Lichen grows on the surface of hornblende granite and causes it to breakdown at a faster rate. This is an example of. _______ Weathering

Answer:

W=45J

Explanation:

W=Fd

W=15(3)=45

W=45J

Answer: A

Explanation:

Answer:

A

Explanation:

Kinetic energy must be moving. Potential energy has the ability to move but is not doing so at the moment.

A is likely the answer. But there's lots involved in that kind of motion.

B If the ball is elevated, it implies it is not moving yet. It has potential energy.

C Again, the spring is compressed. It will push something when it moves, but it is not moving yet.

D The load gun's bullet is not moving. It's still potential energy.

E. The mouse trap is set, but it is not moving. When the mouse eats the bait then it's potential energy will transform into kinetic energy.

Answer:

94.25 seconds

Explanation:

Solve for period (T) using: v=(2*pi*r)/T

rearrange: vT=2*pi*r

rearrange: T=(2*pi*r)/v

Plug in values.

T=(2*pi*150)/10

T=94.25 seconds

If a race car goes around a circular track of a radius of 150 m at speed of 10.0 m/s ,then the time taken to complete the one lap would be 94.25 seconds.

The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object. The unit of speed is a meter/second. The generally considered unit for speed is a meter per second.

As given in the problem a race car goes around a circular track of radius 150 m at speed of 10.0 m/s.

vT = 2 × π × r

T = (2 × π × r)/ v

T = (2 × π× 150)/10

T = 94.25 seconds

Thus, the time taken to complete the one lap would be 94.25 seconds.

To learn more about speed here, refer to the link given below ;

https://brainly.com/question/7359669

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Answer:

The correct option is b: 30 N.

Explanation:

First, we need to find the acceleration due to gravity (a):

[tex] y_{f} - y_{0} = v_{o}t - \frac{1}{2}a(\Delta t)^{2} [/tex]   (1)

Where:

[tex]y_{f}[/tex]: is the final vertical position (obtained from the graph)

[tex]y_{0}[/tex]: is the initial vertical position (obtained from the graph)

v₀: is the initial speed = 0 (it is released from rest)

Δt: is the variation of time (from the graph)

From the graph, we can take the following values of height and time:

t₀ = 0 s → [tex]t_{f}[/tex] = 5 s

y₀ = 300 m → [tex]y_{f}[/tex] = 225 m

Now, by entering the above values into equation (1) and solving for "a" we have:

[tex] a = 2\frac{y_{0} - y_{f}}{(t_{f} - t_{0})^{2}} = 2\frac{300 m - 225 m}{(5 s - 0)^{2}} = 6 m/s^{2} [/tex]

Finally, the weight of the object is:

[tex] W = ma = 5 kg*6 m/s^{2} = 30 N [/tex]

Therefore, the correct option is b: 30 N.

I hope it helps you!                                                                                                            

Answer:

voltage divider,  R₂ = 1000 R₁

measuring the output in the resistance R₁

Explanation:

Let's analyze the situation, in an op amp in open gain loop, the gain is maximum G = 10⁶ V / V

in this case the signal generator gives a minimum wave of 1 10⁻³ V, after passing through the amplified it becomes 10³ V which saturates the oscilloscope.

To solve this problem we must use a simple voltage divider, for this we use the fact that in a series circuit the voltage is the sum of the voltages of each element.

If we use two resistors whose relationship is

            R₂ / R₁ = 10³

            R₂ = 1000 R₁

When measuring the output in the resistance R₁ we have the desired divider, with a tolerance range, for the minimum output of the generator (1 10⁻³V) we have a reading of V = 1 V in the oscilloscope, for which we can use voltage up to 10V on the generator

Answer:

Venus

Explanation:

Answer:

1.matter is any substance that has mass and takes up space by having volume.

2.The three fundamental phases of matter are solid, liquid, and gas (vapour),

3.In gases the particles move rapidly in all directions, frequently colliding with each other and the side of the container. With an increase in temperature, the particles gain kinetic energy and move faster. ... In liquids, particles are quite close together and move with random motion throughout the container.

3.In gases the particles move rapidly in all directions, frequently colliding with each other and the side of the container. With an increase in temperature, the particles gain kinetic energy and move faster. ... In liquids, particles are quite close together and move with random motion throughout the container.

Answer:

gravity pulls down on student the chair pushes up on the student's body with the same force gravity is pulling down on the student

Answer:

83.2 m/s to the West

Explanation:

From the question given above, the following data were obtained:

Mass of plane = 180 Kg

Mass of pilot = 70 Kg

Momentum = 20800 Kg∙m/s West

Velocity =?

Next, we shall determine the total mass. This can be obtained as follow:

Mass of plane = 180 Kg

Mass of pilot = 70 Kg

Total mass =?

Total mass = Mass of plane + Mass of pilot

Total mass = 180 + 70

Total mass = 250 Kg

Finally, we shall determine the velocity. This can be obtained as follow:

Total mass = 250 Kg

Momentum = 20800 Kg∙m/s West

Velocity =?

Momentum = mass × Velocity

20800 = 250 × Velocity

Divide both side by 250

Velocity = 20800 / 250

Velocity = 83.2 m/s West

Thus, the velocity of both plane and pilot is 83.2 m/s to the West

Answer:

20mm per second

Explanation:

The IDEAL mechanical advantage of this ramp is (25m / 5m) = 5 .

But it's only giving you a real MA of (4500N/1000N) = 4.5 .

The friction between the crate and the surface of the ramp is robbing some of the work you do as you slide the crate up the ramp, which degrades the mechanical advantage.

Answer:

I'm gonna say d

Explanation:

bc they all seem very important

hope this helped

Answer:

Force = 0.49 N (Approx)

Explanation:

Given:

Mass = 50 grams = 0.05 Kg

Acceleration = 9.81 m/s²

Find:

Force

Computation:

Force = Mass x Acceleration

Force = 0.05 x 9.81

Force = 0.4905

Force = 0.49 N (Approx)

Explanation:

Given that,

Each vertical line on the graph is 1 millisecond (0.001 s) of time.

We need to find the period and the frequency of the sound wave. The period of a wave is equal to the each vertical line on graph i.e. 0.001 s.

Let f be the frequency of the sound wave. So,

f = 1/T

i.e.

[tex]f=\dfrac{1}{0.001 }\\\\f=1000\ Hz[/tex]

So, the period and the frequency of the sound waves is 1 milliseond and 1000 Hz respectively.

Answer:

v = 3.74 m/s

Explanation:

Given that,

The acceleration of the object in circular path, a = 0.4 m/s²

The radius of the circle, r = 35 m

We need to find the velocity of the object. The acceleration of an object on the circular path is given by :

[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{0.4\times35}\\\\v=3.74\ m/s[/tex]

So, the velocity of the object is equal to 3.74 m/s.

Answer:

MgS

Explanation:

Answer:

A wave can be thought of as a disturbance or oscillation that travels through space-time, accompanied by a transfer of energy. The direction a wave propagates is perpendicular to the direction it oscillates for transverse waves. A wave does not move mass in the direction of propagation; it transfers energy.

Answer:

what situation?

Explanation:

Answer:

2)

Explanation:

Answer:

a. 20 s

b. 0 m/s  

c. right

d.no its inelastic because when the car b was at rest and a was coming in at it, since b had no force what so ever car a swept it away with it moving to the right

Explanation:

im not sure though

By applying conservation of linear momentum, the answers are:

1. Time = 2 s

2. 3 m/s

3. same direction

4. Inelastic collision

There are for types of collision. They are;

Given that a local church is hosting a carnival which includes a bumper car ride. Bumper car A and its driver have a mass of 300 kg; bumper car B and its driver have a mass of 200 kg. Bumper car A has a velocity to the right of 2 m/s and bumper car B is at rest. At t = 0 s, bumper car A and B are separated by 10 m. Bumper car A accelerates at 1 m/s2 to a velocity of 4 m/s and continues at this constant speed until colliding with bumper car B.

1. The time required for bumper car A to travel the 10 m to collide with bumper car B can be calculated by using first equation of linear motion.

V = U + at

Where

V = 4 m/s

U = 2 m/s

a = 1 m/[tex]s^{2}[/tex]

Substitute all the parameters into the formula

4 = 2 + t

t = 4 - 2

t = 2s

2. To calculate the speed of bumper car A following the collision with bumper car B, which now has a velocity to the right of 3 m/s, we will apply conservation of linear momentum

[tex]m_{1}u_{1}[/tex] = [tex]m_{1}v_{1}[/tex] + [tex]m_{2}v_{2}[/tex]

300 x 4 = 300V + 200 x 3

1200 = 300V + 300

300V = 1200 - 300

300V = 900

V = 900/300

V = 3 m/s

3. Since the final velocity of car A is positive, the direction of motion for bumper car A follows the collision with bumper car B to the right.

4. Since the both move at the same velocity, the collision inelastic.

Learn more about Collision here: https://brainly.com/question/25121535

Answer:

THE CORRECT ANSWER FOR THIS KS

Explanation:

STOP USING THIS APP FOR ANSWERS U KNOW NOTHING

What a fan does is create a wind chill effect. ... By blowing air around, the fan makes it easier for the air to evaporate sweat from your skin, which is how you eliminate body heat. The more evaporation, the cooler you feel.

Answer:

a)       [tex]- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{sin \theta}{R }[/tex] =  [tex]\frac{d^2 \theta}{d t^2}[/tex]

b)     f = 2π  [tex]\sqrt{ \frac{R}{ g - \frac{q}{m} \frac{\sigma }{2 \epsilon_o} } }[/tex]  

Explanation:

a) To have the equations of motion, let's use Newton's second law.

Let's set a reference system where the x-axis is parallel to the path and the y-axis is in the direction of tension of the rope.

For this reference system the tension is in the direction of the y axis, we must decompose the weight and the electrical force.

Let's use trigonometry for the weight that is in the vertical direction down

             sin θ = Wₓ / W

             cos θ = W_y / w

             Wₓ = W sin θ

             W_y = W cos θ

we repeat for the electric force that is vertical upwards

              F_{ex} = F_e sin θ

              F_{ey} = F_e cos θ

the electric force is

               F_e = q E

where the field created by an infinite plate is

               E = [tex]\frac{ \sigma}{2 \epsilon_o}[/tex]  

let's write Newton's second law

Y  axis  

           T - W_y = 0

            T = W cos θ

X axis

            F_{ex} - Wₓ = m a                   (1)

we use that the acceleration is related to the position

            a = dv / dt

            v = dx / dt

where x is the displacement in the arc of the curve

substituting

            a = d² x /dt²

we substitute in 1

           q E sin θ - mg sin θ = m [tex]\frac{d^2 x}{dt^2}[/tex]

we have angular (tea) and linear (x) variables, if we remember that angles must be measured in radians

           θ = x / R

           x = R θ

we substitute

           sin θ (q E - mg) = m \frac{d^2 R \ theta}{dt^2}  

          [tex]- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{sin \theta}{R }[/tex] =  [tex]\frac{d^2 \theta}{d t^2}[/tex]

this is the equation of motion of the system

b) for small oscillations

         sin θ = θ

therefore the solution is simple harmonic

      θ = θ₀ cos (wt + Ф)

if derived twice, we substitute

- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o}  ) \frac{\theta}{R } θ₀ cos (wt + Ф) = -w² θ₀ cos (wt + Ф)

       w² =  [tex]\frac{g}{R}[/tex] - [tex]\frac{q}{m} \frac{ \sigma }{2 \epsilon_o} \frac{1}{R}[/tex]  

angular velocity is related to frequency

       w = 2π f

        f = 2π / w

        f = 2π/w

        f = 2π  [tex]\sqrt{ \frac{R}{ g - \frac{q}{m} \frac{\sigma }{2 \epsilon_o} } }[/tex]  

Answer:

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