The pressure inside a hydrogen-filled container was 2.10 atm at 21 ∘C. What would the pressure be if the container was heated to 88 ∘C ?

Answer:

the force experienced by both particles will be of same magnitude as the magnitude of charge is same for both particles but with different sign, the acceleration of both particles will be same in magnitude but opposite in direction, the direction of free electron will be opposite to the direction of electric field whereas the direction of proton will be in the direction of electric field,

Explanation:

the force experienced by both particles can be calculated by the formula

F=|q|E

the acceleration can be calculated by Newton's 2nd law i.e F=ma

the direction of electric field is always towards the positive charge so proton will be directed towards electric field and free electron will be directed away from the electric field.

Answer:

V = 145 cm³

Explanation:

given,

mass of the solid = 2.8 Kg

                             = 2800 g

density of the gold = 19.3 g/cm³

volume of the water displaced = ?

volume of water displaced is equal to the volume of statue.

Volume of statue = ?

[tex]Volume = \dfrac{mass}{density}[/tex]

[tex]Volume = \dfrac{2.8\times 1000}{19.3}[/tex]

        V = 145 cm³

volume of the water displaced will be equal to 145 cm³

The volume of water displaced by the 2.8 kg solid gold statue to verify its purity should be 145.08 cm3, considering the density of pure gold is 19.3 g/cm3.

To determine whether a 2.8 kg solid gold statue is made of pure gold using its density, we need to calculate the volume of gold that mass would occupy and compare it with the volume of water displaced. Since density (d) is mass divided by volume, we can rearrange this formula to solve for volume (V) where V = m/d. First, we need to convert the mass of the gold statue from kilograms to grams because the density of gold is given in grams per cubic centimeter (19.3 g/cm3).

2.8 kg = 2,800 g. Using the density of gold, the volume (in cm3) the statue should displace if it is pure gold is calculated as follows:

Volume of gold (V) = Mass of gold (m) / Density of gold (d)

V = 2,800 g / 19.3 g/cm3

V = 145.08 cm3 (rounded to two decimal places)

In conclusion, to verify that the statue is pure gold, the volume of water displaced should be 145.08 cm3. If the volume of displaced water matches this calculation, it suggests that the statue is indeed made of pure gold.

Answer:

Moment of inertia of the flywheel, [tex]I=1.82\ kg-m^2[/tex]          

Explanation:

Given that,

The maximum energy stored on  flywheel, [tex]E=4\ MJ=4\times 10^6\ J[/tex]

Angular velocity of the flywheel, [tex]\omega=20000\ rev/s=2094.39\ rad/s[/tex]

We need to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :

[tex]E=\dfrac{1}{2}I\omega^2[/tex]

I is the moment of inertia of the flywheel

On rearranging we get :

[tex]I=\dfrac{2E}{\omega^2}[/tex]

[tex]I=\dfrac{2\times 4\times 10^6}{(2094.39)^2}[/tex]

[tex]I=1.82\ kg-m^2[/tex]

So, the moment of inertia of the flywheel is [tex]I=1.82\ kg-m^2[/tex]. Hence, this is the required solution.

The moment of inertia of the flywheel [tex]I= 1.82\ kg-m^2[/tex]

It is given that,

The maximum energy stored on the flywheel is given as

E=4 MJ=[tex]4\times 10^6\ J[/tex]

Angular velocity of the flywheel is

[tex]w=20000\ \dfrac{Rev}{Sec} =2094.39 \frac{rad}{sec}[/tex]

So to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :

[tex]E= \dfrac{1}{2} Iw^2[/tex]

By rearranging the equation

[tex]I=\dfrac{2E}{w^2}[/tex]

[tex]I=1.82\ kg-m^2[/tex]

Thus the moment of inertia of the flywheel [tex]I= 1.82\ kg-m^2[/tex]

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Answer:

0.04865 m

Explanation:

k = Spring Constant

m = Mass

d = Distance

g = Acceleration due to gravity = 9.81 m/s²

Angular frequency is given by

[tex]\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{k}{m}=\omega^2\\\Rightarrow \dfrac{k}{m}=14.2^2[/tex]

At equilibrium we have

[tex]kd=mg\\\Rightarrow d=\dfrac{mg}{k}\\\Rightarrow d=\dfrac{g}{\omega^2}\\\Rightarrow d=\dfrac{9.81}{14.2^2}\\\Rightarrow d=0.04865\ m[/tex]

The distance by which the spring stretches from its unstrained length is 0.04865 m

Answer:

[tex]F_a=1470\ N[/tex]

Explanation:

Friction Force

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.  

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

[tex]\displaystyle F_a-F_{r1}-F_{r2}=m.a=0[/tex]

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

[tex]\displaystyle F_a=F_{r1}+F_{r2}.....[1][/tex]

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

[tex]\displaystyle F_{r2}-T=0[/tex]

The friction forces are computed by

[tex]\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g[/tex]

[tex]\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g[/tex]

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

[tex]\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g[/tex]

Simplifying

[tex]\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)[/tex]

Plugging in the values

[tex]\displaystyle F_{a}=0.25(9.8)[400+2(100)][/tex]

[tex]\boxed{F_a=1470\ N}[/tex]

Final answer:

To make the 100 kg block slip, a horizontal force of at least 245 N must be applied to the 400 kg lower block, calculated based on the maximum static friction force determined by the coefficient of friction and the weight of the 100 kg block.

Explanation:

To find the horizontal force F applied to the lower block that is necessary for the 100 kg block (upper block) to start slipping, we need to calculate the force required to overcome the static friction between the 100 kg block and the 400 kg block. The coefficient of friction (μ) is given as 0.25 for both contacts.

First, calculate the maximum static friction force that can act on the 100 kg block:

The weight of the 100 kg block (W) = 100 kg × 9.8 m/s² = 980 N.

Maximum static friction force (Ffriction) = μ × Normal force = 0.25 × 980 N = 245 N.

To initiate slipping, the horizontal force (F) applied must at least be equal to the maximum static friction force, which is 245 N. However, since this force will be applied to the combined system of both blocks, we must consider the total mass involved when finding the acceleration caused by the applied force.

The total mass of the system is 500 kg (400 kg + 100 kg). To move this mass with an acceleration that would cause the upper block to slip, we calculate:

F = ma, where m is the total mass and a is the acceleration. Since the force to overcome static friction is 245 N, this gives us the minimum force needed to initiate slipping when applied to the 400 kg block.

Answer:

0.15756 m/s

Explanation:

There are 10 blades and 10 gaps

To move through one blade or gap the windmill has to rotate

[tex]\dfrac{2\pi}{20}=0.31415\ rad[/tex]

This divided by the angular velocity is gives us the time

[tex]\dfrac{0.31415}{1.1}=0.28559\ s[/tex]

When the ball moves it does in a way that the ball must travel a distance of its own diameter which is [tex]4.5\times 10^{-2}\ m[/tex]

[tex]Speed=\dfrac{Distance}{Time}[/tex]

[tex]v=\dfrac{4.5\times 10^{-2}}{0.28559}\\\Rightarrow v=0.15756\ m/s[/tex]

The minimum linear speed is 0.15756 m/s

Answer:

37.96usd , 9.49usd

Explanation:

From the data given we have

cost of electricity= $0.13,

and annual cost of electricity is express as

Annual cost= power *hours*days*cost of electricity

note the unit of power must be in kilowatt(Kw),

a. to determine the cost of running 100w we first convert to kw

[tex]1w=0.001kw\\100w=0.1kw\\[/tex]

Hence the annual cost of running it 8.00 hours per day is calculated as

[tex]annual cost=0.1*8*365*0.13\\annual cost=37.96 USD[/tex]

b. to determine the cost of running 25w we first convert to kw

[tex]1w=0.001kw\\25w=0.025kw\\[/tex]

Hence the annual cost of running it 8.00 hours per day is calculated as [tex]annual cost=0.025*8*365*0.13\\annual cost=9.49 USD[/tex]

Hence from the answers we can conclude that less amount of money required to run the 25w incandescent bulb compare to the 100w

Answer:

Angle at which object is launched is 17.15°

Explanation:

We have given initial velocity at which object is launched u = 15 m/sec

It rises to a height of 1 m

So height h = 1 m

We have to find the angle of projection [tex]\Theta[/tex]

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

We know that maximum height is given by [tex]h=\frac{u^2sin^2\Theta }{2g}[/tex]

So [tex]1=\frac{15^2\times\ sin^2\Theta }{2\times 9.8}[/tex]

[tex]sin^2\Theta =0.0871[/tex]

[tex]sin\Theta =0.295[/tex]

[tex]\Theta =sin^{-1}0.295=17.15^{\circ}[/tex]

So the angle at which object is launched is 17.15°

Answer:

Avoidant attachment pattern

Explanation:

The Attachment theory proposes that there are 4 attachment styles:

Secure attachment

Anxious attachment

Avoidant attachment

Anxious-Avoidant attachment

A child who does not seek their mother when they leave and are unbothered by the fact that the mother is not around shows the avoidant attachment pattern. The child is independent and self-directed, thus does not seek intimacy. These types of people are assumed to be fearful or avoidant when it comes to intimacy and commitment, as the theory suggests.

Differentiating it from the other style:

Secure attachment pattern show comfort in intimacy. They easily display interest to other parties, but at the same time, they do not necessarily seek intimacy as they are capable of being alone as well.

Anxious attachment on the other hand, is more like a needy and clingy type of behavior. They seek companionship more desperately and always needs reassurance.

Anxious-avoidant attachment is a more aggressive (in a defensive sense) type. They totally avoid commitment or any form of intimacy. What makes them different is the distrust. They do not trust anyone that tries to get close to them and can sometimes emotionally vent out, (verbally or physically) if anyone attempts to get to know them.

Answer:

option B

Explanation:

given,

Satellite B has an orbital radius nine times that of satellite A.

R' = 9 R

now, orbital velocity of the satellite A

[tex]v_a =\sqrt{\dfrac{GM}{R}}[/tex]........(1)

now, orbital velocity of satellite B

[tex]v_b=\sqrt{\dfrac{GM}{R'}}[/tex]

[tex]v_b=\sqrt{\dfrac{GM}{9R}}[/tex]

[tex]v_b=\dfrac{1}{3}\sqrt{\dfrac{GM}{R}}[/tex]

from equation 1

[tex]v_b=\dfrac{v_a}{3}[/tex]

hence, the correct answer is option B

The speed of Satellite B is one-third of the speed of Satellite A (vA/3), based on Kepler's laws of planetary motion and centripetal force requirements for circular orbits. So the correct option is B.

The question asks for the speed of Satellite B assuming that Satellite A has speed vA and that the orbital radius of Satellite B is nine times that of Satellite A. To find the speed of Satellite B, we can use Kepler's third law and the fact that the centripetal force required for circular motion is provided by the gravitational force.

The orbital speed equation reveals a relation v ≈ r^-1/2, meaning that if the radius increases, the speed decreases. Specifically, if the radius becomes nine times larger, then the speed will be the square root of 1/9 times the initial speed, which is 1/3. So, the correct answer is that the speed of Satellite B is Satellite A's speed divided by 3, which is vA/3.

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Answer:

Thalamus

Explanation:

Thalamus is a mass of gray matter located just above the brain stem, between the cerebral cortex and the midbrain. It plays a role in pain sensation, attention and alertness. It consists of four parts: the hypothalamus, the epythalamus, the ventral thalamus and the dorsal thalamus.

It (Thalamus) serves as the major relay station for most sensory impulses that reach the primary sensory areas of the cerebral cortex from the spinal cord and brain stem.

The thalamus serves as the major relay station for most sensory impulses to reach the cerebral cortex. It acts as a clearinghouse and relay station, conducting sensory signals to their respective processing areas in the cortex.

The thalamus serves as the major relay station for most sensory impulses that reach the primary sensory areas of the cerebral cortex from the spinal cord and brain stem. It is a structure in the forebrain that acts as a clearinghouse and relay station for sensory signals, except for smell. When the sensory signals exit the thalamus, they are conducted to the specific area of the cortex dedicated to processing that particular sense.

Final answer:

The spring can produce 0.270 kJ of work after it has been compressed 3 cm from its unloaded length, with a spring constant of 6 kN/cm.

Explanation:

The work done by a spring is given by the expression ½ kx², where k is the spring constant and x is the displacement from the spring's equilibrium position. In this case, we are given a spring constant of 6 kN/cm and a compression of 3 cm. First, convert the spring constant to N/m by multiplying by 10³ (since 1 kN = 10³ N and 1 cm = 10⁻² m), which gives us 600,000 N/m. The work can then be calculated as:

W = ½ kx² = ½ (600,000 N/m) (0.03 m)² = 270 J

Convert joules to kilojoules, 1 kJ = 1,000 J, gives:

W = 0.270 kJ

Therefore, the spring can produce 0.270 kJ of work after being compressed 3 cm from its unloaded length.

Answer:

η = 40 %  

Explanation:

Given that

Qa ,Heat addition= 1000  J

Qr,Heat rejection= 600 J

Work done ,W= 400 J

We know that ,efficiency of a engine given as

[tex]\eta=\dfrac{W(net)}{Q(heat\ addition)}[/tex]

Now by putting the values in the above equation ,then we get

[tex]\eta=\dfrac{400}{1000}[/tex]

η = 0.4

The efficiency in percentage is given as

η = 0.4  x 100 %

η = 40 %

Therefore the answer will be 40%.

The efficiency of the heat engine is calculated using the formula Efficiency = W/Q(in), which gives an efficiency of 40% given the provided values.

To determine the efficiency (e) of a heat engine, we use the formula:

where W is the work done by the engine and Q(in)is the heat absorbed from the high-temperature reservoir.

Using the values:

Thus, the efficiency of the heat engine is 40%.

[tex]\rm 220^oC[/tex] is the surface temperature of the rod. The fuel must be kept below [tex]\rm 300^oC[/tex] to avoid reaching the cooling water from the critical heat flux and the temperature below that it is reasonable.

The average kinetic energy of particles in a medium, such as a gas, liquid, or solid, is measured as temperature. It is a basic physical characteristic that enables us to comprehend how hot or cold something is. Depending on the location, temperature is often measured using different scales, such as Celsius or Fahrenheit, or in scientific situations, Kelvin (K). Temperature, said simply, tells us how quickly the particles in a substance are moving. The substance is hotter or cooler depending on how quickly the particles are moving.

Heat generated = Heat transferred by convection + Heat conducted

[tex]\rm q_{gen }= q'' \times V\\\rm q_{gen} = 150 \times 10^{6 }W/m^3 \times \pi \times r^2\\\rm q_{conv} = h \times A \times (T_{s }- T )\\\rm q_{conv} = 5000 W/m^2\timesK \times2 \times \pi \times r \times L \times (T_s - T)\\\rm q_{cond} = k \times A \times (T_{s} - T) / L\rm 150 \times 10^6 \times \times \times r^2= 5000 \times 2 \times \pi \times r \timesL \times (T_{s} - T) + k \times 2 \times \times \times r \timesL \times (T_{s} - T) / L\\\rm T_{s }=220^oC[/tex]

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Answer:

B. the stars to come back to the same positions in the sky.

Explanation:

In fact, the solar day is equivalent to more than a rotation, because when the point has turned completely, it is not, as it should, in the same position with respect to the Sun.

The reason for this is that while performing the rotation, the Earth simultaneously moved following its orbit around the Sun.

When the reference point completed its rotation, the Earth already moved almost 2,500,000 km., So that to see the Sun again it will be necessary to turn a little more.

Solar day is more than a rotation. The sidereal or sidereal day, commonly used by astronomers, is also based on the rotation of the Earth; but in this case a distant star is taken as a reference (sidereal comes from the Latin sidus which means "star").

Answer:

I₁ > I₃ > I₂

Explanation:

Taking the pic shown, we have

m₁ = 10m₀

m₂ = 2m₀

m₃ = m₀

r₁ = r₀

r₂ = 2r₀

r₃ = 3r₀

We apply the formula

I = mr²

then

I₁ = m₁r₁² = (10m₀)(r₀)² = 10m₀r₀²

I₂ = m₂r₂² = (2m₀)(2r₀)² = 8m₀r₀²

I₃ = m₃r₃² = (m₀)(3r₀)² = 9m₀r₀²

finally we have

I₁ > I₃ > I₂

Final answer:

To rank the objects according to their moments of inertia, calculate I = mr² for each, where m is mass and r is the distance from the axis. Rank from largest to smallest moment of inertia value.

Explanation:

The student's question involves ranking objects according to their moments of inertia, which requires an understanding of rotational motion in physics. The moment of inertia (I) of an object is calculated as I = Σmr², where m represents the mass of the object and r the perpendicular distance from the object to the axis of rotation. This equation states that for point masses, the moment of inertia is a product of the mass and the square of its distance from the rotation axis.

The student would need to calculate the moment of inertia for each object by multiplying its given mass by the square of its specified distance from the rotation axis (I = mr² for point masses). The objects should be ranked from the highest value of the moment of inertia to the lowest.

The concepts of rotational inertia and the parallel-axis theorem may also be relevant if the objects are not point masses, but the question provided seems to simplify the scenario to treat the masses as points. Additionally, the question posed hints that integration could be required for more complex shapes, but for the simple case of point masses or uniformly distributed mass like a hoop or rod, the formulas are straightforward.

Final answer:

To find the speed of the bullet before striking the block, we can use the law of conservation of momentum. By setting the initial momentum equal to the final momentum, we can solve for the initial velocity of the bullet.

Explanation:

To solve this problem, we can use the law of conservation of momentum. Momentum is defined as the product of an object's mass and velocity. In this case, we know the mass and velocity of the wooden block and bullet after the collision, and we want to find the velocity of the bullet before the collision.

Using the law of conservation of momentum, we have:

Initial momentum = Final momentum

Mass of bullet * Initial velocity of bullet + Mass of block * Initial velocity of block = Mass of bullet * Final velocity of bullet + Mass of block * Final velocity of block

Substituting the given values, we have:

(0.0107 kg)(Initial velocity of bullet) + (4.588 kg)(0 m/s) = (0.0107 kg)(1.74 m/s) + (4.588 kg)(1.74 m/s)

Simplifying the equation and solving for the initial velocity of the bullet, we get:

Initial velocity of bullet = 1.74 m/s - (4.588 kg)(1.74 m/s) / 0.0107 kg

Calculating the value, we find that the initial velocity of the bullet was 271.36 m/s.

Answer:W = -5.808J

Explanation:

Given:

Mass of hoop,m=120kg

Centre of mass speed of hoop,v= 0.220m/s

Rotational inertia,I= mr^2

I=120r^2

Kinetic energy of hoop= Linear kinetic energy + Rotational kinetic energy

K= (1/2mv^2) + (1/2Iw^2)

But w= (v/r)^2

K= (1/2mv^2) + (1/2×120r^2(v/r)^2)

K = (1/2×120×0.220^2) + (1/2×120×0.220^2)

K= 5.808J

To stop the hoop,its final kinetic energy must be zero,

Workdone = Kfinal-Kinitial = 0 - 5.808

W= -5.808J

Explanation:

We have equation of motion v = u + at

     Initial velocity, u = 8.5 m/s

     Final velocity, v = 0 m/s    - At maximum height

     Time, t = ?

     Acceleration , a = -9.81 m/s²

     Substituting

                      v = u + at  

                      0 = 8.5 + -9.81 x t

                      t = 0.867 s

  It takes 0.867 seconds to get to the top of its motion

Answer:

0.87 s

Explanation:

initial velocity, u = 8.5 m/s

Let it takes time t to reach to maximum height. At maximum height the velocity is zero, so, v = 0

Use first equation of motion

v = u - gt

where, g be the acceleration due to gravity

0 = 8.5 - 9.8 t

t = 0.87 s

Thus, the time taken to reach at top is 0.87 s.

Answer: 16N

Explanation:

According to Coulomb's law, the force existing between two charges is directly proportional to the product of their charges and inversely proportional to the distance between the charges. Mathematically,

F = kq1q2/r²

Where F is the force between them

q1 and q2 are the charges

r is the distance between them.

If the force between them is 10N, the formula becomes;

10 = kq1q2/r²... (1)

If the distance between them is now halved, the force will become

F = kq1q2/(r/2)²

F = kq1q2/(r²/4)

F = 4kq1q2/r² ... (2)

Dividing equation 1 by 2 we have;

4/F = (kq1q2/r²)÷ (4kq1q2/r²)

4/F = kq1q2/r²×r²/4k1q1q2

4/F = 1/4

Cross multiplying we have;

F = 4×4

F = 16N

Therefore the new force between the charges is 16N

Explanation:

De Broglie related the wavelength to the magnitude of the momemtum of the particle, as follows:

[tex]\lambda=\frac{h}{p}=\frac{h}{mv}[/tex]

Here h is the planck constant. The momemtum is directly proportional to the particle's mass. So, the wavelength of a particle is inversely proportional to its mass. Thus, as mass increases its wavelength decreases.

In a gravitational field, the net work done by the field when an object moves from rest at point A to rest at point B depends only on the positions of A and B, not the path taken or the object's velocity while moving.

When an object moves from rest at point A to rest at point B in a gravitational field, the net work done by the field is dependent on the positions of A and B only (option 5). This is based on the principle of conservation of energy in physics. If we consider the gravitational field to be conservative, the work done is independent of the path taken, and depends only on the initial (A) and final points (B).

The mass of the object and the gravitational field strength will determine the gravitational potential energy at these points, hence affecting the work done when the object is moved. It does not depend on the nature of the external force moving the object or the velocity of the object while moving. These factors could affect the time it takes to move the object, but not the net work done by the gravitational field.

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The net work done by the field when an object is moved from rest at point A to rest at point B in a gravitational field depends on the nature of the external force moving the object from A to B, and both the positions of A and B and the path taken between them.

The net work done by the field when an object is moved from rest at point A to rest at point B in a gravitational field depends on the nature of the external force moving the object from A to B, and both the positions of A and B and the path taken between them.

If the object is moved vertically upwards or downwards, the work done by the gravitational field will be positive or negative respectively. However, if the object is moved in a horizontal direction, there will be no work done by the field as there is no change in height.

Therefore, option 3, both the positions of A and B and the path taken between them, is the correct answer.

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Answer:

Time period, T = 0.5 seconds

Explanation:

Given that,

One complete wave passes a putrefying pile of poached pickerel in 0.50 seconds, T = 0.5 s

The piled pickerel protrude 20 cm along the beach, A = 20 cm

We need to find the period of the wave. The time period of the wave is defined as the time taken by the wave to complete one oscillation. So, the time period of the wave will be 0.5 seconds.

Hence, this is the required solution.      

Answer:

493.02 g.

Explanation:

Density = mass/volume

mass = density×volume

m = D×V ............................ Equation 1

m = mass of the aluminum foil, D = Density of the aluminum foil, V = volume of the aluminum foil.

V = L×W×H................... Equation 2

Where L = length of the aluminum foil, W = wide of the aluminum foil, H = thickness of the aluminum foil.

Given: L = 59.3 yd = 59.3×0.9144 m = 54.224 m, W = 13 in = 13×0.0254 m = 0.3302 m, H = 0.00040 in = 0.00040×0.0254 m = 0.0000102 m.

Substituting into equation 2

V = 54.224×0.3302×0.0000102

V = 0.0001826 m³ = (0.0001826×1000000) cm³

V = 182.6 cm³

Also given D = 2.70 g/cm³

Substitute into equation 1

m = 2.7×182.6

m = 493.02 g.

The mass in grams of the foil = 493.02 g.

Answer:

Explanation:

Neurons communicate via both electrical signals and chemical signals. The electrical signals are action potentials, which transmit the information from one of a neuron to the other; the chemical signals are neurotransmitters, which transmit the information from one neuron to the next.

The electrical signal travels down the axon to the axon terminals where it tells the vesicles to release the neurotransmitters into the synaptic cleft which travel to the receptors of the receiving cell which releases the second messengers

Neurons convey information using both electrical and chemical signals. Electrical signaling occurs through the generation of action potentials, while chemical signaling occurs at the synapse through the release of neurotransmitter molecules.

Neurons, specialized cells in the nervous system, convey information using both electrical and chemical signals. Electrical signaling occurs through the generation of action potentials, which are electrical impulses that travel along the neuron's axon. When a neuron is stimulated, a wave of depolarization is generated, leading to the opening of ion channels and a change in the neuron's membrane potential. This electrical event allows the action potential to propagate down the axon.

Chemical signaling, on the other hand, occurs at the synapse, the junction between two neurons. When an action potential reaches the axon terminal of the presynaptic neuron, it triggers the release of neurotransmitter molecules into the synapse. These neurotransmitters diffuse across the synaptic gap and bind to receptors on the postsynaptic neuron, generating a chemical signal. This signal can either excite or inhibit the postsynaptic neuron, depending on the type of neurotransmitter and receptor involved.

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Answer:

The magnitude of the average force exerted on the water by the blade is 960 N.

Explanation:

Given that,

The mass of water per second that strikes the blade is, [tex]\dfrac{m}{t}=30\ kg/s[/tex]

Initial speed of the oncoming stream, u = 16 m/s

Final speed of the outgoing water stream, v = -16 m/s

We need to find the magnitude of the average force exerted on the water by the blade. It can be calculated using second law of motion as :

[tex]F=\dfrac{\Delta P}{\Delta t}[/tex]

[tex]F=\dfrac{m(v-u)}{\Delta t}[/tex]

[tex]F=30\ kg/s\times (-16-16)\ m/s[/tex]

F = -960 N

So, the magnitude of the average force exerted on the water by the blade is 960 N. Hence, this is the required solution.

Answer:

It will take 667.239 sec to cover 0.7 miles

Explanation:

We have given the men's world record in swimming is 1500 m in 1 min 34.56 sec

So distance d = 1500 m

Time t = 14 min 34.56 sec

As 1 minute = 60 sec

So time t = [tex]14\times 60+34.56=888.56sec[/tex]

So speed [tex]v=\frac{distance}{time}=\frac{1500}{888.56}=1.688m/sec[/tex]

Now new distance d = 0.7 miles

As 1 miles = 1609 m

So 0.7 miles =[tex]0.7\times 1609=1126.3m[/tex]

So time taken [tex]t=\frac{distance}{velocity}=\frac{1126.3}{1.688}=667.239sec[/tex]

Answer:

E = 4.01×10^6 N/C

direction is toward q2

Explanation:

q1 = q2 = 12micro-C = 12×10^-6C

r1 = 18cm = 0.18m

r2 = 40cm = 0.40m

k = 9×10^9

E1 = kq1/(r1)²

E1 = 3.33×10^6 N/C

E2 = kq2/(r2)²

E2 = 6.75×10^5 N/C

Enet = E1 + E2

= 3.33×10^6 + 6.75×10^5

= 4.01×10^6 N/C

direction is toward q2