According to roosevelt, what are the for freedom to which every one in the world is entitle?

How will the force of friction affect a wood block being pushed over a table?

olasank [31]

Answer:

D

friction acts in the opposite direction of motion but does not affect the motion of the object

5 0

2 years ago

to start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 290 m/s at 53.0° above the ho

kap26 [50]

So this is easy to calculate when you split the velocity into x and y components. The x component is going to equal cos(53) * 290 and the y component is going to equal sin(53)*290.

The x location therefore is 290*cos(53)*35 = 6108.4m

The y location needs to factor in the downwards acceleration of gravity too, which is 9.81m/s^2. We need the equation dist. = V initial*time + 0.5*acceleration*time^2.

This gives us d=290*sin(53)*35 + (0.5*-9.81*35^2)=2097.5m

So your (x,y) coordinates equals (6108.4, 2097.5)

5 0

3 years ago

This is a velocity versus time graph of a car starting from rest. If the area under the line is 10 meters, what is the correspon

adelina 88 [10]

So the area under a velocity time graph is distance or displacement, if you have done calculus yet you will understand that if you take the integral of a velocity function then you end up with displacement. Thats for later understanding however.

So this appears to be a right triangle so we can find the area of a triangle as:

0.5bh = A

Since our area is 10 meters lets alter our formula a bit to fit the situation:

Our base here is time and our height is velocity so:

0.5tv = Δx

So we can read off the graph that our velocity at the end, or our final velocity appears to be near 2.0 m/s

So we have v, and Δx so lets isolate for time by dividing by v and 0.5

t = Δx / 0.5v

Now lets plug all that in:

t = 10 / 0.5(2)

t = 10 seconds

Hope this helped!

8 0

3 years ago

For an object with a given mass on Earth, calculate the weight of the object with the mass equal in magnitude to the number repr

leonid [27]

Answer: The weight of the object is 29.4 N

Explanation:

To calculate the weight of the object, we use the equation:

$W=m\times g$

where,

m = mass of the object = 3 kg

g = acceleration due to gravity = $9.8m/s^2$

Putting values in above equation, we get:

$W=3kg\times 9.8m/s^2\\\\W=29.4N$

Hence, the weight of the object is 29.4 N

6 0

3 years ago

For each of the following pairs of gas properties, describe the relationship between the properties, describe a simple system th

asambeis [7]

Answer:

For each pair of properties of a gas, the relationships are (see the explanation for the description of the systems):

(a) Volume and pressure: The relationship between them is inversely proportional.
(b) Pressure and temperature: They have a directly proportional relationship.
(c) Volume and temperature: They relationship is directly proportional.
(d) Number of gas particles and pressure: The relationship is directly proportional between them.

Explanation:

1. Volume and pressure (temperature and amount of particles constant):

They have an inversely proportional relationship, because if volume is reduced, the pressure increases, or if the volume increases, the pressure decreases.

A simple system could be one similar to the one used by Boyle to test this relationship:

Seal the short extreme of a translucent J tube. It could be glass or plastic.
Put some water on it. As much as needed to have both sides of the tube filled.
Using a syringe, and a flexible small tube,inject a determined volume of air in the bottom in a way that the bubble is trapped in the seal side of the J tube.
Then if more water is added to the tube, it will increase the pressure (from the pressure definition is possible to in the trapped air, and is possible to measure the compression of the air bubble. The same is possible if using the syringe, and the flexible tube, some water is removed, and the increasing of volume could be observed.

2. Pressure and temperature (volume and amount of particles of the gas remains constant)

They have a directly proportional relationship, because if temperature is reduced, the pressure decreases, or if the temperature increases, the pressure would increase, also.

A simple system to show this is two cans of soda.

The can is rigid, so the volume is always constant, and the amount of gas inside the soda is the same.
Put one can under the sun, and the other in the cooler.
After a while, take it out the can in the cooler, and open both cans.
The one that was under the sun will "explode", in other words, it will liberate a lot of foam of gas and soda, meaning that the pressure inside the can was high.
The one that was in te cooler, won't liberate any foam, meaning that the pressure was low.

3. Volume and temperature (pressure and amount of particles of the gas remains constant)

They have a directly proportional relationship, because if temperature is reduced, the pressure decreases, or if the temperature increases, the pressure will increase, also.

A simple system to show this is a party balloon.

Fill the party balloon with some air, not enough to be close to explode, but enough to have it of a medium size. Tie the filling hole of the balloon.

The air inside the balloon would be at the same pressure than the atmosphere around it, so always will be at this pressure, and the close hole ensure that it has always the same amount of air inside.

Now is possible to use some heat source, for example as a hair dryer to increase the temperature of the balloon and its contents. The size of the balloon will increase. Then using water is possible to cool it down and watch how its size decreases.

4. Number of gas particles and pressure (volume and temperature of gas remains constant)

They have a directly proportional relationship, because if the amount of gas particles is reduced, the pressure decreases, or if quantity of gas particles increases, the pressure will increase, also.

A simple system to show this would be a bicycle tire:

The tire is rigid, so its volume is essentially constant, and the temperature would remains the same if not moving or driving it.

Using a tire gauge, it is possible to know the manometric pressure inside the tire, that is the difference between the actual pressure inside the tire and the atmospheric pressure.

Then each time that using an air pump some air is injected in the tire, it si possible to check the pressure inside it using the gauge, and observe how is increasing.

Also, is possible to open the valve, to allow some air to escape, then use the gauge to observe how the pressure decreases.

7 0

4 years ago