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mamaluj [8]
3 years ago
14

A bicycle racer rides from a starting marker to a turnaround marker at 10 m/s. She then rides back along the same route from the

turnaround marker to the starting marker at 16 m/s. What is her average speed for the whole race?
Physics
1 answer:
vitfil [10]3 years ago
8 0

Answer:

12.31 m/s

Explanation:

If we recall from the previous knowledge we had about speed,

we will know that:

speed = distance/ time.

As such:

The average speed of the rider bicycle is

average speed = total distance/ total time

Mathematically, it can be computed as:

v_{avg} = \dfrac{d+d}{\dfrac{d}{v_1}+ \dfrac{d}{v_2}}

v_{avg} = \dfrac{2d}{\dfrac{d}{10 \ m/s}+ \dfrac{d}{16 \ m/s}}

v_{avg} = \dfrac{2}{\dfrac{1}{10 \ m/s}+ \dfrac{1}{16 \ m/s}}

v_{avg} = \dfrac{2}{\dfrac{13}{80 \ m/s}}

\mathbf{v_{avg} =12.31 \ m/s}

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Suppose a firm is producing 2,475 units of output by hiring 50 workers (W = $20 per hour) and 25 units of capital (R = $10 per h
Neko [114]

Answer

given,

firm is producing  = 2,475 units

output by hiring 50 workers W = $20 per hour

25 units of capital R = $10 per hour

marginal product of labor = 40

marginal product of capital = 25

\dfrac{MP_l}{MP_c}=\dfrac{40}{25}

\dfrac{MP_l}{MP_c}=\dfrac{8}{5}

\dfrac{W}{R}=\dfrac{20}{10}

\dfrac{W}{R}=2

\dfrac{MP_l}{MP_c} < \dfrac{W}{R}

Firm is not minimizing the cost because the firm use more capital and less labor.

3 0
4 years ago
A closely wound circular coil has a radius of 6.00 cmand carries a current of 2.65 A. How many turns must it have if the magneti
ad-work [718]

Answer:

Given:

radius of the coil, R = 6 cm = 0.06 m

current in the coil, I = 2.65 A

Magnetic field at the center, B = 6.31\times 10^{4} T

Solution:

To find the number of turns, N, we use the given formula:

B = \frac{\mu_{o}NI}{2R}

Therefore,

N = \frac{2BR}{\mu_{o}I}

N = \frac{2\times 6.31\times 10^{4}\times 0.06}{4\pi \times 10^{- 7}\times 2.65}

N = 22.74 = 23 turns (approx)

 

8 0
3 years ago
Please can anybody tell me what are these lab equipments ​
olga nikolaevna [1]

Answer:

5 is the tripoid stand

Thanks have a bangtastic day

4 0
2 years ago
Read 2 more answers
At what speed does a typical electron pass by any given point in the wire?
attashe74 [19]
A 5.00 A current runs through a 12 gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5*10^28 free electrons per cubic metre.

a) How many electrons pass through the light bulb each second?

b) What is the current density in the wire? (answer in A/m^2)

<span>c) At what speed does a typical electron pass by any given point in the wire? (answer in m/s)

</span>a) 5.0 A = 5.0 C/s 
. Number of electrons in 5.0C = 5.0 / 1.60^-19 = 3.125^19
. 5.0 A ►= 3.125^19 electrons/s

b) A/m² = 5.0 / π(1.025^-3 m)² .. .. ►= 1.52^6 A/m²

c) Charge density (q/m³) = 8.50^28 e/m³ x 1.60^-19 = 1.36^10 C/m³

(q/m³)(m²)(m/s) = q/s (current i in C/s [A])
(m²) = Area
(m/s) = mean drift speed

(q/m³)(A)(v) = i

v = i.[(q/m³)A]ˉ¹

<span>v = 5.0 [1.36^10 * π(1.025^-3 m)²]ˉ¹.. .. ►v = 1.10^-4 m/s</span>
7 0
3 years ago
Next weekend you are going to Eleanor Rigby’s house to fix her grandfather clock that is a bit slow. Based on what you have lear
Bumek [7]

Answer:

By pushing the pendulum Bob up so it moves faster

Explanation:

In pendulum physics the length of the pendulum Bob determines the speed of the clock. So since the grandfather's clock is slow it means the Bob is has moved down so to move it up you have to achieve this by adjusting the but upwards thereby making the clock move faster.

8 0
4 years ago
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