please help on magnitude !

Answer:

0.5 m

Explanation:

The following data were obtained from the question:

Force (F) applied = 5 N

Work done (W) = 2.5 J

Distance (s) =?

Work done (W) can be defined as the product of force (F) and distance (s) moved in the direction of the force. From the above definition, work done (W) can be represented mathematically as:

Work done (W) = Force (F) × Distance (s)

W = F × s

With the above formula, we can obtain the distance moved by the book as shown below:

Force (F) applied = 5 N

Work done (W) = 2.5 J

Distance (s) =?

W = F × s

2.5 = 5 × s

Divide both side by 5

s = 2.5/5

s = 0.5 m

Therefore, the book moved a distance of 0.5 m.

Answer:

the charge of each small sphere, which is + Q / 2

Explanation:

For this exercise we must use the fact that a charged object induces charges on nearby bodies

Induced charge comes from the fact that charges of the same sign repel and charges of different signs attract.

In this case the large, fixed ball with a -Q charge induces a positive charge in the nearest part and the negative charges are repelled to the furthest point, but the net charge on the metallic sphere remains zero. It should be emphasized that since the charges are of different signs, there is an attractive force between the two spheres.

This first metallic sphere now has a negative charge on the back, this charge induces a positive charge on the second sphere, as the charges are of a different sign, they attract each other, which is why the force is attractive.

When the first sphere stops the second sphere hits it, at this moment the charge of the two spheres is equal, therefore the induced charge in the two spheres is + Q. When the two spheres are separated, the charge on each of them is half, that is, the sphere has a charge + Q / 2 and the second sphere has a charge + Q / 2.

Therefore the first sphere is subjected to two forces: an attractive force with the large sphere of charge -Q and a repulsive force with the second sphere of charge + Q / 2.

So the first sphere must approach the big ball and the second sphere must move away from the big sphere.

This is the process of the movement of this exercise, unfortunately the statements with which to compare this process do not appear, but one of the most common questions of what is the charge of each small sphere, which is + Q / 2

Answer:

W= 0.319992 J  and distance is 3.75 cm

Explanation:

The energy needed to stretch the spring from 30 cm to 45 cm = 3 J

Now we required to find the requirement of energy when the spring is stretched from 35 cm to 37 cm.

So first find the work done to stretch the spring from 35 cm to 45 cm.

Work done, w = (1/2) kx^2

3 = (1/2)k(0.45 – 0.30)^2

k = 266.66 N/m

now, x1 = 0.35 – 0.30 = 0.05m

x2 = 0.37 – 0.30 = 0.07m

Now the amount of work done to stretch from 35cm to 37.

w = (1/2) k (x2^2 – x1^2)

w = (1/2) (266.66) (0.07^2 – 0.05^2 )

w= 0.319992 J

(b). Given F = 10 N

F = kx

x = F / k

x = 10 / 266.66

x = 0.0375m

x = 3.75 cm

Thus, distance is 3.75cm

Answer:

40 m/s

Explanation:

Answer:

mass of the block is 7.2Kg

Tension on string A = 72N

Tension on the string with an angle is =  108N

Explanation:

The Horizontal string will have the cos component of the force...

Hence, mg cos60 = 36

by keeping acceleration due to gravity = 10 m/s^2, we get,

m = 7.2Kg

The tension in String A = mg = 7.2*10 = 72N

since the whole string holding the block is pulled by another string with 36 N

the resultant force will be the tension of the string with an angle.

Hence, 72 +36 = 108 N

Please let me know whether I got this right or not...

Momentum = (mass) x (speed)

We don't know the mass of the bowling ball, but we know how much it weighs.  So if we knew the acceleration of gravity in the place where the ball is, we could calculate its mass.  The question doesn't tell us what planet the bowling ball is on.  So if we only use the information given in the question, there's no way to find the answer, and we're stuck here.

But I urgently need the points, so I'm going to go out on a limb here and make a big assumption:  I'll assume that the alley where this ball is rolling and bowling is located on planet Earth, where the acceleration  of gravity is around  9.8 m/s²  everywhere on the planet's surface.  NOW I can go ahead and answer the question that I just invented.

weight = m  g

Mass = (weight) / (g)

Mass = (78.0 N) / (9.8 m/s²)

Mass = 7.96 kilograms

Momentum = (mass) x (speed)

Momentum = (7.96 kg) x (8.00 m/s)

Momentum = 63.7 kg-m/s

Answer:

1.) 19.21 m/s

2.) 57 degree

Explanation: Given that a model airplane is flying north with a velocity of of 15m/s. A strong wind blowing east at 12m/s

The airplanes resultant speed can be calculated by using pythagorean theorem

R = sqrt ( 15^2 + 12^2 )

R = sqrt( 225 + 144 )

R = sqrt( 369 )

R = 19.21 m/s

The magnitude of the resultant vector is 19.21 m/s

The direction of velocity vector will be:

Tan Ø = 15 /12

Tan Ø = 1.25

Ø = tan^-1(1.25)

Ø = 57.04

Ø = 57 degree.

Therefore, the airplanes is heading 57 degree in the horizontal direction of velocity vector

Answer:

removal of heat by cooling towers

Answer:

500 watts

Explanation:

Recall that the definition of power is the amount of energy delivered per unit of time.

In our case, the energy delivered is potential energy which we can estimate as the product of the weight of the object times the distance it is lifted above ground:

200 N x 10 m = 2000  Nm

then the power is the quotient of this potential energy divided the time it took to lift the object to that position:

Power = 2000 / 4   Nm/s = 500 Nm/s = 500 watts

Answer:

90 m

Explanation:

There are two unknowns: the amount of time the squirrel spent running, and the length of the race.  Let's call these t and x, respectively.

The average velocity is the total distance divided by the total time.

5.0 m/s = x / (t + 3.0)

The total distance is the time she spent running times the speed she ran at.

x = (6.0 m/s) t

Substitute and solve:

5 = 6t / (t + 3)

5 (t + 3) = 6t

5t + 15 = 6t

t = 15

She ran for 15 seconds (not including the 3 seconds she stopped).  So the length of the race is:

x = (6.0 m/s) (15 s)

x = 90 m

Answer:

Y(1s) = [tex]10\sqrt{3}[/tex] - 10.1

X(1s) = 10m/s

Explanation:

In annex I've done the explanation for the equations that I will just present here.

Assuming that the arrow stars from the position (0 ; 0) in the Cartesian Graphic, and with Xo and Yo the initial speeds:

[tex]Yo =\frac{\sqrt{3} }{2} . Vr\\Yo = \frac{\sqrt{3} }{2} . 20\\Yo = 10\sqrt{3} m/s \\Xo = \frac{1}{2} Vr\\Xo = 10m/s[/tex]

Ignoring friction with air, Xo = Xf

So, Xo is the same during all the movement.

X(1s) = 10m/s

For Yo is different. That component is suferring reductions from gravity.

We can find Yo(1s) with one the basic functions of cinematics:

Vf = Vo + at

Vf = Final Velocity

Vo = Start Velocity

a = aceleration - gravity (g) is negative here

t = time

Yf = Yo + gt

Yf = [tex]10\sqrt{3}[/tex] - 10.1

If you prefere, can be: Yf = 10. ([tex]\sqrt{3} - 1[/tex])

Answer: 131.14km per day

Explanation: since the second half of the terns migration takes 122 days we can assume that the full migration would take 244 days. using this we can divide the total distance by the total amount of days it takes (because speed = distance/time) which is 32,000/244, which would be 131.14

Answer:

F = 0.012 N

Explanation:

Given that,

Mass of the object, m = 6 g

Acceleration, a = 2 m/s²

1 kg = 1000 grams

6 g = 0.006 kg

Force, F = ma

So,

[tex]F=0.006\ kg\times 2\ m/s^2\\F=0.012\ N[/tex]

So, the force is 0.012 N.

Answer:

Open and globular

Explanation:

Answer: A. a straight line inclined to the time axis

Explanation:

Answer:

Explanation:

we know that

s=vt here v is the speed and s is distance covered by the signals

given data

v=3*10^8

t=10 min we have to convert it into seconds

1 minute=60 seconds

so

10 minutes =10*60/1 =600 seconds

now putting the value of v and t we can find the value of s

s=vt

s=3*10^8*600

s=1.8*10^11m

i hope this will help you

Speed is the rate of distance over time.

The signal will travel [tex]1.8 \times 10^{11}[/tex] meters in the first 10 minutes

The given parameters are:

[tex]s = 3 \times 10^8\ ms^{-1}[/tex] ---the speed of light

[tex]t = 10\ min[/tex] -- the time of travel

The relationship between speed, distance (d) and time is:

[tex]s = \frac dt[/tex]

Make d the subject

[tex]d = s \times t[/tex]

Substitute values for t and s

[tex]d = 3 \times 10^8\ ms^{-1} \times 10\ min[/tex]

Convert minutes to seconds

[tex]d = 3 \times 10^8\ ms^{-1} \times 10 \times 60s[/tex]

So, we have:

[tex]d = 3 \times 10^8\ m \times 10 \times 60[/tex]

[tex]d = 18 \times 10^{10}m[/tex]

Rewrite as:

[tex]d = 1.8 \times 10^{11}m[/tex]

Hence, the signal will travel [tex]1.8 \times 10^{11}[/tex] meters in the first 10 minutes

Read more about speed at:

https://brainly.com/question/21791162

Answer:

Explanation

He runs at 5m/s, so in 30 s he should be at 150m.  So you have to do 125m - 150m and you'll get -25m, this is his initial position. They want to know the time when he hits 75m, so you would do 75 + 25, and get 100. Then do 100m / 5m/s, and you will get 25 seconds.

Answer:

their is always a animal or bug in nature

Explanation:

Answer:

17.6 m/s², 26.2° above x axis

Explanation:

Apply Newton's second law.

Sum of forces in the x direction:

∑Fₓ = ma

40 N + 55.0 N cos 45° = (5.00 kg) aₓ

aₓ = 15.8 m/s²

Sum of forces in the y direction:

∑Fᵧ = ma

55.0 N sin 45° = (5.00 kg) aᵧ

aᵧ = 7.78 m/s²

Use Pythagorean theorem to find the magnitude of the resultant.

a² = aₓ² + aᵧ²

a² = (15.8 m/s²)² + (7.78 m/s²)²

a = 17.6 m/s²

Use trigonometry to find the angle.

tan θ = aᵧ / aₓ

tan θ = (7.78 m/s²) / (15.8 m/s²)

θ = 26.2°

Answer:

D. 803 lbs

Explanation:

In order to find the compressive stress on all three blocks we first need to find the normal surface area of each:

Surface Area of 1 Block = 3.5 x 3.5

Surface Area of 1 Block = 12.25

Surface Area of all 3 Blocks = A = 3 x 12.25

Area = 36.75

Now, the stress is given by the following formula:

Stress = Force/Area

Stress = 29500 lbs/36.75

Stress = 802.72 lbs

Hence, the correct option will be:

D. 803 lbs

Answer:

The tangential speed is [tex]v = 7.5 m/s [/tex]

The centripetal acceleration is [tex]a = 154 \ m/s^2[/tex]

Explanation

Generally the angular velocity is mathematically represented as

[tex]w = e * \frac{ 2 \ pi \ rad }{s}[/tex]

substituting 3.27 rev/s for e we have that

[tex]w = 3.27 * \frac{ 2 \ pi \ rad }{s}[/tex]

=> [tex]w = 20.55 \ rad /s[/tex]

The tangential speed is mathematically represented as

[tex]v = w * r[/tex]

substituting 0.365 m for r we have that

[tex]v = 20.55 * 0.365 [/tex]

=> [tex]v = 7.5 m/s [/tex]

Generally the centripetal acceleration is mathematically represented as

[tex]a = \frac{v^2}{r}[/tex]

=>      [tex]a =  \frac{7.5^2}{ 0.365}[/tex]

=>      [tex]a =  154 \  m/s^2[/tex]

Answer:

0 m/s²

Explanation:

Acceleration is the change in velocity over change in time.  If the velocity is constant, then the acceleration is 0.

Answer:

B

Explanation:

because when the air rises the density increases

We want to predict what will happen to the density of the air on the surface of the Earth when it warms up.

We will see that the correct option is D: "It decreases, then increases"

We know that the temperature of a given object (in this case a mass of air) is related to the kinetic energy of the particles that conform it.

As the temperature increases, the kinetic energy also increases, thus, the amount of motion of each particle increases, thus, the volume of the object increases.

Now remember that:

density = Mass/Volume.

So if the volume of something increases, we will see that the density decreases (as the volume is in the denominator).

Then if the temperature of the air increases, we will see that the density of the air in the surface decreases.

But it does not end there, as only the air near the surface suffers this change of density, we will have a denser mass of air (colder air) above it. And because it is denser (has more mass in the same volume) we can say that it is heavier.

Then eventually the hot air will rise, and the cold air will fall down, thus the density of the air in the surface increases again, as the colder and denser air comes near the surface.

This is one way of how wind currents are born.

Concluding we can see that the correct option is D: "It decreases, then increases"

If you want to learn more, you can read:

https://brainly.com/question/14482731

Answer:

8

Explanation:

Answer:

Chemical Properties of Oxygen

At standard temperature and pressure (STP), two atoms of the element bind to form dioxygen, a colorless, odorless, tasteless diatomic gas with the formula O2. Oxygen is a member of the chalcogen group on the periodic table and is a highly reactive nonmetallic element.

Explanation:

Answer:

C

Explanation:

I think it's C, because at that point, you are going fastest. Sorry if im wrong, hope this helps.

Answer:

In between and the middle one

Explanation:

Answer:

m = 81281.5 pounds.

Explanation:

Given that,

Force, F = 73 kN

Acceleration of the space shuttle, a = 16000 mi/h²

1 miles/h² = 0.0001241 m/s2

16000 mi/h² = 1.98 m/s²

We need to find the mass of the spacecraft.

According to Newton's second law,

F = ma

m is mass of the spacecraft

[tex]m=\dfrac{F}{a}\\\\m=\dfrac{73\times 10^3\ N}{1.98\ m/s^2}\\\\m=36868.68\ kg[/tex]

Since, 1 kg = 2.20462 pounds

m = 81281.5 pounds

Hence, the mass of the spacecraft is 81281.5 pounds.

Answer:

speed = 7.9 m/s

Explanation:

speed = total distance / time taken

speed = 300 / 38

speed = 7.89473684 m/s

to the nearest tenth

speed = 7.9 m/s

Answer:

 y = y₀ (1 - ½ g y₀ / v²)

Explanation:

This is a free fall problem. Let's start with the ball that is released from the window, with initial velocity vo = 0 and a height of the window i

          y = y₀ + v₀ t - ½ g t²

          y = y₀ - ½ g t²

for the ball thrown from the ground with initial velocity v₀₂ = v

         y₂ = y₀₂ + v₀₂ t - ½ g t²

in this case y₀ = 0

         y₂2 = v t - ½ g t²

at the point where the two balls meet, they have the same height

         y = y₂

         y₀ - ½ g t² = vt - ½ g t²

         y₀i = v t

         t = y₀ / v

since we have the time it takes to reach the point, we can substitute in either of the two equations to find the height

         y = y₀ - ½ g t²

         y = y₀ - ½ g (y₀ / v)²

         y = y₀ - ½ g y₀² / v²

        y = y₀ (1 - ½ g y₀ / v²)

with this expression we can find the meeting point of the two balls