horizontal surface is 0.50. and the coefficient of kinetic friction is 0.30.

(a) if the system is released from rest what will be its acceleration

g = acceleration due to gravity = 9.81 m/s²

Coefficient of Kinetic Friction = μk = 0.30

m1 = 10kg

m2 = 4.0kg

The formula to solve question a is given as:

a = acceleration at rest

m2g- μk m1g = (m1+ m2) a

Making a the subject of the formula:

a = (m2g- μk×m1g )/(m1+ m2)

a = [(4.0 kg × 9.81m/s²) – (0.30 ×9.81 × 10) ]/(10+4)

a = 0.7 m/s²

A 30%-efficient car engine accelerates the 1300 kg car from rest to 10 m/s . How much energy is transferred to the engine by burning gasoline

Answers

Answer:

The Energy transferred to the engine by burning gasoline = 216.67 KJ

Explanation:

The parameters given are:

The efficiency of the car engine, E = 30% = 0.3

Mass, m = 1300 kg

Initial velocity, u = 0, since the car is from rest

The final velocity, v = 10 m/s

Since the car was moving, we calculate its kinetic energy.

kinetic energy = ((1/2) (m) (v^2)

((1/2) (1300 kg) (10 m/s^2)

= 65,000 j

The Energy, Q transferred to the engine by burning gasoline in this case

= potential energy / The efficiency of the car engine, E

Q = 65,000 j / 0.3

= 216,666.66 J

Converting Joule to kilojoule

where 1KJ = 1000j

216,666.66 J = 216.67 KJ

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