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3 years ago

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How could you decrease the total energy of a system consisting of a jar full of water? O A. Heat the jar in a microwave oven. O

B. Place a lid on the jar and shake it. O C. Place the jar in a freezer. O D. Place the jar in sunlight.

2 answers:

denpristay [2]3 years ago

8 0

Answer: C

Explanation: when you freeze the water molecules can’t really move if they are frozen therefore creating less energy.

Vaselesa [24]3 years ago

6 0

Answer:

A

Explanation:

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Represent a vector of 100 N in North-East direction

andreyandreev [35.5K]

Answer:

please find the attachment to this question.

Explanation:

In this question, we represent the 100N in the North-East direction, but first, we define the vector representation:

It is generally represented through arrows, whose length and direction reflect the magnitude and direction of the arrow points. In this, both size and direction are necessary because the magnitude of a vector would be a number that can be compared to one vector.

Please find the attachment:

8 0

3 years ago

A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru

r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

= 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

$s = u t +\dfrac{1}{2}at^2$

$-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2$

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

$t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}$

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

6 0

3 years ago

What is the current in a wire of radius R = 2.02 mm if the magnitude of the current density is given by (a) Ja = J0r/R and (b) J

Sloan [31]

Explanation:

For this problem we have to take into account the expression

J = I/area = I/(π*r^(2))

By taking I we have

I = π*r^(2)*J

(a)

For Ja = J0r/R the current is not constant in the wire. Hence

$I(r) = \pi r^{2} J(r) = \pi r^{2} J_{0}r/R = \pi r^{3} (3.74*10^{4}A/m^{2})/(2.02*10^{-3}m)$

and on the surface the current is

$I(R) = \pi r^{2} J(R) = \pi r^{2} J_{0}R/R = \pi(2.02*10^{-3})^{2} (3.74*10^{4}) = 0.47 A$

(b)

For Jb = J0(1 - r/R)

$I(r)=\pi r^{2}J(r) =\pi r^{2} J_{0}(1 - r/R)=\pi r^{2}J_{0}(1-\frac{r}{2.02*10^{-3}} )$

and on the surface

$I(R)=\pi r^{2}J_{0}(1-R/R)=\pi r^{2}J_{0}(1-1)= 0$

(c)

Ja maximizes the current density near the wire's surface

Additional point

The total current in the wire is obtained by integrating

$I_{T}=\pi\int\limits^R_0 {r^{2}Ja(r)} \, dr = \pi \frac{J_{0}}{R}\int\limits^R_0 {r^{3}} \ dr =\pi \frac{J_{0}R^{4}}{4R}=\frac{1}{4}\pi J_{0}R^{3}=2.42*10^{-4} A$

and in a simmilar way for Jb

$I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]$

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

7 0

3 years ago

A satellite is to be launched into an orbit of radius,r Show that v = 2gr, where V is the

Snezhnost [94]

Explanation:

We start by using the conservation law of energy:

$\Delta{K} + \Delta{U} = 0$

or

$\dfrac{1}{2}mv^2 - G\dfrac{mM}{r} = 0$

Simplifying the above equation, we get

$v^2 = 2G\dfrac{M}{r}$

We can rewrite this as

$v^2 = 2\left(G\dfrac{M}{r^2}\right)r$

Note that the expression inside the parenthesis is simply the acceleration due to gravity $g$ so we can write

$v^2 = 2gr$

where $v$ is the launch velocity.

6 0

3 years ago

The temperature of a smelting furnace is found to be 2000 ℃. Find the

IRISSAK [1]

Answer:

a) The equivalent temperature of 2000 ºC is 3632 ºF.

b) The equivalent temperature of 2000 ºC is 4091.67 R.

c) The equivalent temperature of 2000 ºC is 2273.15 K.

d) The equivalent temperature of 2000 ºC is 1600 ºRe.

Explanation:

a) The equivalent temperature on the Fahrenheit scale is defined by the following formula:

$T_{F} = \frac{9}{5}\cdot T_{C}+32$ (Eq. 1)

Where:

$T_{C}$ - Temperature, measured in degrees Celsius.

$T_{F}$ - Temperature, measured in degrees Fahrenheit.

If we know that $T_{C} = 2000\,^{\circ}C$ , then the temperature is:

$T_{F} = \frac{9}{5}\cdot (2000\,^{\circ}C)+32\,^{\circ}F$

$T_{F} = 3632\,^{\circ}F$

The equivalent temperature of 2000 ºC is 3632 ºF.

b) From result in a) we determine the equivalent temperature on the Rankine scale by using the following formula:

$T_{R} = T_{F}+459.67$ (Eq. 2)

Where:

$T_{F}$ - Temperature, measured in degrees Fahrenheit.

$T_{R}$ - Temperature, measured in Rankine.

If we know that $T_{F} = 3632\,^{\circ}F$ , then the temperature is:

$T_{R} = 3632\,^{\circ}F+459.67\,R$

$T_{R} = 4091.67\,R$

The equivalent temperature of 2000 ºC is 4091.67 R.

c) The equivalent temperature on the absolute scale is calculated by using this expression:

$T_{K} = T_{C}+273.15$ (Eq. 3)

Where:

$T_{C}$ - Temperature, measured in degrees Celsius.

$T_{K}$ - Temperature, measured in Kelvin.

If we know that $T_{C} = 2000\,^{\circ}C$ , then the temperature is:

$T_{K} = 2000\,^{\circ}C+273.15\,K$

$T_{K} = 2273.15\,K$

The equivalent temperature of 2000 ºC is 2273.15 K.

d) The equivalent temperature on the Réaumur scale is calculated by applying this expression:

$T_{Re} = \frac{4}{5}\cdot T_{C}$ (Eq. 4)

Where:

$T_{C}$ - Temperature, measured in degrees Celsius.

$T_{Re}$ - Temperature, measured in degrees Réaumur.

If we know that $T_{C} = 2000\,^{\circ}C$ , then the temperature is:

$T_{Re} = \frac{4}{5}\cdot (2000\,^{\circ}C)$

$T_{Re} = 1600\,^{\circ}Re$

The equivalent temperature of 2000 ºC is 1600 ºRe.

5 0

3 years ago