a shiny object reflects light. will a shiny object make a shadow when light shine on it? explain your answer​

Answer:

1) true 2) true 3) true 4) true 5) false 6) false

Explanation:

let the volume be x

19 = 38/x

x=38/19

x=2

volume is 2 cm^3

Answer:

What’s the question??

Explanation:

The density of water is approximately 1 g/cm^3. Therefore, the volume of 2800 g of water would be 2800 cm^3 because density is mass/volume, and so volume is mass/density.

Since this volume is inside a cubic box, the length of each side of the cube (a, for instance) could be found by taking the cubic root of the volume. This is because the volume of a cube is calculated by a^3 (length of one side cubed). Hence, a = cube root of 2800 cm^3 ≈ 14.1 cm.

Converting centimeters to meters (as 1 meter is equal to 100 centimeters), we get approximately 0.141 meters.

So the filled cubic box has a side length of approximately 0.141 m.

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

\(a = \frac{f}{m} \\ \)

where

f is the force

m is the acceleration

From the question we have

\(a = \frac{6.2}{3.2} \\ = 1.9375\)

We have the final answer as

Hope this helps you

Given data:

* The speed of the water coming out of the pipe is,

* The diameter of the left side of pipe is,

* The diameter of the right side of the pipe is,

Solution:

(a). The area of the right side of pipe is,

The radius of the right side of pipe is,

Thus, the area of the right side of pipe is,

The volume of the water coming out from the right side of the pipe is,

Where t is the time taken, v_1 is the velocity of the water, and A_1 is the area of right side of pipe,

Substituting the known values,

Thus, the volume of water flows into the atmosphere is 1.97 meter cube.

Answer:

when it has a fewer amount of electrons than protons

The horizontal component of the force pushing him forward is 37 N.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Frictional force due to ice, F =   μN

Normal force N = mg = 75*9.8 = 735 N

Frictional force = 0.05*735 = 36.75 N

Coefficient of friction, μ for ice is 0.05

The horizontal component of the force pushing him forward is 37 N.

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Answer:

I = 0.483 kgm^2

Explanation:

To know what is the moment of inertia I of the boxer's forearm you use the following formula:

\(\tau=I\alpha\)  (1)

τ: torque exerted by the forearm

I: moment of inertia

α: angular acceleration = 125 rad/s^2

You calculate the torque by using the information about the force (1.95*10^3 N) and the lever arm (3.1 cm = 0.031m)

\(\tau=Fr=(1.95*10^3N)(0.031m)=60.45J\)

Next, you replace this value of τ in the equation (1) and solve for I:

\(I=\frac{\tau}{\alpha}=\frac{60.45Kgm^2/s ^2}{125rad/s^2}=0.483 kgm^2\)

hence, the moment of inertia of the forearm is 0.483 kgm^2

To convert 813 into grams if it is already in grams, then the value remains as 813 grams. If it is in pounds, 813 pounds would be approximately 368,646.696 grams. If it is in kilograms, 813 kilograms would be equal to 813,000 grams.

To convert 813 into grams, we need to know the unit of measurement you're referring to. Grams are typically used to measure the weight of objects, but without additional context, it's difficult to provide an accurate answer.

If you are referring to 813 grams, then the conversion is straightforward, as it is already in grams.

If you are referring to another unit, such as pounds or kilograms, we can convert it to grams:

If you meant 813 pounds, one pound is approximately equal to 453.592 grams. Therefore, 813 pounds would be approximately 368,646.696 grams.

If you meant 813 kilograms, one kilogram is equal to 1,000 grams. Therefore, 813 kilograms would be equal to 813,000 grams.

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Answer:

Explanation:

It's graph A because the pressure in the tire is increasing as the amount of air going into it increases. B says the pressure drops exponentially as air goes in, and C says that the pressure stays the same as air goes in. Pressure in a tire increases proportionally to the amount of air in it.

In space, there is no air to carry sound waves. Sound waves are vibrations that travel through a medium, such as air, water, or solid objects. Without a medium to carry the vibrations, sound cannot be transmitted. This is why astronauts in space cannot hear one another or any other sounds. They must use radios or other communication devices to communicate with one another.

Answer:

Below

Explanation:

Unless they have an electronic intercom in their spacesuits. there is no air between them to conduct the soundwaves.

If the current is held constant, an increase in voltage will result in an increase in resistance.

hope it helps...!!!

The average acceleration of the bus is −6.00 m/s².

a = ? ; Δt = 1.50 s ; v1 = 9.00 m/s ; v2 = 0.00m/s

The formula to use in this case is the definition of acceleration stating that the acceleration a = Δv / Δt or a = (v2- v1) / Δt where v2 is the final velocity and v1 is the initial velocity.

Solving for the acceleration a

a = (v2- v1) / Δt

a = (0 - 9.00 m/s) / 1.50 s

a = - 9.00 m/s / 1.50 s

a = −6.00 m/s^2

The average acceleration of the bus is −6.00 m/s^2.

The complete question is,
When a bus comes to a sudden stop to avoid hitting a dog, it slows from 9.00 m/s to 0.00m/s in 1.50 seconds. What is the average acceleration of the bus?

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Answer:

(a) 22 kN

(b) 36 kN, 29 kN

(c) left will decrease, right will increase

(d) 43 kN

Explanation:

(a) When the truck is off the bridge, there are 3 forces on the bridge.

Reaction force F₁ pushing up at the first support,

reaction force F₂ pushing up at the second support,

and weight force Mg pulling down at the middle of the bridge.

Sum the torques about the second support.  (Remember that the magnitude of torque is force times the perpendicular distance.  Take counterclockwise to be positive.)

∑τ = Iα

(Mg) (0.3 L) − F₁ (0.6 L) = 0

F₁ (0.6 L) = (Mg) (0.3 L)

F₁ = ½ Mg

F₁ = ½ (44.0 kN)

F₁ = 22.0 kN

(b) This time, we have the added force of the truck's weight.

Using the same logic as part (a), we sum the torques about the second support:

∑τ = Iα

(Mg) (0.3 L) + (mg) (0.4 L) − F₁ (0.6 L) = 0

F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.4 L)

F₁ = ½ Mg + ⅔ mg

F₁ = ½ (44.0 kN) + ⅔ (21.0 kN)

F₁ = 36.0 kN

Now sum the torques about the first support:

∑τ = Iα

-(Mg) (0.3 L) − (mg) (0.2 L) + F₂ (0.6 L) = 0

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.2 L)

F₂ = ½ Mg + ⅓ mg

F₂ = ½ (44.0 kN) + ⅓ (21.0 kN)

F₂ = 29.0 kN

Alternatively, sum the forces in the y direction.

∑F = ma

F₁ + F₂ − Mg − mg = 0

F₂ = Mg + mg − F₁

F₂ = 44.0 kN + 21.0 kN − 36.0 kN

F₂ = 29.0 kN

(c) If we say x is the distance between the truck and the first support, then using our equations from part (b):

F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L − x)

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (x)

As x increases, F₁ decreases and F₂ increases.

(d) Using our equation from part (c), when x = 0.6 L, F₂ is:

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L)

F₂ = ½ Mg + mg

F₂ = ½ (44.0 kN) + 21.0 kN

F₂ = 43.0 kN

Answer:

Explanation:

given:

weight of bridge = 44 kN

weight of truck = 21 kN

a) truck is off the bridge

since the bridge is symmetrical, left support is equal to right support.

Left support = Right support = 44/2

Left support = Right support = 22 kN

b) truck is positioned  as shown.

to get the reaction at left support, take moment from right support = 0

∑M at Right support = 0

Left support (0.6) - weight of bridge (0.3) - weight of truck (0.4) = 0

Left support = 44 (0.3) + 21 (0.4)  

                                  0.6

Left support = 36 kN

Right support = weight of bridge + weight of truck - Left support

Right support = 44 + 21 - 36

Right support = 29 kN

c)

as the truck continues to drive to the right, Left support will decrease

as the truck get closer to the right support,  Right support will increase.

d) truck is directly under the right support, find reaction at Right support?

∑M at Left support = 0

Right support (0.6) - weight of bridge (0.3) - weight of truck (0.6) = 0

Right support = 44 (0.3) + 21 (0.6)  

                                  0.6

Right support = 43 kN

The electron number cannot be used instead because electrons can be removed from or added to an atom (option C)

The element of an atom is determined by its proton count, while the electron count can exhibit variability. Take, for instance, a sodium atom, which encompasses 11 protons and 11 electrons. However, it has the capacity to relinquish one electron, transforming into a sodium ion housing only 10 electrons.

This occurs due to the relatively loose binding of electrons to the nucleus, enabling their removal through the influence of an electric field or alternative mechanisms.

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The total distance is 105 kilometers and the total displacement is 15 kilometers east. Option C

To calculate the total distance, we add the distances traveled in each leg of the journey: 60 kilometers (from A to B) + 45 kilometers (from B back to A) = 105 kilometers.

However, displacement refers to the change in position of an object in a straight line from its starting point to its ending point. In this case, since the boat starts and ends at the same point (A), the total displacement is zero.

Hence The total distance is 105 kilometers and the total displacement is 15 kilometers east.

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Answer:

F_{total}= 8.25 \(\frac{G m^2}{L^2}\),      θ’= 194º

Explanation:

To solve this problem we must use the law of universal gravitation and vectorly add the forces

         F = \(G \frac{m_{1} m_{2} }{r^{2} }\)

Let us call the mass m with the subscript 1, the mass 3m with the subscript 3 and the mass 5m with the subscript 5, the total force on particle 1 is

          F_total = F₁₅ + F₁₃

The bold are vectors, in the attachment we can see a diagram of the angles and the forces, the distance between the masses is

           r = L

let's find the force between m1 and m5

           F₁₅ = G m₁ m₅ / r²

           F₁₅ = G m 5m / L²

           F₁₅ = G 5m² / L²

this force is on the line that joins the two masses, let's use trigonometry to decompose this force

           cos 30 = F₁₅ₓ / F₁₅

           sin 30 = \(\frac{Fx_{15y} }{F_{15} }\)

           F₁₅ₓ = F₁₅ cos 30

           \(F_{15y}\) = F₁₅ sin 30

equally with the force between mass 1 and mass 3

            F₁₃ = -G 3 m² / L²

            F₁₃ₓ = F₁₃ cos 30

            F_{13y} = F₁₃ sin 30

to find the total force we can add each component independently, see attached

X axis

           F_total x = -F₁₅ₓ + F₁₃ₓ

           F_total x = -G 5m2 / L² + G 3m² / L²

           F_total x = - G 2m² / L²

Y axis  

           F_total y = - F_{15y} - F_{13y}

           F_total y = - G 5m² / L² - G 3 m² / L²

           F_toal y = - G 8 m² / L²

We can give the result in two ways

1) F_total = - G m ^ 2 / L² (2 i ^ + 8 j ^)

2) in the form of module and angle.

Let's use the Pythagorean theorem

       \(F_{total}^{2} = F_{total x}^{2} + F_{total y}^2\)

       \(F_{total}\) = \(\frac{G m^{2} }{L^2} \sqrt{(2^2 + 8^2)}\)

       F_{total}= 8.25 \(\frac{G m^2}{L^2}\)

with trigonometry

        tan θ = \(\frac{F_{total y} }{F_{total x} }\)

        tan θ = 8/2

        θ = tan⁻¹ 4

        θ = 76º

if we measure this angle from the positive side of the x axis in a counterclockwise direction

       θ’= 270 -76

       θ’= 194º

The particle B will pass particle A in 1.10s.

Using kinematic formula,

d = Vini x t + 1/2 at²

where,

d is distance

Vini is velocity initial

a is acceleration

t is time

So, putting values

7.7 t + 1/2 x 1.8 t^² = 2.9 t + 1/2 x 5.7 t²

7.7 t + 2.7 t² = 2.9 t + 8.55 t²

t + t² / t + t² = 2.9 + 8.55 / 7.7 + 2.7

t = 1.10s

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Answer:

The speed after 2 seconds is 14.4 m/s

Explanation:

Vertical Launch Upwards

In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.

The speed changes from an initial value vo to a final value vf, given by:

\(v_f=v_o-g.t\)

Where g is the gravitational acceleration (\(9.8\ m/s^2\))

The tennis ball is shot up with an initial speed of vo=34 m/s. We are required to find the final speed at t=2 s:

\(v_f=34-9.8*2\)

\(v_f=14.4\ m/s\)

The speed after 2 seconds is 14.4 m/s

the velocity after 2 seconds of launch is 14.4 m/s

the initial velocity(a) is 34m/s

the time(t) is 2 seconds

acceleration(a) due to gravity is 9.8g

the formula is

v= u - at

v= 34 -(9.8)(2)

= 34 - 19.6

= 14.4 m/s

hence the final velocity is 14.4 m/s

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Answer:

\(2.4375\ \text{m/s}\)

Explanation:

\(m_1\) = Mass of cannon ball = 39 kg

\(m_2\) = Mass of performer = 65 kg

\(u_1\) = Initial horizontal component of cannon ball's velocity = 6.5 m/s

\(u_2\) = Initial horizontal component of performer's velocity = 0

v = Velocity of combined mass

As the momentum of the system is conserved we have

\(m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{39\times 6.5+0}{39+65}\\\Rightarrow v=2.4375\ \text{m/s}\)

The speed of the performer immediately after catching the cannon ball is \(2.4375\ \text{m/s}\).

Answer:

NO

Explanation:

because it just doesn't.

A piece of copper at 100°C is transferred to a copper calorimeter with a liquid at 30°C. The final temperature is 50°C. By applying the principle of conservation of energy, the specific heat capacity of the liquid is calculated to be approximately 2100 J/kg/°C.

To calculate the specific heat capacity of the liquid, we can apply the principle of conservation of energy. The heat lost by the copper piece will be equal to the heat gained by the liquid and calorimeter.

The heat lost by the copper piece can be calculated using the formula:

Heat lost = Mass of copper × Specific heat capacity of copper × Temperature change

Given:

Mass of copper = 400 g

Specific heat capacity of copper = 390 J/kg/°C (assuming it remains constant)

Temperature change of copper = 100°C - 50°C = 50°C

Heat lost = 400 g × 390 J/kg/°C × 50°C

Heat lost = 7,800,000 J

The heat gained by the liquid and calorimeter can be calculated using the formula:

Heat gained = (Mass of liquid + Mass of calorimeter) × Specific heat capacity of liquid × Temperature change

Given:

Mass of liquid = 100 g

Mass of calorimeter = 10 g

Temperature change of liquid = 50°C - 30°C = 20°C

Heat gained = (100 g + 10 g) × Specific heat capacity of liquid × 20°C

Now, by equating the heat lost and heat gained:

7,800,000 J = (110 g) × Specific heat capacity of liquid × 20°C

Specific heat capacity of liquid = 7,800,000 J / (110 g × 20°C)

Specific heat capacity of liquid ≈ 3545.45 J/kg/°C

Therefore, the specific heat capacity of the liquid is approximately 3545.45 J/kg/°C.

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umm actually is 10 points

Classical mechanics describes the motion of objects on a macroscopic scale, while quantum mechanics deals with the behavior of particles on a microscopic scale. Classical mechanics is deterministic, meaning that it predicts precise outcomes based on initial conditions, while quantum mechanics is probabilistic, providing probabilities of different outcomes. Classical mechanics follows the principle of causality, where every effect has a specific cause, whereas quantum mechanics introduces inherent uncertainty and wave-particle duality. Classical mechanics is well-suited for describing everyday objects, while quantum mechanics is necessary to explain the behavior of particles at the atomic and subatomic levels.

~~~Harsha~~~

Answer:

a)  W = 1,639 10⁶ J, b)     v = 38.60 m / s, c)    k = 3.78 10⁵ N / m

Explanation:

a) In this part they ask us the work done by the motor, let's use the work energy theorem

            W = ΔEm = Em_f - Em₀

            W = mh y₂ - mgy₁

            W = mg (y₂ -y₁)

            W = 2200 9.8 (80 -4)

            W = 1,639 10⁶ J

b) for this part let's use conservation of energy

starting point. Highest point

            Em₀ = U = m g y₂

final point . Punoit in contact with the spring

           Em_f = K + U

           Em_f = ½ m v² + m g y₁

           Em₀ = Em_f

           m g y₂ = ½ m v² + mg y₁

           v = √(2g (y₂ - y₁))

           v = √(2 9.8 (80-4))

           v = 38.60 m / s

c) the stiffness of the spring

starting point. Just when it comes into contact with the spring

           Em₀ = K + U

           Em₀ = ½ m v² + mgy₁

Final point. With the spring at maximum compression x = 3 m

           Em_f = Ke + U

           Em_f = ½ k x² + m g y₃

           Em₀ = Em_f

           ½ m v² + mgy₁ = ½ k x² + m g y₃

           ½ k x² = ½ m v² + m g (y₁ - y₃)

let's calculate

            ½ 3² k = ½ 2200 38.60² + 2200 9.8 (4 -1)

            4.5 k = 1.639 10⁶ +6.468 10⁴ = 1.7036 10⁶

            k = 1.7036 10⁶ / 4.5

             k = 3.78 10⁵ N / m