If an ice cube with the mass of 5.0 grams melts in a closed system such as a closed glass jar what is the mass of the liquid water after the ice cube completely melts

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

Answer:

5766.7 K

Explanation:

We are given that

Radius of Sun , R=[tex]6.96\times 10^{8} m[/tex]

Distance between the Sun and the Earth, D=[tex]1.50\times 10^{11}m[/tex]

Irradiance arriving on the Earth is the value for AMO=[tex]1350W/m^2[/tex]

We have to find the temperature at the surface of the Sun.

We know that

Temperature ,T=[tex](\frac{K_{sc}D^2}{\sigma R^2})^{\frac{1}{4}}[/tex]

Where [tex]K_{sc}=1350 W/m^2[/tex]

[tex]\sigma=5.67\times 10^{-8}watt/m^2k^4[/tex]

Using the formula

[tex]T=(\frac{1350\times (1.5\times 10^{11})^2}{5.67\times 10^{-8}\times (6.96\times 10^{8})^2})^{\frac{1}{4}}[/tex]

T=5766.7 K

Hence, the temperature at the surface of the sun=5766.7 K

Answer:

a) Average speed is 24.04 m/s

b) the rms speed is 25.84 m/s

c) the most probable speed is 16 m/s

Explanation:

Given the data in the question;

a) Determine the average speed.

To determine the average speed, we simply divide total some of speed by number of particles;

Average speed =  [(2×11 m/s)+(7×16 m/s)+(4×19 m/s)+(3×26 m/s)+(6×31 m/s)+(1×37 m/s)+(2×45 m/s)] / 25    

= 601 / 25

= 24.04 m/s

Therefore, Average speed is 24.04 m/s

b) Determine the rms speed

we know that  (rms speed)² = sum of square speed / total number of particles

so

(rms speed)² =  [(2×11²)+(7×16²)+(4×19²)+(3×26²)+(6×31²)+(1×37²)+(2×45²)] / 25

(rms speed)² =  16691 / 25

(rms speed)² =  667.64

(rms speed) = √ 667.64

(rms speed) = 25.84 m/s

Therefore, the rms speed is 25.84 m/s

c) Determine the most probable speed.

Most particles (7) have velocity 16 m/s

i.e 7 is the maximum number of particle for a particular speed ,

Therefore, the most probable speed is 16 m/s

Answer:

If the speed does not change at all, the object would be moving at a constant speed.

Answer:

it will option option A hope it helps

Answer:

3kg

Explanation:

Given parameters:

Force  = 9N

Acceleration  = 3m/s²

Unknown:

Mass = ?

Solution:

From Newton's second law of motion:

        Force  = mass x acceleration

So;

             9  = mass x 3

             mass  = 3kg

Answer:

P.E = 98 Joules

Explanation:

Given the following data;

Mass = 5kg

Speed = 4m/s

Height = 2m

We know that acceleration due to gravity is equal to 9.8m/s²

To find the potential energy;

Potential energy can be defined as an energy possessed by an object or body due to its position.

Mathematically, potential energy is given by the formula;

[tex] P.E = mgh[/tex]

Where, P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Substituting into the equation, we have;

[tex] P.E = 5*9.8*2[/tex]

P.E = 98 Joules

Answer:

- 21⁰C .

Explanation:

Speed of jet = 2.05 x 10³ km /h

= 2050 x 1000 / (60 x 60 ) m /s

= 569.44 m / s

Mach no represents times of speed of sound , the speed of jet

1.79 x speed of sound = 569.44

speed of sound = 318.12 m /s

speed of sound at 20⁰C = 343 m /s

Difference = 343 - 318.12 = 24.88⁰C

We know that 1 ⁰C change in temperature changes speed of sound

by .61 m /s

So a change in speed of 24.88 will be produced by a change in temperature of

24.88 / .61

= 41⁰C  

temperature = 20 - 41 = - 21⁰C .  

Answer:

2577 K

Explanation:

Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.

So, T = ⁴√(P/σεA)

Since P = 60 W, we substitute the vales of the variables into T. So,

T = ⁴√(P/σεA)

= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)

= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)

= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)

= ⁴√(0.00441 × 10¹⁶K⁴)

= 0.2577 × 10⁴ K

= 2577 K

Answer:

30J

Explanation:

Given parameters:

Mass of hamster  = 0.104kg

Velocity  = 24m/s

Unknown:

Kinetic energy  = ?

Solution:

Kinetic energy is the energy due to the motion of a body. It is mathematically derived by;

  Kinetic energy  = [tex]\frac{1}{2}[/tex] m v²  

m is the mass

v is the velocity

  Kinetic energy  = [tex]\frac{1}{2}[/tex] x 0.104 x 24²   = 30J

Answer:

in fact, kinetic energy is directly proportional to mass: if you double the mass, then you double the kinetic energy. Second, the faster something is moving, the greater the force it is capable of exerting and the greater energy it possesses. ... Thus a modest increase in speed can cause a large increase in kinetic energy.

Explanation:

Answer: The more mass of an object has, the more Kinetic energy it has.

Explanation:

Kinetic energy is comparable to mass. If you double the mass then you double the kinetic energy. The faster the object is moving the greater the energy possesses. A large increase in speed can have a large increase in kinetic energy.

Answer:

hello your question has a missing information

The other is Δ-connected with an impedance of (60 - j45) ohm per phase.

answer : A) 5A ∠0° ,

               p( real power )  = 1800 and  Q ( reactive power ) = 0 VAR

 B) 193.64 v

C) current at load 1 = 2.236 A , current at load 2 = 4.472 A

 D) Load 1 : 450 watts(real power ) , 600 VAR ( reactive power )

      Load 2 : 1200 watts ( real power ), -900 VAR ( reactive power )

Explanation:

First convert the Δ-connection to Y- connection attached below is the conversion and pre-solution

A) determine the current, real power and reactive power delivered by the sending-end source

current power delivered (Is)  =  5A ∠0°

complex power delivered ( s ) = 3vs Is  

                                                  = 3 * 120∠0° * 5∠0° = 1800 + j0 ---- ( 1 )

also s = p + jQ  ------ ( 2 )

comparing equation 1 and 2

p( real power )  = 1800 and  Q ( reactive power ) = 0 VAR

B) determine Line-to-line voltage at the load

Vload = √3 * 111.8

           = 193.64 v

c) Determine current per phase in each load

[tex]I_{l1} = Vl1 / Zl1[/tex]

     = [tex]\frac{111.8<-10.3}{50<53.13}[/tex] = 2.236∠ 63.43° A   hence current at load 1 = 2.236 A

[tex]I_{l2} = V_{l2}/Z_{l2}[/tex]  

     = [tex]\frac{111.8<-10.3}{25<-36.87}[/tex]  = 4.472 ∠ 26.57° A hence current at load 2 = 4.472 A

D) Determine the Total three-phase real and reactive powers absorbed by each load

For load 1

3-phase real power = [tex]3I_{l1} ^{2} R_{l1}[/tex] = 3 * 2.236^2 * 30 = 450 watts

3-phase reactive power = [tex]3I_{l1} ^{2} X_{l1}[/tex] = 3 * 2.236^2 * 40 = 600 VAR

for load 2

3-phase real power = [tex]3I_{l1} ^{2} R_{l2}[/tex]  = 1200 watts

3-phase reactive power = [tex]3I_{l1} ^{2} X_{l2}[/tex] = -900 VAR

The sum of load powers and line losses, 1800 W+ j0 VAR and The line voltage magnitude at the load terminal is 193.64 V.

(a) The impedance per phase of the equivalent Y,

[tex]\bar{Z}_{2}=\frac{60-j 45}{3}=(20-j 15) \Omega[/tex]

The phase voltage,

[tex]\bold { V_{1}=\frac{120 \sqrt{3}}{\sqrt{3}}=120 VV }[/tex]

Total impedance from the input terminals,

[tex]\bold {\begin{aligned}&\bar{Z}=2+j 4+\frac{(30+j 40)(20-j 15)}{(30+j 40)+(20-j 15)}=2+j 4+22-j 4=24 \Omega \\&\bar{I}=\frac{\bar{V}_{1}}{\bar{Z}}=\frac{120 \angle 0^{\circ}}{24}=5 \angle 0^{\circ} A\end{aligned} }[/tex]

The three-phase complex power supplied  [tex]\bold {=\bar{S}=3 \bar{V}_{1} \bar{I}^{*}=1800 W}[/tex]  

P =1800 W and Q = 0 VAR delivered by the sending-end source.

(b) Phase voltage at load terminals will be,  

[tex]\bold {\begin{aligned}\bar{V}_{2} &=120 \angle 0^{\circ}-(2+j 4)\left(5 \angle 0^{\circ}\right) \\&=110-j 20=111.8 \angle-10.3^{\circ} V\end{aligned} }[/tex]  

The line voltage magnitude at the load terminal,  

[tex]\bold{\left(V_{ LOAD }\right)_{L-L}=\sqrt{3} 111.8=193.64 V(V }[/tex]    

(c) The current per phase in the Y-connected load,  

[tex]\bold {\begin{aligned}&\bar{I}_{1}=\frac{\bar{V}_{2}}{\bar{Z}_{1}}=1-j 2=2.236 \angle-63.4^{\circ} A \\&\bar{I}_{2}=\frac{\bar{V}_{2}}{\bar{Z}_{2}}=4+j 2=4.472 \angle 26.56^{\circ} A\end{aligned} ​}[/tex]

The phase current magnitude,  

[tex]\bold {\left(I_{p h}\right)_{\Delta}=\frac{I_{2}}{\sqrt{3}}=\frac{4.472}{\sqrt{3}}=2.582 }[/tex]

(d) The three-phase complex power absorbed by each load,

[tex]\bold {\begin{aligned}&\bar{S}_{1}=3 \bar{V}_{2} \bar{I}_{1}^{*}=430 W +j 600 VAR \\&\bar{S}_{2}=3 \bar{V}_{2} \bar{I}_{2}^{*}=1200 W -j 900 VAR\end{aligned}}[/tex]

The three-phase complex power absorbed by the line is  

[tex]\bold{\bar{S}_{L}=3\left(R_{L}+j X_{L}\right) I^{2}=3(2+j 4)(5)^{2}=150 W +j 300 VAR }[/tex]

Since, the sum of load powers and line losses,  

[tex]\bold {\begin{aligned}\bar{S}_{1}+\bar{S}_{2}+\bar{S}_{L} &=(450+j 600)+(1200-j 900)+(150+j 300) \\&=1800 W +j 0 VAR\end{aligned} }[/tex]

To know more about voltage,

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Answer:

Thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.

Explanation:

From the concept of Escape Velocity, derived from Newton's Law of Gravitation, definition of Work, Work-Energy Theorem and Principle of Energy Conservation, which is the minimum speed such that rocket can overcome gravitational forces exerted by the Earth, and according to the Tsiolkovski's Rocket Equation, which states that thrust done by the rocket is equal to the change in linear momentum of the rocket itself, we conclude that thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.

Complete Question

The question image is in the first uploaded image

Answer:

[tex]E=\frac{KQ*4xa}{(x^2-a^2)^2}[/tex]

Explanation:

From the question we are told that

Distance b/w Q mid point and P is given as x

Generally the equation for magnitude of the electric field at the point P is given as

[tex]E=\frac{kQ}{d^2}[/tex]

where

[tex]k=\frac{1}{4\pi e_0}[/tex]

[tex]d=x^2-a^2[/tex]

Therefore

[tex]E= \frac{1}{4\pi e_0} \frac{Q}{(x^2-a^2)^2}- \frac{1}{4\pi e_0} \frac{Q}{(x^2+a^2)^2}[/tex]

[tex]E= \frac{Q}{4\pi e_0} (\frac{1}{(x^2-a^2)^2}- \frac{1}{(x^2+a^2)^2})[/tex]

Therefore equation for magnitude of the electric field at the point P is

[tex]E=\frac{KQ*4xa}{(x^2-a^2)^2}[/tex]

Answer:

Explanation:

C 200÷100=2

Output ÷ Input= MA

Answer:

v = 8.1 m/s

θ = -36.4º (36.4º South of East).

Explanation:

        [tex]p_{ox} = p_{fx} (1)[/tex]

         ⇒ [tex]m_{tr} * v_{tr}= (m_{olds} + m_{tr}) * v_{fx} (2)[/tex]

       [tex]v_{fx} = \frac{m_{tr}*v_{tr} }{(m_{tr} + m_{olds)} } = \frac{4146kg*9.7m/s}{2028kg+4146 kg} = 6.5 m/s (3)[/tex]

        [tex]p_{oy} = p_{fy} (4)[/tex]

        ⇒[tex]m_{olds} * v_{olds}= (m_{olds} + m_{tr}) * v_{fy} (5)[/tex]

       [tex]v_{fy} = \frac{m_{olds}*v_{olds} }{(m_{tr} + m_{olds)} } = \frac{2028kg*(-14.5)m/s}{2028kg+4146 kg} = -4.8 m/s (6)[/tex]

       [tex]v_{f} = \sqrt{v_{fx} ^{2} +v_{fy} ^{2} }} = \sqrt{(6.5m/s)^{2} +(-4.8m/s)^{2}} = 8.1 m/s (7)[/tex]

       [tex]\frac{v_{fy} }{v_{fx} } = tg \theta (8)[/tex]

      ⇒ tg θ = vfy/vfx = (-4.8m/s) / (6.5m/s) = -0.738 (9)

      ⇒ θ = tg⁻¹ (-0.738) = -36.4º

Answer:

hello your question has some missing parts

A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.

answer : c) 0.39 sec

               d)  2.25 m

               e) 1.92 m/sec

Explanation:

The initial velocity of the first ball = 7.67 m/sec ( calculated )

Time required for first ball to reach ceiling = 0.78 secs ( calculated )

Determine how long after the second ball is thrown do the two balls pass each other

Distance travelled by first ball downwards when it meets second ball can be expressed as : d = 1/2 gt^2 =  9.8t^2 / 2

hence d = 4.9t^2  ----- ( 1 )

Initial speed of second ball = first ball initial speed = 7.67 m/sec

3 - d = 7.67t - 4.9t  ---- ( 2 )

equating equation 1 and 2

3 = 7.67t   therefore t = 0.39 sec

Determine how far the balls are above the Juggler's hands ( when the balls pass each other )

form equation 1 ;

d = 4.9 t^2 = 4.9 *(0.39)^2 = 0.75 m

therefore the height the balls are above the Juggler's hands is

3 - d = 3 - 0.75 = 2.25 m

determine their velocities when the pass each other

velocity = displacement / time

velocity = d / t = 0.75 / 0.39 sec  = 1.92 m/sec

Answer:

Explanation:

For resistance of a wire , the formula is as follows .

R = ρ L/S

where ρ is specific resistance , L is length and S is cross sectional area of wire .

for first wire resistance

R₁ =  ρ 3L/3a = ρ L/a

for second wire , resistance

R₂ = ρ 3L/6a

= .5 ρ L/a

For 3 rd wire resistance

R₃ = ρ 6L/3a

= 2ρ L/a

For fourth wire , resistance

R₄ = ρ 6L/6a

=  ρ L/a

So the smallest resistance is of second wire .

Its resistance is .5 ρ L/a

Answer:

2.59m/s

Explanation:

Using the equation of motion

v = u+at

v is the final velocity = 35ms

u is the initially velocity = 9m/s

t is the time = 13.5s

a is the acceleration

Substitute into the formula

35 = 0+13.5a

a = 35/13.5

a = 2.59m/s²

Hence the acceleration of the motorcycle is 2.59m/s

Answer:

A

Explanation:

Ke = 1/2 MV^2

Answer:

2.88m/s

Explanation:

Given parameters:

Displacement  = 7.2m

Time taken  = 2.5s

Unknown:

Velocity of the plane  = ?

Solution:

Velocity is the displacement divided by the time taken.

  Velocity  = [tex]\frac{displacement}{time taken}[/tex]  

 So;

   Velocity  = [tex]\frac{7.2}{2.5}[/tex]    = 2.88m/s

Answer:

the distance moved by the box is 70.03 m.

Explanation:

Given;

mass of the box, m = 35 kg

initial velocity of the box, u = 10 m/s

frictional force, F = 25 N

Apply Newton's second law of motion to determine the deceleration of the box;

-F = ma

a = -F / m

a = (-25 ) / 35

a = -0.714 m/s²

The distance moved by the box is calculated as follows;

v² = u² + 2ad

where;

v is the final velocity of the box when it comes to rest = 0

0 = 10² + (2 x - 0.714)d

0 = 100 - 1.428d

1.428d = 100

d = 100 / 1.428

d = 70.03 m

Therefore, the distance moved by the box is 70.03 m.

Answer:

c

Explanation:

a vector quantity has both magnitude and direction

The value of  20m/s, east is a vector quantity is Hence, option (C) is correct.

A physical quantity that has both directions and magnitude is referred to as a vector quantity.

A lowercase letter with a "hat" circumflex, such as "û," is used to denote a vector with a magnitude equal to one. This type of vector is known as a unit vector.

Given values  200V, 100kg/m, 50J/(kg°C) are denoting magnitude of different physical quantity. Hence, they and scalar quantity ( Physical quantities with merely magnitude and no direction are referred to as scalar quantities. These physical quantities can be explained just by their numerical value without any further guidance.).

But The value of  20m/s, east has a magnitude of 20 m/s and a direction along east. Hence, 20m/s, east denotes a  vector quantity is Hence, option (C) is correct.

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I believe the answer would be protentional because they have the potential energy in them to lift the hammer.