A student is driving through a mountainous region where the road is at some times flat, at some times inclined upward, and at some time inclined downward. The student maintains a speed of 20 m/s on the roadway, but is required to make an emergency stop on the three sepearte occasions. On levels roadway, it takes 25 m to stop. On a downward-sloping roadway, it takes 40 m to stop. On an upward-sloping roadway, it takes 18 m to stop. Explain why the stopping distances are different. (Focus answer using work and energy, other concepts may be used as well but be sure work and energy are included.)

Answer:

4.79m/s

Explanation:

According to law of conservation of momentum;

The sum of momentum of the bodies before collision is equal to the momentum after collision.

m1u1 + m2u2 = (m1+m2)v

Given;

m1 = 0.3kg

u1 = 2.30m/s

m2 = 0.0225kg

u2 = 38m/s

Required

speed of the arrow after passing through the target v

Substituting the given data into the formula

0.3(2.3) + 0.0225(38) = (0.3 + 0.0225)v

0.69 + 0.855 = 0.3225v

1.545 = 0.3225v

v = 1.545/0.3225

v = 4.79m/s

Hence the speed of the arrow after passing through the target is 4.79m/s

The sky changes colors at sunset because the atmosphere ABSORBS some colors of light more than other colors.

The second choice ("The atmosphere bends light") is a correct statement, but it's not the reason that the sky changes colors at sunset.

Answer:

The atmosphere bends light.

Explanation:

I got the test, and passed! =)

Answer:

Explanation:

It takes 3.1 s for the boat to travel from its highest point to its lowest, so the period of oscillation

T = 2 x 3.1 = 6.2 s

frequency of wave n = 1 / T = .1613 per sec

Amplitude of oscillation = .55/2 = .275 mm

The fisherman sees that the wave crests are spaced 4.2 mm apart. so wavelength of wave λ = 4.2 mm .

A ) velocity of wave v = n λ

.1613 x 4.2 = .677 mm /s

B ) Amplitude of wave = .275 mm

C ) The vertical distance determines only the amplitude which does not affect the velocity , so velocity will remain unchanged .

D ) Amplitude of wave depends only on the vertical displacement .

The amplitude will become .5 / 2 = .25 mm .

Distance and time I think

Answer:

y-coordinate where the monkey catches the coconut is 13.664 m.

Explanation:

Given;

time taken for the monkey to catch the coconut, t = 1.4 s

initial upward speed of the coconut, Uy = 2.9 m/s

acceleration due to gravity, g = 9.8 m/s²

The y-coordinate where the monkey catches the coconut is calculated as;

[tex]h_y = U_yt +\frac{1}{2} gt^2\\\\h_y = (2.9\times 1.4) +\frac{1}{2} (9.8)(1.4^2)\\\\h_y = 4.06 + 9.604\\\\h_y = 13.664 \ m[/tex]

Therefore, y-coordinate where the monkey catches the coconut is 13.664 m.

Answer:

cognitive, humanistic, behavioral, sociocultural

Explanation:

Behavioral, sociocultural, cognitive, and humanistic are approaches to psychological science.

Psychology is a term to refer to the discipline that focuses on the study of various topics related to human thought such as:

Due to the above, several subdisciplines have emerged that focus on the study of each of the topics. For example:

Behavioral psychology: focused on the study of human behavior.

Sociocultural psychology: focused on the study of human behavior and thought in different social situations.

Cognitive psychology: focused on mental processes related to learning.

Humanistic psychology: focused on the study of human thought from a comprehensive approach.

According to the above, options A, E, F, and G are correct because they mention different sub-disciplines of psychology while the other options mention terms that are not related to sub-disciplines or psychological sciences.

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Answer:

electrons are located around the nucleus of an atom.

Explanation:

Electron configurations describe where electrons are located around the nucleus of an atom. For example, the electron configuration of lithium, 1s²2s¹, tells us that lithium has two electrons in the 1s subshell and one electron in the 2s subshell.

Answer:

3.33m

Explanation:

Given parameters:

Force applied  =500N

Elastic constant  = 150N/m

Unknown:

Amount of stretch or extension  = ?

Solution:

To solve this problem use the expression below:

    F  = k e

F is the force applied

k is the elastic constant

e is the extension

  So;

           500  = 150 x e

             e  = [tex]\frac{500}{150}[/tex]    = 3.33m

Answer:

(a) the unknown inertia is 0.388 kg

(b) the average acceleration of the heavier cart is 317 m/s²

(c) the average acceleration of the lighter cart is 539 m/s²

Explanation:

Given;

mass of the first cart, m₁ = 0.66 kg

initial speed of the first cart, u₁ = 1.85 m/s

let the mass of the cart with unknown inertia be m₂

initial velocity of the second cart, u₂ = 2.17 m/s to the left

velocity of the first cart after collision, v₁ = 1.32 m/s to the left

velocity of the second cart after collision, v₂ = 3.22 m/s

time of collision, t = 0.010 s

(a) What is the unknown inertia?

Apply the principle of conservation of linear momentum, to determine the unknown inertia.

let leftward direction be negative direction

let rightward direction be positive direction

m₁u₁ + m₂u₂ = m₁v₁  + m₂v₂

0.66(1.85) + m₂(-2.17) = 0.66(-1.32) + m₂(3.22)

1.221 - 2.17m₂ = -0.8712 + 3.22m₂

1.221 + 0.8712 = 3.22m₂ + 2.17m₂

2.0922 = 5.39m₂

m₂ = 2.0922 / 5.39

m₂ = 0.388 kg

The unknown inertia is 0.388 kg

(b) the average acceleration of the heavier cart

the heavier cart has a mass of 0.66 kg

[tex]a = \frac{v_1 - u_1}{t} \\\\a = \frac{-1.32 - 1.85}{0.01} \\\\a = -317 \ m/s^2\\\\|a| = 317 \ m/s^2[/tex]

(c) the average acceleration of the lighter cart;

the lighter cart has a mass of 0.388 kg

[tex]a = \frac{v_2 - u_2}{t} \\\\a = \frac{3.22 - (-2.17)}{0.01} \\\\a =\frac{3.22 \ +\ 2.17}{0.01} \\\\a= 539\ m/s^2[/tex]

Answer:

hello your question has some missing parts

A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.

answer : c) 0.39 sec

               d)  2.25 m

               e) 1.92 m/sec

Explanation:

The initial velocity of the first ball = 7.67 m/sec ( calculated )

Time required for first ball to reach ceiling = 0.78 secs ( calculated )

Determine how long after the second ball is thrown do the two balls pass each other

Distance travelled by first ball downwards when it meets second ball can be expressed as : d = 1/2 gt^2 =  9.8t^2 / 2

hence d = 4.9t^2  ----- ( 1 )

Initial speed of second ball = first ball initial speed = 7.67 m/sec

3 - d = 7.67t - 4.9t  ---- ( 2 )

equating equation 1 and 2

3 = 7.67t   therefore t = 0.39 sec

Determine how far the balls are above the Juggler's hands ( when the balls pass each other )

form equation 1 ;

d = 4.9 t^2 = 4.9 *(0.39)^2 = 0.75 m

therefore the height the balls are above the Juggler's hands is

3 - d = 3 - 0.75 = 2.25 m

determine their velocities when the pass each other

velocity = displacement / time

velocity = d / t = 0.75 / 0.39 sec  = 1.92 m/sec

Answer:

My biggest reason is to make it a habit. Even if the ball goes into the endzone it is a live ball and the offensive players must down the ball. Don't leave any room for "I thought he downed it" or "I thought I heard the whistle" just run to the ball always.

If the players slow down and the returner takes it out of the end zone it could be a big return. Players are on a full sprint for 40+ yards sometimes and instead of breaking down, they choose to contine through the goal line to slow down at a decreased rate (possibly limiting a muscle pull injury).

Answer:

A. Gamma decay

Explanation:

A form of nuclear decay in which the atomic number is unchanged is a gamma decay.

The atom has undergone a gamma decay.

In a gamma decay, no changes occur to the mass and atomic number of the substance.

Answer:

6m/s

Explanation:

Using the law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the momentum after collision.

Using the expression

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the final velocity after collision

Substitute the given values in the formula

0.35(10)+0.35(2) = (0.35+0.35)v

3.5+0.7 = 0.7v

4.2 = 0.7v

v = 4.2/0.7

v = 6m/s

Hence the magnitude of the velocity for Ball B after the collision is 6m/s

Answer: If you increase the voltage across a component, there will be more current in the component .

Answer:

Explanation:

C 200÷100=2

Output ÷ Input= MA

Answer:

argon (Ar)

Explanation:

Argon is the element from the given choices with a complete valence electron shell.

The valence electron shell is the outermost shell of an atom.

Answer:

argon (Ar)

Explanation:

Answer:

B

Explanation:

Answer:

a) vx = -12.7 m/s vy = -56.7m/s

b) v= 58.1 m/s

c) θ = 77.4º S of W

Explanation:

a)

       [tex]v_{ps} = -44m/s (1)[/tex]

       [tex]v_{we} = - 18m/s * cos (45) = -12.7 m (2)[/tex]

       [tex]v_{ws} = - 18m/s * cos (45) = -12.7 m (3)[/tex]

        [tex]v_{e} = v_{x} = -12.7 m/s (4)[/tex]

b)

        [tex]v = \sqrt{(-12.7m/s)^{2} + (-56.7m/s)^{2}} = 58.1 m/s (6)[/tex]

c)

       [tex]tg_{\theta} =\frac{v_{y} }{v_{x} } = \frac{(-56.7m/s)}{(-12.7m/s} = 4.46 (7)[/tex]

        tg⁻¹ θ = tg⁻¹ (4.46) = 77.4º S of W. (8)

Answer:

27.2m approx

Explanation:

Given data

Work done= 4000J

mass= 15kg

W= mg

W= 15*9.81

W= 147.15N

We know that

Work done= W*d

substitute

4000=147.15*d

divide both sides  by 147.15

d= 4000/147.15

d=27.18

Hence the distance is 27.2m approx

This question is incomplete, the complete question is;

According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula; h= (0.04 to 0.09)(D/d)⁴V²/2g

where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity.

Do you think this equation is valid in any system of units

Answer:

YES, the equation is a general equation that is valid in any system of units

Explanation:

Given the data in the question;

h = (0.04 to 0.09)(D/d)⁴ × [tex]\frac{V^{2} }{2g}[/tex]

so

[ N.m/N ] = (0.04 to 0.09) ( m/m)² × (m²/s²)1/2 × (s²/m)

[ N.L/N ] = (0.04 to 0.09) ( L⁴/L⁴) × (L²/T²)1/2 × (T²/L)

∴ [ L ] = (0.04 to 0.09) [L]

So as each term in the equation must have the same dimensions, the constant term (0.04 to 0.09) must be without dimension.

Therefore, YES, the equation is a general equation that is valid in any system of units

A computer is a programmable device that can automatically perform a sequence of calculations or other operations on data once programmed for the task. It can store, retrieve, and process data according to internal instructions. A computer may be either digital, analog, or hybrid, although most in operation today are digital. Digital computers express variables as numbers, usually in the binary system. They are used for general purposes, whereas analog computers are built for specific tasks, typically scientific or technical. The term "computer" is usually synonymous with a digital computers, and computers for business are exclusively digital.

Answer:

The core, computing part of a computer is its central processing unit (CPU), or processor. ... A computer system, therefore, is a computer combined with peripheral equipment and software so that it can perform desired functions.

Explanation:

Hope the answer was helpful

Answer:

Explanation:

In the whole process , potential energy of the cart is converted into kinetic energy . At the top of the vertical loop , the whole of potential energy is regained and kinetic energy becomes zero if we release the cart from a height of 2R because difference of height between lowest and highest point of motion  is 2R .  In that case kinetic energy at top = 0 , velocity v = 0

At the top , weight mg is acting which is providing centripetal force . So cart must have some velocity at the top . If it be v

mv²/R = mg

v = √ gR .

For that purpose , the cart must be released from a height greater than 2R .

The extra height beyond 2R will make the velocity at the top non-zero.

Answer:

It determines how long you do a certain workout.

Answer:

Explanation:

To find the angular velocity of the tank at which the bottom of the tank is exposed

From the information given:

At rest, the initial volume of the tank is:

[tex]V_i = \pi R^2 h_i --- (1)[/tex]

where;

height h which is the height for the free surface in a rotating tank is expressed as:

[tex]h = \dfrac{\omega^2 r^2}{2g} + C[/tex]

at the bottom surface of the tank;

r = 0, h = 0

∴

[tex]h = \dfrac{\omega^2 r^2}{2g} + C[/tex]

0 = 0 + C

C = 0

Thus; the free surface height in a rotating tank is:

[tex]h=\dfrac{\omega^2 r^2}{2g} --- (2)[/tex]

Now; the volume of the water when the tank is rotating is:

dV = 2π × r × h × dr

Taking the integral on both sides;

[tex]\int \limits ^{V_f}_{0} \ dV = \int \limits ^R_0 \times 2 \pi \times r \times h \ dr[/tex]

replacing the value of h in equation (2); we have:

[tex]V_f} = \int \limits ^R_0 \times 2 \pi \times r \times ( \dfrac{\omega ^2 r^2}{2g} ) \ dr[/tex]

[tex]V_f = \dfrac{ \pi \omega ^2}{g} \int \limits ^R_0 \ r^3 \ dr[/tex]

[tex]V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{r^4}{4} \Big]^R_0[/tex]

[tex]V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{R^4}{4} \Big] --- (3)[/tex]

Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.

Then [tex]V_f = V_i[/tex]

Replacing equation (1) and (3)

[tex]\dfrac{\pi \omega^2}{g}( \dfrac{R^4}{4}) = \pi R^2 h_i[/tex]

[tex]\omega^2 = \dfrac{4g \times h_i }{R^2}[/tex]

[tex]\omega =\sqrt{ \dfrac{4g \times h_i }{R^2}}[/tex]

[tex]\omega = \sqrt{\dfrac{4 \times 9.81 \ m/s^2 \times 0.7 \ m}{(0.5)^2} }[/tex]

[tex]\omega = \sqrt{109.87 }[/tex]

[tex]\mathbf{\omega = 10.48 \ rad/s}[/tex]

Finally, the angular velocity of the tank at which the bottom of the tank is exposed  = 10.48 rad/s

Answer:

The Carnot engine has zero power

Explanation:

Although theoretically the Carnot engine has more efficiency than the real engine. In practice however they tend to have zero power.

This is because all its processes are reversible (that is isothermic and adiabatic).

So the system equilibrates with its surroundings at every point in time. This makes work done very slow and the power generated is zero.

Carnot cycles requires attaining isothermal heat transfers which is quite difficult and take a long time. Also a pump that can handle liquid-vapour phase mixture will be required.

This is not practical.

Answer:

40 N

Explanation:

According to the scenario, computation of given data are as follows:

Mass (m) = 20 kg

Initially moving (v) = 3

Actual distance (d) = 2.25 m

So, we can calculate friction (f) by using following formula,

f × d = [tex]\frac{1}{2} mv^{2}[/tex]

By putting the value, we get

f × 2.25 = [tex]\frac{1}{2}[/tex] × 20 × [tex]3^{2}[/tex]

f × 2.25 = 10 × 9

f = 90 ÷ 2.25

= 40 N.

Answer:

a. Capacitance

b. Charge on the plates  

e. Energy stored in the capacitor

Explanation:

Let A be the area of the capacitor plate

The capacitance of a capacitor is given as;

[tex]C = \frac{Q}{V} = \frac{\epsilon _0 A}{d} \\\\[/tex]

where;

V is the potential difference between the plates

The charge on the plates is given as;

[tex]Q = \frac{V\epsilon _0 A}{d}[/tex]

The energy stored in the capacitor is given as;

[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} (\frac{\epsilon _0 A}{d} )V^2[/tex]

Thus, the physical variables listed that will change include;

a. Capacitance

b. Charge on the plates  

e. Energy stored in the capacitor

Answer:

43955.12N and 7325.85N

Explanation:

Step one:

Given data

Weight of car= 9800N

therefore the mass of the car is

w=mg

m= w/g

m= 9800/9.81

m= 998.98 kg

Speed= 22m/s

duration of impact= 0.5 seconds

we know that

impulse, Ft= mv

F= mv/t

F= 998.98*22/0.5

F=21977.56/0.5

F=43955.12N

Hence the force is 43955.12N

if the time is 3 seconds

then the force will be

F= mv/t

F= 998.98*22/3

F=21977.56/3

F=7325.85N

Answer:

F = 2.40 × [tex]10^{-6}[/tex]  N

Explanation:

given data

charge q1 = 3.95 nC

x= 0.198 m

charge q2 = 4.96 nC

x= -0.297 m

solution

force on a point charge kept in electric field F = E × q       ................1

here E is the magnitude of electric field and q is the magnitude of charge

and

first we will get here electric field at origin

So net field at origin is

E = (Kq2÷r2²) - (kq1÷r1²)           ...............2

put here value

E = 9[(4.96÷0.297²)-(3.95÷0.198²)]

E = 400.72 N/C        ( negative x direction )

so that force will be

F = 6 × [tex]10^{-9}[/tex] × 400.72

F = 2.40 × [tex]10^{-6}[/tex]  N

The net force on the third charge is 2.404 x 10⁻⁶ N.

The given parameters:

The force on the third charge due to first charge is calculated as follows;

[tex]F_{13} = \frac{kq_1 q_3}{r^2} \\\\F_{13} = \frac{9\times 10^9 \times 3.95 \times 10^{-9} \times 6 \times 10^{-9} }{(0.198)^2} (+i)= 5.44 \times 10^{-6} \ N \ (+i)[/tex]

The force on the third charge due to second charge is calculated as follows;

[tex]F_{23} = \frac{kq_2q_3}{r^2} \\\\F_{23} = \frac{9\times 10^9 \times 4.96 \times 10^{-9}\times 6 \times 10^{-9} }{(0.297)^2} (-i)\\\\F_{23} = (3.036 \times 10^{-6} ) \ N \ (-i)[/tex]

The net force on the third charge is calculated as follows;

[tex]F_{net} = 5.44 \times 10^{-6} - 3.036 \times 10^{-6} \\\\F_{net} = 2.404 \times 10^{-6} \ N[/tex]

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