Adults sometimes lump all "screen time" together—social media, music videos, games and so on. Is this valid, or are there distinct differences among different types?

The tension in the string marked A is determined as 331.35 N.

tanθ = Δy/Δx

tanθ = (6 - 1)/(7 - 1) = 0.8333

θ = arc tan(0.8333) = 39.81⁰

with respect to horizontal = 90 - 39.81 = 50.19⁰

tanθ = Δy/Δx

tanθ = (9 - 6)/(7 - 5) = 1.5

θ = arc tan(1.5) = 56.31 ⁰

tanθ = Δy/Δx

tanθ = (6 - 4)/(14 - 7) = 0.2857

θ = arc tan(0.2857) = 15.95 ⁰

Bcosθ + Aosθ = Ccosθ

Bcos(56.31) + A[cos(50.19)] = 56.3cos(15.95)

0.55B + 0.64A = 54.13 ----- (1)

Bsinθ + Asinθ = Csinθ

Bsin(56.31) + A[sin(50.19)] = 56.3sin(15.95)

0.832B + 0.77A = 15.47---- (2)

Solve (1) and (2)

0.2A = 66.27

A = 66.27/0.2

A = 331.35 N

Thus, the tension in the string marked A is determined as 331.35 N.

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Considering that the moment you provide the action force, the car will certainly apply the response force. You will consequently come into an opposing force.

According to Newton's third law of motion, every action has an equal and opposite reaction. As well as the action and reaction are always acted in pairs.

You manually move a big automobile. You are then pushed back by the automobile with an opposing but equivalent force.

Because the automobile will undoubtedly apply the response force when you apply the action force. As a result, you will encounter an opposing force.

Hence, when you push a heavy car by hand. The car, in turn, pushes back due to the action-reaction analogy.

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The altimeter reading is 29.17 from the Kollsman window is 29.17 in Hg

The given parameters are:

The formula to calculate the pressure altitude is:

Altitude = (29.92 – Altimeter reading) * 1,000 + Ground level

Substitute the known values

1450 = (29.92 - Altimeter reading) * 1000 + 700

Evaluate the like terms

750 = (29.92 - Altimeter reading) * 1000

Divide both sides by 1000

0.75 = 29.92 - Altimeter reading

Evaluate the like terms

Altimeter reading = 29.17

Hence, the altimeter reading is 29.17 from the Kollsman window is 29.17 in Hg

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Answer:

Current = 5.45amps

Power = 654 watts

Explanation:

E =120V

I =?

R = 22.02 Ohms

I= E/R

I= 120/22.02

I = 5.45AmPs

P = ?                    P= E x R

E = 120V              P= 120x5.45

I = 5.45 AmPs      P=  654W

(a) The object moves with uniform velocity from A to B.

(b) The object moves with constant velocity from B to C.

(c) The object moves with increasing velocity from C to D.

V(A to B) = (6 - 0)/(4 - 0) = 1.5 m/s

V(B to C) = (6 - 6)/(11 - 4) = 0 m/s

V(C to D) = (7 - 6)/(12 - 11) = 1 m/s

final velocity = 1 + 1.5 m/s = 2.5 m/s

Thus, we can conclude the following;

The object moves with uniform velocity from A to B.

The object moves with constant velocity from B to C.

The object moves with increasing velocity from C to D.

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Answer:The object moves with

positive accelaration

from A to B. It

speeds up

from B to C. It moves with

constant velocity

from C to D.

Explanation:I took the test

The average acceleration of snowboarder's is 0.19 m/s² and distance travel by snowboarder is 6.70 metres.

We have given that snowboarder start moving from rest means

Initial speed (u) = 0 m/s

Final speed (v) = 1.6 m/s

time taken (t) = 8.4 s

Applying first equation of motion

v = u + at

1.6 = 0 + a×8.4

1.6 = 8.4a

a = 1.6/8.4 m/s²

a = 0.19 m/s²

So average acceleration is 0.19 m/s².

Now Applying second equation of motion

s = ut + 1/2at²

s = (0) × (8.4) + (1/2)(0.19)(8.4)² metre

s = (1/2)(0.19)(8.4)²  metre

s = 6.70 metre

So that distance covered by snowboarder is 6.70 metre.

So we find out the average acceleration by using first equation of motion and the average acceleration came out to be 0.19 m/s² and to find out the distance covered by snowboarder we use second equation of motion and the distance came out to be 6.70 metre.

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Answer:

5

Explanation:

Use the Pythagorean theorem. 3^2 + 4^2 = 25, which is the square of 5.

Brainliest, please :)

The direction of the vector would eventually depend on the horizontal component and vertical component which are 4 and 3 respectively. So, it would be calculated as 5 according to the Pythagoras theorem.

In physics, the direction of the vector may be defined as the orientation of the vector, that is, the angle it makes with the X-axis. A vector is drawn by a line with an arrow on the top and a fixed point at the other end.

The direction in which the arrowhead of the vector is directed gives the direction of the vector. The direction of the vector measures the angle it makes with a horizontal line.

According to the question,

The horizontal component of the vector = 4

The vertical component of the vector = 3

The direction of the vector =  square of the horizontal component + square of the vertical component.

                                             = [tex]4^2+3^2[/tex] = 16 + 9 = 25 i.e. = [tex]5^2.[/tex]

Therefore, the direction of the vector would eventually depend on the horizontal component and vertical component which are 4 and 3 respectively. So, it would be calculated as 5 according to the Pythagoras theorem.

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The mouse will be 4.61 meters distant horizontally. and the owl won't fall into the nest.

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

Diameter of nest,d = 30 cm

The velocity of flying, V =  3.50 m/s toward the east

The angle of flying below the horizontal, Θ = 32°

Position of owl,12.0 m above the middle of the 30 cm-diameter nests, at 4.00 m west.

Height of owl above the ground,h = 12 m

Distance from the west direction, a = 4.00 m

The vertical component of velocity,vₐ

The horizontal component of velocity,vₓ

After covering a vertical distance of 12 m, the vertical component of the mouse's velocity, v, is given by

vₐ-u²=2gh

u = asinΘ

Substitute the above value;

vₐ²- (a sin Θ )² = 2gh

Substitute the given value;

vₐ² - (4 sin 32)² = 2 ×9.81 ×12

vₐ² - (4 × 0.53)² = 235.44

vₐ² - 4.494 = 235.44

vₐ² = 235.44 + 4.494

vₐ² = 239.934

vₐ = 15.49 m/s

The time required to descend the aforementioned 12 meters;

[tex]\rm t = \frac{v_{rel}}{g}[/tex]

t = (vₐ - u )/g

t =( vₐ - a sin Θ )/g

Substitute the given value;

t =(15.49 - 4 ×sin 32°) / 9.81

t= (15.49 - 4 * 0.53) / 9.81

t =(15.49 - 2.12) / 9.81

t = 13.37 / 9.81

t= 1.36 s

Horizontal distance traveled for the given period;

x=uₓ × t

x =a cosΘ × t

Substitute the given value;

x = 4×(cos 32°)×1.36

x= 4×0.848×1.36

x= 4.61 m

Hence, the mouse will be 4.61 meters distant in the horizontal direction. and the owl won't fall into the nest,

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(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The  weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

rT - Rf = Iα  ----- (1)

T - f = Ma  -------- (2)

a = Rα

where;

Torque equation for frictional force;

[tex]f = (\frac{r}{R} T) - (\frac{I}{R^2} )a[/tex]

solve (1) and (2)

[tex]a = \frac{TR(R - r)}{I + MR^2}[/tex]

since the yoyo is pulled in vertical direction, T = mg [tex]a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2[/tex]

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

W = mg

W = 0.15 x 9.8 = 1.47 N

T = mg = 1.47 N

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

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Since the tension in the string over the top of the circle is negligible, the tension in the string at the bottom of the circle will be zero.

Circular motion is the same as rotational motion. That is, motion moving in a circle.

Given that a small ball is tied to a string and set rotating with negligible friction in a vertical circle. If the ball moves over the top of the circle at its slowest possible speed (so that the tension in the string is negligible)

Since the tension in the string over the top of the circle is negligible, the tension in the string at the bottom of the circle, assuming there is no additional energy added to the ball during rotation will be zero.

The tension is negligible at the top in order to allow the ball rotate. And since the tension is negligible, there no tension force pulling inward

Therefore, the numerical value of the answer is Zero.

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Answer:

And so calculating we get the maximum speed at each proton will reach To be 1.36 Times 10 to the four m the second

Answer:

H = L sin θ / 2    height of center of mass

E = m g H = m g L sin θ / 2

I = m L^2 / 3        moment of inertia about end of rod

1/2 I ω^2 = m g L sin θ / 2      KE of rod when it hits surface

1/2 m L^2 / 3  ω^2 = m g L sin θ / 2

ω^2 = 3 g sin θ / L

ω = (3 g sin θ / L)^1/2      angular speed of rod striking surface

P = I α       if P = torque    

α = m g L/2 / (m L^2 / 3) = 3 g / (2 L)       angular acceleration of rod

Rightward and upward

Whenever any object moves in one direction only, the motion is known as one-dimensional motion.

Two-dimensional motion is the movement of the object in two directions simultaneously.
Example: A ball thrown at an angle is a two-dimensional motion.

In two-dimensional motion, there are two axes used, generally the x-axis and the y-axis.

Generally, the x-axis is positive towards the rightward direction and negative towards the leftward direction.

Similarly, the y- axis is positive towards the upward direction and negative toward the downward direction.

So, the rightward and upward directions are chosen as positive.

Therefore:

When working with two-dimensional motion, choose the direction of  , the rightward and upward to be a positive direction.

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Velocity and displacement are the two quantities that increases as the other decreases during simple harmonic motion.

If a body's velocity at any given moment is inversely proportional to its displacement from the mean position, its motion is said to be simple harmonic.

Angular harmonic frequency and time t are functions that can be used to explain simple harmonic motion:

Displacement from equilibrium has amplitude A:

x = Asin(ωt)

and velocity has amplitude ωA

v = dx/dt = ωAcos(ωt)

At ωt = 0, x = 0 while v = Aω (max).

At ωt = π/2, x = A (max) while v = 0.

As a result, v reaches it's maximum /2 (90°) before x does. Therefore, displacement follows velocity at a 90° angle.

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The work done on  both Elastic X and Elastic Y is 25.2 J.

The work done on the Elastic X is calculated as follows;

W = ¹/₂fx

where;

W = ¹/₂(42)(1.2)

W = 25.2 J

The work done on the Elastic Y is calculated as follows;

W = ¹/₂fx

where;

W = ¹/₂(24)(2.1)

W = 25.2 J

Thus, the work done on  both Elastic X and Elastic Y is 25.2 J.

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Answer:

frequency and amplitude increases

The boom makes an angle of θ = 60.2° with the vertical wall and the tension in the horizontal rope is mathematically given as

[tex]T_1=730.85 \mathrm{~N}[/tex]

[tex]T_1'=\frac{1980.51 \mathrm{~N}}{}[/tex]

Generally, the equation for is  mathematically given as

1) Tension in vertical rope,

[tex]T_{1}=74.5 \times 9.81\\T_1=730.85 \mathrm{~N}[/tex]

2) Tension in horizontal rope,

[tex]\sum{Mg} =0\\Mg\frac{l}{2} \sin \theta+T_{1} \frac{l}{2} \sin \theta &=T_{2} l \cos \theta \\[/tex]

[tex]T_{2} &=\frac{M g(l / 2) \sin \theta+T_{1}(l / 2) \sin \theta}{l \cos \theta} \\\\&=\frac{M g \tan \theta}{2}+\frac{T_{1} \tan \theta}{2} \\\\T_{2} &=\frac{142 \times 9.81 \tan 61.8}{2}+\frac{730.85 \times \tan 61.8}{2} \\\\=& 1299+681.51[/tex]

[tex]T_1=\frac{1980.51 \mathrm{~N}}{}[/tex]

In conclusion, the tension in the horizontal rope is

[tex]T_1=\frac{1980.51 \mathrm{~N}}{}[/tex]

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(a) The magnitude of the static friction force is 145 N

(b) The minimum possible value of static friction is 0.44.

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

it is defined as the product of the coefficient of friction and normal reaction.

Given data;

The horizontal force exerted, F =145 N

Mass of crate,m =33.2 kg.

The velocity of the crate,v =0 m/sedc

The magnitude of the static friction force, [tex]\rm f_s[/tex] =? N

The minimum possible value of the coefficient of static friction between the crate and the floor is,

Normal reaction,N = mg

a)

The forces acting on the crate are balanced since it is at rest; as a result, the horizontal force is equal to the frictional force, f:

F = [tex]\rm f_s[/tex] = 145 N

(b)

The following equations provide the greatest allowable force of friction between the floor and the crate:

[tex]\rm f_{min}= \mu_s N \\\\ \rm f_{min}= \mu_s mg \\\\[/tex]

The force applied to the box must be less than or equal to the maximum force of friction in order for it to stay at rest.

[tex]\rm f \leq f_{min} \\\\ f \leq \mu_s mg \\\\ 145 \leq \mu_s (33.2 \times 9.81 ) \\\\ \mu_s \geq 0.44[/tex]

Hence, the magnitude of the static friction force and minimum possible value of the coefficient of static friction between the crate and the floor will be 145 N and 0.44.

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The two forces are 12N and 10N (option B)

The net force is the force that has the same effect in magnitude and direction as the two forces acting together.

This problem can be solved using a right angled triangle;

Let the forces be P and Q

sin 40 = P/16

P= 16 sin 40

P = 10 N

Cos 40 = Q/16

Q = 16 cos 40

Q = 12 N

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Answer:

122.17 m/s

Explanation:

x cos 40 = horizontal velocity

1500 m / x cos 40   = time in the air = 1958.11 / x

x sin 40 = vertical velocity

 find when shell vertical velocity = 0 (this is max height....1/2 way through its flight)  , the time when it hits the ground will be twice this...  

   0 =  x sin 40 - 9.8 t

  t =  x sin40 / 9.8        time in the air is twice this = .13118 x

Equate the two times from above to solve for x

1958.11/ x  =  .13118 x

x = 122.17 m/s

The frequency the student detect from waves omitted from the fork and waves coming from the wall is 348.1 Hz.

The observed frequency is determined by applying Doppler effect;

f' = f₀(v + v₀)/(v)

where;

f' = 342(6 + 336)/(336)

f' = 348.1 Hz

Thus, the frequency the student detect from waves omitted from the fork and waves coming from the wall is 348.1 Hz.

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The reverberation time with 800 audiences is 0.875 seconds.

R₁V₁ = R₂V₂

where;

R₂ = R₁V₁/V₂

R₂ = (1.4 x 500) / 800

R₂ = 0.875 seconds

Thus, the reverberation time with 800 audiences is 0.875 seconds.

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The height of the building in meters, given the data is 108 m

The height of the building can be obtained as illustrated below:

h = ½gt²

h = ½ × 9.8 × 4.7²

h = 108 m

Thus, the height of the building is 108 m

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A. The force that causes the water on the lettuce to come off the lettuce and go to the walls of the bowl is centrifugal force.

Centrifugal force is an inertial force that appears to act on all objects when viewed in a rotating frame of reference.

This force is directed away from the center around which the body is moving.

This is force that acts on a body moving in a circular path and is directed towards the center around which the body is moving.

While centripetal force is directed towards to the center, the centrifugal force is directed away.

Thus, the force that causes the water on the lettuce to come off the lettuce and go to the walls of the bowl is centrifugal force.

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Centripetal force causes the water on the lettuce to come off the lettuce and go to the walls of the bowl.

Centripetal force is the force that is necessary to keep an object moving in a curved path and that is directed inward toward the center of rotation. This force can move the body in the circular direction.

So we can conclude that Centripetal force causes the water on the lettuce to come off the lettuce and go to the walls of the bowl.

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(a) The maximum potential difference across the resistor is 339.41 V.

(b) The maximum current through the resistor is 0.23 A.

(c) The rms current through the resistor is 0.16 A.

(d)  The average power dissipated by the resistor is 38.4 W.

Vrms = 0.7071V₀

where;

V₀ = Vrms/0.7071

V₀ = 240/0.7071

V₀ = 339.41 V

I(rms) = V(rms)/R

I(rms) = (240)/(1,540)

I(rms) = 0.16 A

I₀ = I(rms)/0.7071

I₀ = (0.16)/0.7071

I₀ = 0.23 A

P = I(rms) x V(rms)

P = 0.16 x 240

P = 38.4 W

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The work done in dragging an object along a straight path is  173.2 J.

Work done is equal to product of force applied and distance moved.

Given You are trying to drag an object 5.0 m along a straight path. You are pulling with a force of 40 N along a rope that is inclined 30° to the horizontal.

Work = Force x Distance x cos(angle)

W= 40 x 5 x cos 30°

W = 173.2 Joules

Thus, the work done is  173.2 Joules

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Sedimentation ponds are useful for reducing the effects of soil erosion because : ( D ) They provide a place to collect soil sediments so that it doesn't reach ponds, lakes and rivers

Sedimentation ponds are ponds that hold sediments and suspended solids discharged by runoff waters thereby preventing the soil sediments from reaching water bodies such as ponds, lakes and rivers.

The sediment ponds hold the water while allowing the debris and solids contained in the water to settle down.

Hence we can conclude that sedimentation ponds provide a place to collect soil sediments so that it doesn't reach ponds, lakes and rivers

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The middle ear provides impedance matching between air and fluid by the reduction in area between the tympanic membrane and the stapes footplate.

Middle ear is part of ear that separates the outer ear from the inner ear.

The middle ear consists of the following bones:

The middle ear provides impedance matching between air and fluid by

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Answer:

All people with the observant trait are often a steadying force that always wants to get things done. Their energy is very elegant in the sense of working on real things in real time.

Explanation:

Hope I helped