A 0. 45 kg soccer ball changes its velocity by 20. 0 m/s due to a force applied to it in 0. 10 seconds. What force was necessary for this change in velocity? N.

Answer:it equals 0

Explanation:hope this is a good answer

Angular velocity is the ratio of angular displacement to time. Then the number of turns by the shaft is zero.

The rate of change of angular displacement is called angular velocity. It is a vector quantity.

We know that the angular velocity formula

[tex]\rm Angular \ velocity = \dfrac{Angular \ displacement}{time}[/tex]

The motor shaft turns through between the time when the current is reversed and the instant when the angular velocity is zero.

Then the number of turns will be

[tex]\rm Number \ of\ turns = Angular \ velocty \times time \\\\Number \ of\ turns = 0 *time \\\\Number \ of\ turns = 0[/tex]

The number of turns by the shaft is zero.

More about the Angular velocity link is given below.

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Answer:

[tex]4 ms^{-2}[/tex]

Explanation:

Normal force - should be the reaction from the surface it's on - nullifies the weight of the object. At this point the only force is of [tex]22-14 = 8N[/tex] from whoever is pushing the square and the friction, towards the right. At this point we divide the force by the mass of the object to obtain an acceleration of [tex]8/2 = 4 ms^{-2}[/tex]

Answer:

physical science

Explanation:

i guess??

Please find attached photograph for your answer.

Hope it helps.

Do comment if you have any query.

Answer:

D

Unbalanced or net forces acting on the car will cause it to move

The correct answer is option D. The net force on the car is zero​.

Since the car is not accelerating, the force of air resistance and the friction force must add to give the exact opposite of the applied force. The net force here would be zero, as well.

Newton's First Law is the law of inertia. An object with no net forces acting on it which is initially at rest will remain at rest. If it is moving, it will continue to move in a straight line with constant velocity.

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#SPJ2

Using the Hubble law v = H₀d where v = recessional speed = 70,000 km per second H₀ = hubble constant = 70 km/s/Mpc and d = distance of galaxy.

Making d subject of the formula, we have

d = v/H₀

Substituting the values of the variables into the equation, we have

d = v/H₀

d = 70000 km/s/70 km/s/Mpc

d = 1000 Mpc

So, the galaxy is 1000 Mpc away from us.

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Answer:

stored Energy = Power x time = 65 x 25x 60 = 105.6 Kjoules

Answer:

a)object is in uniform motion

Explanation:

velocity is constant.and a=0

body is in uniform motion.

BRAINLIST PLS

Answer:

Explanation:

Givens

m = 15000 kg

t = 12.1 seconds

vi = 86 km / hr

vf =0

Formula

F = m*a

a = (vf - vi)/ t

vi has to be converted to m/s

86 km/hr [1000 m / 1 km] * [1 hour / 3600 seconds]

86 * 1000 / 3600 = 23.89 m/s

Solution

a = (vf - vi) / t

a = (0 - 23.89)/12.1

a = - 1.97 m/s^2 The minus means that the trains is slowing down.

F = 15000 kg * - 1.97 m/s^2

F = -29615.7 Newtons.

39°C=1.28L

8°C=? Less

=8°C/39°C×1.28L

=8/39×1.28L

=10.24/39

=512/195

=2.6L

Answer:

whats the question

Explanation:

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

[tex] E_{t} = KE_{i} + PE_{i} [/tex]

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

[tex] E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J [/tex]

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

[tex] PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J [/tex]

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

[tex] KE_{A} + PE_{A} = KE_{B} + PE_{B} [/tex]

[tex] KE_{B} = KE_{A} + PE_{A} - PE_{B} [/tex]

Since

[tex] KE_{A} + PE_{A} = KE_{i} + PE_{i} [/tex]

we have:

[tex] KE_{B} = KE_{i} + PE_{i} - PE_{B} =  19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J [/tex]

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

[tex] PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J [/tex]

5) The kinetic energy of the roller coaster at point C is:

[tex] KE_{i} + PE_{i} = KE_{C} + PE_{C} [/tex]            

[tex] KE_{C} = KE_{i} + PE_{i} = 19820 J [/tex]      

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

[tex] KE_{C} = \frac{1}{2}mv_{C}^{2} [/tex]

[tex] v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s [/tex]

Hence, the velocity of the roller coaster at point C is 19.91 m/s.

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I hope it helps you!

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]

m is the mass in kg

v is the velocity in m/s

From the question

m = 20 kg

v = 2.5 m/s

From the question we have

[tex]k = \frac{1}{2} \times 20 \times {2.5}^{2} \\ = 10 \times 6.25 \\ = 62.5\: \: \: \: \: \: \: \: \: \: [/tex]

We have the final answer as

Hope this helps you

Answer:

this is an example of Newton's Second Law, acceleration happens when a force acts on a mass.

the force that makes the bicycle move is your foot pushing against the pedal

Explanation:

Shorter wavelength since UV light has more energy than visible light

365 nm - What is the peak energy wavelength of UV light?  It allows both infrared daylight and ultraviolet night-time communications by being transparent between 320 nm and 400 nm and also the longer infrared and just-barely-visible red wavelengths. Its maximum UV transmission is at 365 nm

Temperature inversion leads to an increase in temperature as altitude increases.

The term temperature inversion refers to a situation in which a layer of warm air lies over a layer of cool air. This is also referred to as thermal inversion. This occurs when the air below to loose heat rapidly.

One of the effects of temperature inversion is reduction in visibility. So, thermal inversion leads to an increase in temperature as altitude increases.

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Answer:

Od- Wooden

Explanation:

It would not make sense if it was non-wood because that make it no a wood base product, OC is not right because its not a a plant that produces wood as its structural tissue and thus has a hard stem.   An OA justT o take or get a supply of wood. so its Od

If the spring constant is known, the extension or compression of the spring can be used to calculate the applied force (pull or push) on the spring.

According to Hook's law, the force applied to an elastic material is directly proportional to the extension of the material.

The force applied;

F = kx

where;

The applied force can be in form of push or pull. A push force can result in compression of the spring while a pull force can result in extension of the spring.

If the spring constant is known, the extension or compression of the spring can be used to calculate the applied force (pull or push) on the spring.

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Answer:

Ink

mud

hcl

ocean water

co2

h20

Answer:

It should be mentioned that an efficient way to work this vector addition problem is with the cosine law for general triangles (and since  

a

,

b

 and  

r

 from an isosecles triangle, the angles are easy to figure). However, in the interest of reinforcing the usual systematic approach to vector addition, we note that the angle  

b

 makes with the +x axis  is 30

0

+105

0

=135

0

(a) The x component of  

r

x

​

​

=(10.0m)cos30

0

+(10.0m)cos135

0

=1.59m

(b)  the y component of  

r

 is r

y

​

=(10.0m)sin135

0

 = 12.1 m.

(c) the magnitude of  

r

 is r =  ∣

r

∣=

(1.59m)

2

+(12.1m)

2

​

=12.2m

(d) The angle between  

r

 and the +x direction is tan

−1

[(12.1m)/(1.59m)]=82.5

0

Explanation:

Answer:

the direction of motion of the pulse

Answer: friction :)

Explanation:

I took the test and got 100%

Answer:

a balance and a beaker of water

Explanation:

The mass can easily be measured with balance. If you drop the key in a beaker of water, the water inside the beaker will increase and this increases the volume of water that will be equal to volume of key.

Answer:

Pushing a car and a truck

Explanation:

Hi there!

We can use the following:

U (Gravitational Potential Energy) = mgh

m = mass of object (kg)

g = acceleration due to gravity OR gravitational field strength

h = height, or displacement (m)

ΔU = Uf - Ui, so:

Uf = 3 × 1.5 × 10 = 45 J

Ui = 3 × 0 × 10 = 0

ΔU = Uf - Ui = 45 - 0 = 45 J

Explanation:

u=0, v=?, a=5m/s², t=8sec

So, by the formula,

v=u+at

v=0+5×8

v=40m/s.

hope this helps you .

Answer: beta particles

Explanation:

that's the answer

Answer:

Choice C: Approximately [tex]1.64[/tex].

(Assuming that light entered into this glass block from air.)  

Explanation:

Let [tex]n_{1}[/tex] denote the refractive index of the first medium (in this example, air.) Let [tex]\theta_{1}[/tex] denote the angle of incidence.

Let [tex]n_{2}[/tex] denote the refractive index of the second medium (in this example, the glass block.) Let [tex]\theta_{2}[/tex] denote the angle of refraction.

By Snell's Law:

[tex]n_{1} \, \sin(\theta_{1}) = n_{2}\, \sin(\theta_{2})[/tex].

Rearrange the equation to find an expression for [tex]n_{2}[/tex], the refractive index of the second medium (the glass block.)

[tex]\begin{aligned}n_{2} = \left(\frac{\sin(\theta_{1})}{\sin(\theta_{2})}\right)\, n_{1}\end{aligned}[/tex].

The refractive index of air is approximately [tex]1.00[/tex]. Substitute in the values and solve for [tex]n_{2}[/tex], the refractive index of the glass block:

[tex]\begin{aligned}n_{2} &= \left(\frac{\sin(\theta_{1})}{\sin(\theta_{2})}\right)\, n_{1} \\ &= \left(\frac{\sin(46^{\circ})}{\sin(26^{\circ})}\right)\times 1.00 \\ &\approx 1.64\end{aligned}[/tex].