A 50 kg child is riding on a carousel (merry-go-round) at a constant speed of 5 m/s. What is the magnitude of the change in the child's momentum ∣Δp⃗ ∣∣ in going all the way around (360∘)?In going halfway around (180∘)? It is very helpful to draw a diagram, and to do the vector subtraction graphically.

Answer:

  y /y₀ = 1 /√2 = 0.70

Explanation:

To close the door, a torque must be applied in such a way that there is an angular acceleration and after a period of time the angle of the door moves.

If the force is doubled keeping the other parameters without altering angular acceleration is

                 F r = I α

                 α = F r / I

if we double the force

           α = 2F r / I

            α=2 α₀

let's use the angular kinematics relations

            θ = w₀ t + ½ α t²

we assume the door was still before pushing it

            θ = ½ a t²

            t = √ (2θ / a)

as we saw the angular acceleration is double

            t = √ (2θ / 2a₀)

            t = 1 / √2  a₀

the relationship between the two times is

            y /y₀ = 1 /√2 = 0.70

This time that has been reduced closes the door.

Answer: 3.38 km per hour

Explanation:

Answer:

Magnetic flux is zero

Explanation:

We know that

magnetic flux (B) =

magnetic field x area x cosစ

And စ being the angle between magnetic field and area vector

So we are given စ= 90°

So

B = magnetic field x area x cos(90)

B= 0

Answer:

Loudness and pitch are the characteristics of a sound related to amplitude and frequency.

Explanation:

Loudness and pitch are the characteristics of a sound that are related with the frequency and amplitude of a sound. If any change occurs in the frequency and amplitude of a sound, change also occurs in the loudness and pitch of a sound. If a source has larger amplitude so it produces large loudness or volume of a sound. On the other hand when the frequency is higher than the pitch will also be higher and the sound produce will be high.

Answer:

v_wind = 101.46 km / h   ,  θ = 61.8

Explanation:

This is a velocity composition exercise.

Let's do the problem in parts. Let's start by knowing the speed of the plane without air.

           v = d / t

           v = 750 / 3.14

           v = 238.85 km / h

This is the speed of the plane relative to the Earth and it does not change.

In the second part, when there is wind, the travel time is greater than when there is no wind, therefore the wind delays the plane. To be more general, suppose that the wind has two components vₓ and [tex]v_{y}[/tex]

Let's use trigonometry to find the components of the plane's speed

          cos θ = v_N / v

          sin θ  = v_W / v

          v_N = v cos θ

          v_W = v sin θ

let's calculate

          V _N = 238.85 cos 22 = 221.46 km / h

           v_W = -238.85 sin 22 = -89.47

the negative sign is because the plane is going west and the positive sign is the east direction.

As it indicates that the destination of the avine is towards the north, the x component of the wind must be

              vₓ - v_W = 0

              vₓ = v-w

              vₓ = 89.47 km / h

in the direction to the East.

Now let's analyze the component of the wind in the Nort-South direction,

Indicate the travel time, let's calculate the speed that the component must have the speed of the plane

             v_total = d / t

             v_total = 750 / 4.32

             v_total = 173.61 km / h

This is the final speed of the plane, which can be written

              v_total = v_n - vy

               vy = v_n - v_total

               vy = 221.46 - 173.61

               vy = 47.85 km

this component is directed towards the south

Let's use the Pythagorean Theorem, to find the magnitude

             v_wind² = vₓ² + vy²

             v_wind = √ (89.47² + 47.85²)

             v_wind = 101.46 km / h

the address will then be found using trigonometry

             θ = Vy / vx

             θ = tan⁻¹ (vy / vx)

             θ = tan⁻¹1 (47.85 / 89.47)

             θ = 28.14

Therefore, the magnitude of the wind speed is 101.5 km / h and its direction is 28º south of the East, to give this value

                  90- θtea = 90- 28.2

                  θ = 61.8

East of South

Answer:

the work done is 1192.5Nm

Explanation:

work done = force × distance

w.d = 265N × 4.5m

w.d = 1192.5Nm

mark me as brainliest pllyyyzzz

Answer:

The health hazard of directly going to these places to take direct readings and observations.

Explanation:

The Chernobyl accident was a catastrophic radioactive accidents, with an immediate fatal effect on some victims (mostly firemen and law enforcement officers). Scientists investigating the effects of the radiation left on the venues of the accident face a health risk if they intend to carry out a full scale on-field testing and experiment. Some of the relocated occupants of these places, that have long been relocated, have shown some health effect of the radiation. And scientist studying and working on these places for too long stand a risk of developing health complications if they are exposed to the radiations for too long.

Complete question is;

A particle moves along a line so that its velocity at time t is

v(t) = t² − t − 6

(measured in meters per second).

(a) Find the displacement of the particle during the time period 1 ⩽ t ⩽ 4.

(b) Find the distance traveled during this time period.

Answer:

A) -4.5 m

B) 10.17 m

Explanation:

We are given;

v(t) = t² − t − 6

We know that v = ds/dt

Thus,

S = v dt

s(4) - s(1) = (4,1)∫t² − t − 6 dt

= (4,1)[(t³/3) - (t²/2) - 6t]

= ((4³/3) - (4²/2) - 6(4)) - ((1³/3) - (1²/2) - 6(1))

= 64/3 - 8 - 24 - 1/3 + 1/2 + 6

= -4.5 m

B) Since v(t) = t² − t − 6, then factorizing we can write it as;

v(t) = (t + 2)(t - 3)

Thus, v(t) ⩽ 0 at the interval (1, 3) and v(t) ≥ 0 at the interval (3, 4)

Thus;

-x1 = (1, 3)∫t² − t − 6 dt

-x1 = (1,3)(t³/3 - t²/2 - 6t)

-x1 = (3³/3 - 3²/2 - 6(3)) - (1³/3 - 1²/2 - 6(1))

-x1 = 9 - 9/2 - 18 - 1/3 + 1/2 + 6

-x1 = -3 - 4 - 1/3

-x1 = -22/3 m

x1 = 22/3 m

x2 = (3, 4)∫t² − t − 6 dt

x2 = (3, 4)(t³/3 - t²/2 - 6t)

x2 = ((4³/3) - (4²/2) - 6(4)) - (3³/3 - 3²/2 - 6(3))

x2 = 64/3 - 8 - 24 - 9 + 9/2 + 18

x2 = -23 + 64/3 + 9/2

x2 = (-138 + 128 + 27)/6

x2 = 17/6 m

Thus, total distance = x1 + x2 = 22/3 + 17/6 = 10.17m

Answer:

1m/s should be the answer

Answer:

[tex]\Huge \boxed{\mathrm{1 \ m/s }}[/tex]

[tex]\rule[225]{225}{2}[/tex]

Explanation:

We will convert 1 m/s into mph.

1 m/s = 2.237 mph

2.237 mph > 1 mph

1 m/s is faster than 1 mph.

[tex]\rule[225]{225}{2}[/tex]

Answer:

Young's Modulus = 6 x 10⁶ Pa = 6 MPa

Explanation:

First we need to find the stress on tendon:

Stress = Force/Area

Stress = 11.2 N/πr²

Stress = (11.2 N)/π(4.45 x 10⁻³ m)²

Stress = 180 x 10³ Pa

Now, we calculate the strain:

Strain = Change in length/Original Length

Strain = (9 x 10⁻³ m)/(0.3 m)

Strain = 0.03

Now, we calculate the Young's Modulus:

Young's Modulus = Stress/Strain

Young's Modulus = (180 x 10³ Pa)/0.03

Young's Modulus = 6 x 10⁶ Pa = 6 MPa

Answer:

Faraday's law , he direction of the magnetic field changes by 180º, in the polarity inversion processes, induces a voltage.  

Explanation:

For this exercise let's use Faraday's law

          E = - dФ / dt

          Ф = B.A = B A cos θ

where B is the magnetic field, A is the area and θ is the angle between the field line and the normal to the area.

We can see that an electromotive force (voltage) is indexed when there is a variation of the field B, a variation of the area and change of the angle or when there is a combinational of them.

In this case, the magnitude of the field is constant, as the wire is rigid metal, the area is constant, but the direction of the magnetic field changes by 180º, in the polarity inversion processes, for which reason each change induces a voltage.

If a voltage is created in the ring, which has a resistance, a current is also generated in it.

Therefore the answer is If a current is created in the hoop

Answer:

40000÷40=1000 joules is required to work in 40 seconds

Answer:

motion of an object for which the direction of the acceleration is the same as the direction of motion of the object occurs when the body or object travels with an increase in velocity

Explanation:

Because for direction of object to be the same with direction of acceleration velocity must be increasing

Answer:

λ = 3.62 x 10⁻⁷ m = 362 nm

Explanation:

The grating equation gives the relationship between the wavelength, the diffraction line order and the diffraction angle. The grating equation is written as follows:

mλ = d Sinθ

where,

m = order of diffraction = 6

λ = longest wavelength = ?

d = 1/(460 rulings/mm)(1000 mm /1 m) = 2.17 x 10⁻⁶ m/ruling

θ = Diffraction angle = 90° (for longest wavelength)

(6)λ = (2.17 x 10⁻⁶ m/ruling) Sin 90°

λ = (2.17 x 10⁻⁶ m/rulings)/6

λ = 3.62 x 10⁻⁷ m = 362 nm

Answer:

[tex]$ 1.81 \times 10^{-7} \ T$[/tex]

Explanation:

Given :

r = 1.50 m

R = 1  m

[tex]$\frac{dE}{dt}$[/tex] = [tex]$ 4.88 \times 10^{10} $[/tex]  V / m s

Therefore the displacement current is

[tex]$ I_d = \epsilon_0. \frac{dE}{dt} . A $[/tex]

    = [tex]$ \epsilon_0. \frac{dE}{dt} . \pi R^2 $[/tex]

Now according to law

[tex]$ B= \frac{\mu_0I_d.\frac{dE}{dt}.\pi R^2}{2 \pi r}$[/tex]

  = [tex]$ \frac{\mu_0I_d.\frac{dE}{dt}. R^2}{2 r}$[/tex]

  = [tex]$ \frac{4 \pi \times 10^{-7} \times 8.85 \times 10^{-12} \times 4.88 \times 10^{10} \times 1^2}{2 \times 1.5} $[/tex]

  = [tex]$ 1.81 \times 10^{-7} \ T$[/tex]

Therefore, the magnetic field strength at 1.50 m from the center of the ring is [tex]$ 1.81 \times 10^{-7} \ T$[/tex].

Answer:

Explanation:

Rx = -28.2 units

Ry = 19.6 units

magnitude of R = √  [( - 28.2 )² + ( 19.6 ) ]

= √ ( 795.24 + 384.16 )

= 34.34 units

If θ  be the angle measured counterclockwise from the +x-direction

Tanθ = 19.6 / - 28.2 = -0.695

θ = 180 - 34.8

= 145.2° .

[tex]\star[/tex][tex]\blue\star[/tex]

Star coding

Hope it helps

Answer:

66.8°

Explanation:

Calculating for tthe first leg

dNx = 0;

dNy = + 0.8 mi ......................(0 , 0.8)

then for the second

dWx = - 0.3 mi;

dWy = 0 ...................(-0.3 , 0)

then the third leg

dSx = 0;

dSy = - 0.1 mi ......................(0 , -0.1)

So the displacement vector is

db = ( -0.3 , (0.8 - 0.1) )

db = ( -0.3 , 0.7)

Finding the magnitude i= √(0.3^2 + 0.7^2 )

= 0.76

So the direction is

စ= tan^-1( 0.7 / 0.3 ) = 66.8°

Answer:

The question is incomplete. Let me explain the relationship between electric field and equipotential lines.

The electric field lines point symmetrically away from the positive point charge, this clearly indicates that they are perpendicular to the equipotential lines.

Explanation:

The potential for a point charge is V = [tex]\frac{kQ}{r}[/tex]

where k is a constant with value of 8.99 x 109 N m2/ , and r is the distance from point charge.

This shows that the potential for a point charge is the same anywhere on an imaginary sphere of radius r surrounding the charge.

The attached image is an isolated positive point charge Q and its electric field lines.

In the image, the electric field lines radiate out from a positive charge and terminate on negative charges.

The blue lines indicates the magnitude and the direction of the electric field.

This clearly helps us to see that the electric field lines point away from the positive point charge, which means that they are perpendicular to the equipotential lines.

Answer: Provided in the explanation

Explanation:

I have understood that I have been influenced by the 'affinity bias' for quite a while. It caused me to feel fascination and feel better and right about individuals who had comparable intrigue and thought designs. I sort of began to feel this is the one for me based on those likenesses yet later used to be miserable when I comprehended that those similitudes are not many and insufficient consistently. I have begun to beat this inclination by rehearsing self reflection. I have begun to think about what causes me to feel pulled in to somebody and on the off chance that I introspect that it's exclusively founded on likenesses, at that point I cause myself to get that and monitoring that helps in escaping the inclination.

I make an effort not to get into the snare of paradoxes in my own announcements yet I have been forced to bear deceptions during contentions. One deception which I have encountered most is the 'Foul play' error as there have been a great deal of occurrences when individuals began to tear down me and my family when they couldn't win on a contention regarding balanced and rationale.

Basic reasoning causes us in understanding our defects and those issues of our own which keeps us from taking better choices and furthermore forestalls us tackling issues in more successful manners. Through basic reasoning, one can reflect and recognize utilization of deceptions in the contentions and that will help him in deciphering the circumstance from an alternate perspective. He will assess the circumstance better and through appropriate induction, he will have the option to get over those issues in thinking and subsequently, have the option to take better and more powerful choices for taking care of an issue.

The most fascinating idea which I have learned is that of the utilization of reflection or explicitly self reflection to comprehend ourselves better and furthermore to display basic reasoning appropriately. Being said that, I am a still somewhat confused about the manners by which we can reflect appropriately at each circumstance in a successful manner. This disarray remains in light of the fact that most deceptions and predispositions are oblivious in nature while the basic considering aptitudes reflection is cognizant and I meander whether a cognizant ability will have the option to break down and distinguish each oblivious inclinations or whether a few guards will keep a few inclinations covered up.

Answer:

a teacher that ive enjoyed is my 5th grade reading teacher. I enjoyed her because she always made class fun

Explanation:

Answer:

Tension on the string is 48 N

Explanation:

Mass of A = 6 kg

mass of B = 4 kg

If the 9 kg accelerates downwards then the equation for the 6 kg mass side of the pulley will be

6g - T = 6a    ....1

where

g is acceleration due to gravity = 10 m/s^2

a is the acceleration of the body

T is the tension in the string.

For the 4 kg mass side of the string, we have

T - 4g = 4a    ....2

transposing, we have

T = 4a + 4g    ....3

substitute equation 3 in equation 1, we have

6g - (4a + 4g) = 6a

6g - 4a - 4g = 6a

6g - 4g = 6a + 4a

2g = 10a

but g = 10 m/s^

2 x 10 = 10a

20 = 10a

acceleration a = 20/10 = 2 m/s^2

substitute value of a in equation 1

6g - T = 6a

6(10) - T = 6(2)

60 -T = 12

60 - 12 = T

T = 48 N

Answer:

-2.5

Explanation:

Edge 2020

By observing the given graph The acceleration of the object would be -2.5  m/s², in other words, one can say the deacceleration of the object is 2.5 m/s²

The rate of change of the velocity with respect to time is known as the acceleration of the object. Generally, the unit of acceleration is considered as meter/seconds².

acceleration =change in velocity/change in time

The slope of the velocity time diagram represents the acceleration of the moving object

As we can see from the graph the Y axis is representing the velocity and the X axis is representing the time

The Y coordinate changes from 40 m/s to 10 m/s and the X coordinates change from 0 s to 12 seconds

slope = (10 -40 )/(12-0)

acceleration = 30/12

acceleration =  -2.5 m/s²

Thus, By looking at the provided graph In other terms, the object's acceleration would be -2.5 m/s2, or its deacceleration would be 2.5 m/s².

Learn more about acceleration from here

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Answer:

its B

Explanation:

i just took the test

The correct option is B. The magnitude of the force on the charge is 12N and the direction is downward.

The electric field is a region of space where electric force can be felt. The formula for calculating the magnitude of a force in an electric field is expressed as:

[tex]\overline F=qE[/tex] where:

Given the following parameters:

q = -0.06C

E = 200N/C

Substitute the given parameters into the formula as shown:

[tex]\overline F = -0.06 * 200\\\overline F = -12N\\[/tex]

Since the modulus of the force is in the negative sense, hence the direction will be downward.

Learn more here: https://brainly.com/question/22735564

Answer:

Some of the most severe weather occurs when a single thunderstorm affects one location for an extended time. Thunderstorms can bring heavy rains (which can cause flash flooding), strong winds, hail, lightning, and tornadoes. Severe thunderstorms can cause extensive damage to homes and property.

Explanation:

Answer:

1) the new power coming from the amplifier is 19.02 W

2) The distance away from the amplifier now is 5.50 m

3) u₁ = 69.24 m

Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther

Explanation:

Lets say that I am at a distance "u" from the TV,

Let I₁ be the corresponding intensity of the sound at my location when sound level is 125dB

SO

S(indB) = 10log (I₁/1₀)

we substitute

125 = 10(I₁/10⁻¹²)

12.5 = log (I₁/10⁻¹²)

10^12.5 = I₁/10^-12

I₁ = 10^12.5 × 10^-12

I₁ = 10^0.5 W/m²

Now I₂ will be intensity of sound when corresponding sound level is 107 dB

107 = 10log(I₂/10⁻²)

10.7 = log(I₂/10⁻¹²)

10^10.7 = I₂ / 10^-12

I₂ = 10^10.7  ×  10^-12

I₂ = 10^-1.3 W/m²

Now since we know that

I = P/4πu² ⇒ p = 4πu²I

THEN P₁ = 4πu²I₁ and P₂ =4πu²I₂

Therefore

P₁/P₂ = I₁/I₂

WE substitute

P₂ = P₁(I₂/I₁) = 1200 × ( 10^-1.3 / 10^0.5)

P₂ = 19.02 W

the new power coming from the amplifier is 19.02 W

2)

P₁ = 4πu²I₁

u =√(p₁/4πI₁)

u = √(1200/4π × 10^0.5)

u = 5.50 m

The distance away from the amplifier now is 5.50 m

3)

Let I₃ be the intensity corresponding to required sound level 85 dB

85 = 10log(I₃/10⁻¹²)

8.5 = log (I₃/10⁻¹²)

10^8.5 = I₃ / 10^-12

I₃ = 10^8.5  × 10^-12

I₃ = 10^-3.5 w/m²

Now, I ∝ 1/u²

so I₂/I₃ = u₁²/u²

u₁ = √(I₂/I₃) × u

u₁ = √(10^-1.3 / 10^-3.5) ×  5.50

u₁ = 69.24 m

Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther

Answer:

3.5E-15m

Explanation:

Using

R= Ro A^ 1/3

Where Ro= 1.2fm

and A= 7

= 1.2*10^-15 x (7)^1/3

= 3.5*10^-15m

Answer:

60 V

The direction will be anticlockwise from the right side of the magnet.

Explanation:

The change in magnetic flux = ∆Φ = 0.3 T.m^2

The change in time = ∆t = 0.2 s

number of turns = 40 turns

The induced emf E = N∆Φ/∆t =

E = (40 x 0.3)/0.2 = 60 V

If the magnet is moved away from the coil, the induced current on the coil will try to oppose the motion of the magnet by attracting the magnet towards the coil. For the magnet to be attracted towards the coil, it must possess the equivalent of a magnetic south pole. For the equivalent of a magnetic south pole, the current on the coil will flow in the clockwise direction when viewed from the left side of the magnet. When viewed from the right side of the magnet, the direction will appear as anticlockwise.

Answer:

2.04 minutes

Explanation:

He travels the first mile at 50 mph so:

[tex]60/50=1.2[/tex]

so it takes 1.2 minutes to travel each mile when going 50 miles per hour

and for the second mile he travels at 71 mph so:

[tex]60/71=0.84[/tex] (this answer is rounded)

so it takes him 0.84 minutes to travels each mile while going 71 mph.

If you ad them together his total times while biking two miles was 2.04 minutes.