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Salsk061 [2.6K]
2 years ago
13

A student stays at her initial position for a bit of time, then walks slowly in a straight line for a while, then stops to rest

a while, and finally runs quickly back to her initial position along a straight line. Which of the following statements is true about the average speed and the magnitude of the average velocity of the student during her trip? Her average speed is greater than the magnitude of her average velocity. Her average speed is the same as the magnitude of her average velocity. Her average speed is less than the magnitude of her average velocity. The student's average speed and velocity cannot be compared without knowing the farthest point she reached.
Physics
1 answer:
IRINA_888 [86]2 years ago
5 0

Answer:Average speed is greater than average velocity.

Explanation  :

Given

student walks slowly along a straight line for a while ,then stops to rest a while, and finally runs quickly back to her initial position

Let x be the length of track and the whole process takes t time

For average speed =\frac{distance\ traveled}{time\ taken}

Average speed=\frac{2x}{t}

For average velocity =\frac{Displacement}{time\ taken}

Since displacement is zero as she returns to its initial position.

Average velocity=0

Therefore Average speed is greater than average velocity.

You might be interested in
would an elephant standing on one leg exert a higher force on a scale than an elephant on four legs. why​
zlopas [31]

Answer:

no becaus force is mass multiplied by acceleration. the mass of the elephant does not change

7 0
2 years ago
: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at
ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

The  expression is   F_{No} =   A [P_{No} - \frac{P_{sea}}{2}]

b

F_{No}= 7771.125 \ N

c

 F_p = 2.2*10^{6} N

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

          The altitude of container in Denver is  d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m

           The surface area of the container lid is A = 0.0155m^2

           The altitude of container in New Orleans  is sea-level

           The air pressure in Denver is  P_D = 79000 \ Pa

            The air pressure in new Orleans is P_{ro} = 100250 \ Pa

Generally force is mathematically represented as

            F_{No} = \Delta P A

  So we are told the pressure inside is  is half the pressure the at sea level so the  the pressure acting on the container would

   The  pressure at sea level is a constant with a  value of  

               P_{sea} = 101000 Pa

So the \Delta P which is the difference in pressure within and outside the container is  

           \Delta P = P_{No} - \frac{P_{sea}}{2}

Therefore

                F_{No} =   A [P_{No} - \frac{P_{sea}}{2}]

Now substituting values

                F_{No} =   0.0155 [100250 - \frac{101000}{2}]

                       F_{No}= 7771.125 \ N

The force to remove the lid in Denver is  

           F_p = \Delta P_d A

So we are told the pressure inside is  is half the pressure the at sea level so the  the pressure acting on the container would

 The  pressure at sea level is a constant with a  value of  

               P_{sea} = 101000 Pa    

 At  sea level the air pressure in Denver is mathematically represented as

              P_D = \rho g h

     =>     g = \frac{P_D}{\rho h}      

Let height at sea level is h = 1

  The air pressure at height d_D

             P_d__{D}} = \rho gd_D

    =>     g = \frac{P_d_D}{\rho d_D}

  Equating the both

                 \frac{P_D}{\rho h}  = \frac{P_d_D}{\rho d_D}

                 P_d_D =  P_D * d_D

Substituting value  

                   P_d__{D}} = 1828.2 * 79000

                    P_d__{D}} = 1.445*10^{8} Pa

    So

              \Delta P_d  = P_{d} _D - \frac{P_{sea}}{2}

=>          \Delta P_d  = 1.445 *10^{8} - \frac{101000}{2}    

                        \Delta P_d = 1.44*10^{8}Pa

  So

               F_p = \Delta P_d A

                  = 1.44*10^8 * 0.0155

              F_p = 2.2*10^{6} N

               

                 

             

             

6 0
2 years ago
Ben starts walking along a path at 3 3 mi/h. One and a half hours after Ben leaves, his sister Amanda begins jogging along the s
Drupady [299]

Answer:

3 hours

Explanation:

Given:

- The speed of Ben v_b = 3 mi/h

- The speed of Amanda v_a = 6 mi/h

- The total time taken to cover distance(d) by ben = t_b

Find:

How long will it be before Amanda catches up to Ben?

Solution:

- The distance d traveled by Ben:

                                 d = v_b*t_b

                                 d = 3*t_b

- The distance d traveled by Amanda:

                                 d = v_a*t_a

                                 d = 6*t_a

- Equate the distance as when they meet:

                                 3*t_b = 6*t_a

- Where ,

                                  t_b = t_a + 1.5

                                  t_a = t_b - 1.5

- Substitute the time relationship in distance relationship:

                                  3*t_b = 6*(t_b - 1.5)

                                  3*t_b = 6*1.5

                                      t_b = 2*1.5 = 3 h

- Hence, It would take 3 hours since Ben starts walking that amanda catches up.

4 0
2 years ago
A thin layer of oil with index of refraction no = 1.47 is floating above the water. The index of refraction of water is nw = 1.3
garik1379 [7]

Answer:

A thin layer of oil with index of refraction ng = 1.47 is floating above the water. The index of refraction of water is nw = 1.3. The index of refraction of air is na= 1. A light with wavelength λ = 775 nm goes in from the air to oil and water.

 Part (a) Express the wavelength of the light in the oil,  λ₀, in terms of  λ  and n⁰ (b) Express the minimum thickness of the film that will result in destructive interference, t min, in terms of  λ o

(c) Express tmin in terms of  λ  and no.

(d) Solve for the numerical value of tmin in nm.

Explanation:

n₀ = 1.47

refraction of water = 1.3

refraction of air = 1

wavelength λ = 775 nm

(a) wavelength of light in water ⇒  λ₀ = λ / n₀

(b) minimum thickness of the film that will result in destructive interference

t(min) = λ₀ / 2

(c) the express t(min)

t =  λ /2n₀

(d) the thickness is

t = 775 / 2(1.47)

= 263.61 nm

4 0
1 year ago
Mr. and Mrs. Tew attended this summer band concert to hear their son play his violin. They sat in the back row. They noticed tha
Hitman42 [59]

Option (a) is correct.

Change in volume during the band concert might have been caused by the constructive and destructive interference of sound waves.

Interference is the process of redistribution of energy when two or more waves superimpose on each other.When two sound waves which are in phase superimpose on each other, constructive interference takes place. During constructive interference , the amplitude of resulting waves increases.Thus the loudness of sound increases.

When two sound waves which are out of phase superimpose on each other, destructive interference takes place. During destructive interference , the amplitude of resulting waves decreases.Thus the loudness of sound decreases.

3 0
1 year ago
Read 2 more answers
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