True or false if the temperature does not change but the mass of the object increases the thermal energy in the object decreases

A future use of space stations may be to provide hospitals for severely burned persons. It is very painful for a badly burned pe

inessss [21]

Answer:

1.5min

Explanation:

To solve the problem it is necessary to take into account the concepts related to Period and Centripetal Acceleration.

By definition centripetal acceleration is given by

$a_c = \frac{V^2}{r}$

Where,

V = Tangencial velocity

r = radius

With our values we know that

$a_c = \frac{V^2}{r}$

$\frac{V^2}{r} = \frac{1}{10}g$

Therefore solving to find V, we have:

$V = \sqrt{\frac{1}{10}g*r}$

$V = \sqrt{\frac{9.81*200}{10}}$

$V = 14m/s$

For definition we know that the Time to complete are revolution is given by

$t = \frac{Perimeter}{Speed}$

$t = \frac{2\pi R}{V}$

$t = \frac{2\pi * 200}{14}$

$t = 1.5min$

6 0

3 years ago

_____ is very corrosive and can cause rusting so metal tanks shouldn't be used

bagirrra123 [75]

Answer:

Iron

Explanation:

5 0

3 years ago

A roller coaster starts at the top of a hill of height h, goes down the hill, and does a circular loop of radius r before contin

jeka94

a) See free-body diagram in attachment

b) Net force in the y-direction: $F_y=mg+N$ [/tex]

c) The velocity at which the roller coaster will fall is $[tex]v=\sqrt{gr}$ [/tex]

d) The speed of the roller coaster must be 17.1 m/s

e) The roller coaster should start from a height of 90 m

f) The roller coaster should start from a height of 100 m

Explanation:

a)

See the free-body diagram in attachment. There are only two forces acting on the roller coaster at the top of the loop:

The weight of the roller coaster, acting downward, indicated by $mg$ (where m is the mass of the roller coaster and g is the acceleration of gravity)
The normal reaction exerted by the track on the roller coaster, acting downward, and indicated with N

The two forces are represented in the diagram as two downward arrows (the length is not proportional to their magnitude, in this case)

b)

Since there are only two forces acting on the roller coaster at the top of the loop, and both forces are acting downward, then we can write the vertical net force as follows (we take downward as positive direction):

$F_y = mg + N$

where

mg is the weight

N is the normal reaction

Since the roller coaster is in circular motion, this net force must be equal to the centripetal force, therefore

$m\frac{v^2}{r}=mg+N$

where v is the speed of the car at the top of the loop and r is the radius of the loop.

c)

For this part of the problem, we start from the equation written in part b)

$m\frac{v^2}{r}=mg+N$

where the term on the left represents the centripetal force, and the terms on the right are the weight and the normal reaction.

We now re-arrange the equation making v (the speed) as the subject:

$v=\sqrt{gr+\frac{Nr}{m}}$

However, the velocity at which the roller coaster will fall is the velocity at which the normal reaction becomes zero (the roller coaster loses contact with the track), so when

N = 0

And as a result, the minimum velocity of the cart is

$v=\sqrt{gr}$

d)

In this part, we are told that the radius of the loop is

r = 30 m

And the mass of the cart is

m = 50 kg

Moreover, the acceleration of gravity is

$g=9.8 m/s^2$

We said that the minimum velocity that the cart must have in order not to fall at the top is

$v=\sqrt{gr}$

And substituting, we find

$v=\sqrt{(9.8)(30)}=17.1 m/s$

e)

According to the law of conservation of energy, the initial gravitational energy of the roller coaster at the starting point must be equal to the sum of the kinetic energy + gravitational potential energy at the top of the loop, therefore:

$mgh = \frac{1}{2}mv^2 + mg(2r)$

where

h is the initial height at the starting point

(2r) is the height of the roller coaster at the top of the loop

We can re-arrange the equation making h the subject,

$h=\frac{v^2}{2g}+2r$

And substituting the minimum speed of the cart,

$v=\sqrt{gr}$

this becomes

$h=r+2r=3r$

And since r = 30 m, we find

$h=3(30)=90 m$

f)

In this case, 10% of the initial energy is lost during the motion of the roller coaster. We can rewrite the equation of the previous part as

$0.90mgh = \frac{1}{2}mv^2 + mg(2r)$

Because only 90% (0.90) of the initial energy is converted into useful energy (kinetic+potential) when the cart reaches the top of the loop.

Re-arranging the equation, this time we get

$h=\frac{\frac{v^2}{2g}+2r}{0.90}$

Again, by substituting $v=\sqrt{gr}$ , we get

$h=\frac{3r}{0.90}$

And therefore, the new initial height must be

$h=\frac{3(30)}{0.9}=100 m$

Learn more about kinetic and potential energy:

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#LearnwithBrainly

4 0

3 years ago

Why do blades come in different lengths in a jig saw sander?

trasher [3.6K]

Answer:

To determine the minimum blade length, add 1" to the workpiece thickness. One type of material, and some materials can be cut by more than one type of blade. No matter the material, there's likely a jigsaw blade designed specifically for. Armed with the right blade, follow these pointers to make your work go (and cut) .

Explanation:

6 0

3 years ago

In the film, 2001: A Space Odyssey, astronauts travel through space and experience artificial gravity created by a spinning, cyl

katrin2010 [14]

Artificial gravity is created by rotating the cylinder

So here we can say that the acceleration is due to centripetal acceleration

So here we know that the formula of centripetal acceleration is given by

$a_c = \frac{v^2}{R}$