A 1-kg iron frying pan is placed on a stove. The pan increases from 20°C to 250°C. If the same amount of heat is added to a pan with a greater specific heat, what can you predict about the temperature of this second pan? A) The second pan would reach a lower temperature. B) The second pan would reach a higher temperature. C) The second pan would reach the same temperature. D) No conclusion can be made from this information.

Answer

6.6 km

The description of the problem is shown in the attached figure, where the line "d" represents the final displacement vector.

First, the trekker walked 3.3km in a 40 ° direction, as shown in the figure. We can write this vector in its Cartesian coordinates:

[tex]-3.3sin (40)x + 3.3cos (40)y[/tex]

Then the hiker walked 3.4 km in a 60 degree northwest direction.

We can write this as a vector in its Cartesian coordinates:

[tex]-3.4sin (60)x + 3.4cos (60)y[/tex].

When adding this two vectors we will obtain the final displacement "d"

[tex]d = [- 3.3sin(40) -3.4sin (60)]x + [3.3cos(40) + 3.3cos (60)]y\\[/tex]

[tex]d = -5.07x +4.23y\\[/tex]

To obtain the magnitude of this vector we calculate its module:

[tex]\sqrt{5.07 ^2 +4.23 ^ 2}[/tex]

Then the magnitude of the final displacement was:

6.6 km

10 N/m is the answer to "A spring extends by 10 cm when a mass of 100 g is attached to it. What is the spring constant."

A spring extends by 10 cm when a mass of 100 g is attached to it.  Spring constant will be 10.

The force a spring applies to items fastened to its ends is proportional to the distance the spring travels from its equilibrium length and is always pointed in the direction of equilibrium.

Consider a spring that has one end attached to a wall or ceiling and the other end being pulled or pushed by an object. The spring is pulled by the object, and the object is pulled by the spring.

The spring applies a force F to the object that is in the opposite direction as the free end's displacement. The equilibrium point of the spring's free end is at x = 0, and if the x-axis of a coordinate system is selected to be parallel to the spring.

Therefore, A spring extends by 10 cm when a mass of 100 g is attached to it.  Spring constant will be 10.

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The pet store would be the reference point because it is where he started and it will not move. Hope this helped.

The pet store would be the reference point because it is where he started motion and it will not move.

The phenomenon of an item changing its position with respect to time is known as motion in physics. In mathematics, displacement, distance, velocity, and acceleration are used to explain motion.

A reference point is a location or object that is used as a point of comparison to ascertain whether something is moving. When an object shifts in relation to a fixed point, it is said to be in motion. Good reference points are things that are fixed in relation to Earth, like a building, a tree, or a sign.

The pet store would be the reference point because it is where he started motion and it will not move.

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Well, there aren't really any choices on the list you provided that I would consider to be an acceptable answer to this question.  So I just have to make something up:

From the given information, I can conclude:

-- The car's velocity is constant, and points East.

-- The car's acceleration is zero.

-- The forces on the car are balanced.

-- The centripetal force acting on the car is zero.

-- The force delivered from the engine to the wheels is EXACTLY EQUAL to the sum of the forces of air resistance and friction with the road.

Yes, the mass of the astronaut’s bathroom is same but its weight will be less on the moon.

Further explanation:

First of all we will know about the mass and weight,

Mass: Mass of an object is its quantity or amount of inertia.

It remains constant and does not affected by change in the gravity.

Weight: Weight of an object is the amount of force experienced by the object when it comes in a gravitational field. If there is no gravitational field then there will be no weight of the object but the mass of the object will remain same.  

It is also a type of force.

Weight of the object can be calculated as,

[tex]\boxed{W = mg}[/tex]  

Here, [tex]W[/tex] is the weight of the object, [tex]m[/tex] is the mass of the object and [tex]g[/tex] is the acceleration due to gravity.

Consider that the mass of the astronaut’s bathroom is [tex]20{\text{ kg}}[/tex].

So, its weight on the earth will be,

[tex]{W_e} = 20{g_e}[/tex]

Here, [tex]{g_e}[/tex] is the value of acceleration due to gravity of earth and its value is [tex]9.8{\text{ }}{{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}^2}[/tex] .

Substitute this value in above equation.

[tex]\begin{aligned}{W_e}&=20\times9.8\\&=196{\text{N}}\\\end{aligned}[/tex]

Now, we will calculate it weight on the moon.

[tex]{W_m} = m{g_m}[/tex]

Here, [tex]{g_m}[/tex] is the value of moon’s gravity and its value is given as [tex]1.6{\text{ }}{{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}^2}[/tex].

Substitute this value of [tex]{g_m}[/tex] in above equation.

[tex]\begin{aligned}{W_m}&=20\times 1.6\\&=32{\text{N}}\\\end{aligned}[/tex]

Here, we can see the weight of the astronaut’s bathroom is more on earth as compare to moon.

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Answer detail:

Grade: High School

Subject: Physics

Chapter: Gravitation

Keywords:

Astronaut, bathroom, weight, moon, earth, mass, change, unchanged, constant, gravitation, force, quantity, same, acceleration due to gravity, force, acceleration.

When wire is coiled on the plastic tube and current flow through that wire then the system will behave like a solenoid in which current flows through the coiled wires and then it produce magnetic field along the axis of solenoid

So here it will also produce magnetic field along the axis of solenoid and then due to this magnetic field the compass placed inside the tube will experience torque on its needle due to which it will have tendency to oriented along the direction of magnetic field.

So here we can say that compass needle will lie along the axis of the plastic tube.

here magnetic field along the axis of tube will be given same as solenoid which is given as

[tex]B = \mu_0 ni[/tex]

so here direction of compass needle is axis of the tube

The compass inside the tube would point in the direction tangential to the circular loop of the wire at the location of the compass, following the right-hand rule with respect to the direction of the current in the wire.

When a current-carrying wire is wound around a tube to form a circular loop, it creates a magnetic field. According to Ampere's circuital law, the magnetic field inside a solenoid (which is essentially what the wire-wrapped tube resembles) is uniform and parallel to the axis of the solenoid. However, in this case, since the wire is wound only once around the tube, the situation is more akin to a single loop of wire carrying a current.

The magnetic field due to a current-carrying loop can be determined using the Biot-Savart law or by considering Ampere's law in the context of a loop. The magnetic field lines inside the loop are concentric circles that are parallel to the plane of the loop. The direction of the magnetic field at any point inside the loop can be found using the right-hand rule: if you point the thumb of your right hand in the direction of the current, your fingers will curl in the direction of the magnetic field.

Therefore, a compass placed inside the tube will align itself with the magnetic field created by the current in the wire. The compass needle, being a magnetic dipole, will point in the direction of the magnetic field lines, which is tangential to the circular path of the wire at the location of the compass. This direction is consistent with the right-hand rule applied to the current in the wire.

It's important to note that the strength of the magnetic field will vary depending on the distance from the wire, with the field being strongest near the wire and weakening as one moves towards the center of the tube. However, the direction of the field will remain the same, following the right-hand rule, as long as the compass is inside the loop formed by the wire.

SOLUTION is given in attachment below.

First write down all your known variables:

vi = 25m/s

a = 7.0m/s^2

t = 6.0s

vf = ?

Then choose the kinematic equation that relates all the variables and solve for the unknown variable:

vf = vi + at

vf = (25) + (7.0)(6.0)

vf = 67m/s

The final velocity of the motorcycle is 67m/s.

90 km/hr  

Unit conversion 1000 m/km 60 min/hr 60 s/min  

Answer: The speed of car is 90 km/hr

Explanation:

We are given:

Speed of car = 25 m/s

To convert this speed into km/hr, we use the conversion factors:

1 km = 1000 m

1 hr = 3600 s

Converting the speed into km/hr, we get:

[tex]\Rightarrow (\frac{25m}{s})\times (\frac{1km}{1000m})\times (\frac{3600s}{1hr})\\\\\Rightarrow 90km/hr[/tex]

Hence, the speed of car is 90 km/hr

Ummm thats not a question buddy...

A vector in a given plane can have any direction

In XY plane we can say that its direction is

North = + Y direction

South = - Y direction

East = + X direction

West = - X direction

so all of the above directions can be considered as direction of a given vector as a vector can incline in all above direction as well and at any angle with all also

so here it is also possible to have a direction which is along 45 degree North of East which will incline between X and Y direction both

So here all four options may be the possible direction of a vector

The x-component of a vector can be calculated using the magnitude of the vector and the angle it forms with the positive x-axis.

In a rectangular coordinate system in a plane, the x-component of a vector is given by the dot product of the vector with the unit vector Î (î). The x-component can be calculated as:

Ex = E * cos(θ)

where E is the magnitude of the vector and θ is the angle the vector forms with the positive x-axis.

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The x-component of a vector describes the part of the vector in the x direction. In the case of vector E⃗, it can be computed based on the angle θ it makes with the x-axis and its magnitude E through the relationship E cosθ.

The x-component of a vector describes the effect of the vector in the 'x' direction. This can be visualized by understanding that any vector can be viewed as a resultant of its components along the various axes available (here, along 'x'). Thus, when a given vector is expressed in rectangular or Cartesian components, the respective projections of the vector on 'x' (and 'y') directions essentially give the x-component (and y-component).

In the case of vector E⃗, the x-component can be computed based on the angle θ (theta) it makes with the 'x' axis and its overall magnitude E, using simple trigonometric principles. The x-component (Ex) is given by E cosθ because cosθ gives the fraction of the magnitude E that lies along the 'x' direction.

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The answer is:

The water level by rock affected by wave as:

if the direction of the wave is going to the rock naturally by the effect of the tide which comes in, the water level will rise and increase if the wave is escaping or receding the water level will decrease and will be lower.

when the tsunami is  a tidal wave which is a series of waves in a water body caused by the displacement of a large volume of water.

Waves, especially tsunamis, can cause significant changes to the water level by rocks due to their large energy. The water level can rise or fall along the rock's surface, a process known as wave run-up. The amount of change depends on the wave's characteristics and the rock's features.

In physics, the interaction between waves and shoreline features can significantly affect the water level by rocks. Waves, especially large ones like tsunamis, carry a significant amount of energy. When a wave crashes into a land formation such as a rock, some of this energy is transferred to the water, causing it to rise or fall along the rock's surface. This is known as wave run-up. The amount of change to the water level depends on factors such as the wave's size and speed, and the rock's shape and orientation. In the case of a tsunami, the water level by a rock could rise dramatically due to the enormous energy carried by the tsunami wave.

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Answer:

Physical evidence can be defined as the real evidence or material that plays a major role in the matter that gave rise to the litigation, which can be introduced in the court as a evidence in the judicial proceedings.

It can be any object or material which was present at the time of the incident. Example: A knife, table, footwear, chair, box, et cetera.

It is a strong support to favor the argument or to reject the argument in the courtroom.

Physical evidence refers to tangible or intangible material objects that can establish a crime has been committed, or link the crime to its victim or perpetrator. Examples can include biological materials such as blood or hair, weapons found at the crime scene, or documents proving relevant aspects of the offense.

Physical evidence refers to any material object that can establish that a crime has been committed, or link a crime and its victim or its perpetrator. Such evidence can be both tangible or intangible.

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Answer:

  She will make the jump.

Explanation:

We have equation of motion , [tex]s= ut+\frac{1}{2} at^2[/tex], s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

First we will consider horizontal motion of stunt women

   Displacement = 77 m, Initial velocity = 28 cos 15 = 27.05 m/s, acceleration = 0

Substituting

   [tex]77= 27.05t+\frac{1}{2} *0*t^2\\ \\ t=77/27.05=2.85 seconds[/tex]

So she will cover 77 m in 2.85 seconds

 Now considering vertical motion, up direction as positive

    Initial velocity = 28 sin 15 = 7.25 m/s, acceleration =acceleration due to gravity = -9.8 [tex]m/s^2[/tex], time = 2.85

    Substituting

           [tex]s=7.25*2.85-\frac{1}{2}*9.8*2.85^2=20.69-39.80 =-10.11 m[/tex]

  So at time 2.85 stunt women is 10.11 m below from starting position, far side is 25 m lower. So she will be at higher position.

  So she will make the jump.

Final answer:

The stuntwoman will not make the jump across the canyon. The calculation of horizontal and vertical motion shows that she will fall short of the ramp on the far side.

Explanation:

To determine if the stuntwoman will make the jump, we can analyze the horizontal and vertical components of the motion separately.

For the horizontal motion, we can use the equation:

horizontal distance = velocity x time

Since the stuntwoman leaves the ramp with a velocity of 28 m/s and the width of the canyon is 77 m, we can calculate:

time = horizontal distance / velocity = 77 m / 28 m/s = 2.75 s

Now, for the vertical motion, we can use the equation:

vertical distance = initial vertical velocity x time + (1/2) x acceleration x time^2

Since the ramp on the far side of the canyon is 25 m lower than the takeoff ramp, we can calculate:

vertical distance = -25 m

Using the acceleration of gravity of 9.8 m/s², we can rearrange the equation to solve for the initial vertical velocity:

initial vertical velocity = (vertical distance - (1/2) x acceleration x time^2) / time = (-25 m - (1/2) x 9.8 m/s² x (2.75 s)^2) / 2.75 s = -33.86 m/s

Since the initial vertical velocity is negative, it means the stuntwoman will not clear the canyon. She will fall short of the ramp on the far side.

The tiny sports car would accelerate faster, because it requires less kinetic energy to move due to it's mass being less than the truck

Hello!

To find the speed of the train, we have to divide the distance by the time.

Speed = [tex]\frac{distance}{time}[/tex]

600 / 4 = 150

The speed of the train is 150 km per hour.

Its average would be 35km in 30 minutes Hope this helps :D

C.  

The force of friction = coefficient of friction * normal force.  

Adding the book to the boy increases the normal force and the component of the gravitational force directed down the slide.  This in turn increases the force of friction as can be seen by the relationship from the above equation.  For a stationary object, the force of static friction is always equal to the force applied (in this case, it is the component of the gravitational force directed down the slide).  That means that so long as the boy is not moving and his mass increases, the frictional force is increasing also to balance the increased downward gravitation force directed down the slide.

The correct awnser is true i belive

TRUE

Answer

It is false

Explanation:

In old days it was assumed that if we place wooden pallets on concrete. The casing of pallets may break down and their may be leakage of battery. So to  protect the case, people placed the pallets on sand or gravel.

Here they use the concept that if you place a glass jar on concrete with a force it may got a crack and if you place this glass jar with same force on sand it will not break.

But it is not true because pallets have strong casing which cant break on placing it on concrete.

Hence When using wooden pallets to store batteries it can be placed anywhere.

The velocity of an object dropped and falling for 3.0 seconds under the influence of gravity and ignoring air resistance will be approximately 29.4 m/s downward.

The velocity of an object dropped and falling under the influence of gravity increases with time due to the acceleration caused by gravity. In the absence of air resistance, an object in freefall near the surface of the Earth accelerates downwards at a rate of 9.8 m/s². This means that for every second that passes, the object's downward speed will increase by about 9.8 m/s.

So, if an object has been falling for 3.0 seconds, we can find its velocity using the formula v = gt, where v is velocity, g is the acceleration due to gravity (9.8 m/s² near the surface of the Earth), and t is time. Substituting in our values we get: v = 9.8 m/s² * 3.0 s = 29.4 m/s.

Therefore the velocity of a dropped object after it has fallen for 3.0 seconds would be approximately 29.4 m/s downward.

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Cody could have drawn several conclusions if a flake of cereal wasn't attracted to a weak magnet: the cereal may not contain any magnetic materials, the magnet might be too weak, or a barrier or absence of liquid might have interfered with the attraction.

If Cody used a weak magnet and noticed that the flake of cereal was not attracted to it, he could have drawn a few possible conclusions.

Firstly, he might conclude that the cereal does not contain any magnetic materials, such as iron. Many cereals are fortified with iron, and this fortification is often in a form that is magnetic.

Secondly, it's also possible that the magnet was too weak to attract the iron in the cereal. Therefore, a stronger magnet could still cause attraction.

Lastly, the cereal might have been too far away from the magnet or there might have been a barrier (like container's wall) between them which prevented the attraction.

Another scenario is that the cereal may not have been in liquid, which often helps distribute the magnetic particles and makes them respond more effectively to the magnet.

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Answer: B) The kinetic energy is increases proportional to the mass of the bowling ball.

Explanation: Kinetic energy is the energy possessed by an object by virtue of its motion.

Kinetic energy is calculated by the formula:

[tex]K.E=\frac{1mv^2}{2}[/tex]

K.E= Kinetic energy

m= mass of object

v = velocity of object

As Kinetic energy is directly proportional to the mass of the object, Kinetic energy increases as the mass of the object increases.

Kinetic energy increases proportional to the square of the velocity.

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

[tex]x=v_0\cos20.0^\circ t+\dfrac12a_xt^2[/tex]

[tex]y=v_0\sin20.0^\circ t+\dfrac12a_yt^2[/tex]

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

[tex]x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ[/tex]

[tex]y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ[/tex]

So we have enough information to solve for the components of the acceleration vector, [tex]a_x[/tex] and [tex]a_y[/tex]:

[tex]x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}[/tex]

[tex]y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}[/tex]

The acceleration vector then has direction [tex]\theta[/tex] where

[tex]\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ[/tex]

kangaroo is capable of jumping to a height of 2.62 m

It means it will reach maximum height of 2.62 m when it will jump off with some maximum capable speed from the ground

So here as it will reach to its maximum height the final speed of the kangaroo will be zero

and also we know that during the motion of kangaroo the acceleration of kangaroo is due to gravity which is given by g = 9.8 m/s^2

now we can use kinematics equation to find out take off speed

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

here we know that

[tex]v_f = 0[/tex]

a = - 9.8 m/s^2

d = 2.62 m

now we will have

[tex]0^2 - v_i^2 = 2 * (-9.8)* 2.62[/tex]

[tex] - v_i^2 = - -51.352 [/tex]

[tex]v_i = \sqrt{51.352}[/tex]

[tex]v_i = 7.17 m/s[/tex]

so the take off speed of the kangaroo will be 7.17 m/s and correct answer is "option d"

Answer:

vi = 7.17 m/s

Explanation:

(4225 m2/s2)/(6 m/s2) = d

d = 704 m

Return to Problem 8

Given:

vi = 22.4 m/s

vf = 0 m/s

t = 2.55 s

Find:

d = ??

d = (vi + vf)/2 *t

d = (22.4 m/s + 0 m/s)/2 *2.55 s

d = (11.2 m/s)*2.55 s

d = 28.6 m

Return to Problem 9

Given:

a = -9.8 m/s2

vf = 0 m/s

d = 2.62 m

Find:

vi = ??

vf2 = vi2 + 2*a*d

(0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(2.62 m)

0 m2/s2 = vi2 - 51.35 m2/s2

51.35 m2/s2 = vi2

vi = 7.17 m/s

A test can be conducted using a light meter and a lamp where each lens from a pair of sunglasses is placed between the lamp and light meter one after the other. If the light intensity readings after each test are the same for both lenses, it indicates they are of equal effectiveness.

To design a test to ascertain if the two lenses in a pair of sunglasses are equally effective in reducing daylight, a light meter and a lamp can be used. The test would involve these steps:

Note that while sunglasses dim light, they also have other attributes such as polarization and photochromic properties. These features also contribute to the effectiveness of lenses in sunglasses. However, their effects cannot be properly measured using this simple test with a light meter and a lamp.

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To measure the effectiveness of both lenses in a pair of sunglasses, use a light meter and a lamp. Measure the amount of light that passes through each lens from the lamp, keeping the exposure consistent. Differences in readings indicate different effectiveness.

In order to design a test to determine the effectiveness of both lenses of a pair of sunglasses, you would first need a light meter and a lamp to simulate sunlight. You would then measure the amount of light that passes through each lens individually. Ensure to use the same angle while measuring light.from the lamp passing through each lens. This way, the light exposure would remain consistent for both lenses.

Start with measuring the initial light intensity from the lamp without the sunglasses. Then place one lens of the sunglasses between the lamp and light meter and record the reading. Repeat the same process with the second lens. If the readings for both lenses are equal, it indicates they are equally effective.

Moreover, sunglasses may have different coatings and treatments like polarization or photochromic lenses that react differently to light. These features can also affect the amount of light that passes through the lenses.Thus it's important to know what kind of sunglasses lens you are testing.

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After 28 seconds, the velocity of the object will be 151.2 m/s.

The velocity of an object after a given time can be found using the equation:

In physics, the equation used to calculate the velocity of an object when accelerated from rest is v = at, where 'v' is velocity, 'a' is acceleration, and 't' is time passed. In this question, we have an object accelerating from rest at a constant acceleration of 5.4m/s2 for a duration of 28seconds. By substituting the given values into the equation, we get v = 5.4 m/s2 * 28s = 151.2 m/s. Therefore, the velocity of the object after 28 seconds would be 151.2 m/s.

v = u + at

Where:

Substituting the known values into the equation:

v = 0 + (5.4 m/s2) * (28s)

Simplifying the equation:

v = 151.2 m/s

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x is the displacement.

Answer:

In hooks law,The x represents the extension of elastic material.

The mass of the astronaut is the same on both, but weight is actually a force and it depends on the acceleration due to gravity.  On the moon, the acceleration due to gravity is 1/6 of the Earth’s so the astronaut’s weight will be 1/6 lighter on the moon.

Answer:

The weight of the astronaut on the moon is less than the weight of the astronaut on the Earth.            

Explanation:

Matter contained in a body is known as mass. It remains constant even if one goes to another celestial body. Weight is the force due to gravity acting on a mass. Since, gravitational force varies at each celestial body, the weight also changes.

The acceleration due to gravity on the Moon is [tex]\frac{1}{6}^{th}[/tex] the acceleration due to gravity on the Earth. Therefore, the astronaut would weigh less on moon than on the Earth. His weight would be [tex]\frac{1}{6}^{th}[/tex] of that on Earth.

Volume=0.09m³

Density=4000kg/m³

Force=?

Density=mass/volume ⇒mass=volume×density

m=0.09×4000=360kg

Force=mass×accelaration

F=360×9.8

F=3528N

Answer

3,528 N

Explanation

density = mass/volume

ρ = m/v

∴ m = ρv

      = 4000 × 0.09

        = 360 Kg

Weight is defined as the force that acts on  a body and is directed towards the center of the earth.

Weight = mass × acceleration due to gravity

W = mg          Where g = 9.8 m/s²

W = 360 × 9.8

  = 3,528 N