Which friction static friction, kinetic friction, or air friction? Can be more than one response.
Gripping friction.
kinetic friction
air friction
static friction

Answer:

The gravitational potential energy is 2,750 Joules

Explanation:

Given the following data;

Mass of brick = 1kg

Height of Eiffel Tower = 275m

Potential energy can be defined as the energy possessed by an object due to its position.

Mathematically, potential energy is;

[tex] P = mgh[/tex]

Also, we know acceleration due to gravity = 10m/s2

[tex] P = mgh[/tex]

[tex] P = 1 * 275 * 10 [/tex]

P = 2750 Joules.

Hence, the gravitational potential energy the brick acquired is 2750 Joules.

Answer:

5 seconds

Explanation:

Answer:

5 seconds

Explanation:3=1   6=2 9=3 12=4 15=5       15 equal 5 mean 15 meters in 5 sec

Answer:Accelerating your car to get up to speed on a freeway

An airplane accelerating to take off

Accelerating out of the starting blocks at a track meet

Explanation:

Answer:

1.More number of natural disasters will occur.

2. Disease will spread.

Explanation:

An ecosystem is a geographic area where plants, animals, and other organisms live in and interact with each other in an environment.

Forests are homes to orangutans, elephants, tigers, and other animals.

As more than 8.6 million acres of forest were destroyed in Indonesia, Malaysia, and Papua New Guinea to build palm oil plantations, this may lead to the following consequences:

1.More number of natural disasters will occur.

2. Disease will spread.

Answer:

45m/s^1

Explanation:

v=u+at

v=(30)+(2.5)(6)

v=45m/s^1

Answer:

Power is the rate which work is done.

Explanation:

Power is the rate which work is done. Power is measured in watts.

Work is the use of force to move an object. Work is measured in joules

Answer:

The rocket above the ground is in 44 sec.

Explanation:

Given that,

Initial velocity = 92 m/s

Acceleration = 4 m/s²

Altitude = 1200 m

Suppose, How long was the rocket above the ground?

We need to calculate the time

Using equation of motion

[tex]s=ut-\dfrac{1}{2}at^2[/tex]

Put the value into the formula

[tex]1200=92t+\dfrac{1}{2}\times4t^2[/tex]

[tex]2t^2+92-1200=0[/tex]

[tex]t=10\ sec[/tex]

We need to calculate the velocity

Using equation of motion

[tex]v=u+at[/tex]

Put the value into the formula

[tex]v=92+4\times10[/tex]

[tex]v=132\ m/s[/tex]

When the rocket hits the ground,

Then, h'=0

We need to calculate the time

Using equation of motion

[tex]h'=h+ut-\dfrac{1}{2}at^2[/tex]

Put the value into the formula

[tex]0=1200+132t-\dfrac{1}{2}\times9.8t^2[/tex]

[tex]4.9t^2-132t-1200=0[/tex]

[tex]t=34\ sec[/tex]

When the rocket is in the air it is the sum of the time when it reaches 1000 m and the time when it hits the ground

So, the total time will be

[tex]t'=34+10[/tex]

[tex]t'=44\ sec[/tex]

Hence, The rocket above the ground is in 44 sec.

Answer:

5 Kg force I think try

Explanation:

the force is equal to the weight divided by two. It will require a 5 Kg force to lift a 10Kg weight.

Answer:

Explanation:

A: wrong. Mass and weight are different.

B: Wrong. The mass here and the mass on the moon are the same. The weight, which is Mass * the acceleration  is equal to weight.

C: Correct.

D: Wrong. Weight is not measured in Kg. Mass is.

Dennis, 60 feet in 0.5 seconds is the average speed of a pitcher's fastball. So, the correct option is (A).

Average speed is defined as the total distance covered by the object in a particular time interval. Average speed is a scalar quantity which is represented by the magnitude and not the direction.

The formula for average speed is found by calculating the ratio of the total distance traveled by the body to the time taken to cover that distance which is expressed as:

Average speed= Total distance travelled/ Time taken

The unit of the average speed is meter/second.

In the above given information, the pitcher's fastball average speed of Dennis is 120

Thus, Dennis, 60 feet in 0.5 seconds is the average speed of a pitcher's fastball. So, the correct option is (A).

Learn more about Average speed, here:

https://brainly.com/question/12322912

#SPJ5

Explanation:

When Two Samples of Water are Mixed, what Final Temperature Results?

Go to Mixing Two Amounts of Water: Problems 1 - 10

Go to calculating the final temperature when mixing water and a piece of metal

Worksheet #2

Back to Thermochemistry Menu

Example #1: Determine the final temperature when 32.2 g of water at 14.9 °C mixes with 32.2 grams of water at 46.8 °C.

This is problem 8a from Worksheet #2.

First some discussion, then the solution. Forgive me if the points seem obvious:

1) The colder water will warm up (heat energy "flows" into it). The warmer water will cool down (heat energy "flows" out of it).

2) The whole mixture will wind up at the SAME temperature. This is very, very important.

3) The energy which "flowed" out (of the warmer water) equals the energy which "flowed" in (to the colder water)

This problem type becomes slightly harder if a phase change is involved. For this example, no phase change. What that means is that only the specific heat equation will be involved

Solution Key Number One: We start by calling the final, ending temperature 'x.' Keep in mind that BOTH water samples will wind up at the temperature we are calling 'x.' Also, make sure you understand that the 'x' we are using IS NOT the Δt, but the FINAL temperature. This is what we are solving for.

The warmer water goes down from to 46.8 to x, so this means its Δt equals 46.8 − x. The colder water goes up in temperature, so its Δt equals x − 14.9.

That last paragraph may be a bit confusing, so let's compare it to a number line:



To compute the absolute distance, it's the larger value minus the smaller value, so 46.8 to x is 46.8 − x and the distance from x to 14.9 is x − 14.9.

These two distances on the number line represent our two Δt values:

a) the Δt of the warmer water is 46.8 minus x

b) the Δt of the cooler water is x minus 14.9

Solution Key Number Two: the energy amount going out of the warm water is equal to the energy amount going into the cool water. This means:

qlost = qgain

However:

q = (mass) (Δt) (Cp)

So:

(mass) (Δt) (Cp) = (mass) (Δt) (Cp)

With qlost on the left side and qgain on the right side.

Substituting values into the above, we then have:

(32.2) (46.8 − x)(4.184) = (32.2) (x − 14.9) (4.184)

Solve for x

Answer:

The first object is positively charged, and the second object is negatively charged. (D)

Explanation:

Answer:

Explanation:

Find the diagram of the scenario in the attached file.

From the diagram, the weight = mass * acceleration due to gravity

mass of sled = 30.0kg

acc. due to gravity = 9.81m/s

weight of the sled = 30.0*9.81

weight of the sled = 294.3N

The normal reaction force R in the diagram will be gotten by resolving the 12N force along the vertical

Ry = 12sin45°

R = 12 * 1/√2

R = 12√2/√2

R = 6√2 N

R = 8.49N ≈ 9.0N

The normal force exerted on the sled is approximately is 9.0N

Get the acceleration

Using the formula [tex]\sum Fx = ma_x[/tex] where;

m is the mass of the sled = 30.0kg

[tex]a_x[/tex] is the acceleration of the sled

[tex]\sum Fx = 8 + 12cos45^0\\\sum Fx = 8 + 12(0.7071)\\\sum Fx = 8 + 8.49\\\sum Fx = 16.49N[/tex]

Substitute into the formula;

[tex]a_x = \frac{\sum Fx }{m} \\ a_x = \frac{16.49}{30}\\ a_x = 0.55 m/s^2[/tex]

Hence the acceleration of the sled rounded to the nearest hundredth is 0.55m/s²

Answer:

omg, dude you're  giving out this many free points??

TYSM!!!!

God bless!

Answer:

7357.5N

Explanation:

Weight is the force used to pull a mass of object towards the centre of the Earth. The force is 9.81 N/kg.

so, Weight = mass × 9.81

the weight of the space probe should be: 750 × 9.81 = 7357.5 N

Answer:

7350N

Explanation:

Data

m-750kg

g-9.8

W=mg

=750 x 9.8

=7350N

Answer:

v=5.86 m/s

Explanation:

Given that,

Length of the string, l = 0.8 m

Maximum tension tolerated by the string, F = 15 N

Mass of the ball, m = 0.35 kg

We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :

[tex]F=\dfrac{mv^2}{r}[/tex]

v is the maximum speed

[tex]v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{15\times 0.8}{0.35}} \\\\v=5.86\ m/s[/tex]

Hence, the maximum speed of the ball is 5.86 m/s.

Answer:

Contact forces can be classified according to six types: tensional, spring, normal reaction, friction, air friction, and weight. Noncontact forces: Forces that take place when two objects do not touch. These forces can be classified according to three types: gravitational, electrical, and magnetic.

Explanation:

Answer:

1.66*10^-3N

Explanation:

Answer:

This chemical equation represents a synthesis reaction. This identification is based on the fact that the reactant chemical compounds combine to form a single chemical substance that is different from the initial chemical substance.

Explanation:

Answer:

a.42.5 km/h or .71 km/m

b. 66.7 km/h or 1.1repeating km/m

c.41.5 km/h or .69 km/m

Explanation:

These are all average speed problems, meaning you divide the distance traveled by the time [tex]s=\frac{d}{t}[/tex]

A. 170km/4hrs or 170km/240 mins

B. 100km/1.5hrs or 100km/90mins

C. The only tricky thing here is the question seems like it wants the speed for all of the information given, so you do include the one hour that he stops

270km/6.5 hrs or 270km/390 mins

Answer:

90 seconds

Explanation:

Answer:

metallic bond

Explanation:

Answer:

9) a = 25 [m/s^2], t = 4 [s]

10) a = 0.0875 [m/s^2], t = 34.3 [s]

11) t = 32 [s]

Explanation:

To solve this problem we must use kinematics equations. In this way we have:

9)

a)

[tex]v_{f}^{2} = v_{i}^{2}-(2*a*x)\\[/tex]

where:

Vf = final velocity = 0

Vi = initial velocity = 100 [m/s]

a = acceleration [m/s^2]

x = distance = 200 [m]

Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.

0 = (100)^2 - (2*a*200)

a = 25 [m/s^2]

b)

Now using the following equation:

[tex]v_{f} =v_{i} - (a*t)[/tex]

0 = 100 - (25*t)

t = 4 [s]

10)

a)

To solve this problem we must use kinematics equations. In this way we have:

[tex]v_{f} ^{2} = v_{i} ^{2} + 2*a*(x-x_{o})[/tex]

Note:  The positive sign of the equation means that the car increases his speed.

5^2 = 2^2 + 2*a*(125 - 5)

25 - 4 = 2*a* (120)

a = 0.0875 [m/s^2]

b)

Now using the following equation:

[tex]v_{f}= v_{i}+a*t\\[/tex]

5 = 2 + 0.0875*t

3 = 0.0875*t

t = 34.3 [s]

11)

To solve this problem we must use kinematics equations. In this way we have:

[tex]v_{f} ^{2} = v_{i} ^{2} + 2*a*(x-x_{o})[/tex]

10^2 = 2^2 + 2*a*(200 - 10)

100 - 4 = 2*a* (190)

a = 0.25 [m/s^2]

Now using the following equation:

[tex]v_{f}= v_{i}+a*t\\[/tex]

10 = 2 + 0.25*t

8 = 0.25*t

t = 32 [s]

Answer:

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Explanation: