A typical elevator car with people has a mass of 1500.0 kg. Elevators are currently approaching speeds of 20.0 m/s - faster than the speed.

Required:
What is the upward force required if the elevator moves upward 200.0 meters before reaching 20.0 m/s?

Answer:

The normal force exerted on the ant is 0.75 N.

Explanation:

Given;

diameter of the ball, D = 40 cm = 0.4m

radius of the ball, r = 0.2m

mass of the beach ball, m₁ = 300 g = 0.3 kg

mass of the ant, m₂ = 4 x 10⁻⁶ kg

speed of the ball, v = 4 m/s

The area of the ball, assuming spherical ball is given by;

A = 4πr²

A = 4π(0.2)² = 0.5027 m²

The drag force (resistance) experienced by the spherical ball is given as;

[tex]F_D = \frac{1}{2}C\rho Av^2[/tex]

where;

C is the drag coefficient of the spherical ball = 0.45

ρ is density of air = 1.21 kg/m³

[tex]F_D = \frac{1}{2}C\rho Av^2\\\\F_D = \frac{1}{2}(0.45)(1.21) (0.5027)(4)^2\\\\F_D = 2.19 \ N[/tex]

The downward force of the ball due to its weight and that of the ant is given by;

[tex]F_g = mg\\\\F_g =g(m_{ant} + m_{ball})\\\\F_g = g(4*10^{-6} \ kg\ + \ 0.3\ kg)\\\\F_g = g(0.300004 \ kg) \ \ \ (mass \ of \ the \ ant \ is \ insignificant)\\\\F_g = 9.8(0.3)\\\\F_g = 2.94 \ N[/tex]

The net downward force experienced by the ball is given by;

[tex]F_{net} = F_g - F_D\\\\F_{net} = 2.94 \ N - 2.19 \ N\\\\F_{net} = 0.75 \ N[/tex]

This downward force experienced by the ball is equal to the normal reaction it exerts on the ant.

Thus, the normal force exerted on the ant is 0.75 N.

Answer:

Mercury is the only one in liquid state at room temperature. It's used in thermometers because it has high coefficient of expansion.

Explanation:

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Mercury is the only one in liquid state at room temperature. It's used in thermometers because it has high coefficient of expansion. they still use mercury even though it is the poorest conductor of heat.

Answer:

Explanation:

These are the laws of Newton

Answer:

the first law, an object will not change its motion unless a force acts upon it. the 2nd one, the force of an object is equal to its mass times it acceleration. the 3rd one is when 2 objects interact, they apply forces to each other of equal magnitude and opposite direction.

Complete Question:

A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.

They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet

Answer:

aw = 3 i + 6 j m/s2

Explanation:

       [tex]a_{c} = \omega^{2} * r (1)[/tex]

       ∴ ωp = ωw (2)

           ⇒ [tex]a_{p} = \omega_{p} ^{2} * r_{p} (3)[/tex]

               [tex]a_{w} = \omega_{w}^{2} * r_{w} (4)[/tex]

        [tex]\frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}[/tex]

        [tex]a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2[/tex]

Explanation:

Speed is the rate of change of distance with time. It is a scalar quantity with magnitude both no direction.

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

The unit is m/s or km/hr or mile/hr

Distance covered is simply the length of the path traveled.

 The unit is m or km or miles

Time taken is the duration of an event.

  The unit is seconds or minutes or hour.

Answer:

106.24 kJ.

Explanation:

Given that,

Mass of sample of sand, m = 8 kg

Specific heat of sand, c = 664 J/kg-°C

The temperature changes from 20° C to 40° C. We need to find the change in thermal energy. It is given by :

[tex]Q=mc\Delta T\\\\Q=8\times 664(40-20)\\\\=106240\ J\\\\=106.24\ kJ[/tex]

So, the change in thermal energy is 106.24 kJ.

Answer:

Thermal energy

Explanation:

The internal energy of a system is widely known as thermal energy. Now, thermal energy is also called heat energy and it is an internal energy of a component which is produced when an increase in temperature causes atoms and molecules within the component to move faster and start colliding with one other.

Therefore, the more heat the is applied to the component, the hotter the substance and the more its particles move which in turn leads to a higher thermal energy.

Answer:

its 0.5 for all i beleive

Explanation:

Answer:

a. V = 19.1m/s

b. The mass of the car does not matter

Explanation:

A.

KE = 1/2mv² = fd --------(1)

Fd = umgd ---------(2)

Therefore,

1/2mv² = umgd ---------(3)

M will cancel itself out from both sides of equation 3.

Then we will have:

1/2v² = ugd

Then we cross multiply to make v² the subject of the formula

V² = 2ugd

V = √2ugd -------(4)

U = 0.38

g = 9.81

d = 98

When we input these values into equation 4, we will have:

V = √2x0.38x9.81x98

V = √730.6488

V = 27.03m/s

B.

The mass of the car does not actually matter as the mass was cancelled out on the both sides of equation 3

The time taken by the car to achieve the final speed is 6.25 seconds.

The equations of motion can be defined as the equation that represents the relationship between the time, velocity, acceleration, and displacement of a moving object.

The mathematical expressions for the equations of motions can be written as:

[tex]v= u+at\\S=ut+(1/2)at^2\\v^2-u^2=2aS[/tex]

Given, the initial speed of the car, u = 15 m/s

The final speed of the given car, v = 25m/s

The distance covered by car, S = 125 m

From the third equation of motion: v² = u²+ 2aS

(25)² = (15)² + 2×a× 125

a = 1.6 m/s²

From the first equation of motion we can find the time to achieve the final speed:

v = u+ at

25 = 15 + (1.6) × t

t = 6.25 sec

Therefore, 6.25 seconds will be taken by the car to catch the final speed.

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Answer:

I thinks it's

deficit spending

Explanation:

cause When a government spends more than it collects in taxes, it is said to have a budget deficit.

Answer:

t = 1.32 s

Explanation:

We are given;. Frequency of C4 note; F_c = 262 Hz

In conversions, we know that 1 Hz = 1 cycle/s

Thus, F_c = 262 cycles/s

Now, we want to find out how much time it takes for 346 air pressure maxima to pass a stationary listener.

346 air pressure maxima denotes that the air pressure maxima is 346 cycles.

Thus, time will be;

t = 346 cycles/262 cycles/s

t = 1.32 s

The time taken for the musical note to pass the stationary listener is 1.32 s.

The given parameters:

The frequency of a sound wave is defined as the number of cycles completed per second by the wave.

[tex]F = \frac{n}{t}[/tex]

where;

The time taken for the musical note to pass the stationary listener is calculated as follows;

[tex]262 = \frac{n}{t} \\\\t = \frac{n}{262} \\\\t = \frac{346}{262} \\\\t = 1.32 \ s[/tex]

Thus, the time taken for the musical note to pass the stationary listener is 1.32 s.

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Answer:

x = 704 [m]

Explanation:

To solve this problem we must use the following equation of kinematics.

[tex]v_{f} ^{2} =v_{o} ^{2} +2*a*x[/tex]

where:

Vf = final velocity = 65 [m/s]

Vo = initial velocity = 0 (starts from rest)

a = acceleration = 3 [m/s²]

x = distance [m]

Now replacing we have:

65² = 0 + 2*3*x

4225 = 6*x

x = 704 [m]

Answer:

2.93 m  (which agrees with answer "C" on the list)

Explanation:

Recall that the speed of the wave equals the product of the wave's length times its frequency. Therefore, the wavelength is going to be the quotient of the speed of the signal divided its frequency:

Wavelength = 2.997  10^8 / 1.023  10^8 =  2.93 m

Complete question:

Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire feels a repulsive force of 2.4 x 10^-4 N created by the interaction of the 12.5 A current and the magnetic field created by the bottom current (I). Find the magnitude and direction of the bottom current, if the distance between the two wires is 40cm.

Answer:

The bottom current is 12.8 A to the right.

Explanation:

Given;

length of the wires, L = 3.0 m

current in the top wire, I₁ = 12.5 A

repulsive force between the two wires, F = 2.4 x 10⁻⁴ N

distance between the two wires, r = 40 cm = 0.4 m

The repulsive force between the two wires is given by;

[tex]F = \frac{\mu_oI_1I_2L}{2\pi r}\\\\I_{2} = \frac{2F\pi r}{\mu_oI_1L}[/tex]

Where;

I₂ is the bottom current

The direction of the bottom current must be in the same direction as the top current since the force between the two wires is repulsive.

[tex]I_{2} = \frac{2F\pi r}{\mu_oI_1L}\\\\I_{2} = \frac{2(2.4*10^{-4})(\pi)(0.4)}{(4\pi*10^{-7})(12.5)(3)}\\\\I_{2} = 12.8 \ A[/tex]

Therefore, the bottom current is 12.8 A to the right.

Answer:

Less precipitation, droughts9: How might agriculture in southern Europe change by the end of the century if conditions follow the RCP8.

Explanation:

Precipitation and droughts are the specific changes in two climate variables that are expected to lead to major decreases in soil moisture.

Drought is defined as a period of protracted water scarcity, whether it is due to atmospheric surface water, or groundwater constraints.

Droughts can last months or years, although they can be proclaimed in as little as 15 days.

It has the potential to have a significant influence on the afflicted region's ecology and agriculture as well as harm the local economy.

Precipitation and droughts are the specific changes in two climate variables that are expected to lead to major decreases in soil moisture in southern Africa and the Mediterranean region.

Hence Precipitation and droughts are the specific changes in two climate variables.

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Answer:

no deflection

Explanation:

current is flowing from west to east. As the magnetic field of a long wire carrying current is circular, its direction will be north below the wire and south above the wire (according to the right hand rule). So, when the compass is placed underneath the wire, it will still point towards the north direction.

Answer:

If the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will increase because the toy car will have a greater acceleration.

Explanation:

I hope this helped

Hypothesis: If the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will increase because the toy car gains more potential energy at the higher starting point.

When a toy car moves along a sloped racetrack, it converts its potential energy (due to its height above the ground) into kinetic energy (energy of motion). The higher the starting height of the racetrack, the more potential energy the toy car possesses initially.

As the toy car moves down the sloped track, it will accelerate due to the force of gravity. The potential energy is converted into kinetic energy, and the car's speed increases. According to the law of conservation of energy, the total mechanical energy (sum of potential and kinetic energy) remains constant as long as no external forces, such as friction, act on the car.

Therefore, if the starting height of the racetrack is increased, the toy car will have more potential energy to start with. As it moves down the track, it will convert this increased potential energy into kinetic energy, resulting in a higher speed compared to when it starts from a lower height.

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Answer:

a. 250kg I think it's the right answer. hope it helps:)

Answer:

C.10

Explanation:

because when you divide 50 divided by 5 = 10

Answer:

a) s = 79 m W

b) s = 2 km S

c) s = 250 m W

d) s = 0 km  

Explanation:

We take the following sign convention for the directions:

North (N)  ---> positive

South (S)  ---> negative

East (E) ---> negative

West (W)  ---> positive

a)

45 m W, 34 m W

s = 45 m + 34 m

s = 79 m W

b)

5 km N, 7 km S

s = 5 km - 7 km

s = - 2 km

s = 2 km S

c)

350 m E , 800 m W, 200 m E

s = -350 m + 800 m - 200 m

s = 250 m

s = 250 m W

d)

850 km N, 850 km S

s = 850 km - 850 km

s = 0 km

Answer:

The change in the internal energy of the gas 1,595 J

Explanation:

The first law of thermodynamics establishes that in an isolated system energy is neither created nor destroyed, but undergoes transformations; If mechanical work is applied to a system, its internal energy varies; If the system is not isolated, part of the energy is transformed into heat that can leave or enter the system; and finally an isolated system is an adiabatic system (heat can neither enter nor exit, so no heat transfer takes place.)

This is summarized in the expression:

ΔU= Q - W

where the heat absorbed and the work done by the system on the environment are considered positive.

Taking these considerations into account, in this case:

Replacing:

ΔU= 2,092 J - 500 J

ΔU= 1,592 J  whose closest answer is 1,595 J

The change in the internal energy of the gas 1,595 J

Answer:

The correct answer is - phototropism, thigmotropism, hydrotropism, and gravitropism in order ( already match correctly).

Explanation:

phototropism is a phenomenon in which plants grow towards the light or sun which is accomplished by the hormone auxin in the cells far from the light.

Thigotropism is a type of plant growth that occurs around the tree to support itself which is a touch stimuli response.

The movement of the plant in the direction of the source of the water is known as hydrotropism. In which stimuli is humidity or the water concentration.

The movement of the plant or roots of the plants towards the soil or earth is known as gravitropism here gravity is the stimuli.

Answer:

Use the drop-down menus to complete each sentence.

A plant grows toward a sunny window. This response is an example of

phototropism

.

Sometimes a plant grows around a tree for support. This response is an example of

thigmotropism

.

The roots of a plant grow toward a water source. This response is an example of

hydrotropism

.

The roots of a plant grow down into the soil. This response is an example of

gravitropism

.

Explanation:

Answer:

 x = 10.75 m

Explanation:

For this problem we will solve it in two parts, the first using energy and the second with kinematics

Let's use the energy work relationship to find the velocity of the block as it exits the ramp

       W = [tex]Em_{f}[/tex] - Em₀

Starting point. Higher

       Em₀ = U = m g h

the height from the edge of the ramp of the graph has a value

        h = 9-3 = 6 m

Final point. At the bottom of the ramp

       Em_{f} = K = ½ m v²

Friction force work

      W = - fr  d

The friction force has the formula

      fr = μ N

On the ramp, we can use Newton's second law

         N - W cos θ = 0

         N = W cos θ

where the angle is obtained from the graph

         tan θ = (9-3) / (0.5-4) = -6 / 3.5

         θ = tan⁻¹ (-1,714)

         θ = -59.7º

the distance d is

         d = √ (Δx² + Δy²)

         d = √ [(0.5-4)² + (9-3)²]

         d = 6.95 m

for which the work is

       W = - μ mg cos 59.7 d

we substitute

        W = Em_{f} -Em₀

        - μ mg cos 59.7 d = ½ m v² - m g h

In the graph o text the value of the friction coefficient is not observed, suppose that it is μvery = 0.2

        - μ g cos 59.7 d = ½ v² - g h

         v² = 2g (h - very d coss 59.7)

let's calculate

         v² = 2 9.8 (6 - 0.2 6.95 cos 59.7)

         v = √ 103.8546

         v = 10.19 m / s

in the same direction as the ramp

in the second part we use projectile launch kinematics

let's look for the components of velocity

         v₀ₓ = vo cos -59.7

         [tex]v_{oy}[/tex] = vo sin (-59,7)

         v₀ₓ = 10.19 cos (-59.7) = 5.14 m / s

         v_{oy} = 10.19 if (-59.7) = -8.798 m / s

Let's find the time to get to the floor (y = o)

          y = y₀ + v_{oy} t - ½ g t²

to de groph y₀=3 m

          0 = 3 - 8.798 t - ½ 9.8 t²

          t² - 1.796 t - 0.612 = 0

we solve the quadratic equation

          t = [1.796 ±√(1.796² + 4 0.612)] / 2

          t = [1,795 ± 2,382] / 2

          t₁ = 2.09 s

          t₂ = -0.29 s

since time must be a positive quantity the correct value is t = 2.09 s

we calculate the horizontal displacement

          x = v₀ₓ t

          x = 5.14 2.09

          x = 10.75 m

The motion of the box, after it exits the incline is the motion and trajectory

of a projectile.

Horizontal distance from the right-hand edge of the incline to the point of

contact with the floor is approximately 1.24613 m.

Reasons:

Mass of the block,  m = 3.5 kg

Coefficient of kinetic friction, μ = 1.2

Location of the = 0.5 m from the origin

Required:

Horizontal distance between the block's point of contact with the floor and

the bottom right-hand edge of the incline.

Solution:

Let θ represent the angle the incline make with the horizontal.

The normal reaction of the incline on the block, [tex]F_N[/tex] = m·g·cos(θ)

Work done on friction = [tex]F_N[/tex]×μ×Length of the incline, L

Rise of the incline = 10 - 3 = 7

Run of the incline = 4

L = √(6.125² + 3.5²) = [tex]\dfrac{7 \times \sqrt{65} }{8}[/tex]

Let ΔP.E.₁  represent the potential energy transferred to kinetic energy

and work along the incline, we have;

Energy of the block at the bottom of the incline, M.E.₂, is found as follows;

K.E.₂ = mgh - m·g·μ·cos(θ)·L

[tex]K.E. =\frac{1}{2} \times 3.5 \times v^2 = 3.5 \times 9.81 \times 6.125 - 3.5 \times 9.81 \times 1.2 \times \dfrac{4}{\sqrt{65} } \times \dfrac{7 \times \sqrt{65} }{8}[/tex]

v ≈ 6.1456 m/s

The vertical component of the velocity is therefore;

[tex]v_y = v \cdot sin(\theta)[/tex]

[tex]v_y = 6.1456 \times \dfrac{7}{\sqrt{65} } \approx 5.33588[/tex]

From the equation, h = u·t + 0.5·g·t² derived from Newton's Laws of motion, we have;

ΔP.E.₁ = 3.5×9.81×7

3 = 5.33588·t + 0.5×9.81·t²

Factorizing, the above quadratic equation, we get;

The time it takes the block to reach the floor, t ≈ 0.40869 seconds

Horizontal component of the velocity is [tex]v_x \approx 6.1456 \times \dfrac{4}{\sqrt{65} } \approx 3.04908[/tex]

The horizontal distance, x = vₓ × t

∴ x = 3.04908 × 0.40869 ≈ 1.08194

Horizontal distance from the right-hand edge of the incline to the point of

contact with the floor, x ≈ 1.24613 m.

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Answer:

The pH scale measures of how acidic or basic water is.

The pH scale also measures whether there is more hydronium or hydroxide in a solution.

Explanation:

The range goes from 0-14, with 7 being neutral. Less than 7 indicates acidity and more than seven indicates the substance is a base.