Concept Simulation 2.3 offers a useful review of the concepts central to this problem. An astronaut on a distant planet wants to determine its acceleration due to gravity. The astronaut throws a rock straight up with a velocity of 17.4 m/s and measures a time of 12.4 s before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet

Answer:

Required energy Q = 231 J

Explanation:

Given:

Specific heat of copper C = 0.385 J/g°C

Mass m = 20 g

ΔT = (50 - 20)°C = 30 °C

Find:

Required energy

Computation:

Q = mCΔT

Q = 20(0.385)(30)

Required energy Q = 231 J

Answer:

The energy an object has because it is moving

Explanation:

It has been a while since I have talked about kinetic energy so I can't give you an explanation why that answer is right but it is.

Answer:

30 m/s

Explanation:

Speed is distance over time. 60 meters / 2 seconds, = 30 m/s.

Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

         Em₀ = U = m g y₁

final point. At the exit of the ramp

         Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

          Em₀ = Em_f

         m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

        v = w r

the moment of inertia of a sphere is

         I = [tex]\frac{2}{5}[/tex] m r²

we substitute

         m g (y₁ - y₂) = ½ m v² + ½ ([tex]\frac{2}{5}[/tex] m r²) ([tex]\frac{v}{r}[/tex])²

         m g h = ½ m v² (1 + [tex]\frac{2}{5}[/tex])

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

         mg h = [tex]\frac{7}{5}[/tex] ([tex]\frac{1}{2}[/tex] m v²)

         v = √5/7  √2gh

This is the exit velocity of the vertical movement of the sphere

         v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

   starting point

          Em₀ = U = mg y₁

   final point

          Em_f = K + U = ½ m v² + m g y₂

          Em₀ = Em_f

          mg y₁ = ½ m v² + m g y₂

          m g (y₁ -y₂) = ½ m v²

          v = √2gh

this is the speed of the box

          v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

          v² = v₀² - 2 g y

at the highest point v = 0

           y = vo₀²/ 2g

for the sphere

           y_sphere = 5/7 2gh / 2g

           y_esfera = 5/7 h

for the block

           y_block = 2gh / 2g

            y_block = h

therefore the block reaches higher than the sphere

         [tex]\frac{y_{sphere}} {y_bolck} = 5/7[/tex]

Answer:

17⁰ north of west

Explanation:

We shall represent forces by vector notations . i will represent unit vector towards east and j will represent unit vector towards north .

Jane: 80 pounds, straight east

F₁ = 80 i

Brittany: 100 pounds, north (90 degrees relative to east)

F₂ = 100 j

Robyn: 80 pounds, southwest (225 degrees relative to east ) or 45⁰ south of west

F₃ = - 80cos 45  i - 80 sin45 j = - 80/√2 x ( i + j )

= - 40√2 ( i + j )

Scott: 100 pounds, northwest (135 degrees relative to east) or 45⁰ west of north .

F₄ = - 100 cos 45 i + 100 sin45 j =  50√2 ( - i + j )

Resultant force

= F₁ + F₂ + F₃ + F₄

F =  80 i + 100 j + - 40√2 ( i + j ) + 50√2 ( - i + j )

F = 80 i - 90√2 i + 110√2 j

80 i - 127.26 i + 114.14 j

F = - 47.26 i + 114.14 j

Direction of force

TanФ = 114.14 / 47.26 = 2.4

Ф = 67⁰

67⁰ north of west .

Answer:

C. zero for both

Explanation:

In case of solid metal sphere , when it is given any charge , all the charges are transferred on the surface and within surface no charge exists . In case of hollow metal sphere , all charges reside on surface . In this way , in both solid and hollow sphere , all charge reside on the surface and no charge resides inside it . Hence due to absence of any charge inside  , there is no electric field inside the sphere in both the cases .  

Hence in both the case electric field is zero .

option C is correct .

This question involves the concepts of compression and rarefraction.

This wave should be classified as "A) a longitudinal wave".

A surface wave is a mechanical wave which propagates on the boundary interface between two different media.

A longitudinal wave is the wave in which the direction of travel of the wave and medium are the same. This wave consists of compressions and rarefactions.

A transverse wave is the wave in which the direction of travel of the wave and medium are perpendicular to each other. It consists of crests and troughs.

An electromagnetic wave is the one which consists of electric field and magnetic field vibrating perpendicular to each other.

Hence, the wave with compressions and rarefactions is classified as a longitudinal wave.

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Answer:

120.237 seconds

Explanation:

Given that:

V = 144 V

I = 10.5 A

H = 60.6 kJ

Using the formula:

H = I²RT

From H = I²RT; making T the subject, we have:

[tex]T = \dfrac{H}{I^2R}[/tex]

where;

[tex]R = \dfrac{V}{I}[/tex]

∴

[tex]T = \dfrac{H}{V \times I}[/tex]

[tex]T = \dfrac{60.6 \times 10^3 }{144 \times 10.5}[/tex]

T = 40.079

[tex]T_{neq} = 3T[/tex]

[tex]T_{neq} =3 \times 40.079[/tex]

[tex]\mathbf{T_{neq} =120.237 \ sec}[/tex]

Answer:

Convection

Explanation:

Convection refers to the transfer of heat by the actual movement of the heated molecules from the hot parts to the cooler parts.

Thus, the two diagrams show illustrate transfer of heat by the movement of molecules of a fluid.

This mode of heat transfer is known as convection. It is the concept illustrated by both diagrams.

Answer:

W = 2.3 10² [tex]F_{e}[/tex]

Explanation:

The force of the weight is

         W = m g

let's use the concept of density

         ρ= m / v

the volume of a sphere is

         V = [tex]\frac{4}{3}[/tex] π r³

         V = [tex]\frac{4}{3}[/tex] π (1.0 10⁻³)³

         V = 4.1887 10⁻⁹ m³

the density of water ρ = 1000 kg / m³

          m = ρ V

          m = 1000 4.1887 10⁻⁹

          m = 4.1887 10⁻⁶ kg

therefore the out of gravity is

          W = 4.1887 10⁻⁶ 9.8

          W = 41.05 10⁻⁶ N

now let's look for the electric force

           F_e = q E

           F_e = 12 10⁻¹² 15000

           F_e = 1.8 10⁻⁷ N

the relationship between these two quantities is

          [tex]\frac{W}{F_e}[/tex] = 41.05 10⁻⁶ / 1.8 10⁻⁷

           \frac{W}{F_e} = 2,281 10²

             W = 2.3 10² [tex]F_{e}[/tex]

therefore the weight of the drop is much greater than the electric force

Answer:

The joule (symbol: J) is the basic SI unit of energy. A joule is equal to the kinetic energy of a kilogram mass moving at the speed of one meter per second (one joule is a kg⋅m2⋅s−2). ... Because it is named for a person, the first letter of the symbol is uppercase (J instead of j)

Explanation:

Answer:

the answer is 32

Explanation:

he was riding pretty far

Answer:

0.76m

Explanation:

Given data

Frequency= 230Hz

speed= 350m/s

Since we are told that the frequency is the fundamental frequency n= 1

For a standing wave

Fn= nv/2L

n= 1

230= 1*350/2*L

230= 350/2L

cross multiply

2L= 350/230

2L=1.521

L=1.521/2

L=0.76m

Hence the length is 0.76m

Answer:

(a) 1.21 m/s² (b) 1.75 m/s²

Explanation:

The initial speed of the car, u = 17.8 m/s

Case 1.

Final speed of the car, v = 23.5 m/s

Time, t = 4.68-s

Acceleration = rate of change of velocity

[tex]a=\dfrac{23.5 -17.8 }{4.68}\\\\a=1.21\ m/s^2[/tex]

Case 2.

Final speed of the car, v = 15.3 m/s

[tex]a=\dfrac{23.5 -15.3}{4.68}\\\\a=1.75\ m/s^2[/tex]

Hence, this is the required solution.

Acceleration is the change in the velocity of a body with time.

The term acceleration is the change in the velocity of a body with time. We have to find the acceleration in two separate cases;

Case 1;

a = -(23.5 m/s - 17.8 m/s)/4.68-s

a = -1.2 m/s^2

Case 2;

a = -(15.3 m/s - 17.8 m/s)/4.68-s

a = 0.53 ms-2

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Answer:

Explanation:

Horizontal component of initial velocity of throw = 20 cos 53 = 12 m /s

Vertical component = 20 sin 53 = 15.97 m /s

Distance to be travelled horizontally = 6 m .

time taken by ball to travel this distance = 6 / 12 = 0.5 s

vertical displacement during this period can be calculated as follows .

Initial vertical velocity = 15.97 m /s

time of travel = .5 s

acceleration = - 9.8 m /s²

s = ut - 1/2 g t²

= 15.97 x .5 - .5 x 9.8 x 0.5²

= 7.985 - 1.225

= 6.76 m

Goal post is 6 m high , so ball will cross the goal post .76 m or 76 cm above cross bar  .

b ) vertical component of ball when it crosses the goal post . Let it be v .

v = u - gt

Applying this formula for vertical movement ,

v = 15.97 - 9.8 x .5

= 15.97 - 4.9

= 11.07 m /s .

Answer:

66.85 m

Explanation:

We are given that

Acceleration ,a=[tex]2.7m/s^2[/tex]

Speed of truck, v=9.5 m/s

We have to find the distance beyond which the traffic signal will the automobile overtake the truck.

Initial speed of automobile, u=0

We know that

[tex]s=ut+\frac{1}{2}at^2[/tex]

Using the formula

[tex]s=0+\frac{1}{2}(27)t^2=\frac{27}{2}t^2[/tex]

For constant speed

Acceleration, a=0

Again

[tex]s=vt+0=9.5t[/tex]

[tex]9.5t=\frac{27}{2}t^2[/tex]

[tex]t=\frac{9.5\times 2}{2.7}=7.037s[/tex]

Substitute the value of t

[tex]x=9.5(7.037)=66.85m[/tex]

Hence, the distance beyond which the traffic signal will the automobile overtake the truck=66.85 m

Answer:

#1 ( Balanced ) ( Rest)

#2 ( Unbalanced)  ( Accelerating)

Explanation:

Answer:

14.1seconds approx

Explanation:

Given data

Distance= 59.1m

Your velocity= 2.35m/s

Walkway velocity= 1.85m/s

Total velocity= 2.35+1.85= 4.2m/s

We know that

Velocity= distance/time

time= distance/velocity

substitute

time= 59.1/4.2

time= 14.07

time=14.1seconds approx

Hence the time is 14.1seconds approx

Answer:

The correct answer is "2524 W".

Explanation:

The given values are:

Ia = 20 A

Ea = 103 V

Ra = 0.19 Ω

La = 4 mH

According to KVL,

⇒ [tex]V_0=E_a-I_aR_a[/tex]

On substituting the given values in the above equation, we get

⇒      [tex]=103-(20)(0.19)[/tex]

⇒      [tex]=103-3.8[/tex]

⇒      [tex]=126.2 \ V[/tex]

Now,[tex]=126.2\times 20[/tex]

The average power flow into the converter will be:

⇒ [tex]P=V_0 I_a[/tex]

On substituting the given values, we get

⇒     [tex]=(126.2\times 20)[/tex]

⇒     [tex]=2524 \ W[/tex]

Answer:

  t = 781.25 s

Explanation:

This is an exercise in velocity composition, if we set a reference system where the x-axis is perpendicular to the river and the y-axis is parallel to the river.

The swimmer has a velocity on the x axis

           vx = 1.60 m / s

a velocity on the y axis, created by the current of the river

           vy = 0.549 m / s

time is a scalar, therefore the time it takes to cross the river is the same time it creates the displacement in e; Axis y

X axis

            vₓ = x / t

            t = x / vₓ

            t = 1250 / 1.6

            t = 781.25 s

in this time a distance has descended

            y = v_y t

            y = 0.549 781.25

            y = 428.9 m

Answer:

27872.2 miles per hour

Explanation:

Given that :

Radius of Jupiter is approximately = 44,360 miles

Revolution is 10 hours ;

Jupiter makes one revolution in 10 hours :

Using the relation to obtain the velocity :

V = re

r = radius

w = 2π/T

Hence,

V = r * 2π/ T

V =44360 * 2 * π/10

V = 88720 * π/10

V = 278722.10 / 10

V = 27872.210

V = 27872.2 miles per hour

Answer:

A concave lens is exactly the opposite with the outer surfaces curving inward, so it makes parallel light rays curve outward or diverge. That's why concave lenses are sometimes called diverging lenses.

Answer: A concave lens

A concave lens is exactly the opposite with the outer surfaces curving inward, so it makes parallel light rays curve outward or diverge. That's why concave lenses are sometimes called diverging lenses.

Hopes this helps :)

Answer:

Explanation:

The enclosure will behave as black body . For black body , the formula for radiant energy is given by Stefan's law as follows

E = σ A T⁴ where σ = 5.67 x 10⁻⁸ W m⁻² T⁻⁴

Area A = .01 m²

E = 48 W

48 = .01 x 5.67 x 10⁻⁸ T⁴

T⁴ =  846.56 x 10⁸

T= 539 K

Following are the calculation to the temperature of the interior enclosure wall:

Given:

[tex]\to A_s=50\ m^2\\\\\to A_0=0.01\ m^2\\\\\to q_{rad}=48\ w\\\\\to \sigma = 5.67 \times 10^{-8}\ \frac{W}{m^{-2} T^{-4}} \\\\[/tex]

To find:

[tex]T_s=?[/tex]

Solution:

Using the value to calculate the surface is black radiates power:

[tex]\to q_{rad}=A_0 E_0 T_s\\\\\to q_{rad}=A_0 \sigma T_{s}^4\\\\\to 48=0.01 \times 5.67 \times 10^{-8} \times T_{s}^4\\\\[/tex]

[tex]\to T_{s}^4 =\frac{48}{0.01 \times 5.67 \times 10^{-8}} \\\\[/tex]

        [tex]=\frac{48\times 100}{ 5.67 \times 10^{-8}} \\\\=\frac{48\times 100\times 10^{8} }{ 5.67 } \\\\=\frac{48\times 10^{10} }{ 5.67 } \\\\=8.46\times 10^{10} \\\\[/tex]

[tex]\to \bold{T_s=\pm 539.31 \ K}[/tex]

Therefore the final answer is "[tex]\bold{\pm 539.31 \ K}[/tex]".

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Answer:

The answer is a 4.2m

Explanation:

Given data

Please see attached the rough drawing for your reference.

From the drawing, you ran 18m west and 2.4m south

The displacement is

= 1.8+2.4

=4.2m

Answer:it’s A. right for ape x

Explanation:

Based on data, digital waves are more reliable because they are less likely to change when copied (Option A).

A wave is a periodic distortion capable of traveling through a suitable media (either air or water).

In conclusion, based on data, digital waves are more reliable because they are less likely to change when copied (Option A).

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Answer:

a horizontal force of 120N to Push the block into the floor is called net force

Answer:

A star's brightness also depends on its proximity to us. The more distant an object is, the dimmer it appears.

Answer:

Explanation:

We shall apply Bernoulli's formula to solve the problem . It is as follows .

P + ρ gh + 1/2 ρ v² = constant .

P₁ + ρ gh + 1/2 ρ v₁²  = P₂ + ρ gh + 1/2 ρ v₂²

P₁ +  1/2 ρ v₁²  = P₂  + 1/2 ρ v₂²

P₁ - P₂ = 1/2 ρ (v₂² - v₁² )

= .5 x 1,1 ( 30² - 20² )

= 275 N / m²

velocity over moon roof is high , pressure will be lower there by 275 N / m²

Given pressure difference already existing = 90500 - 90000 = 500 N / m²

Additional pressure difference due to velocity difference = 275 N / m²

Total pressure difference = 275 + 500 = 775 N / m²

Area of roof = .5 m²

Total force acting upwards on the roof

= .5 x 775 N

= 387.5 N .