If the same force is applied to a tiny sports car and a huge dump truck, which one would experience greater acceleration and why?

Answer: A sealed calorimeter is a closed system because thermal energy is not transferred to the environment.

A closed system is an isolated system where there is no transfer of energy with its surroundings. Also the system is not subjected to any external force whose source is external to the system.

Answer:C. Because thermal energy has not transferred to the environment.

Explanation:

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.

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.

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

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

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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Ummm thats not a question buddy...

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

A) gravity is the attraction of any mass to any other other mass .

D) gravity is a force that exist between any two masses in the universe

both are ok ...

x is the displacement.

Answer:

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

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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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.

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.

To learn more about motion refer to the link:

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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.

10% of the income.

Answer:

20% of their income

Explanation:

I got a five on Plato

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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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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We are standing still inside an elevator

So our speed in the frame of elevator is zero

we can write this relative speed as

[tex]v_{pE} = v_p - v_E[/tex]

also we know that speed of the elevator is 55 mph

[tex]v_E = 55 mph[/tex]

now the velocity in relative frame is zero as he is  still in the moving elevator while in the ground frame or observing from the ground its velocity will be same as velocity of elevator.

So whenever an object is at rest in a moving frame then its relative velocity with respect to that frame must be ZERO

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.

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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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

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

ANSWER:

Abseiling rope is tested beyond twice the maximum possible weight when in use because of safety purposes.

EXPLANATION:

Every rope has its elastic limit.Abseiling rope is used basically for climbing or descending purposes so this type of rope must be of quite good quality and  it must bear maximum possible weight. We test it  just to check the elastic limit of the rope so that the rope shouldnot increase its elastic limit and it doesnot break when we use it.

The displacement of the rock will be the same as the total horizontal distance traveled. Here the rock's horizontal position is given by

[tex]x=\left(15\,\dfrac{\mathrm m}{\mathrm s}\right)\cos30^\circ\,t[/tex]

so to find the horizontal distance it traversed, we need to know the time it took for the rock to return to the ground. We use the rock's vertical position over time to figure that out:

[tex]y=\left(15\,\dfrac{\mathrm m}{\mathrm s}\right)\sin30^\circ\,t-\dfrac g2t^2=0[/tex]

where [tex]g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}[/tex] is the acceleration due to gravity. Then we find that [tex]t\approx1.5\,\mathrm s[/tex], at which point we find [tex]x\approx20\,\mathrm m[/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

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]

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.

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.

(A) The resistance of the block of aluminum is given by:

[tex]R=\frac{\rho L}{A}[/tex]

where

[tex]\rho[/tex] is the aluminum resistivity

L is the length of the cylinder

A is the cross-sectional area of the cylinder

We already know the aluminum resistivity ([tex]\rho=2.65 \cdot 10^{-8} \Omega m[/tex]) and the length of the cylinder (L=20 m), so we must find the cross-sectional area A, which is given by the difference between the area of the larger cylinder and the area of the radial hole:

[tex]A=\pi (R^2 -r^2)[/tex]

where [tex]R=2.5 cm=0.025 m[/tex] and [tex]r=20 mm=0.02 m[/tex] (assuming that the 20 mm removed radially refers to the radius of the hole).

Therefore, the cross-sectional area is

[tex]A=\pi ((0.025 m)^2-(0.020 m)^2)=7.06 \cdot 10^{-4} m^2[/tex]

Substituting into the initial formula of the resistance, we find:

[tex]R=\frac{\rho L}{A}=\frac{(2.65 \cdot 10^{-8} \Omega m)(20 m)}{7.06 \cdot 10^{-4} m^2}=7.51 \cdot 10^{-4} \Omega[/tex]

(B) The resistivity of the tungsten is [tex]\rho=5.6 \cdot 10^{-8} \Omega m[/tex], so a cylinder of tungsten of the same size would have a resistance of

[tex]R=\frac{\rho L}{A}=\frac{(5.6 \cdot 10^{-8} \Omega m)(20 m)}{7.06 \cdot 10^{-4} m^2}= 1.59 \cdot 10^{-3} \Omega[/tex]

The behaviour of the resistance versus temperature is given by:

[tex]R=R_0 (1 + \alpha \Delta T)[/tex]

where [tex]\alpha[/tex] is a coefficient that for aluminum is equal to [tex]\alpha = 0.004308[/tex], R0 is the resistance of the piece of aluminum we found at point (A), and R is the resistance of the tungsten. Re-arranging the formula and substituting, we find

[tex]\Delta T = \frac{R/R_0 -1}{\alpha}=\frac{\frac{1.59 \cdot 10^{-3} \Omega}{7.51 \cdot 10^{-4} \Omega}-1}{0.004308}=259.3^{\circ}C[/tex]

So, the temperature must increase by 259.3 degrees.

(C) The power dissipated is given by:

[tex]P=I^2 R[/tex]

where I=30 A is the current. Substituting the numbers into the formula, we find

[tex]P=(30 A)^2 (7.51 \cdot 10^{-4} \Omega)=0.68 W[/tex]