1. Opposite charges
O repel
attract

Answer:

None of them: the direction of the pressure fluctuations is parallel to the direction of motion of the wave

Explanation:

Answer:

Q = 425 kJ

Explanation:

Given that,

Mass, m = 25 kg

The clown doll head that heats up from 25°C to 35°C

The specific heat is 1700 J/kg°C

We need to find the internal energy of it. The heat required to raise the temperature is given by the formula as follows :

[tex]Q=mc\Delta T\\\\Q=25\times 1700\times (35-25)\\\\Q=425000\ J\\\\Q=425\ kJ[/tex]

So, 425 kJ of thermal energy is severed.

Answer:

The correct answer is option B

Explanation:

Step one:

given data

a. force F= 110N

distance s= 3meters

we know that work= Force* distance

work= 110*3

Work= 330Joules

Step two:

data

Force= 60 pounds

distance= 30 ft

convert pounds to Newton

1 pound= 4.44822N

60 pounds= 60*4.44822

=266.9N

convert ft to meteres

1 ft = 0.3048meter

30ft= 0.3048*30

=9.144N

we know that work= Force* distance

work= 266.9N*9.144N

Work= 2440.53Joules

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:

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:

See connections below

Explanation:

1   [tex]\Rightarrow[/tex] b

2   [tex]\Rightarrow[/tex] a

3   [tex]\Rightarrow[/tex] d

4   [tex]\Rightarrow[/tex] i

5   [tex]\Rightarrow[/tex] g

6   [tex]\Rightarrow[/tex]  h

7   [tex]\Rightarrow[/tex]  c

8   [tex]\Rightarrow[/tex]  e

9   [tex]\Rightarrow[/tex] f

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:

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:

 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.

Answer:

The magnetic field is [tex]B = 2.319 *10^{-3} \ T[/tex]

Explanation:

From the question we are told that

   The radius of the wire is [tex]r = 0.8 \ cm = 0.008 \ m[/tex]

    The current is [tex]I = 106 \ A[/tex]

    The position considered is  d = 0.07 cm = 0.0007 m

 Generally the magnetic field is mathematically represented as

         [tex]B = \frac{\mu_o * I}{2\pi * \frac{r^2}{d} }[/tex]

Here [tex]\mu_o[/tex] is the permeability of free space with value [tex] 4\pi * 10^{-7} N/A^2[/tex]

So    

        [tex]B = \frac{ 4\pi * 10^{-7} * 106 }{2 * 3.142 * \frac{0.008^2}{0.0007} }[/tex]

=>     [tex]B = 2.319 *10^{-3} \ T[/tex]

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:

please mark me brainlist

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:

B I believe

Explanation:

Answer:

Speed = 100 km/h

Explanation:

Given:

Speedometer reading = 100 kph for one lap

Assume;

Time taken to complete one lap = 1 hour

Computation:

Speed =  Distance / Time

Speed =  100 / 1

Speed = 100 km/h

Answer:

Explanation:

Force = mass * acceleration

Given

Mass = 1500kg

Get the acceleration using the equation of motion;

v² = u²+2aS

20² = 0+2s(200)

400 = 400a

a = 400/400

a = 1m/s²

Get the upward force required

F = 1500 * 1

F = 1500N

Hence the upward force required if the elevator moves upward 200.0 meters before reaching 20.0 m/s is 1500N

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:no sure sorry

Explanation:

Answer:

45.9m

Explanation:

Given parameters:

Final velocity  = 30m/s

Initial velocity  = 0m/s

Unknown:

Time it takes for the object of fall  = ?

Height of fall = ?

Solution:

For the first problem, we use the equation below to solve for t;

      V = U + gt

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

t is the time taken

          30  = 0 + 9.8 x t

          30  = 9.8t

            t  = [tex]\frac{30}{9.8}[/tex]  = 3.1s

Now, height of fall;

  V² = U²  + 2gH

   30² = 0² + 2 x 9.8 x H

   900  = 19.6H

      H  = 45.9m

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:

Reverse check the answer

Explanation:

I believe it is very important that once someone is done with any calculation, the person ought to go over the calculations again. And even, recheck the answer in inverted form.

This is so because while doing the calculations, we can possibly make errors that we won't notice until after submission. Knowing 2 * 3 = 6, but writing 2 * 3 = 5 in the course of calculations can happen to anybody. So therefore, cross checking and reverse checking is needed

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

Answer:

more than 51.45 N

__________________________________________________________

We are given:

Mass of the television = 15 kg

Coefficient of Static friction = 0.35

Minimum force required to move the television:

Normal Force:

We know that the normal force is equal and opposite to the Weight of the television

Weight of the television = Mg            

[where m is the mass and g is the acceleration due to gravity]

Weight = 15 * 9.8

Weight = 147 N

Force of Friction:

We are given the coefficient of Friction = 0.35

We know that coefficient of Friction = Force of friction / Normal Force

replacing the variables

0.35 = Force of Friction / 147

Force of Friction = 147 * 0.35                     [multiplying both sides by 147]

Force of Friction = 51.45 N

Since a force of 51.45 N is will be applied opposite to the direction of application of Force, the television will only move when we apply more force than 51.45 N

Answer:

it is C

Explanation:

Answer: When two objects in contact with each other are at different temperatures, they are said to be in thermal equilibrium.

Explanation: . When two objects not in contact with each other are at the same pressure, they are said to be in thermal equilibrium.

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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