A hiker walks 7.7 miles to the east in 3.6 hours, then turns around and walks 2.1 miles to the west in 2.4 hours. What was the magnitude of her average velocity during the trip?

Answer:

2m/s

Explanation:

v=f×wavelength

v=2×1

=2m/s

Answer:

Gravitational

Explanation:

gravitational

Answer:

Gravitational

Explanation:

In order from strongest to weakest.

Strong nuclear

Electromagnetic

Weak nuclear

Gravitational

Answer:

30 m/s

Explanation:

Speed is distance over time. 60 meters / 2 seconds, = 30 m/s.

net force = 30 N

mass = 8.16 kg

acceleration = 3.68 m/s²

Given

80 N force applied

mass of object = 10 kg

Friction force = 50 N

Required

Net force

mass

acceleration

Solution

Net force = force applied(to the right) - friction force(to the left)

Net force = 80 - 50 = 30 N

Gravitational force(downward) : F = mg

m = F : g

m = 80 : 9.8

m = 8.16 kg

a = F net / m

a = 30 / 8.16

a = 3.68 m/s²

Answer:

a) v₀ₓ = 62.76 m / s, b)   θ₁ = 17.6º,   θ₂ = 67.0º

Explanation:

We can solve this exercise using the projectile launch ratios

a) Let's find the time it takes for the bullet to reach the water level

       y = y₀ + v_{oy} t - ½ g t²

when it reaches the water its height is zero y = 0, as the bullet is fired horizontally its initial vertical velocity is zero

       0 = y₀ + 0 - ½ g t²

       t =[tex]\sqrt{2y_o/g}[/tex]

       t = [tex]\sqrt{2 \ 7 /9.8}[/tex]          

       t = 1,195 s

now we can calculate the speed with the horizontal movement

        x = v₀ₓ t

        v₀ₓ = x / t

        v₀ₓ = 75.0 / 1.195

        v₀ₓ = 62.76 m / s

b) if the speed of the bullets is half of that found

         v₀ = 62.76 / 2 = 31.38 m / s

let's write the expressions for the distance

          x = v₀ cos θ t

          y = y₀ + v_{oy} sin θ t - ½ g t²

          t = [tex]\frac{x}{v_o \ cos \theta}[/tex]

we substitute

          [tex]0 = y_o + v_o sin \theta \ \frac{x}{v_o \cos \thetay} - 1/2 g \ (\frac{x}{v_o \ cos \theta})^2[/tex]

          [tex]0 = y_o + x tan \theta - \frac{1}{2} g \ \frac{x^2}{ v_o^2 \ cos^2 \theta}[/tex]    

let's use the identified trigonometry

          sec² θ = 1 + tan² θ

         sec θ = 1 / cos θ

we substitute

          [tex]0 = y_o + x tan \theta - \frac{g x^2}{2 v_o^2} ( 1 + tan^2 \theta)[/tex]

          [tex]\frac{g x^2}{2v_o^2} tan^2 \theta - x tan \theta + \frac{gx^2}{2v_o^2} - y_o = 0[/tex]

we change variable

         tan θ = H

         [tex]\frac{gx^2}{2 v_o^2 } H^2 - x H + \frac{gx^2}{2v_o^2}-y_o =0[/tex]

we subtitle the values

         [tex]\frac{9.8 \ 75^2}{2 \ 31.38^2} H^2 - 75 H + \frac{9.8 \ 75^2}{2 \ 31.38^2}-7 =0[/tex]

         27.99 H² - 75 H + 20.99 = 0

         H² - 2.679 H + 0.75 = 0

we solve the quadratic equation

         H = [2.679 ± [tex]\sqrt{2.679^2 - 4 0.75}[/tex]] / 2

         H = [2,679 ± 2,044] / 2

         H₁ = 0.3175

         H₂ = 2.3615

now we can find the angles

          H₁ = tan θ₁

          θ₁ = tan⁻¹ H₁

          θ₁ = tan⁻¹ 0.3175

          θ₁ = 17.6º

          θ₂ = 67.0º

for these two angles the bullet hits the boat

Answer:

Answer:

15.67 seconds

Explanation:

Using first equation of Motion

Final Velocity= Initial Velocity + (Acceleration * Time)  

v= u + at

v=3

u=50

a= - 4 (negative acceleration or deceleration)  

3= 50 +( -4 * t)

-47/-4 = t

Time = 15.67 seconds

We have that the speed  must be at the speed below if they want to just hit their target

From the Question we are told that

Distance [tex]d=50m[/tex]

Height [tex]h=70m[/tex]

Constant Velocity [tex]v= 7.0 m/s[/tex]

Generally the equation for the time  is mathematically given as

[tex]T=\frac{h}{v}\\\\T=\frac{70}{7}\\\\T=10s[/tex]

Therefore

The velocity required to make horizontal movement is

[tex]V=\frac{d}{T}\\\\V=\frac{50}{10}\\\\V=5m/s[/tex]

Given that

Velocity on the Vertical axis is

[tex]v_y=7m/s[/tex]

Velocity on the  horizontal axis is

[tex]v_x=5m/s[/tex]

Therefore resultant speed

[tex]v_r=\sqrt{v_x^2+V_y^2}\\\\v_r=\sqrt{(5)^2+(7)^2}[/tex]

[tex]v_r=8.6023m/s[/tex]

In conclusion

[tex]v_r=8.6023m/s[/tex] must be their total speed if they want to just hit their target

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Answer:

Its C

Explanation:

Because I got it wrong for you

Wave B has about 1.6 times more energy than wave A.

Energy is the ability or capability to do tasks, such as the ability to move an item (of a certain mass) by exerting force. Energy can exist in many different forms, including electrical, mechanical, chemical, thermal, or nuclear, and it can change its form.

The amplitude and energy of a mechanical wave. If mechanical wave A has an amplitude of 4 cm and mechanical wave B has an amplitude of 5 cm wave B has about 1.6 times more energy than wave A.

Wave B has about 1.6 times more energy than wave A.

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Answer:

the required height is 0.2449 m only

Explanation:

Given the data in the question;

Initial height = 2m

so speed of the puck before hitting the ground will be;

u² = 2gh

Initial speed u_ball = √2gh

u_ball = √( 2 × 9.8 × 2 )

u_ball =  √39.2

u_ball = 6.26 m/s

given that; FOR THE FROZEN PUCK, coefficient of restitution = 0.35 only

R = - (v_ball - v_ground / u_ball - u_ ground)

so

0.35 = - (v_ball - 0 / 6.26 - 0)

0.35 = -v_ball / - 6.26

-v_ball = 0.35 × (- 6.26)

-v_ball = -2.191 m/s

v_ball = 2.191 m/s

to get the height;

v² = 2gh

h = v² / 2g

we substitute

h = (2.191)² / 2×9.8

h = 4.800481 / 19.6

h = 0.2449 m

Therefore, the required height is 0.2449 m only

Answer:

6 kg

Explanation:

F=ma

F is Force(newtons)

m is mass(kg)

a is acceleration(m/s^2)

Plug in the numbers

3000 = m(500)

divide both sides by 500 to cancel out the 500.

3000/500=6

6 = m

6kg is the mass

Answer:

  λ = 440 nm

Explanation:

The phenomenon of constructive interference is described by the expressions

          d sin θ= m λ                    (1)

where d is the separation of the slits d = 0.0250 mm = 2.50 10⁻⁵ m, lam is the wavelength of the incident radiation and m is an integer indicating the order of interference

let's use trigonometry to find the angle

          tan θ = y / L

where L is the distance to the screen L = 1.25 m

in general interference experiments angles are very small

         tan θ = [tex]\frac{sin \ \theta }{cos \ \theta}[/tex]

         ten θ = sin θ

substituting

         sin θ = y / L

we substitute 1

          d y / L = m λ

          λ = [tex]\frac{ d \ y }{m \ L}[/tex]

in the exercise indicate

        m = 3

         y = 6.6 cm = 6.6 10-2 m

we calculate

          λ = 2.50 10⁻⁵ 6.6 10⁻² /( 3   1.25)

          λ = 4.4 10⁻⁷ m

let's reduce to nm

          λ = 4.4 10⁻⁷ m (10⁹ nm / 1 m)

          λ = 440 nm

The force formula can be rewritten  to solve the acceleration as:

acceleration = force/mass.

Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).

The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

From Newton's 2nd law of motion, we can write that:

Force = mass × acceleration

⇒ acceleration = force/mass.

Hence, the force formula can be rewritten  to solve the acceleration as:

acceleration = force/mass.

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Answer:

The four-season year is typical only in the mid-latitudes. The mid-latitudes are places that are neither near the poles nor near the Equator. The farther north you go, the bigger the differences in the seasons.

Explanation:

hope this helps have a good day :) ❤

At a latitude equal to zero degrees there is little seasonal variation. This phenomenon is due each day the Sun's rays strike the Earth's surface at approximately the same angle near the Equator.

The Equator is the line of 0° (zero degrees) latitude around the middle of the Earth.

Moroever, the intensity of solar radiation and therefore also the temperature at the Earth's surface largely depends on the angle of incidence of the Sun's rays.

At 0° latitude, there is a very little seasonal variation because all days the Sun's rays strike the Earth's surface at approximately the same angle. At the Equator, the Sun's rays strike the Earth's surface at an angle of 90°, causing warmer temperatures compared to higher latitudes.

In additon, at 0° latitude, all days also have the same number of hours of light and dark (i.e., approximately 12 hours of sunlight).

In conclusion, at a latitude equal to zero degrees (i.e., at the Equator) there is little seasonal variation. This phenomenon is due each day the Sun's rays strike the Earth's surface at approximately the same angle near the Equator.

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Answer:

The variables that can be measured in joules are

B. Energy

D. Work

Hope it will help :)

Answer:

Ep = 117600 J

Explanation:

Data:

Use formula:

Replace:

Multiply operations, and units:

What is the potential energy?

The potential energy is 117600 Joules.

Answer:

D

Explanation:

Work = Distance x Mass

work done = 100,000 FT LB

Work is done whenever a force moves something over a distance or The work done by a force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement.

Work done = force * displacement

given :

force = 500 pound

displacement = 200 feet

work done = 500 * 200 = 100,000 FT LB

correct option is d. 100,000 FT LB

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Answer:

Astronomers can learn about the elements in stars and galaxies by decoding the information in their spectral lines. There is a complicating factor in learning how to decode the message of starlight, however. If a star is moving toward or away from us, its lines will be in a slightly different place in the spectrum from where they would be in a star at rest. And most objects in the universe do have some motion relative to the Sun.

Answer:

 E = -1.5 10⁵ N / C

Explanation:

In a capacitor the electric field is uniform between the blades, therefore we use the expression

                   V = - E s

                   E = - V / s

let's calculate

                   E = - 3000 / 0.02

                   E = -1.5 10⁵ N / C

the sign indicates that the field and the potential are opposite, when one increases the gold decreases

Answer:

0.1s,5s,0.017s

Explanation:

T=1÷frequency

Answer:

a =

✔ 6

The period is

✔ 2 seconds.

b =

✔ pi

Explanation:

Graph the function using the graphing calculator. Find the least positive value of t at which the pendulum is in the center.

t =

✔0.5 sec

To the nearest thousandth, find the position of the pendulum when t = 4.25 sec.

d =

✔ 4.243 in.

Answer:

Δr = 0.425 m

Explanation:

This is a sound interference exercise, the expression for destructive interference is

          Δr = (2n + 1) λ / 2

in this case the movement is in the same direction as the sound, therefore the movement is one-dimensional

let's use the relationship between the speed of sound and its frequency and wavelength

          v = λ f

          λ = v / f

the first minium occurs for n = 0

           Δr = λ / 2

            Δr = v / 2f

            Δr = [tex]\frac{340}{2 \ 400}[/tex]

             Δr = 0.425 m

this is the distance from the current position that we assume in the center of the room

Answer:

2880 N/c

Explanation:

Given that:

Charge per unit length ; λ = 4 * 10^-9

radius, r = 10

Radius, R = 0.5m

Using the relation :

2λr / 4πE0R²

Columb's constant, k = 1/4πE0 =. 9* 10^9Nm²/C²

Hence, we have :

2λrk/ R²

(2 * 4 * 10^-9 * 10 * 9 * 10^9) / 0.5^2

(720 ÷ 0.25)

= 2880 N/c

Answer:

am not sure about the answer

Explanation:

you need to find out the amount of force it's going in for example 10n or 100n then you need to times it the distance then devide by the time

Answer: the higher the kinetic energy

Explanation:

Answer:

* Larger mirrors collect more light and therefore fainter and more distant objects can have enough intensity to be detected

* arger mirrors decreases the angle of dispersion giving a better resolution of the bodies

Explanation:

Refracting telescopes get bigger every day for two main reasons.

* Larger mirrors collect more light and therefore fainter and more distant objects can have enough intensity to be detected

* the diffraction process for circular apertures is given by

               θ = 1.22 λ / D

where d is the diameter of the mirror, therefore having larger mirrors decreases the angle of dispersion giving a better resolution of the bodies

Answer:

i think its ancestor

Explanation:

sry if im wrong

Answer:

scientists compare organisms DNA to support the theory that all species share a common Ancestor.

Explanation:

thermodynamics is the study of heat.

Answer The study of heat and its relationship to useful work is called thermodynamics and involves macroscopic quantities such as pressure, temperature, and volume without regard for the molecular basis of these quantitie

Explanation:

Answer:

WE NEED TO ADD ALL 40+2.38 +75+5

Explanation:

PLSE GIVE SOME POINTS DUDE

Answer:

4

2

1

3

Explanation:

Be safe, lovelies <3

Answer:

the radius of curvature of the track for this instant is 266 m

Explanation:

Given that;

The Initial Velocity u = 100 km/h = 100 × [tex]\frac{5}{18}[/tex] = 27.77 m/s

velocity of the train at t=12 s is;

[tex]V_{t=12}[/tex] = 50 km/h = 50 × [tex]\frac{5}{18}[/tex] = 13.89 m/s

now, we calculate the deceleration of the train

[tex]V_{t=12}[/tex]  = u + at

13.89 = 27.77 + [tex]a_{t}[/tex]12

[tex]a_{t}[/tex] = (13.89 - 27.77) / 12

[tex]a_{t}[/tex] = -13.88 / 12

[tex]a_{t}[/tex] = - 1.1566 m/s²

Now, the velocity of the train at 6 seconds is;

[tex]V_{t=6}[/tex]  = u + at

[tex]V_{t=6}[/tex]  = 27.77 + ( - 1.1566 m/s²)6

[tex]V_{t=6}[/tex]  = 27.77 - 6.9396

[tex]V_{t=6}[/tex]  = 20.83 m/s

The acceleration at t=6 s is;

a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]

a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]

we substitute

2m/s² = √[ (- 1.15 )² + ([tex]a_{n}[/tex])²]

4 = (- 1.1566 )² + ([tex]a_{n}[/tex])²

4 = 1.3377 +  ([tex]a_{n}[/tex])²

([tex]a_{n}[/tex])² = 4 - 1.3377

([tex]a_{n}[/tex])² = 2.6623

[tex]a_{n}[/tex] = √2.6623

[tex]a_{n}[/tex]  = 1.6316 m/s²

Now the radius of curve is;

a = V² / p

[tex]p_{t=6}[/tex] = ( [tex]V_{t=6}[/tex] )² /  [tex]a_{n}[/tex]

[tex]p_{t=6}[/tex] = ( 20.83 m/s )² /  1.6316 m/s²

[tex]p_{t=6}[/tex] = 433.8889 / 1.6316

[tex]p_{t=6}[/tex] = 265.9 m ≈ 266 m

Therefore;  the radius of curvature of the track for this instant is 266 m