In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.0342 s, during which time it experiences an acceleration of 186 m/s2. The ball is launched at an angle of 45.9 ° above the ground. Determine the (a) horizontal and (b) vertical components of the launch velocity.

Answer:

a) [tex]V=14.904m/s[/tex]

b) [tex]d = 191.49 m[/tex]

c) [tex]t= 25.696 s[/tex]

Explanation:

From the question we are told that:

Radius [tex]r =378m[/tex]

Acceleration [tex]a=0.580[/tex]

a)

Generally the  equation for speed of the car is mathematically given by

 [tex]a=\frac{v^2}{r}[/tex]

 [tex]V=\sqrt{a*r}[/tex]

 [tex]V=\sqrt{0.58*383}[/tex]

 [tex]V=14.904m/s[/tex]

b)

Generally the  equation for distance traveled of the car is mathematically given by

 [tex]V^2=u^2+2ad[/tex]

 [tex]d=\frac{V^2}{2a}[/tex]

 [tex]d=\frac{14.904^2}{2*0.58}[/tex]  

 [tex]d = 191.49 m[/tex]

c)

Generally the  equation for time of the car is mathematically given by

 [tex]V=u+at[/tex]

 [tex]t=\frac{V}{a}[/tex]

 [tex]t=\frac{14.904}{0.58}[/tex]

 [tex]t= 25.696 s[/tex]

Answer:

Explanation:

Blue: 10/30

Red: 5/30

Yellow: 15/30

The probability of finding the blue red and yellow chips is 1/3, 1/6 and 1/2

The extent to which an event is likely to occur, measured by the ratio of the favorable cases to the whole number of cases possible.

Given that

Blue chips =10

Red chips = 5

Yellow chips = 15

Number of the total samples =10+15+5=30

Probability of choosing Blue chips = [tex]\dfrac{10}{30}= \dfrac{1}{3}[/tex]

Probability of Red chips =[tex]\dfrac{5}{30}=\dfrac{1}{6}[/tex]

Probability of Yellow chips =[tex]\dfrac{15}{30}=\dfrac{1}{2}[/tex]

Thus the probability of finding the blue red and yellow chips is 1/3, 1/6 and 1/2

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Answer:

the answer is B i hope it helps :)

[tex]\huge\color{purple}\boxed{\colorbox{black}{♡Answer}}[/tex]

B. The energy required to force two nuclei to undergo nuclear fusion. ✅

[tex]\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{!}}}}}[/tex]

Answer:

Explanation:

Given that:

Focal length for the objective lens = 20 mm, 4 mm, 1.4 mm

For objective lens of focal length f₁ = 20 mm

s₁' = 120 mm + 20 mm = 140 mm

∴

Magnification [tex]m_1 = \dfrac{s'_1}{f_1}[/tex]

[tex]m_1 = \dfrac{140}{20}[/tex]

[tex]m_1 = 7 \ m[/tex]

For objective lens of focal length f₁ = 4 mm

s₁' = 120 mm + 4 mm = 124 mm

[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]

[tex]m_1 = \dfrac{124}{4}[/tex]

[tex]m_1 = 31 \ m[/tex]

For objective lens of focal length f₁ = 1.4 mm

s₁' = 120 mm + 1.4 mm = 121.4 mm

[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]

[tex]m_1 = \dfrac{121.4}{1.4}[/tex]

[tex]m_1 = 86.71 \ m[/tex]

The magnification of the eyepiece is given as:

[tex]m_e = 5X \ and \ m_e = 15X[/tex]

Thus, the largest angular magnification when  [tex]m_1 \ and \ m_e \ are \ large \ is:[/tex]

[tex]M_{large}= (m_1)_{large} \times (m_e)_{large}[/tex]

= 86.71 × 15

= 1300.65

The smallest angular magnification derived when [tex]m_1 \ and \ m_e \ are \ small \ is:[/tex]

[tex]M_{small}= (m_1)_{small} \times (m_e)_{small}[/tex]

= 7 × 5

= 35

The largest magnification will be 1300.65 and the smallest magnification will be 35.

Magnification is defined as the ratio of the size of the image of an object to the actual size of the object.

Now for objective lens and eyepieces, it is defined as the ratio of the focal length of the objective lens to the focal length of the eyepiece.

It is given in the question:

Focal lengths for the objective lens is = 20 mm, 4 mm, 1.4 mm

now we will calculate the magnification for all three focal lengths of the objective lens.

Also, each objective forms an image 120 mm beyond its second focal point.

(1) For an objective lens of focal length   [tex]f_1=20 \ mm[/tex]

[tex]s_1'=120\ mm +20 \ mm =140\ mm[/tex]

Magnification will be calculated as

[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{140}{20} =7[/tex]

(2) For an objective lens of focal length [tex]f_1= \ 4 \ mm[/tex]

s₁' = 120 mm + 4 mm = 124 mm

[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{124}{4} =31[/tex]

(3) For an objective lens of focal length [tex]f_1=1.4\ mm[/tex]

s₁' = 120 mm + 1.4 mm = 121.4 mm

[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{121.4}{1.4} =86.71[/tex]

Now the magnification of the eyepiece is given as:

[tex]m_e=5x\ \ \ & \ \ m_e=15x[/tex]

Thus, the largest angular magnification when  

[tex]m_1 = 86.17\ \ \ \ m_e=15x[/tex]

[tex]m_{large}= (m_1)_{large}\times (m_e)_{large}[/tex]

[tex]m_{large}=86.71\times 15=1300.65[/tex]

The smallest angular magnification derived when

[tex]m_1=7\ \ \ \ m_e=5[/tex]

[tex]m_{small}=(m_1)_{small}\times (m_e)_{small}[/tex]

[tex]m_{small}=7\times 5=35[/tex]

Thus the largest magnification will be 1300.65 and the smallest magnification will be 35.

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Answer:

The speed of the ball when it reaches equilibrium position is 3.31 m/s

Explanation:

Given;

mass of the object, m = 0.25 kg

initial displacement of the object, h₁ = 0.56 m

spring constant, k = 105 N/m

displacement at equilibrium position, h₂ = 0

initial velocity of the object, v₁ = 0

velocity of the object at equilibrium position = v₂

The change in gravitational potential energy at the equilibrium position is given as;

ΔP.E = mg(h₂ - h₁)

The change in kinetic energy of the object at the equilibrium position is given as;

ΔK.E = ¹/₂m(v₂² - v₁²)  

Apply the principle of conservation of mechanical energy;

ΔK.E  +  ΔP.E = 0

¹/₂m(v₂² - v₁²)  +  mg(h₂ - h₁) = 0

¹/₂m(v₂² - 0)  +  mg(0 - h₁) = 0

¹/₂mv₂²  -  mgh₁  =  0

¹/₂mv₂²  = mgh

¹/₂v₂² = gh

v₂² = 2gh

v₂ = √2gh

v₂ = √(2 x 9.8 x 0.56)

v₂ = 3.31 m/s

Therefore, the speed of the ball when it reaches equilibrium position is 3.31 m/s

Answer:

Explanation:

There are two main ways to de-starch leaves of a plant - the 'Light Exclusion' Method and the 'Carbon Dioxide Deprivation' Method. The 'Light Exclusion' method is a simpler procedure and is used often. Leaves can be destarched by depriving them of light for an extended period of time, usually 24-48 hours.

Answer:

[tex]\lambda=1.39\times 10^{-4}\ s^{-1}[/tex]

Explanation:

Given that,

The half-life of Barium-139 is [tex]4.96\times 10^3[/tex]

A sample contains [tex]3.21\times 10^{17}[/tex] nuclei.

We need to find the decay constant for this decay. The formula for half life is given by :

[tex]T_{1/2}=\dfrac{0.693}{\lambda}\\\\\lambda=\dfrac{0.693}{T_{1/2}}[/tex]

Put all the values,

[tex]\lambda=\dfrac{0.693}{4.96\times 10^3}\\\\=1.39\times 10^{-4}\ s^{-1}[/tex]

So, the decay constant is [tex]1.39\times 10^{-4}\ s^{-1}[/tex].

Answer:

30.5°

Explanation:

Since the light travels horizontally through the prism, it undergoes minimum deviation. So, the angle between the direction of the incident ray and that of the outgoing ray D is gotten from

n = [sin(D + α)/2]/sin(α/2) where n = refractive index of prism = 1.42 and α = angle of prism = 60° (since it is a n equilateral glass prism).

Making D subject of the formula, we have

n = [sin(D + α)/2]/sin(α/2)

nsin(α/2) = [sin(D + α)/2]

(D + α)/2 = sin⁻¹[nsin(α/2)]

D + α = 2sin⁻¹[nsin(α/2)]

D = 2sin⁻¹[nsin(α/2)] - α

So, substituting the values of the variables into the equation, we have

D = 2sin⁻¹[nsin(α/2)] - α

D = 2sin⁻¹[1.42sin(60°/2)] - 60°

D = 2sin⁻¹[1.42sin(30°)] - 60°

D = 2sin⁻¹[1.42 × 0.5] - 60°

D = 2sin⁻¹[0.71] - 60°

D = 2(45.23°) - 60°

D = 90.46° - 60°

D = 30.46°

D ≅ 30.5°

Answer:

Following are the solution to the given points:

Explanation:

For point a:

Find the schematic of the empty body and in attachment. Upon on ball during the pitch only two forces act:

The strength of the pitcher F is applied that operates horizontally. Its gravity force acting on an object is termed weight, which value is where m denotes mass, and g the acceleration of gravity.

For point b:

[tex]160.2\ N[/tex]

First, they must find that ball's acceleration. You can use the SUVAT equation to achieve that

where

[tex]v = 47\ \frac{m}{s} \\\\u = 0 \\\\a =?\\\\d = 1.0 \ m \\\\[/tex]

Solving for a,

[tex]a=\frac{v^2-u^2}{2d}=\frac{47^2-0}{2(1.0)}=1104.5 \ \frac{m}{s^2}[/tex]

Calculating the mass:

[tex]m = 145 g = 0.145 kg[/tex]

Calculating the force:

[tex]F=ma=0.145 \times 1104.5= 160.2 \ N[/tex]

 For point c:

0.195 times the pitcher's weight

[tex]m = 84 \ kg \\\\g = 9.8\ \frac{m}{s^2}\\\\[/tex]

Solving for W:

[tex]W=84 \times 9.8= 823.2 \ N[/tex]

Now the force of Part B could be defined as the fraction of the mass of the pitcher:  

[tex]\frac{F}{W}=\frac{160.2}{823.3}=0.195[/tex]

Answer:

Final velocity = 10 m/s

Time taken to travel 100 meter = 8.16 second (Approx.)

Explanation:

Given:

Acceleration = 2 m/s²

Initial velocity = 0 m/s

Find:

Velocity after 5 seconds

Time taken to travel 100 meter

Computation:

Using first equation of motion

v = u + at

v = 0 + (2)(5)

Final velocity = 10 m/s

Using Second equation of motion

s = ut + (1/2)(a)(t²)

100 = (0)(t) + (1/2)(3)(t²)

100 = (1/2)(2)(t²)

100  = (t²)

t = 10

Time taken to travel 100 meter = 8.16 second (Approx.)

570k or 1050k

270k+290k= 570k

270k•290k = 1050k

Answer: θ would equal approximately 28.7°

This is a kinematics problem, where one is only given the theta value 13.0° in regards to the range; thus, the problem is testing one's understanding of the relationships between the variables.

Range (aka x) = (v₀ sin (2θ₀))/g, where θ₀ = 13.0°

Now if we multiply the range by 2, we get:

2x = 2((v₀ sin (2θ₀))/g) → to verbalize, if range equates to (v₀ sin (2θ₀))/g, and doubling the range equals twice the product value, then:

2θ = sin⁻¹(2sin(2(13.0° )) = sin⁻¹(2(0.76255845048)) = sin⁻¹ (1.52511690096) = 57.35560850015109°/2 = θ

Thus, θ = 28.67780425

It's been awhile since I did this; though I hope it helped!

Answer:

a)  [tex]V=7.5m/s[/tex]

b) [tex]rms=8.4m/s[/tex]

c) Generally the most probable speed is 8m/s as it the most posses by particles being the average

Explanation:

From the question we are told that:

Sample size N=15

  [tex]Speed 1 v_1=2m/s\\\\Speed 2 v_2=3m/s\\\\Speed 3 v_2=5m/s\\\\Speed 4 v_4=8m/s\\\\Speed 3 v_5=9m/s\\\\Speed 2 v_6=15m/s\\\\[/tex]

Generally the equation for Average speed is mathematically given by

 [tex]V_{avg}=\frac{\sum(nv)}{N}[/tex]

Therefore

 [tex]V_{avg}=\frac{(2+2(3)+3(5)+4(8)+3(9)+2(15))}{15}[/tex]

 [tex]V=7.5m/s[/tex]

b)

Generally the equation for RMS speed of the particle is mathematically given by

  [tex]rms=\sqrt{\frac{\sum(nv^2)}{N}}[/tex]

  [tex]rms=\sqrt{\frac{2^2+2(3)^2+3(5)^2+4(8)^2+3(9)^2+2(15)^2}{15}}[/tex]

  [tex]rms=\sqrt{69.73}[/tex]

  [tex]rms=8.4m/s[/tex]

c

Generally the most probable speed is 8m/s as it the most posses by particles being the average

Answer:

a.49 n

b. 63 n

c. 112 n

Explanation:

a.10 times 9.8 from gravity/2 = 49 n

b. 49n times 4.5/8-4.5 = 63 n

c 49n + 63 n = 112 n

Answer:

The work done by the friction force is - 139927.5 J.

Explanation:

mass of diver, m = 45 kg

distance falls, h = 450 m

initial speed, u = 0 m/s

final speed = 51 m/s

According to the work energy theorem,

Work done by the gravity + work done by the friction force = change in kinetic energy

[tex]m g h + W' = 0.5 m ()v^2 - u^2)\\\\45\times 9.8\times 450 + W' = 0.5\times 45\times (51^2 - 0)\\\\198450 + W' = 58522.5\\\\W' = - 139927.5 J[/tex]

Answer:

I don't see anything on your question?

Answer:

Δd = 7.22 10⁻² m

Explanation:

For this exercise we must use the dispersion relationship of a diffraction grating

           d sin θ = m λ

let's use trigonometry

           tan θ = y / L

how the angles are small

           tant θ = sinθ  /cos θ = sin θ

we substitute  

           sin θ = y / L

          d y / L = m λ

          y = m λ L / d

let's use direct ruler rule to find the distance between two slits

If there are 500 lines in 1 me, what distance is there between two lines

         d = 2/500

        d = 0.004 me = 4 10⁻⁶ m

diffraction gratings are built so that most of the energy is in the first order of diffraction m = 1

let's calculate for each wavelength

λ = 656 nm = 656 10⁻⁹ m

         d₁ = 1 656 10⁻⁹ 1.7 / 4 10⁻⁶

         d₁ = 2.788 10⁻¹ m

λ = 486 nm = 486 10⁻⁹ m

         d₂ = 1 486 10⁻⁹ 1.7 / 4 10⁻⁶

         d₂ = 2.066 10⁻¹ m

the distance between the two lines is

         Δd = d1 -d2

         Δd = (2,788 - 2,066) 10⁻¹

         Δd = 7.22 10⁻² m

Answer:

the mass of the bullet is 10.5 g

Explanation:

Given;

initial velocity, u₁ = 280 m/s

final velocity of the bullet, v₁ = 70 m/s

final velocity of the block, v₂ = 0.2 m/s

mass of the block, m₂ = 11 kg

initial velocity of the block, u₂ = 0

let the mass of the bullet = m₁

Apply the principle of conservation of linear momentum for elastic collision to calculate the mass of the bullet.

m₁u₁ + m₂u₂ = m₁v₁  + m₂v₂

280m₁  +  11(0)  = 70m₁  +  11 x 0.2

280m₁ = 70m₁  + 2.2

280m₁ - 70m₁  = 2.2

210m₁ = 2.2

m₁ = 2.2/210

m₁ = 0.0105 kg

m₁ = 10.5 g

Therefore, the mass of the bullet is 10.5 g

Answer:

The product of mass and velocity is the correct answer

Explanation:

Momentum is defined as mass × velocity

p = mv

Answer:

The product of mass and velocity

Explanation:

I  just did it and got it right with a 100%

Posted 1/3/23

Answer: Geothermal energy is produced by the heat of Earth's molten interior. This energy is harnessed to generate electricity when water is injected deep underground and returns as steam (or hot water, which is later converted to steam) to drive a turbine on an electric power generator.

Explanation: Hope this helps!

Answer:

The impulse delivered to the ball is [tex]Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right][/tex].

Explanation:

By Impulse Theorem, the motion of the tennis ball is modelled after the following expression:

[tex]Imp = m\cdot (\vec v_{f} - \vec v_{o})[/tex] (1)

Where:

[tex]m[/tex] - Mass of the ball, in kilograms.

[tex]\vec v_{o}[/tex] - Vector of the initial velocity, in meters per second.

[tex]\vec v_{f}[/tex] - Vector of the final velocity, in meters per second.

[tex]Imp[/tex] - Impulse, in meters per second.

If we know that [tex]m = 0.06\,kg[/tex], [tex]\vec v_{o} = \left(45\,\frac{m}{s} \right)\cdot (\cos 47^{\circ}, \sin 47^{\circ})[/tex] and [tex]\vec v_{f} = \left(35\,\frac{m}{s} \right)\cdot (-1, 0)[/tex], then the impulse delivered to the ball is:

[tex]Imp = (0.06\,kg)\cdot \left[\left(35\,\frac{m}{s} \right)\cdot (-1,0) -\left(45\,\frac{m}{s} \right)\cdot (\cos 47^{\circ}, \sin 47^{\circ})\right][/tex]

[tex]Imp = (0.06\,kg)\cdot (-65.670, -32.911)\,\left[\frac{m}{s} \right][/tex]

[tex]Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right][/tex]

The impulse delivered to the ball is [tex]Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right][/tex].

Answer:

The Sum of mass and energy is always conserved in a nuclear reaction. Mass changes to energy, but the total amount of mass and energy combined remains the same

Explanation:

Every single radioactive decay, every single nuclear collision, every single nuclear reaction will conserve mass number and charge.