A given container of 5.0 kg of water maintains a constant temperature over 24 hours. If its mass doesn't change, what statements would support identifying it as a closed system.

I. Only energy was exchanged with its surroundings.
II. Its mass was constant because the water lost to evaporation was replaced with an equal volume of water.
III. It didn't exchange matter or energy with its environment.
IV. External work was used to maintain its internal energy.

A) III
B) l and III
C) II and IV
D) I and IV​

If the output work of a simple machine equals the input work, the machine is said to have 100% efficiency.

[tex]\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘[/tex]

Answer:

four glorps

Explanation:

We know :

[tex]$y=v_{0y}t + \frac{1}{2}a_yt^2$[/tex]

[tex]$\Rightarrow -1 \text{glorp} = 0 - \frac{g}{2} \times (1 g\text{ gleep})^2$[/tex]

[tex]$\Rightarrow 1 \text{ glorp}= \frac{g}{2} (1 \text{ gleep})^2$[/tex]   .............(i)

Now, t' = 2 gleep

[tex]$y=v_{0y}t + \frac{1}{2}a_yt^2$[/tex]

  [tex]$=0+ \frac{-g}{2} (2 \text{ gleep})^2$[/tex]

  [tex]$=-\frac{4g}{2}(2 \text{ gleep})^2$[/tex]

 [tex]$=4\left[\frac{-g}{2} (\text{gleep})^2\right]$[/tex]

 = 4 (-1 gleep)     (From (i))

So, |y| = 4 glorp

[tex]\huge{ \mathfrak{  \underline{ Answer }\:  \:  ✓ }}[/tex]

Copper wires are used as connecting wires in circuits because they offer very less resistance.

__________________________________________________________

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Answer:

It is example of Closed system

Answer:

(a) Fₓ = 0 N

[tex]F_y[/tex] = 9.08 N

(b)

(a) Fₓ = 0 N

[tex]F_y[/tex] = 9.08 N

(c) [tex]F_y[/tex] = 0 N

Fₓ = 9.08 N

Explanation:

The magnitude of the force will remain the same in each case, which is given as follows:

F = ma (Newton's Second Law)

where,

F = force = ?

m = mass = 4.54 kg

a = acceleration = 2 m/s²

Therefore,

F = (4.54 kg)(2 m/s²)

F = 9.08 N

Now, we come to each scenario:

(a)

Since the motion is in the vertical direction. Therefore the magnitude of the force in x-direction will be zero:

Fₓ = 0 N

For upward direction the force will be positive:

[tex]F_y[/tex] = 9.08 N

(b)

Since the motion is in the vertical direction. Therefore the magnitude of the force in x-direction will be zero:

Fₓ = 0 N

For upward direction the force will be negative:

[tex]F_y[/tex] = - 9.08 N

(c)

Since the motion is in the horizontal direction. Therefore the magnitude of the force in y-direction will be zero:

[tex]F_y[/tex] = 0 N

Fₓ = 9.08 N

Answer:

1. F = 3400 N = 3.4 KN

2. [tex]K.E_f=92832.6\ J = 92.83\ KJ[/tex]

3. v = 14.9 m/s

Explanation:

1.

First, we will calculate the acceleration of Pam by using the third equation of motion:

[tex]2as = v_f^2-v_i^2[/tex]

where,

a = acceleration = ?

s = distance = 27.3 m

vf = final speed = 62 m/s

vi = initial speed = 0 m/s

Therefore,

[tex]2a(27.3\ m) = (62\ m/s)^2-(0\ m/s)^2\\\\a = 70.4\ m/s^2[/tex]

Now, we will calculate the force by using Newton's Second Law of Motion:

F = ma

F = (48.3 kg)(70.4 m/s²)

F = 3400 N = 3.4 KN

2.

Final kinetic energy is given as:

[tex]K.E_f = \frac{1}{2}mv_f^2\\\\K.E_f = \frac{1}{2} (48.3\ kg)(62\ m/s)^2[/tex]

[tex]K.E_f=92832.6\ J = 92.83\ KJ[/tex]

3.

According to the law of conservation of energy:

[tex]Potential\ Energy\ at\ top = Kinetic\ Energy\ at\ bottom\\mgh = \frac{1}{2}mv_2 \\\\v = \sqrt{2gh}[/tex]

where,

v = speed at bottom = ?

g = acceleration due to gravity = 9.81 m/s²

h = height at top = 11.3 m

Therefore,

[tex]v = \sqrt{(2)(9.81\ m/s^2)(11.3\ m)}[/tex]

v = 14.9 m/s

Rt = 3 ohms

Explanation:

Let R1 = 4-ohm resistor

R2 = 12-ohm resistor

For 2 resistors connected in parallel, the total resistance Rt is given by

1/Rt = 1/R1 + 1/R2

or

Rt = R1R2/(R1 + R2)

= (4 ohms)(12 ohms)/(4 ohms + 12 ohms)

= 48 ohms^2/16 ohms

= 3 ohms

Answer:

5 x 10⁻⁵ T

Explanation:

Magnetic field at a point d distance away from a current carrying wire having current of I

B = 10⁻⁷ x 2I / d

Here magnetic field due to current of 20 A

B₁ = 10⁻⁷ x 2 x 20 / 10 x 10⁻²

= 4 x 10⁻⁵ T .

Here magnetic field due to current of 5 A

B₂ = 10⁻⁷ x 2 x 5 / 10 x 10⁻²

=  10⁻⁵ T .

These magnetic fields are acting in the same direction

Total magnetic field

B = B₁ + B₂

= 4 x 10⁻⁵ + 10⁻⁵ = 5 x 10⁻⁵ T .

Answer:

The right approach is "8.1 m/s". A further explanation is provided below.

Explanation:

According to the table,

Speed of Boat

= [tex]\frac{s}{t}[/tex]

[tex]V_b=\frac{140}{20}[/tex]

[tex]V_a = 4 \ m/s[/tex]

[tex]V_B = 7 \ m/s[/tex]

Now,

⇒  [tex](V_{relative})^2 = (7)^2+(4)^2[/tex]

or,

⇒  [tex](V_r)^2=49+16[/tex]

              [tex]=65[/tex]

         [tex]V_r=\sqrt{65}[/tex]

              [tex]=8.1 \ m/s[/tex]

Answer:

check out of phase

Explanation:

this is my answer

Answer:

See Explanation

Explanation:

Sound is a mechanical wave. A mechanical wave requires a material medium for propagation. This means that sound waves must be carried in air. If there are no air molecules, sound waves can not travel.

When air is gradually removed from the jar by the pump, the sound intensity from the bell gradually decreases owing to the fact that air which is the medium through which sound waves are propagated is gradually being removed from the jar.

You really don't need any help on the question.  It's all right there in the picture.  What you need help with is the answer.

The number of times the same thing happens each second is called its "frequency".  The frequency of the dragonfly's flaps is 477 Hz.  (If you're close enough to the dragonfly, you can hear the wings flapping.  It sounds like a raspy tone with a frequency of 477 Hz.)

The "period" is just the length of time it takes to happen once.  That length of time is just  (1 / frequency) .

The dragonfly flaps its wings once every  (1 / 477 Hz) = 0.0021 second (C)

Answer:

[tex]175\ \text{N/C}[/tex]

Explanation:

[tex]E_1[/tex] = Initial electric field = 3.5 N/C

[tex]E_2[/tex] = Final electric field

[tex]r_1[/tex] = Initial distance = 10 m

[tex]r_2[/tex] = Final distance = 20 cm

Electric field is given by

[tex]E=\sqrt{\dfrac{2P}{\pi r^2c\varepsilon_0}}[/tex]

So,

[tex]E\propto \dfrac{1}{r}[/tex]

[tex]\dfrac{E_2}{E_1}=\dfrac{r_1}{r_2}\\\Rightarrow E_2=E_1\dfrac{r_1}{r_2}\\\Rightarrow E_2=3.5\dfrac{10}{0.2}\\\Rightarrow E_2=175\ \text{N/C}[/tex]

The electric field amplitude at the required point is [tex]175\ \text{N/C}[/tex].

Answer:

static friction=0.126

Answer:

yong may check po ayon yong sagot

Answer:

230N-m

Explanation:

Given data

Power P= 75kW

Frequency f= 60Hz

w= 2πf

w=2*π*60

w=120π rad/s

Power= T*w

Power= IV= 415*I

75*10^3= 415*I

I=180.72 A

V= 480V

Power= 180.72*480

Power= 86.746kW

Power= 86.746= T*120π

T= 120π/86.746

T= 230N-m

Answer:

A

Explanation:

because uhhh I'm doing the same test as you kinda LOL

Complete question is;

A small car of mass m and a large car of mass 2m drive around a highway curve of radius R. Both cars travel at the same speed (v). The

centripetal acceleration (Grad) of the large car is the centripetal acceleration of the small car. How does the Force of the small car FS compare to the force of the large car FL as they round the curve.

four times

twice

half

equal to

Answer:

Half

Explanation:

Formula for centripetal force is given as;

F = mv²/R

Where;

v is velocity

R is radius

Now, centripetal acceleration is given by;

a = v²/R

Since they both travel with the same velocity V and radius remains the same, we can say that;

F = ma

For the small car;

FS = ma

For the big car;

FL = 2ma

This means the force of the small car is half of that of the Large car

Thus;

FS = ½FL

Answer:

Cellular respiration takes place in the cells of all organisms. It occurs in autotrophs such as plants as well as heterotrophs such as animals. Cellular respiration begins in the cytoplasm of cells. It is completed in mitochondria

Explanation:

Cellular respiration takes place in the cells of all organisms. It happening in autotrophs such as plantas as well as heterotrophs such as animals. Cellular respiration starts in the cytoplasm of cells.

It is finished in mitochondria.

Answer:

Height of the door = 2m = 2000 cm

1 in = 2.54 cm

So 1 cm = 1/2.54 in

2000 cm = 200000/ 254

=

787.401574803

So no.c is correct

The door is  78.7 inch tall. Hence, option (A) is correct.

Any arbitrarily selected and widely used reference standard for length measurement is referred to as a unit of length. The metric system, which is adopted by every nation on earth, is the most widely utilized in modern times.

The American customary units are also in use in the United States. In the UK and several other nations, British Imperial units are still used sometimes. There are SI units and non-SI units in the metric system.

Given that: the height of the door is = 2 meter

= 2*100 centimeter

= 200 centimeter.

There are 2.54 centimeter in 1 inch.

Hence,  the height of the door is = 2 meter  = 200 centimeter

= (200/2.54) inch

= 78.7 inch.

The door is 78.7 inch tall.

Learn more about length here:

https://brainly.com/question/17139363

#SPJ2

Answer:

H = 30.8 m

Explanation:

Its a lot to write down so I'm not going to but if you want the answer just reply to this and say you need the answer. Thanks.

Answer:

Explanation:

Let after time t , Tina catches up David .

Distance travelled by them are equal ,

Distance travelled by Tina

s = ut + 1/2 a t²

= .5 x 2.10 t²

= 1.05 t²

Distance travelled by David

= 30 t ( because of uniform velocity )

1.05 t² = 30t

t = 28.57 s

Distance travelled by Tina

= 1/2 a t²

= .5 x 2.10 x 28.57²

= 857 m approx.

Answer: [tex]857\ m[/tex]

Explanation:

Given

Speed of David car [tex]v=30\ m/s[/tex]

Tina begins to accelerate [tex]2.1\ m/s^2[/tex] after David pass the tina

Suppose it took t time for tina to catch David

Distance traveled by David in t time

[tex]\Rightarrow s_d=30\times t[/tex]

Using the equation of motion to get the distance of Tina is

[tex]s_t=ut+\dfrac{1}{2}at^2\\\\s_t=0+\dfrac{1}{2}\times 2.1t^2[/tex]

now, [tex]s_d=s_t[/tex]

[tex]30t=\dfrac{2.1}{2}t^2\\\\\Rightarrow 2.1t^2-60t=0\\\Rightarrow t(2.1t-60)=0\\\Rightarrow t=0,28.57\ s[/tex]

Neglecting [tex]t=0[/tex]

Distance traveled by tina in [tex]28.57\ s[/tex] is

[tex]s_t=\dfrac{1}{2}\times 2.1\times (28.57)^2\\\\s_t=857.057\approx 857\ m[/tex]

Answer:

S = Vo t * 1/2 a t^2     equation for distance traveled

w = w0 t + 1/2 a t^2    equivalent circular equation where w equals omega

w = 5.98 * 27.5  + a (27.5)^2/ 2

a = (w2 - w1) / t = -5.98 / 27.5 = -.217

w = 5.98 * 27.5 - 1/2 * .217 * 27.5^2 = 82.4

Since 1 Rev = 2 pi  radians      

Rev = 82.4 / 2 * pi = 13.1 Rev

Explanation:

x-component:

Vx = Vcos(theta)

= (92.5 m)cos(32.0)

= 78.4 m

Answer:

-78.4

Explanation:

For acellus students

[tex]\huge{ \mathfrak{ \underline{question : }}}[/tex]

A body weighing 250 grams was dropped from a helicopter flying at an altitude of 100 meters. Determine the potential energy of this body.

[tex] \huge{ \mathfrak{given : }}[/tex]

[tex]\huge{ \mathfrak{ \underline{ Answer \: \: ✓ }}}[/tex]

[tex] \boxed{ \mathbf{potential \: \: energy = mgh}}[/tex]

potential energy = 250 joules

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The question is incomplete. The complete question is :

In your job as a mechanical engineer you are designing a flywheel and clutch-plate system. Disk A is made of a lighter material than disk B, and the moment of inertia of disk A about the shaft is one-third that of disk B. The moment of inertia of the shaft is negligible. With the clutch disconnected, A is brought up to an angular speed ?0; B is initially at rest. The accelerating torque is then removed from A, and A is coupled to B. (Ignore bearing friction.) The design specifications allow for a maximum of 2300 J of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy?

Solution :

Let M.I. of disk A = [tex]$I_0$[/tex]

So, M.I. of disk B =  [tex]$3I_0$[/tex]

Angular velocity of A = [tex]$\omega_0$[/tex]

So the kinetic energy of the disk A = [tex]$\frac{1}{2}I_0\omega^2$[/tex]

After coupling, the angular velocity of both the disks will be equal to ω.

Angular momentum will be conserved.

So,

[tex]$I_0\omega_0 = I_0 \omega + 3I_0 \omega$[/tex]

[tex]$I_0\omega_0 = 4I_0 \omega$[/tex]

[tex]$\omega = \frac{\omega_0}{4}$[/tex]

Now,

[tex]$K.E. = \frac{1}{2}I_0\omega^2+ \frac{1}{2}3I_0\omega^2$[/tex]

[tex]$K.E. = \frac{1}{2}I_0\frac{\omega_0^2}{16}+ \frac{1}{2}3I_0\frac{\omega_0^2}{16}$[/tex]

[tex]$K.E. = \frac{1}{2}I_0\omega_0^2 \left(\frac{1}{16}+\frac{3}{16}\right)$[/tex]

[tex]$K.E. = \frac{1}{2}I_0\omega_0^2\times \frac{1}{4}$[/tex]

[tex]$\Delta K = \frac{1}{2}I_0\omega_0^2 - \frac{1}{2}I_0\omega_0^2 \times \frac{1}{4} $[/tex]

[tex]$2300=\frac{3}{4}\left(\frac{1}{2}I_0\omega_0^2\right)$[/tex]

[tex]$\frac{1}{2}I_0\omega_0^2=2300 \times \frac{4}{3 } \ J $[/tex]

Therefore, the maximum initial K.E. = 3066.67 J

Answer:

Friction hope this helps <3