a vechile having a mass of 500kg is moving with a speed of 10m/s.Sand is dropped into it at the rate of 10kg/min.What force is needed to keep the vechile moving with uniform speed​

Answers

Answer 1

Answer:

1.67 N

Explanation:

Applying,

F = u(dm/dt)+m(du/dt)................ Equation 1

Where F = force, m = mass of the vehicle, u = speed.

Since u is constant,

Therefore, du/dt = 0

F = u(dm/dt)............... Equation 2

From the question,

Given: u = 10 m/s, dm/dt = 10 kg/min = (10/60) kg/s

Substitute these values into equation 2

F = 10(10/60)

F = 100/60

F = 1.67 N


Related Questions

In 1.0 second, a battery charger moves 0.50 C of charge from the negative terminal to the positive terminal of a 1.5 V AA battery.
Part A:
How much work does the charger do? Answer is 0.75 J
Part B:
What is the power output of the charger in watts?

Answers

Answer:

W = Q * V     work done on charge Q

A. W = .5 C * 1.5 V = .75 Joules

B. P = W / t = .75 J / 1 sec = .75 Watts

Which physical phenomenon is illustrated by the fact that the prism has different refractive indices for different colors

Answers

Answer:

The incoming white light is composed of light of different colors,

Since these different colors have different refractive indices they are refracted at different angles from one another.

The output light is then separated by color creating a color spectrum.

Since n is greater for shorter wavelengths  (violet colors) these wavelengths are refracted thru the larger angles.

The image of the object formed by the lens is real, enlarged and inverted. What is the kind of lens ?​

Answers

Answer:

Converging (convex) lens.

Explanation:

A lens can be defined as a transparent optical instrument that refracts rays of light to produce a real image.

Basically, there are two (2) main types of lens and these includes;

I. Diverging (concave) lens.

II. Converging (convex) lens.

A converging (convex) lens refers to a type of lens that typically causes parallel rays of light with respect to its principal axis to come to a focus (converge) and form a real image. Thus, this type of lens is usually thin at the lower and upper edges and thick across the middle.

Basically, the image of the object formed by a converging (convex) lens. lens is real, enlarged and inverted.

A uniform horizontal bar of mass m1 and length L is supported by two identical massless strings. String A Both strings are vertical. String A is attached at a distance d

Answers

Answer:

a)  T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex] ,  b) T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]

c)  x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex],  d)  m₂ = m₁  ( [tex]\frac{ L}{2d} -1[/tex])

Explanation:

After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act

a) The tension of string A is requested

The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive

      ∑ τ = 0

      T_A d - W₂ x -W₁ L/2 = 0

      T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]

b) the tension in string B

we write the expression of the translational equilibrium

       ∑ F = 0

       T_A - W₂ - W₁ - T_B = 0

       T_B = T_A -W₂ - W₁

       T_ B =   [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]  - g m₂ - g m₁

       T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]

c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system

         T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0

at the point that begins to rotate T_B = 0

          g m₂ (d -x) -  g m₁  (0.5 L -d) + 0 = 0

          m₂ (d-x) = m₁ (0.5 L- d)

          m₂ x = m₂ d - m₁ (0.5 L- d)

          x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex]

 

d) The mass of the block for which it is always in equilibrium

this is the mass for which x = 0

           0 = d - \frac{m_1}{m_2} \  \frac{L}{2d}

         [tex]\frac{m_1}{m_2} \ (0.5L -d) = d[/tex]

          [tex]\frac{m_1}{m_2} = \frac{ d}{0.5L-d}[/tex]

          m₂ = m₁  [tex]\frac{0.5 L -d}{d}[/tex]

          m₂ = m₁  ( [tex]\frac{ L}{2d} -1[/tex])

Container A and container B hold samples of the same ideal gas. The volume and the pressure of container A is equal to the volume and pressure of container B, respectively. If Container A has half as many molecules of the ideal gas in it as Container B does, then which of the following mathematical statements is correct regarding the absolute temperatures TA and TB in Container A and Container B. respectively?
A. TA = TB/2.
B. TA = 4TB.
C. TA = TB/4.
D. TA = 2TB.
E. TA = TB

Answers

Answer:

A. TA = TB/2.

Explanation:

Since container A has half as many molecules of the ideal gas in it as container B. Therefore, container A will have half the volume of gas as in container B:

[tex]V_A = \frac{1}{2}V_B[/tex]

Now, from Charle's Law:

[tex]\frac{V_A}{T_A}=\frac{V_B}{T_B}\\\\\frac{1}{2}\frac{V_B}{T_A}=\frac{V_B}{T_B}\\\\T_A = \frac{T_B}{2}[/tex]

Hence, the correct option is:

A. TA = TB/2.

Which best describes the relationship between heat,intemal energy, and thermal energy?
Internal energy is heat that flows and heat is the part of thermal energy that can be transferred
Internal energy is thermal energy that flows, and thermal energy is the part of heat that can be transferred,
Thermal energy is heat that flows, and heat is the part of intemal energy that can be transferred
Heat is thermal energy that flows, and hennal energy is the part of internal energy that can be transferred.

Answers

Answer:

It is all a thermodynamic system that is highly related to each other.

Explanation:

Because they are in the physics of thermodynamics it is not wrong to say they follow the same thermodynamic rules and has highly the same properties of energy.

Two charged particles exert an electric force of 27 N on each other. What will the magnitude of the force be if the distance between the particles is reduced to one-third of the original separation

Answers

Answer:

243 N

Explanation:

The formula for electromagnetic force is F= Kq1q2/r^2

where r is the distance between the charges, if the distance between the charges is reduced by 1/3 then F will increase by 9 [(1/3r)^2 becomes 1/9r which is 9F] so 27*9 is 243N

The working substance of a certain Carnot engine is 1.50 mol of an ideal monatomic gas. During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles, while during the adiabatic expansion the volume increases by a factor of 5.7. The work output of the engine is 940 J in each cycle. Compute the temperatures of the two reservoirs between which this engine
operates.

Answers

Answer:

The hot temperature is 157.5 K

The cold temperature is 48.8 K

Explanation:

Step 1: Data given

The working substance of a certain Carnot engine is 1.50 mol of an ideal monatomic gas.

The volume increases by a factor of 5.7

The work output of the engine is 940 J in each cycle.

During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles. This means V2 = 2*V1 (and V4 = 2*V3)

Step 2:For a carnot engine:

V2/V1 = V4/V3

Work = nR((T1)ln(V2/V1) - (T2)ln(V4/V3))

⇒with Work = the work done in the cycle = 940J

⇒with n = the number of moles = 1.50 moles

⇒with R = the gas constant = 8.314 J/mol*K

⇒with T1 = the hot temperature

⇒With T2⇒ the cold temperature

where R = 8.31 J/mol K Gas Constant

940J = 1.5moles * 8.314 J/mol*K * (T1*ln(2) - T2*ln(2)))

940 = 1.5 * 8.314 ln(2) * (T1-T2)

(T1-T2) = 940 / (1.5*8.314*ln(2))

(T1-T2) = 108.7K

For the reversible adiabatic expansion: T2 = T1*(V1/V2)^(R/Cv). Where V2/V1 = 5.7 (Because during the adiabatic expansion the volume increases by a factor of 5.7)

For a monatomic ideal gas, Cv = 3/2R

When we combine both, we'll have:

T2 = T1*(1/5.7)^(R/3/2R)

T2 = T1*(1/5.7)^(2/3)

T2= T1 * 0.31

Since we know that (T1-T2) = 108.7K

we have:

T1 - 0.31T1= 108.7K

0.69T1 = 108.7K

T1 = 157.5K

T2 = 157.5*0.31 = 48.8K

A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surface. At a certain time, both engines turn on. The one on the left gives the spacecraft an acceleration component in the x direction of
ax = 5.10 m/s2,
while the one on the back gives an acceleration component in the y direction of
ay = 7.30 m/s2.
The engines turn off after firing for 670 s, at which point the spacecraft has velocity components of
vx = 3670 m/s and vy = 4378 m/s.
What was the magnitude and the direction of the spacecraft's initial velocity before the engines were turned on? Express the magnitude as m/s and the direction as an angle measured counterclockwise from the +x axis.

magnitude m/s
direction ° counterclockwise from the +x-axis

Answers

Answer:

a)    v = 517.99 m / s,  b) θ = 296.3º

Explanation:

This is an exercise in kinematics, we are going to solve each axis independently

X axis

the acceleration is aₓ = 5.10 1 / S², they are on for t = 670 s and reaches a speed of vₓ=  3670 m / s, let's use the relation

           vₓ = v₀ₓ + aₓ t

           v₀ₓ = vₓ - aₓ t

           v₀ₓ = 3670 - 5.10 670

           v₀ₓ = 253 m / s

Y axis  

the acceleration is ay = 7.30 m / s², with a velocity of 4378 m / s after

t = 670 s

          v_y = v_{oy} + a_y t

          v_{oy} = v_y - a_y t

          v_oy} = 4378 - 7.30 670

          v_{oy}  = -513 m / s

to find the velocity modulus we use the Pythagorean theorem

          v = [tex]\sqrt{v_o_x^2 + v_o_y^2}[/tex]

          v = [tex]\sqrt{253^2 +513^2}[/tex]

          v = 517.99 m / s

to find the direction we use trigonometry

         tan θ ’= [tex]\frac{v_o_y}{v_o_x}[/tex]

         θ'= tan⁻¹  [tex]\frac{voy}{voy}[/tex]  

         θ'= tan⁻¹ (-513/253)

         tea '= -63.7

the negative sign indicates that it is below the ax axis, in the fourth quadrant

to give this angle from the positive side of the axis ax

          θ = 360 -   θ  

          θ = 360 - 63.7

          θ = 296.3º

if Petrol diesel etc catches fire one should never try to extinguish in using water why?​

Answers

Answer:

because both petrol and diesel are oil

Explanation:

oil floats on water that's why if we will try to extinguish with water so the fire will float on water

hope u like my answer

please mark methe brainest

A 10.0 L tank contains 0.329 kg of helium at 28.0 ∘C. The molar mass of helium is 4.00 g/mol . Part A How many moles of helium are in the tank? Express your answer in moles.

Answers

Answer:

82.25 moles of He

Explanation:

From the question given above, the following data were obtained:

Volume (V) = 10 L

Mass of He = 0.329 Kg

Temperature (T) = 28.0 °C

Molar mass of He = 4 g/mol

Mole of He =?

Next, we shall convert 0.329 Kg of He to g. This can be obtained as follow:

1 Kg = 1000 g

Therefore,

0.329 Kg = 0.329 Kg × 1000 g / 1 Kg

0.329 Kg = 329 g

Thus, 0.329 Kg is equivalent to 329 g.

Finally, we shall determine the number of mole of He in the tank. This can be obtained as illustrated below:

Mass of He = 329 g

Molar mass of He = 4 g/mol

Mole of He =?

Mole = mass / molar mass

Mole of He = 329 / 4

Mole of He = 82.25 moles

Therefore, there are 82.25 moles of He in the tank.

An electrostatic paint sprayer has a 0.100 m diameter metal sphere at a potential of 30.0 kV that repels paint droplets onto a grounded object. (a) What charge (in C) is on the sphere?(b) What charge must a 0.100-mg drop of paint have to arrive at the object with a speed of 10.0 m/s?

Answers

Answer:

A) q = 1.67 × 10^(-7) C

B) q = 1.67 × 10^(-10) C

Explanation:

We are given;

Potential; V = 30 KV = 30000 V

Radius of sphere; r = diameter/2 = 0.1/2 = 0.05 m

A) To find the charge of the sphere, we will use the formula;

V = kq/r

Where;

q is the charge

k is electric force constant = 9 × 10^(9) N.m²/C²

Thus;

q = Vr/k

q = (30000 × 0.05)/(9 × 10^(9))

q = 1.67 × 10^(-7) C

B) Now, potential energy here is a formula; U = qV

However, for the drop of paint to move, the potential energy will be equal to the kinetic energy. Thus;

qV = ½mv²

q = mv²/2V

Where;

v is speed = 10 m/s

V = 30000 V

m = mass = 0.100 mg = 0.1 × 10^(-6) Thus;

q = (0.1 × 10^(-6) × 10²)/(2 × 30000)

q = 1.67 × 10^(-10) C

Your cell phone typically consumes about 300 mW of power when you text a friend. If the phone is operated using a lithium-ion battery with a voltage of 3.5 V, what is the current (in A) flowing through the cell-phone circuitry under these circumstances

Answers

Answer:

I = 0.0857 A

Explanation:

Given that,

Power consumed by the cellphone, P = 300 mW

The voltage of the battery, V = 3.5 V

Let I is the current flowing through the cell-phone. We know that,

P = VI

Where

I is the current

So,

[tex]I=\dfrac{P}{V}\\\\I=\dfrac{300\times 10^{-3}}{3.5}\\\\I=0.0857\ A[/tex]

So, the current flowing the cell-phone is 0.0857 A.

Drawing a shows a displacement vector (450.0 m along the y axis). In this x, y coordinate system the scalar components are Ax 0 m and Ay 450.0 m. Suppose that the coordinate system is rotated counterclockwise by 35.0, but the magnitude (450.0 m) and direction of vector remain unchanged, as in drawing b. What are the scalar components, Ax and Ay, of the vector in the rotated x, y coordinate system

Answers

Answer:

x ’= 368.61 m,  y ’= 258.11 m

Explanation:

To solve this problem we must find the projections of the point on the new vectors of the rotated system  θ = 35º

            x’= R cos 35

            y’= R sin 35

           

The modulus vector can be found using the Pythagorean theorem

            R² = x² + y²

            R = 450 m

we calculate

            x ’= 450 cos 35

            x ’= 368.61 m

            y ’= 450 sin 35

            y ’= 258.11 m

PLZ help asap :-/
............................ ​

Answers

Explanation:

[16]

[tex]\underline{\boxed{\large{\bf{Option \; A!! }}}} [/tex]

Here,

[tex]\rm { R_1} [/tex] = 2Ω[tex]\rm { R_2} [/tex] = 2Ω[tex]\rm { R_3} [/tex] = 2Ω[tex]\rm { R_4} [/tex] = 2Ω

We have to find the equivalent resistance of the circuit.

Here, [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] are connected in series, so their combined resistance will be given by,

[tex]\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\ [/tex]

[tex]\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\ [/tex]

[tex]\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\ [/tex]

Now, the combined resistance of [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] is connected in parallel combination with [tex]\rm { R_3} [/tex], so their combined resistance will be given by,

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\ [/tex]

Reciprocating both sides,

[tex]\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\ [/tex]

Now, the combined resistance of [tex]\rm { R_1} [/tex], [tex]\rm { R_2} [/tex] and [tex]\rm { R_3} [/tex] is connected in series combination with [tex]\rm { R_4} [/tex]. So, equivalent resistance will be given by,

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\ [/tex]

Henceforth, Option A is correct.

_________________________________

[17]

[tex]\underline{\boxed{\large{\bf{Option \; B!! }}}} [/tex]

Here, we have to find the amount of flow of current in the circuit. By using ohm's law,

[tex] \longrightarrow [/tex] V = IR

[tex] \longrightarrow [/tex] 3 = I × 3.33

[tex] \longrightarrow [/tex] 3 ÷ 3.33 = I

[tex] \longrightarrow [/tex] 0.90 Ampere = I

Henceforth, Option B is correct.

____________________________

[tex] \tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\ [/tex]

Which one is the better material to use for an inexpensive compass? hard iron, soft iron, or any conductor ​

Answers

Answer:

Soft iron

Explanation:

A particle charge of 2.7 µC is at the center of a Gaussian cube 55 cm on edge. What is the net electric flux through the surface?

Answers

Answer:

3.05×10⁵ Nm²C⁻¹

Explanation:

According to Gauss' law,

∅' = q/e₀............... Equation 1

Where ∅' = net flux through the surface, q = net charge, e₀ = electric permittivity of the space

From the question,

Given: q = 2.7 μC = 2.7×10⁻⁶ C,

Constant: e₀ = 8.85×10⁻¹² C²/N.m²

Substituting these values into equation 1

∅' = (2.7×10⁻⁶)/(8.85×10⁻¹²)

∅' = 3.05×10⁵ Nm²C⁻¹

An electric lamp consumes 60W at 220 volts. How many dry cells of 1.5 V and internal resistance 1 Ohm are required to glow the lamp?

Answers

Answer:

1. Number of dry cells of 1.5 V required is 40.

2. Number of internal resistance of 1 ohm required is 807

Explanation:

We'll begin by calculating the resistance. This can be obtained as follow:

Power (P) = 60 W

Voltage (V) = 220 V

Resistance (R) =?

P = V²/R

60 = 220² / R

Cross multiply

60 × R = 220²

60 × R = 48400

Divide both side by 60

R = 48400 / 60

R ≈ 807 Ohm

1. Determination of the number of dry cells of 1.5 V required.

Voltage (V) = 220

Dry Cells = 1.5 V

Number of dry cells (n) =?

n = Voltage / Dry cells

n = 60 / 1.5

n = 40

2. Determination of the number of internal resistance of 1 ohm required.

Resistance (R) = 807 Ohm

Internal resistance (r) = 1 ohm

Number of internal resistance (n) =?

n = R/r

n = 807 / 1

n = 807

SUMMARY:

1. Number of dry cells of 1.5 V required is 40.

2. Number of internal resistance of 1 ohm required is 807

The cable lifting an elevator is wrapped around a 1.1m diameter cylinder that is turned by the elevator's motor. The elevator is moving upward at a speed of 2.6 ms. It then slows to a stop, while the cylinder turns one complete revolution. How long does it take for the elevator to stop? Express your answer to two significant figures and include the appropriate units.

Answers

Answer:

t = 2.7 s

Explanation:

This is a kinematics problem.

How the elevator reduces its speed to zero by a distance equal to the length of the cylinder

        y = 2π r = 2 π d / 2

        y = π d

        y = π 1.1

        y = 3,456 m

now we can look for the acceleration of the system

        v² = v₀² - 2 a y

        0 = v₀² - 2 a y

        a = v₀² / 2y

        a = 2.6² / 2 3.456

        a = 0.978 m / s²

now let's calculate the time

        v = v₀ - a t

        0 = v₀ - at

         t = v₀ / a

         t = 2.6 /0.978

         t = 2.658 s

ask for the result with two significant figures

         t = 2.7 s

A system gains 1500J of heat and 2200J of work is done by the system on its surroundings. Determine the change in internal energy of the system

Answers

Answer:

-700

formula is heat gained - work done

The change in internal energy if A system gains 1500J of heat and 2200J of work is done by the system on its surroundings, is 700 joules.

What is Energy?

Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc.

Additionally, there is heat and work, which is energy being transferred from one body to another. Energy is always assigned based on its nature once it has been transmitted. Thus, heat transmitted may manifest as thermal energy while work performed may result in mechanical energy.

Given:

A system gains 1500J of heat and 2200J of work is done by the system on its surroundings,

Calculate the change in internal energy as shown below,

The change in internal energy = heat gained - work done

The change in internal energy = 1500 - 2200

The change in internal energy = -700 J

Thus, the change in internal energy is 700 joules.

To know more about Energy:

https://brainly.com/question/8630757

#SPJ5

Soap bubble coloring example:
(reflection, interference, refraction, diffraction)​

Answers

Explanation:

Interference is the example of soap bubble colouring

EXTRA INFO:

(LOOK AT THE IMAGE)

An incoming light ray is partly reflected by the top surface of the soap film and partly reflected by the bottom surface. The wave reflected from the bottom surface has traveled further (an extra distance equal to twice the thickness of the film) so emerges out of step with the top wave. When the two waves meet, they add together, and some colors are removed by destructive interference. Where the film is thickest, the bubble appears more blueish; where it's thinner, it will look more violet or magenta.

[tex]\huge\bold\color{black}{ANSWER}[/tex]

Soap bubble coloring example: INTERFERENCE

You walk into a room and you see 4 chickens on a bed 2 cows on the floor and 2 cats in a chair. How many legs are on the ground? (I know this answer just a riddle to see who knows it) (:

Answers

Answer:

18

Explanation:

I'm pretty sure I got it right

You have two identical beakers A and B. Each beaker is filled with water to the same height. Beaker B has a rock floating at the surface (like a pumice stone). Which beaker, with all its contents, weighs more. Or are they equal?

Answers

Answer:

a) if we assume that the water does not spill, Beaker B weighs more than beaker S, or which in this case Beaker A weighs more

b) If it is spilled in water the weight of the two beakers is the same

Explanation:

The beaker weight is

 beaker A

          W_total = W_ empty + W_water

Beaker B

            W_total = W_ empty + W_water + W_roca

a) if we assume that the water does not spill, Beaker B weighs more than beaker S, or which in this case Beaker A weighs more

b) If it is spilled in water, the weight of the two beakers is the same because the amount of liquid spilled and equal to the weight of the stone, therefore the two beakers weigh the same

Part B
What is the approximate amount of thrust you need to apply to the lander to keep its velocity roughly constant? Explain why, using Newton's first
law of motion.

Answers

Answer:

Force is zero.

Explanation:

According to the Newton's second law, when an object is moving with an acceleration the force acting on the object is directly proportional to the rate of change of momentum of the object.

F = m a

if the object is moving with uniform velocity, the acceleration is zero, and thus, the force is also zero.  

Answer: Near the moon’s surface, a thrust over 11,250 N but under 13,500 N would make it travel at a constant vertical velocity.

Explanation: .Newton’s first law of motion states that an object in motion continues to move in a straight line at a constant velocity unless acted upon by an unbalanced force. In accordance with this law, the lunar lander moves in a downward direction toward the surface of the moon under the influence of force due to gravity. A thrust somewhere between 11,250 and 13,500 balances this gravitational force out.

Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 2 m/s, exactly how fast (in m2/s) is the area of the spill increasing when the radius is 39 m?

Answers

Explanation:

The area of a circle of radius r is given by

[tex]A = \pi r^2[/tex]

Taking the derivative of A with respect to time t, we get

[tex]\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}[/tex]

We also know that

[tex]\dfrac{dr}{dt} = 2\:\text{m/s}\:\text{at}\:r = 39\:\text{m}[/tex]

[tex]\dfrac{dA}{dt} = 2\pi (39\:\text{m})(2\:\text{m/s})= 490\:\text{m}^2\text{/s}[/tex]

In Trial II, the same spring is used as in Trial I. Let us use this information to find the suspended mass in Trial II. Use 0.517 ss for the value of the period.
Trial 1 Spring constant is 117N/m, period of oscillations .37s, mass of the block is .400kg .
Trial 2 oscillation period is .52s

Answers

Answer:

[tex]M_2=0.79kg[/tex]

Explanation:

From the question we are told that:

Period [tex]T=0.517s[/tex]

Trial 1

Spring constant [tex]\mu=117N/m[/tex]

Period [tex]T_1=0.37[/tex]

Mass [tex]m=0.400kg[/tex]

Trial 2

Period [tex]T_2=0.52[/tex]

Generally the equation for Spring Constant  is mathematically given by

\mu=\frac{4 \pi^2 M}{T^2}

Since

[tex]\mu _1=\mu_2[/tex]

Therefore

[tex]\frac{4 \pi^2 M_1}{T_1^2}=\frac{4 \pi^2 M_2}{T_2^2}[/tex]

[tex]M_2=M_1*(\frac{T_2}{T_1})^2[/tex]

[tex]M_2=0.400*(\frac{0.52}{0.37}})^2[/tex]

[tex]M_2=0.79kg[/tex]

A car accelerates at 2 meters/s/s. Assuming the car starts from rest how far will it travel in 10 seconds

Answers

Answer:

Distance = velocity x time, so 10 m/s X 10 s = 100 m

Explanation:

If you accelerate at 2 m/s^2 for 10 seconds, at the end of the 10 seconds you are moving at a rate of 20 m/s.

V(f) = V(i) + a*t

Final velocity = initial velocity + acceleration x time

Your average velocity will be half of your final, because you accelerated at a constant rate. So your average velocity is 10 m/s.

Distance = velocity x time, so 10 m/s X 10 s = 100 m

Answer:

100 m

Explanation:

Given,

Initial velocity ( u ) = 0 m/s

Acceleration ( a ) = 2 m/s^2

Time ( t ) = 10 sec s

To find : Displacement ( s ) = ?

By 2nd equation of motion,

s = ut + at^2 / 2

= ( 0 ) ( 10 ) + ( 2 ) ( 10 )^2 / 2

= 0 + ( 2 ) ( 100 ) / 2

= 200 / 2

s = 100 m

a baseball is thrown vertically upward with an initial velocity of 20m/s.
A,what maximum height will it attain? B,what time will elapse before it strike the ground?
C,what is the velocity just before it strike the ground?​

Answers

Answer:

Look at explanation

Explanation:

a)Only force acting on the object is gravity, so a=-g (consider up to be positive)

use: v^2=v0^2+2a(y-y0)

plug in givens, at max height v=0

0=400-19.6(H)

Solve for H

H= 20.41m

b) Use: y=y0+v0t+1/2at^2

Plug in givens

0=0+20t-4.9t^2

solve for t

t=4.08 seconds

c) v=v0+at

v=20-39.984= -19.984m/s

You are helping your friend move a new refrigerator into his kitchen. You apply a horizontal force of 275 N in the positive x direction to try and move the 61 kg refrigerator. The coefficient of static friction is 0.58. (a) How much static frictional force does the floor exert on the refrigerator

Answers

Answer:

f = 347.08 N

Explanation:

The frictional force exerted by the floor on the refrigerator is given as follows:

[tex]f = \mu R = \mu W[/tex]

where,

f = frictional force = ?

μ = coefficient of static friction = 0.58

W = Weight of refrigerator = mg

m = mass of refrigerator = 61 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]f = \mu mg\\f = (0.58)(61\ kg)(9.81\ m/s^2)\\[/tex]

f = 347.08 N

a nano second is what​

Answers

Answer:

one thousand-millionth of a second.

A nanosecond is an SI unit of time equal to one billionth of a second, that is, ​¹⁄₁ ₀₀₀ ₀₀₀ ₀₀₀ of a second, or 10⁻⁹ seconds. The term combines the prefix nano- with the basic unit for one-sixtieth of a minute. A nanosecond is equal to 1000 picoseconds or ​¹⁄₁₀₀₀ microsecond.  

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