Why does water that is frozen in an ice cube tray stay in the shape of a cube when it is

A wire is used as a heating element that has a resistance that is fairly independent of its temperature within its operating ran

dedylja [7]

Answer:

Double the current

Explanation:

The energy delivered by the heater is related to the current by the following relation:

E= $I^{2}R t$

let R * t = k ( ∴ R and t both are constant)

so E= k $I^{2}$

Now let:

E2= k I₂^2

E2= 4E

⇒ k I₂^2= 4* k $I^{2}$

Cancel same terms on both sides.

I₂^2= 4* $I^{2}$

taking square-root on both sides.

√I₂^2 = √4* I^2

⇒I₂= 2I

If we double the current the energy delivered each minute be 4E.

3 0

3 years ago

In the decomposition of water, why is twice as much hydrogen as oxygen formed?

astra-53 [7]

Answer:

I think

D : Hydrogen gas has two atom , and oxygen has one atom

7 0

2 years ago

A bowling ball with a momentum of 18kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 13k

Veronika [31]

Answer:

$14.98\ \text{kg m/s}$

$45.26^{\circ}$

Explanation:

$P_1$ = Initial momentum of the pin = 13 kg m/s

$P_i$ = Initial momentum of the ball = 18 kg m/s

$P_2$ = Momentum of the ball after hit

$55^{\circ}$ = Angle ball makes with the horizontal after hitting the pin

$\theta$ = Angle the pin makes with the horizotal after getting hit by the ball

Momentum in the x direction

$P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}$

Momentum in the y direction

$P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}$

$(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}$

The pin's resultant velocity is $14.98\ \text{kg m/s}$

$P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}$

The pin's resultant direction is $45.26^{\circ}$ below the horizontal or to the right.

4 0

3 years ago

LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.

Svetradugi [14.3K]

Answer:

i) E = 269 [MJ] ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

$E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.$

$W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]$

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

$E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]$