Which is the correct procedure to determine the daily mean temperature?.

How can you measure the distance an object has moved?

Naily [24]

You must observe the object twice.

-- Look at it the first time, and make a mark where it is.

-- After some time has passed, look at the object again, and
make another mark at the place where it is.

-- At your convenience, take out your ruler, and measure the
distance between the two marks.

What you'll have is the object's "displacement" during that period
of time ... the distance between the start-point and end-point.
Technically, you won't know the actual distance it has traveled
during that time, because you don't know the route it took.

8 0

3 years ago

You have a pulley 10.4 cm in diameter and with a mass of 2.3 kg. You get to wondering whether the pulley is uniform. That is, is

madreJ [45]

Answer:

Explanation:

Given

Diameter of Pulley=10.4 cm

mass of Pulley(m)=2.3 kg

mass of book $(m_0)=1.7 kg$

height(h)=1 m

time taken=0.64 s

$h=ut+frac{at^2}{2}$

$1=0+\frac{a(0.64)^2}{2}$

$a=4.88 m/s^2and [tex]a=\alpha r$

where $\alpha$ is angular acceleration of pulley

$4.88=\alpha \times 5.2\times 10^{-2}$

$\alpha =93.84 rad/s^2$

And Tension in Rope

$T=m(g-a)$

$T=1.7\times (9.8-4.88)$

T=8.364 N

and Tension will provide Torque

$T\times r=I\cdot \alpha$

$8.364\times 5.2\times 10^{-2}=I\times 93.84$

$I=0.463\times 10^{-2} kg-m^2$

$I_{original}=\frac{mr^2}{2}=0.31\times 10^{-2}kg-m^2$

Thus mass is uniformly distributed or some more towards periphery of Pulley

4 0

3 years ago

The state of a medium affect the speed of sound

fenix001 [56]

Answer:

no it can not effect the speed of sound not shure tho

5 0

2 years ago

How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C?

Svet_ta [14]

Answer:

229,098.96 J

Explanation:

mass of water (m) = 456 g = 0.456 kg

initial temperature (T) = 25 degrees

final temperature (t) = - 10 degrees

specific heat of ice = 2090 J/kg

latent heat of fusion =33.5 x 10^(4) J/kg

specific heat of water = 4186 J/kg

for the water to be converted to ice it must undergo three stages:

the water must cool from 25 degrees to 0 degrees, and the heat removed would be Q = m x specific heat of water x change in temp

Q = 0.456 x 4186 x (25 - (-10)) = 66808.56 J

the water must freeze at 0 degrees, and the heat removed would be Q = m x specific heat of fusion x change in temp

Q = 0.456 x 33.5 x 10^(4) = 152760 J

the water must cool further to -10 degrees from 0 degrees, and the heat removed would be Q = m x specific heat of ice x change in temp

Q = 0.456 x 2090 x (0 - (-10)) = 9530.4 J

The quantity of heat removed from all three stages would be added to get the total heat removed.

Q total = 66,808.56 + 152,760 + 9,530.4 = 229,098.96 J

6 0

3 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago