A stone is thrown vertically upward with an initial velocity of 40m/s. Taking g = 10 m/s^2 find the maximum height reach by the stone and what is the net displacement and distance covered by the stone.​

Fundamental

forming a necessary base or core; of central importance.

"the protection of fundamental human rights"

Answer:

40 degrees Celsius

Explanation:

Have a great summer :)

Answer:

In a circular motion, the object just moves in a circle. In rotational motion, the object rotates about an axis. ... For example, Earth rotating on its own axis.

Answer:

Its different

Explanation:

it's different because

Answer:

the work done by the woman is 6,000 J

Explanation:

Given;

force applied by the woman, F = 400 N

distance moved by the woman, d = 15 m

The work done by the woman is calculated as follows;

W = F x d

W = 400 N x 15 m

W = 6000 Nm = 6,000 J

Therefore, the work done by the woman is 6,000 J

Explanation:

[tex]C.) \: Conduction \: \\ \\ = > it \: is \: a \: process \: of \: transferring \: \\ \\ of \: heat \: from \: one \: object \: to \: another \\ \\ when \: they \: are \: in \: contact \: [/tex]

By the conduction process the heat reaches the pin.Option C is correct.

The term "heat transfer" refers to the movement of heat. The flow of heat across a system's boundary is due to a temperature differential between the system and its surroundings.

Conduction is the type of heat transfer occurs mainly in solid or by contact.In the given condition the rod is then heated continuously at one end and the pins fall off in the order 4, 3, 2, 1​.

By the conduction process the heat reaches the pin.

Hence option C is correct.

To learn more about the heat transfer refer to the link;

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Answer:

Explanation:

[tex]h=-v^2 /2g[/tex]

[tex]with\\g = 9,8 m/s^2 or 10 m/s^2[/tex]

[tex]h= (-13)^2 / 2 * 9,8 = 8,6[/tex]

A fixed mass of gas has a volume of 25 [tex]cm^3[/tex], the pressure of the gas is 100 kPa, the volume of the gas is slowly decreased by 15 [tex]cm^3[/tex] at a constant temperature, and the change in pressure of the gas is 150 kPa, which is option b.

PV = nRT (P= pressure of the gas, V =volume, n = number of moles of gas, R = gas constant, and T =temperature of the gas in kelvin)

Suppose the gas is an ideal gas and that the temperature is constant,

P1V1 = P2V2

Here P1 = 100 kPa, V1 = 25 [tex]cm^3[/tex], V2 = 10 [tex]cm^3[/tex],

100 kPa x 25 [tex]cm^3[/tex] = P2 x 10 [tex]cm^3[/tex]

P2 = (100 kPa x 25 [tex]cm^3[/tex]) / 10 [tex]cm^3[/tex]

P2 = 250 kPa

the change in pressure of the gas is,

ΔP = P2 - P1 = 250 kPa - 100 kPa = 150 kPa

The reason is that when the volume of a fixed mass of gas is decreased, the pressure of the gas increases proportionally, so here assuming that the temperature is constant it is calculated.

Hence, the volume of the gas is slowly decreased by 15 [tex]cm^3[/tex] at a constant temperature, and the change in pressure of the gas is 150 kPa, which is option b.

Learn more about the calculation of the change in pressure here.

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Answer:

a. 0.41 m

b. 5.72 m/s

c. i. For part (a), I chose the hand as the reference level since the penny was thrown from the hand and the height of the penny at the hand is zero and also, it is easier to calculate from a zero reference level.

ii. For part (b), I chose the ground as the reference level since the height of the penny above the ground is positive and the height of the penny when the penny hits the ground is zero and also, it is easier to calculate from a zero reference level.

d. 5.72 m/s

Explanation:

a. Use energy to find how high the penny goes above your hand before stopping.

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy at the hand, E equals the total mechanical energy when the penny stops in the air, E'.

E = E'

U + K = U' + K' where U = initial potential energy at hand level = mgh where h = height at hand level = 0, K = initial kinetic energy at hand level = 1/2mv² where v = speed at hand level = 2.85 m/s, U' = final potential energy at stopping level = mgh' where h' = height at stopping level, K = final kinetic energy at stopping level = 1/2mv'² where v = speed at stopping level = 0 m/s (since the penny momentarily stops)

So, U + K = U' + K'

mgh + 1/2mv² = mgh' + 1/2mv'²

substituting the values of the variables into the equation, we have

mg(0) + 1/2m(2.85 m/s)² = mgh' + 1/2m(0 m/s)²

0 + 1/2m(8.1225 m²/s²) = mgh' + 0

m(4.06125 m²/s²) = mgh'

h' = 4.06125 m²/s² ÷ g

h' = 4.06125 m²/s² ÷ 9.8 m/s²

h' = 0.41 m

(b) The penny then falls to the floor, 1.26 m below your hand. Use energy to find its speed just before it hits the floor.  

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy when the penny stops in the air, E'  equals the total mechanical energy on the ground, E"

E' = E"

U' + K' = U" + K" where U' = initial potential energy at stopping level = mgh" where h' = height at stopping level = height of penny above hand, h' + height of hand above ground = 0.41 m + 1.26 m = 1.67 m, K = initial kinetic energy at stopping level = 1/2mv'² where v = speed at stopping level = 0 m/s (since the penny momentarily stops), U = final potential energy at ground level = mgh₁ where h₁ = height at ground level = 0, K = final kinetic energy at ground level = 1/2mv"² where v" = speed at ground level,

So, U' + K' = U' + K'

mgh" + 1/2mv'² = mgh₁ + 1/2mv"²

substituting the values of the variables into the equation, we have

mg(1.67 m) + 1/2m(0 m/s)² = mg(0) + 1/2mv"²

1.67mg + 0 = 0 + 1/2mv"²

1.67mg = 1/2mv"²

1.67g = 1/2v"²

v"² = 2(1.67g)

v" = √[2(1.67g)]

v" = √[2(1.67 m × 9.8 m/s²)]

v" = √[2(16.366 m²/s²)]

v" = √[32.732 m²/s²)]

v" = 5.72 m/s

(c) Explain your choice of reference level for parts (a) and (b).

i. For part (a), I chose the hand as the reference level since the penny was thrown from the hand and the height of the penny at the hand is zero and also, it is easier to calculate from a zero reference level.

ii. For part (b), I chose the ground as the reference level since the height of the penny above the ground is positive and the height of the penny when the penny hits the ground is zero and also, it is easier to calculate from a zero reference level.

(d) Choose a different reference level and repeat part (b)

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy when the penny stops in the air, E'  equals the total mechanical energy on the ground, E"

E' = E"

U' + K' = U" + K" where U' = initial potential energy at stopping level = mgh' where h' = height at stopping level = 0.41 m, K = initial kinetic energy at stopping level = 1/2mv'² where v' = speed at stopping level = 0 m/s (since the penny momentarily stops), U = final potential energy at ground level = mgh₁ where h₂ = height of hand above the ground level = height of ground below hand = -1.26 m(it is negative since the ground is below the hand), K = final kinetic energy at ground level = 1/2mv"² where v = speed at ground level,

So, U' + K' = U' + K'

mgh' + 1/2mv'² = mgh₂ + 1/2mv"²

substituting the values of the variables into the equation, we have

mg(0.41 m) + 1/2m(0 m/s)² = mg(-1.26 m) + 1/2mv"²

0.41mg + 0 = -1.26 mg + 1/2mv"²

0.41mg + 1.26mg = 1/2mv"²

1.67mg = 1/2mv"²

1.67g = 1/2v"²

v"² = 2(1.67g)

v" = √[2(1.67g)]

v" = √[2(1.67 m × 9.8 m/s²)]

v" = √[2(16.366 m²/s²)]

v" = √[32.732 m²/s²)]

v" = 5.72 m/s

Answer:

Kayla. You can calculate it using the formula for momentum: momentum=mass×velocity and find the bigger number between the two momentums

Answer:

The iron bolt is attracted to the electromagnet. The force of attraction between the electromagnet's core and the iron bolt is greater than the force of the spring, so the iron bolt is pulled out of its housing in the door

Explanation:

Find the angle θ made by the road. When rounding the curve at 15.0 m/s, the car has a radial acceleration of

a = (15.0 m/s)² / (75.0 m) = 3.00 m/s²

There are two forces acting on the car in this situation:

• the normal force of the road pushing upward on the car, perpendicular to the surface of the road, with magnitude n

• the car's weight, pointing directly downward; its magnitude is mg (where m is the mass of the car and g is the acceleration due to gravity), and hence its perpendicular and parallel components are, respectively, -mg cos(θ) and mg sin(θ)

By Newton's second law, the net forces in the perpendicular and parallel directions are

(perp.) ∑ F = n - mg cos(θ) = 0

(para.) ∑ F = mg sin(θ) = ma

==>   sin(θ) = a/g   ==>   θ = arcsin(a/g) ≈ 17.8°

(Notice that in the paralell case, the positive direction points toward the center of the curve.)

When rounding the curve at 31.8 m/s, the car's radial acceleration changes to

a = (31.8 m/s)² / (75.0 m) ≈ 13.5 m/s²

and there is now static friction (mag. f = µn, where µ is the coefficient of static friction) acting on the car and keeping from sliding off the road, hence pointing toward the center of the curve and acting in the parallel direction. Newton's second law gives the same equations, with an additional term in the parallel case:

(perp.) ∑ F = n - mg cos(θ) = 0

(para.) ∑ F = mg sin(θ) + f = ma

The first equation gives

n = mg cos(θ)

and substituting into the second equation, we get

mg sin(θ) + µmg cos(θ) = ma

==>   µ = (a - g sin(θ)) / (g cos(θ)) = a/g sec(θ) - tan(θ) ≈ 1.12

Answer:

Explanation:

You are in the chapter on Physics about uniform circular motion and gravity. This is a centripetal force problem in particular, and the equation for that is

[tex]F_c=\frac{mv^2}{r}[/tex] where

[tex]F_c[/tex] is the centripetal force needed to keep the car moving in its circular path,

m is the mass of the car,

v is the velocity with which the car is moving, and

r is the radius of the circle that the car is moving around.

For us, the centripetal force is supplied by the friction keeping the car on the road, altering the equation to become

[tex]f=\frac{mv^2}{r}[/tex] and friction is defined by

f = μ[tex]F_n[/tex] (the coefficient of friction multiplied by the weight of the car).

Going on and getting buried even deeper,

[tex]F_n=mg[/tex] which says that the weight of the car is equal to its mass times the pull of gravity. Putting all that together, finally, we have the equation we need to solve this problem:

μ·m·g = [tex]\frac{mv^2}{r}[/tex] and we solve this for μ:

μ = [tex]\frac{mv^2}{mgr}[/tex] and it just so happens that the mass of the car cancels out. (I'll tell you why the mass of the car doesn't matter at the end of this problem). Filling in and solving for the coefficient of friction:

μ = [tex]\frac{31.8^2}{(9.8)(75.0)}[/tex] to 2 significant figures is

μ = 1.4

The mass of the car doesn't affect whether or not the car can stay on the curve. Even though a car with a greater mass will have a greater frictional force, that doesn't mean that it's easier for that car to stay on the road; a larger mass only means that a larger centripetal force is needed to keep it moving in a circle. This makes the gain in friction become offset by the fact that a larger centripetal force is necessary. Thus,

On a flat curve, the mass of the object experiencing circular motion does not affect the velocity at which it can stay on the curve.

Answer:

Gravitational force -an attractive force that exists between all objects with mass; an object with mass attracts another object with mass; the magnitude of the force is directly proportional to the masses of the two objects and inversely proportional to the square of the distance between the two objects.

Explanation:

hope it helped

Answer:

Explanation:

This is a classic Law of Momentum Conservation problem. For us the equation will look like this:

[tex][(m_yv_y+m_bv_b)]_b=[(m_y+m_b)v_{both}]_a[/tex] Filling in with our given info:

[tex][(70.0)(0)+(.40)(10.0)]_b=[(70.0+.40)v_{both}]_a[/tex] and

4.0 = 70.4v and

v = .06 m/s

Explanation:

The voltage of the lamp, V = 240 V

Power of the lamp, P = 40 W

It is running for 62 hours.

The cost of running is $2.50k per KWH

Electric power is,

P = 40×62 Wh

= 2480 Wh

P = 2.48 kWh

At the rate of $2.5 per kWh

P = $6.2

So, the cost of running is $6.2 per kWh.

hope this helps

Answer:

If a person is pushing a desk across the room, then there is an applied force acting upon the object. The applied force is the force exerted on the desk by the person.

Explanation:

Gravitational potential energy is energy an object possesses because of its position in a gravitational field. ... The gravitational potential energy is equal to its weight times the height to which it is lifted. PE = kg x 9.8 m/s2 x m = joules. The 9.8 us the gravitational acceleration constant.

so the answer is "the gravitational potential energy is decreasing"

Answer:

33.33j+6.67i km/hr

Explanation:

From the law of conservation of momentum,

Applying,

mu+m'u' = V(m+m')............... Equation 1

Where m = mass of the truck, m' = mass of the car, u = initial velocity of the truck, u' = initial velocity of the car, V = Final velocity.

Note: let j represent the north, and i  represent the east

From the question,

Given: m = 1500 kg, u = 60j, m' = 1200 kg, u' = 15i

Substitute these values into equation 1

1500*60j+1200*15i = V(1500+1200)

90000j+18000i = 2700V

V = (90000j+18000i)/2700

V = 33.33j+6.67i km/hr

Answer:

Energy is: the ability to do work

The formula for energy is: power x time = energy

Explanation:

Hope this helps (there isn't really an explanation)

By Newton's second law, the net force acting on the crate parallel to the surface is

∑ F = mg sin(10°) - 800 N = ma

where m = 500 kg is the mass of the crate and a is the acceleration.

Solve for a :

a = ((500 kg) (9.80 m/s^2) sin(10°) - 800 N) / (500 kg)

a ≈ 0.102 m/s^2

Answer:2 protons and 2 neutrons

Explanation:In Nuclei, There are 2 forces. 1 force is electrostatic and acts as repulsion between 2 protons. The other is force of attraction called Nuclear force between 2 neutrons.

Answer:

1 is false 2 is true 3 is false

Answer:

313kelvin

Explanation:

40 degree celcius plus 273=313K

Explanation:

Glass is the best sound absorber material.

Answer:

Speed = 1.6 m/s

Explanation:

Formula,

Speed = Distance ÷ Time

Answer:

I am not sure if this is the answer

acceleration: 2.2m/s

Explanation:

here

initial velocity(u): 11m/s

Final velocity(v): 33m/s

time taken(t): 10 s

now

a:v-u/t

or

acceleration:final velocity-initial velocity/time taken

or

a: 33-11/10

or

a:22/10, divide it

: a=2.2m/s#

Answer:

Figure is not there

Explanation: