Two long, straight, parallel wires, 10.0 cm apart carry equal 4.00-A currents in the same direction, as shown in (Figure 1).
a) Find the magnitude of the magnetic field at point P1 , midway between the wires.
b) What is its direction?
c) Find the magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 .
d) What is its direction?
e) Find the magnitude of the magnetic field at point P3 , 20.0 cm directly above P1 .
f) What is its direction?

Two Long, Straight, Parallel Wires, 10.0 Cm Apart Carry Equal 4.00-A Currents In The Same Direction,

Answers

Answer 1

(a) The magnitude of the magnetic field at point P1 , midway between the wires is  1.005 x 10⁻⁴ T and the direction will be out of the page.

(b) The magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 is 2.67 x 10⁻⁶ T and the direction is into the page.

(c) The magnitude of the magnetic field at point P3 , 20.0 cm directly above P1  is 3.88 x 10⁻⁶ T and the direction is downwards.

Magnetic field midway between the wires

B = μ/2π[I₁/0.5r + I₂/0.5r]

B = (μ/2π) x (I/0.5r + I/0.5r)

B = (μ/2π) x (2I/0.5r)

B = μI/0.5r

B = 2μI/r

where;

I is current in the wiresr is the distance between the wires

B = (2 x 4π x 10⁻⁷ x 4)/(0.1)

B = 1.005 x 10⁻⁴ T

The direction of the magnetic field is out of the page.

Magnetic field at 25 cm right of P1

B = μI/2πd

d = 5 cm + 25 cm = 30 cm

B =  (4π x 10⁻⁷ x 4)/(2π x 0.3)

B = 2.67 x 10⁻⁶ T

The direction of the magnetic field is into the page towards P1.

Magnetic field at 20 above P1

B = μI/2πd

d = √(20² + 5²)

d = 20.62 cm

B =  (4π x 10⁻⁷ x 4)/(2π x 0.2062)

B = 3.88 x 10⁻⁶ T

The direction of the magnetic field is downwards towards P1.

Thus, the magnitude of the magnetic field at point P1 , midway between the wires is  1.005 x 10⁻⁴ T and the direction will be out of the page.

The magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 is 2.67 x 10⁻⁶ T and the direction is into the page.

The magnitude of the magnetic field at point P3 , 20.0 cm directly above P1  is 3.88 x 10⁻⁶ T and the direction is downwards.

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Related Questions

Which of the following describes the products of a chemical reaction?
A. The original materials
B. The substances that are changed
C. The chemicals on the left side of a chemical equation
Ο Ο
D. The chemicals on the right side of a chemical equation

Answers

Answer:

D The chemicals on the right side of a chamical equation

A man pushes a block of ice across a frozen pond at a constant velocity. While the coefficients of static and kinetic friction for ice are low, they are not zero. Consider this problem to involve friction. If necessary, use Fs for the force of static friction, and Fk as the force of kinetic friction.

Required:
Draw the Free Body Diagram for the block of ice.

Answers

Answer:

   F₁> F₂

Explanation:

For this exercise Newton's second law is used, in the adjoint we can see the unapplied forces in this exercise.

Y axis y

        N- W = 0

in this axis there is no movement

X axis

          F -fr = m a

as they indicate that the velocity is consonant the acceleration is worth zero

          F - fr = 0

friction force has the expression

           fr = μ N

           fr = μ mg

we substitute

           F = μ m g

by the time the block is stopped the deferred force is

           F₁ = μ_s m g

when it begins to move the force should decrease to

           F₂ = μ_k k m g

as the static coefficient is greater than the dynamic coefficient

             F₁> F₂

The free body diagram consists of applied force (F) and kinetic frictional force acting in opposite direction.

Net force on the block

The net force on the block will result constant speed of the block which is zero acceleration.

[tex]\Sigma F= 0\\\\F - F_f = 0\\\\F - \mu_k F_n= 0\\\\F - \mu_k mg = 0\\\\F - \mu k W = 0\\\\F = F_f\ \ \ or \ \ F = \mu_k W[/tex]

Free body diagram

The free body diagram consists of applied force (F) and kinetic frictional force acting in opposite direction.

                        F →  Ф ← Ff

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when material allow to heat to pass though them rapidly they are known as __________​

Answers

Answer:

They are conductors/conductive. Materials that can transfer thermal energy well are conductive.

Explanation:

Answer:when material allow to heat to pass though them rapidly they are known as Conductors

What are conductors

In physics /electrical engineering They  allow the flow of charge (electrical current) in one or more directions. Materials made of metal are common electrical conductors.

What are examples of conductors

.Silver

.copper

.iron

.aluminum

.brass

Difference between conductors and insulators?

Insulators a material that is a poor conductor (as of electricity or heat)Whereas Conductors are fantastic at allowing the flow of the heat.

Hope this helps

                                                                           -Tobie

Help plz I’ll mark brainliest

Answers

Answer:

It's A

Explanation:

sound waves are longitudinal they need a medium to travel through

Diagram B D c с Which car has: Ke = 100 PE=0? * 1 point A B C D​

Answers

Answer:

The car C has KE = 100, PE = 0

Explanation:

The principle of conservation of energy states that although energy can be transformed from one form to another, the total energy of the given system remains unchanged.

The energy that a body possesses due to its motion or position is known as mechanical energy. There are two kinds of mechanical energy: kinetic energy, KE and potential energy, PE.

Kinetic energy is the energy that a body possesses due to its motion.

Potential energy is the energy a body possesses due to its position.

From the principle of conservation of energy, kinetic energy can be transformed into potential energy and vice versa, but in all cases the energy is conserved or constant.

In the diagram above, the cars at various positions of rest or motion are transforming the various forms of mechanical energy, but the total energy is conserved at every point. At the point A, energy is all potential, at B, it is partly potential partly kinetic energy, However, at the point C, all the potential energy has been converted to kinetic energy. At D, some of the kinetic energy has been converted to potential energy as the car climbs up the hill.

Therefore, the car C has KE = 100, PE = 0

Which of the following will be attracted toward a positively charged cloth?

Positively charged sock
Negatively charged pipe
Sound waves
Light energy

Answers

Postive and negatives attract, positive and positive repel. answer is negatively charged pipe.

sound waves and light energy are not "affected" by static electricity

The next four questions refer to the situation below.
A person is swimming in a river with a current that has speed vR with respect to the shore. The swimmer first swims downstream (i.e. in the direction of the current) at a constant speed, vS , with respect to the water. The swimmer travels a distance D in a time tOut . The swimmer then changes direction to swim upstream (i.e. against the direction of the current) at a constant speed, vS , with respect to the water and returns to her original starting point (located a distance D from her turn-around point) in a time tIn .
What is tOut in terms of vR, vS, and D, as needed?

Answers

Answer:

 t_{out} = [tex]\frac{v_s - v_r}{v_s+v_r}[/tex] t_{in},      t_{out} = [tex]\frac{D}{v_s +v_r}[/tex]

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

         [tex]v_{sg 1} = v_{sr} + v_{rg}[/tex]

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

           [tex]v_{sg1}[/tex] = D / [tex]t_{out}[/tex]

           D = v_{sg1}  t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

        [tex]v_{sg 2} = v_{sr} - v_{rg}[/tex]

         [tex]v_{sg 2}[/tex] = D / [tex]t_{in}[/tex]

         D = v_{sg 2}  t_{in}

with the distance is the same we can equalize

           [tex]v_{sg1} t_{out} = v_{sg2} t_{in}[/tex]

          t_{out} =  t_{in}

           t_{out} = [tex]\frac{v_s - v_r}{v_s+v_r}[/tex] t_{in}

This must be the answer since the return time is known. If you want to delete this time

            t_{in}= D / [tex]v_{sg2}[/tex]

we substitute

            t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

            t_{out} = [tex]\frac{D}{v_s +v_r}[/tex]

What are some technological limitations that currently prevent humans from traveling to distant planets?

Answers

Answer:

Propulsion system, antigravitational tech

Explanation:

Fuel is extremely inefficient and expensive not to mention it weighs a lot. You really only need to reach escape velocity to leave earth. The rest is just a little amount of boosting to alter course and slow down for landing. I couldn't really think of much. Once we have an antigravitational system then you could say the whole rocket is holding you back because the design would be different. Nobody really knows how to defy gravity but that would be a technolgical limitation for sure.

Name the state of matter that diffusion happens the fastest in.

Answers

Answer:

Liquids

Explanation:

Diffusion occurs fastest in liquids.

Consider the air over a city to be a box that measures 100 km per side that reaches up to an altitude of 1.0 km. Wind (clean air) is blowing into the box along one of its sides with a speed of 4 m/s. An air pollutant is emitted into the box at a rate of 10.0 kg/s; the pollutant degrades with a rate constant k = 0.20/hr. a. Find the steady state concentration of the pollutant (µg/m3 ) in the box if the air is assumed to be completely mixed. b. If the wind speed suddenly drops to 1 m/s, estimate the concentration of the pollutant (µg/m3 ) two hours later.

Answers

Answer:

a)  ρ = 6.25 10⁵ μg / m³, b) ρ  = 1 10⁷ μg / m³

Explanation:

Let's analyze the exercise a little before starting, we must know the amount of pollutant in the box, that the one that enters less the one that degrades and with this value find the density or concentration.

Let's start by finding the volume of air that goes into the box

               V = Lh x

Let's find the distance of air that enters per unit of time, as it goes at constant speed

               x = v₀ t

we substitute

               V₀ = Lh v₀ t

At this same time, a quantity of pollutant is distributed

              Q₀ = r t  

the contaminant that is entering reaches the entire box, therefore the total amount of contaminant is

               Q = Qo t

we substitute

               Q = r t²

the net amount of pollutant that remains is that less enters the one that degraded in the same time, as they ask for the steady state

              [tex]Q_{net}[/tex]= Q - k t

 

the pollutant concentration is

              ρ = Q_net / V

              V = L L h

              ρ =[tex]\frac{r \ t^2 - k \ t}{ L^2 h}[/tex]

              ρ = [tex](r \frac{ L^2}{v_o^2} - k \frac{L}{v_o} ) \frac{1}{L^2 h}[/tex]

               ρ = [tex]\frac{r}{ v_o h} -\frac{k}{v_o L h}[/tex]

let's reduce the magnitudes to the SI system

           r = 10 kg / s

           L = 100 km = 100 10³ m

           h = 1 km = 1 10³ m

           k = dq / dt = 0.20 1/h ( 1h/3600 s) = 5.5555 10⁻⁵  1/s

           v₀ = 4 m / s

let's calculate

The volume of the box

             V = (100 100 1) 109

             V = 1 10¹³ m³

            ρ = [tex]\frac{10}{ 4^2 \ 1\ 10^3 } - \frac{5.5556 \ 10^{-5}}{ 4 \ 100 \ 10^3 1 \ 10^3}[/tex]

            ρ = [tex]6.25 10^{-4} - 1.389 ^{-13}[/tex]

            ρ = 6.25 10⁻⁴ kg / m³

       

let's reduce to μg / m³

               ρ = 6.25 10⁻⁻⁴ kg / m³ (10⁹ μg / 1kg)

               ρ = 6.25 10⁵ μg / m³

 

b) in case the air speed decreases to v₀ = 1 m / s

             

             ρ= \frac{10}{ 1^2 \  1\  10^3 } - \frac{5.5556 \ 10^{-5}}{ 1 \ 100 \ 10^3  1 \ 10^3}

             ρ = 1 10⁻² - 5.5556 10⁻¹³

             ρ =  1 10⁻² kg / m³

             ρ  = 1 10⁷ μg / m³

g Suppose that you seal an ordinary 60W lightbulb and a suitable battery inside a transparent enclosure and suspend the system from a very sensitive balance. (a) Compute the change in the mass of the system if the lamp is on continuously for one year at full power. (b) What difference, if any, would it make if the inner surface of the container were a perfect reflector

Answers

Answer:

kekemeeimdeiddnekem

Explanation:

mdjdjdiddmjd jjeneeiej

what type of reaction is being shown in this energy diagram?

X exothermic, because energy is absorbed from the surroundings

O exothermic, because energy is released into the surrounding

X endothermic, because energy is released into the surrounding

X endothermic, because energy is absorbed from the surroundings​

best of luck nerds

Answers

Answer:

O exothermic, because energy is released into the surrounding

Explanation:

From the diagram the energy of the reactant is higher than the energy of the product, thereby making it exothermic. If you study diagram well, exothermic reaction means that the reactions releases energy into the surroundings.

2. One tin for weight control is to:
Eat alone
Eat slowly

Answers

Answer:

Eat slowly

Explanation:

If you eat slower, you'll chew your food better, which leads to better digestion. Digestion actually starts in the mouth, so the more work you do up there, the less you'll have to do in your stomach. This can help lead to fewer digestive problems. Less stress.

Eat slowly is answer

A motorcycle and rider have a total mass equal to 300 kg. The rider applies the brakes, causing the motorcycle to decelerate at a rate of -5 m/s^2. What is the net force on the motorcycle?

Answers

Answer:

Net force = - 1500 N

Explanation:

We calculate the net force acting using Newton's second Law:

[tex]F_{net}=m*a\\F_{net}=(300 \,kg)*(-5\,m/s^2)\\F_{net}=-1500\,N[/tex]

3.
What part of your eye is responsible for regulating the amount of light that enters your eye?

Answers

Answer:

Iris

Explanation:

The iris seems to be the illuminated portion of the eyes which really covers the pupil. It controls the amount of light reaching the eye. The lens is indeed a translucent layer of the retina that serves to concentrate light and objects on the lens.

Answer:

I hope this helps.

Explanation:

2) The track for a racing event was designed so that riders jump off the slope at 37 degrees from a height of 1 m. During a race it was observed that the rider remained in mid air for 1.5 seconds. Determine the speed at which he was traveling off the slope, the horizontal distance he travels before striking the ground and the maximum height he attains. Neglect the size of the bike and rider.

Answers

Answer:

3.277 m

Explanation:

Given :

Maximum Height (Hmax) = (u²sin²θ) / 2g

Xv = Xh + Uv * t + 0.5gt²

Xv and Xh are vertical and horizontal distances

-1 = 0 + sin37 * 1.5 Uv + 0.5*-9.8*1.5^2

-1 = 0 + 0.903Uv - 11.025

-1 + 11.025 = 0.903Uv

10.025 = 0.903Uv

Uv = 10.025 / 0.903

Uv = 11.10 m/s

Hmax = 1 + (u²sin²θ) / 2g

= (11.10^2 * (sin37)^2) / 2*9.8

= 44.624360 / 19.6

= 2.277

Hmax = 1 + 2.277

Hmax = 3.277 m

The dielectric constant of the interior of a protein is considerably smaller than that of water. How would this difference in dielectric constants affect the strength of an electrostatic interaction between two opposite charges with the same distance between them if the charged groups were located in the interior of the protein rather than on its surface

Answers

Answer:

the interaction in the protein is greater than the surface with water

\frac{F_i}{F_s} = \frac{\epsilon_s}{ \epsilon_i} \ > 1

Explanation:

The electric force  for a charge is

          F = [tex]\frac{1}{4\pi \epsilon} \ \frac{q^2}{r^2}[/tex]

In the exercise indicate that the charge is q and the distance r is maintained, the test charge is another  

therefore if we use the index i for the dielectric constant ([tex]\epsilon_i[/tex]) in the protein

         [tex]F_{i} = \frac{1}{4\pi \epsilon_i} \frac{q^2}{r^2}[/tex]  

the electric force in water with dielectric constant ([tex]\epsilon_s[/tex])

           [tex]F_s = \frac{1}{4\pi \epsilon_s} \frac{q^2}{r^2}[/tex]

            [tex]\epsilon_i < \epsilon_s[/tex]

if we look for the relationship between these forces

          [tex]\frac{F_i}{F_s} = \frac{\epsilon_s}{ \epsilon_i} \ > 1[/tex]

therefore the interaction in the protein is greater than the surface with water

A diet is to contain at least 2400 mg vitamin C, 1800mg Calcium, and 1200 calories every day. Two foods, a dairy-based meal and a vegan option are to fulfill these requirements. Each ounce of the dairy-based meal provides 50 mg vitamin C, 30 mg Calcium, and 10 calories. Each ounce of the vegan option provides 20 mg vitamin C, 20 mg Calcium, and 40 calories. If the dairy-based meal costs $0.042 per ounce and the vegan option costs $0.208 per ounce, how many ounces of each food should be purchased to minimize costs? What is that minimum cost (per day)?

Answers

Answer:

The answer is below

Explanation:

Let x represent the number of ounce of dairy based meal and let y represent the number of vegan option in ounce.

Since the diet must contain at least 2400 mg vitamin C, therefore:

50x + 20y ≥ 2400

Since the diet must contain at least 1800 mg Calcium, therefore:

30x + 20y ≥ 1200

Since the diet must contain at least 1200 calories, therefore:

10x + 40y ≥ 1200

Therefore the constraints are:

50x + 20y ≥ 2400

30x + 20y ≥ 1200

10x + 40y ≥ 1200

x > 0, y > 0

The graph was drawn using geogebra online graphing tool, and the solution to the problem is at:

C(30, 45) and D(48, 18)

dairy-based meal costs $0.042 per ounce and the vegan option costs $0.208 per ounce. The cost equation is:

Cost = 0.042x + 0.208y

At C(30, 45);  Cost = 0.042(30) + 0.208(45) = $10.62

At C(48, 18);  Cost = 0.042(48) + 0.208(18) = $5.76

The minimum cost is at (48, 18). That is 48 dairy based meal and 18 vegan

g Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of magnitude 46.0 N is exerted and the magnitude of the astronaut's acceleration is measured to be 0.834 m/s2. Calculate her mass.

Answers

Newton’s second law is F = mass times acceleration.

F divided by acceleration equals her mass.

46/0.834 = 55.156 kg

Explain how a common housecat gets “worms.”eplain(science)

Answers

Answer:

Cats most commonly contract worms after coming into contact with parasite eggs or infected feces. A cat may walk through an area with eggs or infected feces, and since cats are often such fastidious groomers, they will then ingest the eggs or fecal particles as they clean their fur and feet.

Explanation:

this is the only thing in my book hope it helps

How much force is needed to accelerate a Kia Soul with a
mass of 1200 kg to 5 m/s2?

Answers

Answer:

[tex]\boxed {\boxed {\sf 6,000 \ Newtons}}[/tex]

Explanation:

Force is the product of mass and acceleration.

[tex]F=ma[/tex]

The mass of the Kia Soul is 1200 kilograms and its acceleration is 5 meters per square second.

[tex]m= 1200 \ kg \\a= 5 \ m/s^2[/tex]

Substitute the values into the formula.

[tex]F= 1200 \ kg * 5 \ m/s^2[/tex]

Multiply.

[tex]F= 6000 \ kg*m/s^2[/tex]

1 kilgram meter per square second is equal to 1 Newton. Our answer of 6000 kg*m/s² equals 6000 N

[tex]F= 6000 \ N[/tex]

Answer:

Given :-Mass = 1200 kgAcceleration = 5 m/s²To Find :-

Force

Solution :-

We know that

F = ma

F = Force

m = mass

a = acceleration

F = 1200 × 5

F = 6000 N

[tex] \\ [/tex]

If 0.5 C charge passes through a wire in 10 seconds, what will be the value of the current flowing through the wire? *
20 mA
30 mA
50 mA
60 mA​

Answers

Answer:

electric current passing through it will be 50mA

Explanation:

electric current = charge / time

I = Q / TI = 0.5 / 10 I = 0.05 ampere

current = 0.05 A = 50mA

If 0.5C charge passes through a wire in 10 seconds, then 50mA current is flowing through the wire. Thus, the correct option is C.

What is Electric current?

Electric current is the flow of electricity in an electronic circuit. It is the amount of electricity flowing through a electronic circuit. It is generally measured in amperes (A). The larger the value in amperes, the more electricity is flowing in that circuit.

The formula for calculation of Electric current is:

I = Q/T

where, I = electric current,

Q = amount of charge,

T = time required

Therefore, the current flowing in the wire is:

I = 0.5C/ 10 seconds

I = 0.05 A or 50mA (1mA = 10⁻³A)

Therefore, the correct option is C.

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Two blocks collide on a frictionless surface, as shown above. They have a combined mass of 10 kg and a speed of 2.5 m/s. Before the collision, one of the blocks was at rest. This block had a mass of 8 kg. What was the speed of the second block?

Answers

Answer:

12.5 m/s

Explanation:

Excuse my scribbles!

I had to work backwards using the inelastic collision formula for this problem.

Formula: V=(M₁V₁+M₂V₂)/(M₁+M₂)V= Combined SpeedM₁= Block 1's MassV₁= Block 1's Velocity M₂= Block 2's MassV₂= Block 2's Velocity

Step 1: Substitute in the values provided in the problem

Combined mass: 10kgCombined speed: 2.5m/sBlock 1's mass: 8kgBlock 1's speed: 0

2.5=(8*0)+(?*?)/(8+?)

Step 2: Subtract block 1's mass from the combined mass to determine block 2's mass

10-8=2     Block 2's mass is 2.

2.5=(8*0)+(2*x)/(8+2)    

Now simplify.

2.5=(2*x)/(8+2)

2.5=2x/10

Step 3: Multiply both sides by the reciprocal

(5)2.5=2x/10(5)

12.5=x

Answer is checked in the attached images!

Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown to the right. The mass of the left block m1 = 1.4 kg and the mass of the right block m2 = 4.9 kg. The angle between the applied force and the horizontal is θ = 54°. The coefficient of kinetic friction between the blocks and the surface is μ = 0.38. Each block has an acceleration of a = 3.6 m/s2 to the right.

Answers

Answer:

Explanation:The Mass Of The Left Block M1 = 1.3 Kg And The Mass Of The Right Block M2 = 3.1 Kg. The Angle Between The String And The Horizontal Is ... (10%) Problem 8: Two blocks connected by a string are pulled across a horizontal surface by a ... m m, 50% Part (a) Write an equation for the magnitude of the force exerted by the ...

It's time to get a little more specific. Based on the velocity (Vx) graph for the car and the velocity data in the table, divide the total
motion of the car into rough time periods that tell a different "chapter" of the story for this car trip. In each of these time
periods, the car's velocity will be notably different from the previous period. Enter a brief description of the car's motion in each
period. The first one is done for you. Use it as an example to identify and describe the remaining time periods. Note: You can
define as many periods as you think appropriate.
s
B
1
U X
X х.
Font Sizes
А • А
E
E 를 들
E 3
Numbered list
Time Period
Motion Description
0.2 - 4.6 seconds increasing speed in positive direction

Answers

Answer:

0.2 – 4.6 seconds   increasing speed in positive direction

4.6 - 7.8 seconds   decelerating speed in a positive direction

8 - 17.2 seconds  accelerating speed in a negative direction

Explanation:

**Plato** **Edmentum**n~ this question is pretty open ended, so its hard to get it wrong honestly, good luck <3 ~

Answer:

0.2 – 4.6 seconds   increasing speed in positive direction

4.6 - 7.8 seconds   decelerating speed in a positive direction

8 - 17.2 seconds  accelerating speed in a negative direction

Explanation:

Bartender slides a beer mug at 1.1 m/s towards a customer at the end of the bar which is 1.8 m tall. The customer makes a grab for the mug and misses and mug sails at the end of the bar. a) How far away from the end of the bar does the mug hit the floor

Answers

Answer:

Δx = 0.7 m

Explanation:

Once the mug is moving in the horizontal direction, it keeps moving at the same speed of 1.1 m/s, due to no other force acts on it in this direction.Since the horizontal and vertical movements are independent each other (due to they are mutually perpendicular), in the vertical direction, the initial speed is just zero.In the vertical direction, the mug is accelerated by the force of gravity at all times, with a constant value of 9.8 m/s2, aimed downward.So, we can use the following kinematic equation in order to get the time passed from the instant that the mug left the bar, until it hit the floor, as follows:[tex]\Delta y = \frac{1}{2} * g* t^{2} = (1)[/tex]where  Δy = 0-1.8m = -1.8m, g= -9.8m/s2.Replacing these values in (1) and solving for t, we get:

       [tex]t = \sqrt{\frac{2*1.8m}{ 9.8m/s2} } = 0.6 s (2)[/tex]

Now, since the mug obviously finishes its horizontal trip at this same time (hitting ground), we can find the horizontal distance traveled, just applying the definition of average speed, as follows:

       [tex]\Delta x = v_{o} * t = 1.1 m/s* 0.6 s = 0.7 m (3)[/tex]

A large truck and a small car traveling at the same speed have a head-on collision. The vehi-cle to undergo the greater change in velocity will be?

Answers

Answer:

The car ...Explanation:

2.) The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his/her head, forcing them to move quickly away from the net. Suppose that you loft the ball with an initial speed of 15m/s at an angle of 50 degrees from the horizontal. At this moment your opponent is 10m from the ball. They begin to run away from you 0.3 seconds after the ball was launched hoping to reach the ball and hit it back to you at a height of 2.1m above where you hit it. What is the minimum average speed that your opponent must move so that he is in position to hit this ball

Answers

Answer:

The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s

Explanation:

The given parameters of the ball are;

The initial speed of the ball = 15 m/s

The direction in which the ball is launched = 50° above the horizontal

The location of the other tennis player when the ball is launched = 10 m from the ball

The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched

The height at which the ball is hit back = 2.1 m above the height from which the ball is launched

The vertical position, 'y', at time, 't', of a projectile motion is given as follows;

y = (u·sinθ)·t - 1/2·g·t²

When y = 2.1 m, we have;

2.1 = (15·sin(50°))·t - 1/2·9.8·t²

∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0

Solving with the aid of a graphing calculator function, we get;

t = 0.199776187257 s or t = 2.14525782198 s

Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds

The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;

x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m

The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m

Therefore, we have;

The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s

The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m

The minimum average speed the other player has to move with, [tex]v_s[/tex] = d/t₂

∴ [tex]v_s[/tex] = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s

The minimum average speed the opponent must move so that he is in position to hit the ball, [tex]v_s[/tex] ≈ 5.79 m/s.

Imagine a third particle, which we will call a cyberon. It has three times the mass of an electron (3_m). It has a positive charge that is three times the magnitude (3_(qe)) of the charge on an electron. What is the ratio of the speed v_c that the cyberon would have when it reaches the upper plate after being released from rest at position h_0 to the speed ve that the electron would have?

Answers

Answer:

The answer is "The last choice".

Explanation:

Please find the complete question in the attachment.

In an external electric field, its electrical energy at positive charge becomes directed to just the electrical domain. Therefore it will speed towards its base plate whenever cyber one is released to rest at h0. It was never going to reach the top plate. Thus,  the last choice corrects because in this the cyber-on never reaches its upper stage.

Astronauts aboard the ISS move at about 8000 m/s, relative to us when we look upward.How long does an astronaut need to stay aboard the space station to be a full second youngerthan people on the ground? Please show and explain how you would set-up the problem,before you actually try to solve it. If you cannot solve it exactly, please try to offer an estimate.(5 pts)

Answers

Answer:

#_time = 7.5 10⁴ s

Explanation:

In order for the astronaut to be younger than the people on earth, it follows that the speed of light has a constant speed in vacuum (c = 3 108 m / s), therefore with the expressions of special relativity we have.

            t = [tex]\frac{t_p}{ \sqrt{1- (v/c)^2} }[/tex]

where t_p is the person's own time in an immobile reference frame,

           [tex]t_{p} = t \sqrt{1 - (\frac{v}{c})^2 }[/tex]

let's calculate

we assume that the speed of the space station is constant

              [tex]t_p = 1 \sqrt{1 - \frac{8 \ 10^3}{3 \ 10^8} }[/tex]

             [tex]t_p = 1 \sqrt{1- 2.6666 \ 10^{-5}}[/tex]

             t_ =  0.99998666657   s

             

therefore the time change is

             Δt = t - t_p

             Δt = 1 - 0.9998666657                  

              Δt = 1.3333 10⁻⁵ s

this is the delay in each second, therefore we can use a direct rule of proportions. If Δt was delayed every second, how much second (#_time) is needed for a total delay of Δt = 1 s

               #_time = 1 / Δt

               #_time =[tex]\frac{1}{1.3333 \ 10^{-5}}[/tex]

               #_time = 7.5 10⁴ s

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