100 g of water at 25 °C is poured into an insulating cup. 50 g of ice at 0 °C is added to the

Arrange the stars based on their temperature. Begin with the coolest star, and end with the hottest star.

Slav-nsk [51]

Answer:

Uranus, Pluto, Neptune, Saturn , Jupiter, mars, Venus ,mercury and sun

4 0

2 years ago

A diver jumps from a high platform and stays in the air for 1 second. How high was the platform?

ivolga24 [154]

Answer: Depends

Explanation:

Depends on how much the diver weighs.

6 0

2 years ago

At the same temperature, two wires made of pure copper have different resistances. The same voltage is applied at the ends of ea

Semmy [17]

Answer:

Explanation:

As we know that the resistance of the wire is directly proportional to the length of wire and inversely proportional to the area of crossection of the wire.

As the material is copper for both the wires so the resistivity is same and the voltage is also same. As their resistance is different it means either length is different or the area of crossection is different.

6 0

2 years ago

Science not has only achievements but also challenges justify with an example

statuscvo [17]

For example science has discovered many amazing things like atoms,molecules which led to the development of nuclear reactors etc. Challenges would be like trying to prove theories like evolution for 100 of years trying to discover all the stuff we still don’t know exists all the mysteries of life. A lot of science is still theory we can’t prove some things until we have more advanced technology etc.

3 0

2 years ago

Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to

Andrej [43]

(a) $3.1\cdot 10^7 J$

The total mechanical energy of the space probe must be constant, so we can write:

$E_i = E_f\\K_i + U_i = K_f + U_f$ (1)

where

$K_i$ is the kinetic energy at the surface, when the probe is launched

$U_i$ is the gravitational potential energy at the surface

$K_f$ is the final kinetic energy of the probe

$U_i$ is the final gravitational potential energy

Here we have

$K_i = 5.0 \cdot 10^7 J$

at the surface, $R=3.3\cdot 10^6 m$ (radius of the planet), $M=5.3\cdot 10^{23}kg$ (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

$U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J$

At the final point, the distance of the probe from the centre of Zero is

$r=4.0\cdot 10^6 m$

so the final potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J$

So now we can use eq.(1) to find the final kinetic energy:

$K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J$

(b) $6.3\cdot 10^7 J$

The probe reaches a maximum distance of

$r=8.0\cdot 10^6 m$

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

$K_f = 0$

At that point, the gravitational potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J$

So now we can use eq.(1) to find the initial kinetic energy:

$K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J$

3 0

2 years ago