When a drag strip vehicle reaches a velocity of 60 m/s, it begins a negative acceleration by releasing a drag chute and applying its brakes. While reducing its velocity back to zero, its acceleration along a straight line path is a constant -7.5 m/s2 . What displacement does it undergo during this deceleration period

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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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:

a) k= 3232.30 N / m,  b)  f = 4,410 Hz

Explanation:

In this exercise, the car + spring system is oscillating in the form of a simple harmonic motion, as the four springs are in parallel, the force is the sum of the 4 Hocke forces.

The expression for the angular velocity is

          w = √k/m

the angular velocity is related to the period

          w = 2π / T

we substitute

          T = 2[tex]\pi[/tex]  √m/ k

a) empty car

           k = 4π² m / T²

           k = 4 π² 1310/2 2

           k = 12929.18 N / m

This is the equivalent constant of the short springs

           F1 + F2 + F3 + F4 = k_eq x

           k x + kx + kx + kx = k_eq x

           k_eq = 4 k

           k = k_eq / 4

           k = 12 929.18 / 4

            k= 3232.30 N / m

b) the frequency of oscillation when carrying four passengers.

In this case the plus is the mass of the vehicle plus the masses of the passengers

            m_total = 1360 + 4 70

            m_total = 1640 kg

angular velocity and frequency are related

              w = 2pi f

we substitute

             2 pi f = Ra K / m

in this case the spring constant changes us

             k_eq = 12929.18 N / m

             f = 1 / 2π √ 12929.18 / 1640

             f = π / 2 2.80778

             f = 4,410 Hz

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

This question is incomplete, the complete question is;

According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula; h= (0.04 to 0.09)(D/d)⁴V²/2g

where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity.

Do you think this equation is valid in any system of units

Answer:

YES, the equation is a general equation that is valid in any system of units

Explanation:

Given the data in the question;

h = (0.04 to 0.09)(D/d)⁴ × [tex]\frac{V^{2} }{2g}[/tex]

so

[ N.m/N ] = (0.04 to 0.09) ( m/m)² × (m²/s²)1/2 × (s²/m)

[ N.L/N ] = (0.04 to 0.09) ( L⁴/L⁴) × (L²/T²)1/2 × (T²/L)

∴ [ L ] = (0.04 to 0.09) [L]

So as each term in the equation must have the same dimensions, the constant term (0.04 to 0.09) must be without dimension.

Therefore, YES, the equation is a general equation that is valid in any system of units

Answer:

The Carnot engine has zero power

Explanation:

Although theoretically the Carnot engine has more efficiency than the real engine. In practice however they tend to have zero power.

This is because all its processes are reversible (that is isothermic and adiabatic).

So the system equilibrates with its surroundings at every point in time. This makes work done very slow and the power generated is zero.

Carnot cycles requires attaining isothermal heat transfers which is quite difficult and take a long time. Also a pump that can handle liquid-vapour phase mixture will be required.

This is not practical.

Answer:

Explanation:

C 200÷100=2

Output ÷ Input= MA

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:

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:

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:

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:

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.

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:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

Electric Field between two parallel plates = 628 N/C

Separation = 4.22 cm

a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.

Solution:

First of all:

Force on proton due to the Electric field between the plates is:

[tex]F_{p}[/tex] = [tex]q_{p}[/tex]E

and, we know that, F = ma

So,

[tex]m_{p}[/tex]a = [tex]q_{p}[/tex]E

a = [tex]\frac{q_{p}.E }{m_{p} }[/tex]      Equation 1

So,

The distance covered by the electron is:

S = ut + 1/2[tex]at^{2}[/tex]

Here, u = 0.

S = 1/2[tex]at^{2}[/tex]

Put equation 1 into the above equation:

S = 1/2 x ([tex]\frac{q_{p}.E }{m_{p} }[/tex]  )[tex]t^{2}[/tex]      Equation 2

So,  

Similarly, the distance covered by electron will be:

(D-S) = 1/2 x ([tex]\frac{q_{e}.E }{m_{e} }[/tex]  )[tex]t^{2}[/tex]    Equation 3

We know that the charge of electron is equal to the charge of proton so,

[tex]q_{p}[/tex] = [tex]q_{e}[/tex] = q

By dividing the equation 2 by equation 3, we get:

[tex]\frac{S}{D-S}[/tex] = [tex]\frac{m_{e} }{m_{p} }[/tex]

Solve the above equation for S,

S[tex]m_{p}[/tex] = [tex]m_{e}[/tex]D - [tex]m_{e}[/tex]S

So,

S = [tex]\frac{m_{e}.D }{(m_{e} + m_{p}) }[/tex]

Plugging in the values,

As we know the mass of electron is 9.1 x [tex]10^{-31}[/tex] and the mass of proton is 1.67 x [tex]10^{-27}[/tex]

S = [tex]\frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27} }[/tex]

S = 0.002298 cm (Distance from the positive plate at which the two pass each other)

b) In this part, we to calculate distance for Sodium ion and chloride ion as above.

So,

we already have the equation, we need to put the values in it.

So,

S = [tex]\frac{m_{Cl}.D }{(m_{Cl} + m_{Na}) }[/tex]

As we know the mass of chlorine is 35.5 and of sodium is 23

S = [tex]\frac{35.5 . 4.22}{(35.5 + 23)}[/tex]

S = 2.56 cm

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:

A. Gamma decay

Explanation:

A form of nuclear decay in which the atomic number is unchanged is a gamma decay.

The atom has undergone a gamma decay.

In a gamma decay, no changes occur to the mass and atomic number of the substance.

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:

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:

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:

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]

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Answer:

electrons are located around the nucleus of an atom.

Explanation:

Electron configurations describe where electrons are located around the nucleus of an atom. For example, the electron configuration of lithium, 1s²2s¹, tells us that lithium has two electrons in the 1s subshell and one electron in the 2s subshell.

Answer:

La aceleración del autobús es -5.80 m/s².

Explanation:

Podemos encontrar la aceleración del autobús usando la siguiente ecuación:

[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]

Where:

[tex]v_{f}[/tex]: es la velocidad final = 0 (se detiene al final)

[tex]v_{0}[/tex]: es la velocidad inicial = 95 km/h

d: es la distancia recorrida = 60 m

Por lo tanto, la aceleración es:

[tex] a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} = \frac{0 - (95 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2*60 m} = -5.80 m/s^{2} [/tex]

El signo negativo se debe a que el autobús está desacelerando (hasta que se detiene).

Entonces, la aceleración del autobús es -5.80 m/s².

Espero que te sea de utilidad!                      

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:

it will option option A hope it helps

Answer:

40 N

Explanation:

According to the scenario, computation of given data are as follows:

Mass (m) = 20 kg

Initially moving (v) = 3

Actual distance (d) = 2.25 m

So, we can calculate friction (f) by using following formula,

f × d = [tex]\frac{1}{2} mv^{2}[/tex]

By putting the value, we get

f × 2.25 = [tex]\frac{1}{2}[/tex] × 20 × [tex]3^{2}[/tex]

f × 2.25 = 10 × 9

f = 90 ÷ 2.25

= 40 N.

Answer:

average speed is 1 m/s and average velocity is 0 m/s.

Explanation:

Given that :

Length of round trip = 50 m

Time taken = 100 seconds

The average speed :

Total distance / total time taken

Length of complete round trip :

(50 + 50) m, total. Distance = 100 m

100 / 100 = 1m/s

The average velocity :

Total Displacement / total time taken

Total Displacement of round trip = end point - start point = 0

0 / 100 = 0

Average speed is 1 m/s and average velocity is 0 m/s.

The average speed is defined as the ratio of distance to time. Speed is a scalar quantity hence it does not take direction into account while velocity is a vector quantity hence it takes direction into account.

The speed is obtained from;

Speed = Distance/time = 2(50 m)/100 s = 1 m/s.

The velocity is 0 m/s since it is complete round-trip lap.

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