Object A is 8 kg and at rest and Object B is 16 kg and moving at 10 m/s to the Right. If Object B hits Object A and Object B is at rest after the collision, what is the velocity of Object A after the collision? Why?

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

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:

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:

Ray A - incident ray

Ray B - reflected ray

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

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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.

Force is a push or a pull that changes or trends to change the state of rest or uniform motion of an object or changes the direction or shape of an object. It causes objects to accelerate. SI unit is Newton.

1) Can change the state of an object : For example, pushing a heavy stone in order to move it.

2) May change the speed of an object if it is already moving. For example, catching a ball hit by a batsman.

3) May change the direction of motion of an object.

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:

12kWhr

Explanation:

Energy = Power * Time

Power = 4kW

Time = 3hrs

Substitute into the formula

Energy used up = 4kW * 3hrs

Energy used up = 12kWhr

Answer:

coz teaching is their profession.

Answer:

a. [tex]1000\ kg(\frac{1000\ g}{1\ kg}) = 1\ x\ 10^{6}\ g[/tex]

b. [tex]50\ m (\frac{100\ cm}{1\ m} ) = 5000\ cm[/tex]

c. "Nano" is 10⁻⁹ , so there are 10⁻⁹ meter in a nm (OR)  "Nano" is 10⁻⁹ , so there are 10⁹ nm in a meter.

d. micro is 10⁻⁶, so 1kg is 10⁹ ug​  

Explanation:

a.

The conversion factor is written inverted. The correct statement will be:

[tex]1000\ kg(\frac{1000\ g}{1\ kg}) = 1\ x\ 10^{6}\ g[/tex]

b.

The values in the conversion factor used are wrong. The correct statement will be:

[tex]50\ m (\frac{100\ cm}{1\ m} ) = 5000\ cm[/tex]

c.

Change of units is the mistake here. The correct statement will be:

"Nano" is 10⁻⁹ , so there are 10⁻⁹ meter in a nm (OR)  "Nano" is 10⁻⁹ , so there are 10⁹ nm in a meter.

d.

the conversion will be as follows:

[tex]1\ kg(\frac{1000\ g}{1\ kg})(\frac{1\ \mu g}{10^{-6}\ g}) = 10^9 \mu g[/tex]

therefore, the correct statement will be:

micro is 10⁻⁶, so 1kg is 10⁹ ug​

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]

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:

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:

answer 8N 8N

la tensión del ladrillo inferior es 8N ya que los ladrillos son idénticos, decido 24 por 3 para darme 8

la tensión entre el ladrillo superior y el medio también es 8N

Answer:

[tex]F = \frac{m {v}^{2} }{r} \\ = \frac{82 \times { (\frac{590 \times 1000}{3600} )}^{2} }{970} \\ = 2270.6 \: newtons[/tex]

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

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.

[tex]\frac{dx}{dt} = 2.5 \: \frac{cm}{sec} [/tex]

Explanation:

The volume of a cube is V = x^3. Taking the time derivative of this expression, we get

[tex] \frac{dV}{dt} = 3 {x}^{2} \frac{dx}{dt} [/tex]

or

[tex]\frac{dx}{dt} = \frac{1}{3 {x}^{2}} \frac{dV}{dt} [/tex]

We know that dV/dt = 30 cm^3/sec so the value of dx/dt when x = 2 cm is

[tex]\frac{dx}{dt} = \frac{1}{3 {(2 \: cm)}^{2}}(30 \: \frac{ {cm}^{3} }{sec} ) = 2.5 \: \frac{cm}{sec} [/tex]

Answer:

Explanation:

[tex]V=x^3\\\\\frac{dV}{dt}=3x^2\frac{dx}{dt}\\\\30\frac{cm^3}{s}=3x^2\frac{dx}{dt}\\\\\frac{dx}{dt}=\frac{30\frac{cm^3}{s}}{3x^2}~at~x=2cm,~\frac{dx}{dt}=\frac{30\frac{cm^3}{s}}{3*(2cm)^2}=\frac{5}{2}\frac{cm}{s}[/tex]

Answer: hello your question is incomplete below is the missing part

A 69-kg petty thief wants to escape from a third-story jail window. Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg.

answer:

To 2 significant Figures = 1.6 m/s^2

Explanation:

Calculate the magnitude of minimum acceleration at which the thief can descend

we apply the relation below

Mg - T = Ma  --- ( 1 )

M = 69kg

g = 9.81

T = 58 * 9.81

a = ? ( magnitude of minimum acceleration)

From equation 1

a = [ ( 69 * 9.81 ) - ( 58 * 9.81 ) ] / 69

  = 1.5639 m/s^2

To 2 significant Figures = 1.6 m/s^2

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:

Person's body pulling on the earth

Explanation:

The weight of an object is the attraction of a person's body toward the Earth. The weight of an object is given by :

Weight (W) = mass (m) × acceleration due to gravity (g)

We know that, for an action there is an equal and opposite reaction.

So, there must be a reaction force for the weight of an object. The reaction force must be the person's body pulling on the earth.

Explanation:

Let's set the x-axis to be parallel to the and positive up the plane. Likewise, the y-axis will be positive upwards and perpendicular to the plane. As the problem stated, we are going to assume that m1 will move downwards so its acceleration is negative while m2 moves up so its acceleration is positive. There are two weight components pointing down the plane, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex] and two others pointing up the plane, the two tensions T along the strings. There is a normal force N pointing up from the plane and two pointing down, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex]. Now let's apply Newton's 2nd law to this problem:

x-axis:

[tex]m1:\:\:\:\displaystyle \sum_i F_i = T - m_1g \sin \theta = - m_1a\:\:\:\:(1)[/tex]

[tex]m2:\:\:\:\displaystyle \sum_i F_i = T - m_2g \sin \theta = m_2a\:\:\:\:(2)[/tex]

y-axis:

[tex]\:\:\:\displaystyle \sum_i F_i = N - m_1g \cos \theta - m_2g \cos \theta = 0[/tex]

Use Eqn 1 to solve for T,

[tex]T = m_1(g \sin \theta - a)[/tex]

Substitute this expression for T into Eqn 2,

[tex]m_1g \sin \theta - m_1a - m_2g \sin \theta = m_2a[/tex]

Collecting all similar terms, we get

[tex](m_1 + m_2)a = (m_1 - m_2)g \sin \theta[/tex]

or

[tex]a = \left(\dfrac{m_1 - m_2}{m_1 + m_2} \right)g \sin \theta[/tex]

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:

The moon is actually quite dim.

Explanation:

compared to other astronomical bodies. The moon only seems bright in the night sky because it is so close to the earth and because the trees, houses, and fields around you are so dark at night. In fact, the moon is one of the least reflective objects in the solar system.

Answer:

It reflects the light send from the sun.

Explanation:

If the moon is between you and the sun, you will see the back of it which doesnt reflect light.