4. You currently have TWO 9 volt batteries connected to a 82 resistor. How
much current is flowing through the circuit? *
1.125 A
5 A
72 A
2.25 A
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Answer:

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ds

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:

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:

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:

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:

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:

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: 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:

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:

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

To know more about probability follow

https://brainly.com/question/24756209

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.

To know more about magnification follow

https://brainly.com/question/1599771

Answer:

W = 354.9 J

Explanation:

Given that,

The mass of a bale of hay, m = 23 kg

The displacement, d = 3.9 m

The horizontal force exerted on the hay, F = 91 N

We need to find the work done. We know that,

We know that,

Work done, W = Fd

So,

W = 91 N × 3.9 m

W = 354.9 J

So, the required work done is 354.9 J.

Answer:

The number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons

Explanation:

Given;

magnitude of the attractive force, F = 17 mN = 0.017 N

distance between the two objects, r = 24 cm = 0.24 m

The attractive force is given by Coulomb's law;

[tex]F = \frac{1}{4\pi \epsilon _0} \times \frac{Q^2}{r^2} = \frac{kQ^2}{r^2} \\\\Q^2 = \frac{Fr^2}{k} \\\\Q = \sqrt{ \frac{Fr^2}{k}} \\\\Q = \sqrt{ \frac{(0.017)(0.24)^2}{9\times 10^9}} \\\\Q = 3.298 \times 10^{-7} \ C[/tex]

The charge of 1 electron = 1.602 x 10⁻¹⁹ C

n(1.602 x 10⁻¹⁹ C) = 3.298 x 10⁻⁷

[tex]n = \frac{3.298 \times 10^{-7}}{1.602 \times 10^{-19}} = 2.06 \times 10^{12} \ electrons[/tex]

Therefore, the number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons

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]

False

It's not straight up

Newton's third law states that for every action (force) in nature there is an equal and opposite reaction. If object A exerts a force on object B, object B also exerts an equal and opposite force on object A. In other words, forces result from interactions.

Hope it helps

Answer:

Newton's third law state that every action there is equal but opposite reactions

hope this will help you more

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]

570k or 1050k

270k+290k= 570k

270k•290k = 1050k

Answer:

I don't see anything on your question?

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:

a. 59 ev. helpful answer

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:

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