Four 8.5-kg spheres are located at the corners of a square of side 0.90 m.

A. Calculate the magnitude of the gravitational force exerted on one sphere by the other three.

B. Calculate the direction of the gravitational force exerted on one sphere by the other three.

Answer:

  a₃ = -1.08 m/s²,    K = 1.42 J

Explanation:

The particle is in a periodic motion, so the general expression is

           x = A cos (wt + Ф)

let's compare the terms with the expression they give us

           x = 2 cos (t - 2)

the amplitude of motion is A = 2 m, the angular velocity w = 1 rad / s, and the phase is Ф = - 2.

to find the acceleration we use its definition

          v = dx / dt

          a = dv / dt

          a = [tex]\frac{ d^2x}{dt^2}[/tex]

let's perform the derivative

          v = - A w sin (wt + Ф)

          a = - A w² cos wt + Ф)

substituting the values

          a = - 2 1² cos (t-2)

for t = 3 s

          a₃ = 2 cos (3-2)

remember angles are in radians

          a₃ = -1.08 m/s²

To calculate kinetic energy, let's find the velocity for t = 3 s

          v = - 2 sin (t-2)

          v = -2 sin (3-2)

          v = - 1.683 m / s

body mass is m = 1 kg

we calculate

          K = ½ m v²

          K = ½ 1 (-1.683) ²

          K = 1.42 J

Answer:

The minimum distance between two points on the  object that are barely resolved is 0.26 mm

The corresponding distance between the  image points = 0.0015 m

Explanation:

Given  

focal length f = 50 mm and maximum aperture f>2

s =  9.0 m

aperture = 25 mm = 25 *10^-3 m

Sin a = 1.22 *wavelength /D  

Substituting the given values, we get –  

Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m

Sin a = 2.93 * 10 ^-5 rad

Now  

Y/9.0 m = 2.93 * 10 ^-5

Y = 2.64 *10^-4 m = 0.26 mm

Y’/50 *10^-3 = 2.93 * 10 ^-5  

Y’ = 0.0015 m

Answer:

parallel circuit

Explanation:

An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.

Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.

Basically, the components of an electric circuit can be connected or arranged in two forms and these includes;

I. Series circuit

II. Parallel circuit: it's an electrical circuit that has the same potential difference (voltage) across its terminals or ends. Thus, its components are connected within the same common points so that only a portion of current flows through each branch.

Hence, the type of circuit that the above diagram above represents is a parallel circuit.

Answer:

parallel circuit

Explanation:

I got it right on my exam

Answer:

The correct answer is "[tex]\frac{4E}{3}[/tex]".

Explanation:

According to the question,

Energy of satellite,

⇒ [tex]E_s=-\frac{GM_sM_E}{2r}[/tex]

For the very 1st case:

[tex]r = R_E+R_E[/tex]

  [tex]=2R_E[/tex]

or,

⇒ [tex]E=-\frac{GM_sM_E}{4R_E}[/tex]...(1)

For the new case:

[tex]r = R_E+\frac{R_E}{2}[/tex]

  [tex]=\frac{3R_E}{2}[/tex]

then,

⇒ [tex]E'=-\frac{GM_sM_E}{2 \frac{3R_E}{2} }[/tex]

        [tex]=-\frac{GM_sM_E}{3R_E}[/tex]...(2)

From equation (1) and (2), we get

⇒ [tex]E'=\frac{1}{3}(4E)[/tex]

        [tex]=\frac{4E}{3}[/tex]  

Answer:

the moment of inertia of this system of masses about the y-axis is 99 kgm²

Explanation:

Given the data in the question;

mass m₁ = 5.0 kg at point ( 3.0, 4.0 )

mass m₂ = 6.0 kg at point ( 3.0, -4.0 )

Now, Moment of inertia [tex]I[/tex] of this system of masses about the y-axis will be;

Moment of inertia [tex]I[/tex]ₓ = mixi²

Moment of inertia [tex]I[/tex] = m₁x₁² + m₂x₂²

we substitute

Moment of inertia [tex]I[/tex] = [ 5.0 × ( 3 )² ]  + [ 6.0 × ( 3 )² ]

Moment of inertia [tex]I[/tex] = [ 5.0 × 9 ]  + [ 6.0 × 9 ]

Moment of inertia [tex]I[/tex] = 45 + 54

Moment of inertia [tex]I[/tex] = 99 kgm²

Therefore, the moment of inertia of this system of masses about the y-axis is 99 kgm²

Answer:

The location of helicopter is behind the packet.

Explanation:

As the packet also have same horizontal velocity as same as the helicopter, and also it has some vertical velocity as it hits the ground.

The horizontal velocity remains same as there is no force in the horizontal direction. The vertical  velocity goes on increasing as acceleration due to gravity acts.

So, the helicopter is behind the packet.

Answer:

amperes

Ammeter, instrument for measuring either direct or alternating electric current, in amperes. An ammeter can measure a wide range of current values because at high values only a small portion of the current is directed through the meter mechanism; a shunt in parallel with the meter carries the major portion.

Explanation:

hope it helps

Answer:

Ray A - incident ray

Ray B - reflected ray

I don't know about it your answer will give another people

Answer: Let the final velocity be v.

Given,

Initial velocity(u)=8m/s

Acceleration(a)=7m/s2

Time(t)=3 sec

Then,

v=u+at

  =8+7*3 m/s

  =29m/s

Therefore, the final velocity is 29m/s.

Answer:

if I am correct, they indicate less steep terrain. think of it as the steeper the terrain the closer together the lines would be. hope that makes sense for you guys.

Answer:

gentle slopes

Explanation:

Explanation:

m = 19 oz × (28.3 g/1 oz) = 537.7 g

V = 92.8 mL

[tex]\rho = \dfrac{m}{V}= \dfrac{537.7\:g}{92.8\:mL} = 5.79\:\frac{g}{mL}[/tex]

Answer:

a) 10.3 m/s

b) 566 N

Explanation:

[tex]v {}^{2} = {u}^{2} + 2as \\ v {}^{2} = 0 {}^{2} + 2(9.81)(5.4) \\ v = 10.3 \: ms {}^{ - 1} [/tex]

[tex]force \: = \frac{d(mv)}{dt} \\ = 55(10.293) \\ = 566 \: newtons[/tex]

The athelete velocity will be 10.3 and constant force 566 N.

The displacement that an object or particle experiences with respect to time is expressed vectorially as velocity. The meter per second (m/s) is the accepted unit of velocity magnitude (also known as speed).

Alternately, the magnitude of velocity can be expressed in centimeters per second (cm/s). Depending on how many dimensions are included, there are numerous ways to indicate the direction of a velocity vector.

The car's velocity in relation to your body is zero when you are driving. The speed of the car in relation to you if you were to stand by the side of the road is 20 m/s northward.

Therefore, The athelete velocity will be 10.3 and constant force 566 N.

To learn more about velocity, refer to the link:

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Answer:

4592.57 lb

Explanation:

The missing diagram for this question is attached in the image below.

Given that:

the weight of the car = 2890 lb

At point A, the speed of the car [tex](V_A)[/tex] = 58 mi/hr

At point C, the speed of the car [tex](V_C)[/tex] = 18 mi/hr

To ft/s:

[tex](V_A)[/tex]  = 58 mi/hr × 5280 ft/1 mi × 1 hr/3600 s

[tex](V_A)[/tex]  = 85.07 ft/s

[tex](V_C)[/tex] = 18 mi/hr × 5280 ft/1 mi × 1 hr/3600 s

[tex](V_C)[/tex] = 26.4 ft/s

Between A to C, the total distance is;

[tex]S_{AC} = S_{AB}} + S_{BC} \\ \\ S_{AC} = 331 + \dfrac{\pi r}{2} \\ \\ S_{AC}= 331 + \dfrac{\pi \times 207}{2} \\ \\ S_{AC} = 656.154 \ ft[/tex]

Now, we need to determine the deceleration of the car using the formula:

[tex]V_C^2 = V_A^2 + 2 aS_{AC}[/tex]

[tex]26.4^2 = 85.07^2 + 2 a (654.154)[/tex]

[tex]696.96 = 7236.9049+ 2 a (654.154)[/tex]

[tex]696.96-7236.9049 = 2 a (654.154)[/tex]

[tex]-6539.9449 = 2 a (654.154)[/tex]

[tex]a= \dfrac{-6539.9449} {2(654.154)}[/tex]

a = -4.99 ft/s²

The velocity of the car as it passes via B

[tex]v_B^2 = v_A^2 + 2aS_{AB}[/tex]

[tex]v_B^2 = 85.07^2 + 2(-4.99 \times 331)[/tex]

[tex]v_B =\sqrt{ 85.07^2 + 2(-4.99 \times 331)}[/tex]

[tex]v_B =\sqrt{ 85.07^2 +3303.38}[/tex]

[tex]v_B =\sqrt{ 10540.2849}[/tex]

[tex]v_B =102.67 \ ft/s[/tex]

Along B, the car's acceleration is:

[tex]a_B = \sqrt{a^2 + (\dfrac{v_B^2}{r})^2}[/tex]

[tex]a_B = \sqrt{(-4.99)^2 + \dfrac{102.67^2}{207}^2 }[/tex]

[tex]a_B = 51.17 \ ft/s^2[/tex]

Finally, the total horizontal force F exerted = m[tex]a_B[/tex]

[tex]= (\dfrac{2890}{32.2}) \times 51.17[/tex]

= 4592.57 lb

Answer:

(a) 159.84 m

(b) 1.89 m/s²

Explanation:

Applying,

(a)

s = (v+u)t/2.................. Equation 1

Where s = distance traveled by the car, u = initial velocity, v = final velocity, t = time.

From the question,

Given: u = 0 m/s ( from rest), v = 55 mi/h = (55/2.237) m/s = 24.59 m/s, t = 13 s

Substitute these values into equation 1

s = (24.59+0)13/2

s = 159.84 m

(b)

Also applying,

a = (v+u)/t................. Equation 2

Where a = acceleration of the car.

substituting into equation 2,

a = (24.59+0)/13

a = 1.89 m/s²

Answer: High-frequency sound waves are perceived as high-pitched sounds, while low-frequency sound waves are perceived as low-pitched sounds. The audible range of sound frequencies is between 20 and 20000 Hz, with greatest sensitivity to those frequencies that fall in the middle of this range.

Explanation: Obviously explained in the answer

Answer:

A

Explanation:

If the object is moving at a constant speed, the object isn't accelerating as the velocity doesn't change.

Answer: C.

Explanation:  plato users

Answer:

12kWhr

Explanation:

Energy = Power * Time

Power = 4kW

Time = 3hrs

Substitute into the formula

Energy used up = 4kW * 3hrs

Energy used up = 12kWhr

Answer:

46

Explanation:

Answer:

0.53 quart

Explanation:

The volume expansion of the coolant is gotten from ΔV = VγΔθ where ΔV   = change in volume of the coolant, V = initial volume of coolant = 15 quart, γ = coefficient of volume expansion of coolant = 410 × 10⁻⁶ /°C and Δθ = temperature change = θ₂ - θ₁ where θ₁ = initial temperature of coolant = 6 °C and θ₂ = final temperature of coolant = 92 °C. So, Δθ = θ₂ - θ₁ = 92 °C - 6 °C = 86 °C

Since, ΔV = VγΔθ

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

ΔV = VγΔθ

ΔV = 15 × 410 × 10⁻⁶ /°C × 86 °C

ΔV = 528900 × 10⁻⁶ quart

ΔV = 0.528900 quart

ΔV ≅ 0.53 quart

Since the change in volume of the coolant equals the spill over volume, thus the overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92 °C is 0.53 quart.

Complete question:

A 5-m³ container is filled with 900 kg of granite (density of 2400 kg/m3). The rest of the volume is air, with density equal to 1.15 kg/m³. Find the mass of air and the overall (average) specific volume.

Answer:

The mass of the air is 5.32 kg

The specific volume is 5.52 x 10⁻³ m³/kg

Explanation:

Given;

total volume of the container, [tex]V_t[/tex] = 5 m³

mass of granite, [tex]m_g[/tex] = 900 kg

density of granite, [tex]\rho _g[/tex] = 2,400 kg/m³

density of air, [tex]\rho_a[/tex] = 1.15 kg/m³

The volume of the granite is calculated as;

[tex]V_g = \frac{m_g}{ \rho_g}\\\\V_g = \frac{900 \ kg}{2,400 \ kg/m^3} \\\\V_g = 0.375 \ m^3[/tex]

The volume of air is calculated as;

[tex]V_a = V_t - V_g\\\\V_a = 5 \ m^3 \ - \ 0.375 \ m\\\\V_a = 4.625 \ m^3[/tex]

The mass of the air is calculated as;

[tex]m_a = \rho_a \times V_a\\\\m_a = 1.15 \ kg/m^3 \ \times \ 4.625 \ m^3\\\\m_a = 5.32 \ kg[/tex]

The specific volume is calculated as;

[tex]V_{specific} = \frac{V_t}{m_g \ + \ m_a} = \frac{5 \ m^3}{900 \ kg \ + \ 5.32\ kg} = 5.52 \times 10^{-3} \ m^3/kg[/tex]

Answer:

below

Explanation:

The magnitude of the impulse of the truck is equal to 4000 Kg.m/s.

Impulse can be described as the integral of a force over the time interval for which it acts. Impulse is also a vector quantity since force is a vector quantity. Impulse can be applied to an object that generates an equivalent vector change in its linear momentum.

The S.I. unit of impulse is N⋅s and the dimensionally equivalent unit of momentum is kg⋅m/s. A resultant force gives acceleration and changes the velocity of an object for as long as it acts.

Given the mass of the truck, m= 2000 Kg

The initial speed of the truck, u = 6 m/s

The final speed of the truck, v = 4 m/s

The change in the linear momentum is equal to the impulse.

I = ΔP = mv - mu

I = 2000 ×4 - 2000 × 6

I = 8000 - 12000

I = - 4000  Kg.m/s²

Therefore, the magnitude of the impulse is  4000 Kg.m/s².

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Answer:

Road A- dry

Road B- mud

Road C- wet

Explanation:

Surface conditions do affect the ease and speed with which a skateboarder can move, on a muddy surface, the tyres of the skate boards finds it difficult to establish adequate fictional force between the skates trees and the traveling surface. Hence, the muddy surface presents a very slippery travel ground for the skate, hence leading the to skateboarder needing to apply caution.

The speed on a wet surfave is height as the amount of firece that will be applied in other to accelerate is very small. The surface is wet and hence serves as a lubricant between the contact surface.

The dry road also has a high speed but lower than a wet surface, frictional force is high here and this tend to slow the skateboarder down except in sloppy terrains.

Answer:

The number of input force is the same as output. ... If it equals once, then both numbers are equal making it the same.Explanation:

Does this helps

Answer:

The number of input force is the same as output. Formula for MA (Mechanical Advantage) is Input Force/Output Force. If it equals once, then both numbers are equal making it the same. In order to raise MA, you must lower efficiency, something you learn around grade 8. Good luck!

P.S. Direction is the same for both, meaning if you pull something, the object you pull will come towards you.

Answer:

A. Power generated by meteor = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Workdone = 981000 J

Power required = 19620 Watts

Note: The question is incomplete. A similar complete question is given below:

A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5 g and is moving at an impressive 50 km/s, both typical values. What power is generated if the meteor slows down over a typical 2.1 s? Can you see how this tiny object can make a glowing trail that can be seen hundreds of kilometers away? 61. a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed? b. How much power must the motor supply to do this in 50 s at constant speed?

Explanation:

A. Power = workdone / time taken

Workdone = Kinetic energy of the meteor

Kinetic energy = mass × velocity² / 2

Mass of meteor = 1.5 g = 0.0015 kg;

Velocity of meteor = 50 km/s = 50000 m/s

Kinetic energy = 0.0015 × (50000)² / 2 = 1875000 J

Power generated = 1875000/2.1 = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Work done by elevator against gravity = mass × acceleration due to gravity × height

Work done = 1000 kg × 9.81 m/s² × 100 m

Workdone = 981000 J

Power required = workdone / time

Power = 981000 J / 50 s

Power required = 19620 Watts

Therefore, the motor must supply a power of 19620 Watts in order to lift a 1000 kg to a height of 100 m at a constant speed in 50 seconds.

Answer:

Total cost = 56.16 cents

Explanation:

Given the following data;

Power = 45 Watts

Time = 4 hours

Number of days = 30 days

Cost = 10.4 cents

To find how much does it cost her to watch TV for one month;

First of all, we would determine the energy consumption of the TV;

Energy = power * time

Energy = 45 * 4

Energy = 180 Watt-hour = 180/1000 = 0.18 Kwh (1 Kilowatts is equal to 1000 watts).

Energy consumption = 0.18 Kwh

Next, we find the total cost;

Total cost = energy * number of days * cost

Total cost = 0.18 * 30 * 10.4

Total cost = 56.16 cents

Answer:

The energy stored in the spring would be 1 joule.

Explanation:

hope that helps?

Answer:

1.  SG  

true

=2.689

2. The object is probably some sort of minerals and rocks such as Feldspar, Corals, Beryl, etc.

Explanation:

Given:

mass in the air= 0.0675 kg

mass in water= 0.0424 kg

The specific gravity of the object will be 2.6892. It is the ratio of the density of the given fluid and the standard fluid.

Density is specified as the mass divided by the volume. It is represented by the unit of measurement as kg/m³.

The mass of the object in air;

m=Vρ₀

m=0.0675 kg

Buoyant force on the object;

B= Vρₐg

For equilibrium;

N+B=m₀g

n=m₀g-Vρₓg

N/g=m₀-Vρₓ

N/g=0.0424 kg

[tex]\rm \frac{V\rho_0}{V\rho_x} =\frac{0.0675 }{m_0-0.0424 \ kg} \\\\ \frac{\rho_0}{\rho_x} =\frac{0.0675}{0.0675-0.0424} \\\\ \frac{\rho_0}{\rho_x} =2.6892[/tex]

Hence, the specific gravity of the object will be 2.6892.

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Answer:

s₁ = 0.022 m

Explanation:

From the law of conservation of momentum:

[tex]m_1u_1 + m_2u_2 = m_1v_1+m_2v_2[/tex]

where,

m₁ = mass of hockey player = 97 kg

m₂ = mass of puck = 0.15 kg

u₁ = u₂ = initial velocities of puck and player = 0 m/s

v₁ = velocity of player after collision = ?

v₂ = velocity of puck after hitting = 48 m/s

Therefore,

[tex](97\ kg)(0\ m/s)+(0.15\ kg)(0\ m/s)=(97\ kg)(v_1)+(0.15\ kg)(48\ m/s)\\\\v_1 = -\frac{(0.15\ kg)(48\ m/s)}{97\ kg} \\v_1 = - 0.074 m/s[/tex]

negative sign here shows the opposite direction.

Now, we calculate the time taken by puck to move 14.5 m:

[tex]s_2 =v_2t\\\\t = \frac{s_2}{v_2} = \frac{14.5\ m}{48\ m/s} \\\\t = 0.3\ s[/tex]

Now, the distance covered by the player in this time will be:

[tex]s_1 = v_1t\\s_1 = (0.074\ m/s)(0.3\ s)[/tex]

s₁ = 0.022 m