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2 years ago

13

From the point of view of water when the temperature of water increases from room temperature to 90°C the process of heating the

water is

1 answer:

Andrew [12]2 years ago

5 0

its boiling i believe

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Sam is observing the velocity of a car at different times. After two hours, the velocity of the car is 54 km/h. After four hours

MariettaO [177]

With that information you can only suppose a uniformly accelerated motion. This is, acceleration is constant.

Then, acceleration = change in velocity / change in time = (58 -54)km/h / 2 h = 4km/h / 2 h = 2 km/h^2

Then the equation for velocity, V is

V = Vo + a*t = Vo + 2 (km/h^2) * t = Vo + 2t

Vo is the initial velocity, which you can find using V = 54km/h and t = -2

Vo = V after 2 hours - a*(2hours) = 54km/h - 2(km/h^2)*2h = 54km/k - 4km/h = 50km/h

Then, the equation is: V = 50 km/h + 2t

Valid for constant acceleration.

5 0

2 years ago

Which of newtons laws is illustrated by a squid moving forward by shooting water out behind it? Why?

Dimas [21]

I believe its newtons 3rd law for every action there is an equal but opposite reaction since the squid is moving foward by shooting the water its pushing the squid back as its reaction. Hope this helped !

5 0

3 years ago

How far does a car travel in 10 s if its initial velocity is 3 m/s and its acceleration

victus00 [196]

To learn more about Acceleration please visit -
brainly.com/question/13563385
#SPJ4

4 0

1 year ago

A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel

Dafna1 [17]

Answer:

$1.69\cdot 10^{10}J$

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

$E=-G\frac{mM}{2r}$

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

$M=5.98\cdot 10^{24}kg$ is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

$r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m$

So the initial total energy is

$E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J$

When the satellite hits the ground, it is now on Earth's surface, so

$r=R=6370 km=6.37\cdot 10^6 m$

so its gravitational potential energy is

$U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J$

And since it hits the ground with speed

$v=1.90 km/s = 1900 m/s$

it also has kinetic energy:

$K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J$

So the total energy when the satellite hits the ground is

$E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J$

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

$\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J$

8 0

2 years ago

The magnetic field is 0.0 2M from the wire is 0.1 T what is the magnitude of the magnetic field 0.0 1M from the same wire

Kisachek [45]

Answer:

0.2 T

Explanation:

Magnetic field is inversely proportional to the distance from wire since the distance is halved therefore magnetic field will be doubled.

4 0

2 years ago