1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
umka21 [38]
3 years ago
12

A person walks first at a constant speed of 5.10 m/s along a straight line from point circled a to point circled b and then back

along the line from circled b to circled a at a constant speed of 2.75 m/s. (a) what is her average speed over the entire trip? 3.57 correct: your answer is correct. m/s
Physics
1 answer:
Alika [10]3 years ago
4 0
Average speed = total distance / time

1) Distance from A to B

Distance = velocity * time  = V1 * t1 = 5.10 t1

2) Distance from B to A

Distance = velocity * time  = V2*t 2 = 2.75 t2

Distance from A to B = distance from B to A => 5.10t1 = 2.75t2

=> t2 = 5.10t1 / 2.75

3) Average speed = total distance / total time =

total distance = 2 * 5.10 t1 = 10.20 t1

Average speed = [10.20 t1] / [t1 + 5.10 t1 / 2.75]

As you see t1 is factor for the three terms, so you can simplify it

=> Average speed = [10.20 ] / [1 + 5.10/2.75] = 3.57 m/s

Answer: 3.57 m/s

 


You might be interested in
How to use allele In a sentence
disa [49]
<span>Here is an example, the allele for blue eyes and the allele for brown eyes are different versions of the gene for eye color. Alleles are located at the same genetic locus . </span>
6 0
3 years ago
An hydrogen molecule consists of two hydrogen atoms whose total mass is 3.3×10−27 kg and whose moment of inertia about an axis p
dlinn [17]

Answer:

6.9631\times 10^{-11}\ m

Explanation:

I = Moment of inertia = 4\times 10^{-48}\ kg m^2

m = Mass of two atoms = 2m = 3.3\times 10^{-27}\ kg

r  = distance between axis and rotation mass

Moment of inertia of the system is given by

I=mr^2\\\Rightarrow I=2mr^2\\\Rightarrow 4\times 10^{-48}=3.3\times 10^{-27}\times r^2\\\Rightarrow r=\sqrt{\frac{4\times 10^{-48}}{3.3\times 10^{-27}}}\\\Rightarrow r=3.48155\times 10^{-11}\ m

The distance between the atoms will be two times the distance between axis and rotation mass.

d=2r\\\Rightarrow d=2\times 3.48155\times 10^{-11}\\\Rightarrow d=6.9631\times 10^{-11}\ m

Therefore, the distance between the two atoms is 6.9631\times 10^{-11}\ m

3 0
3 years ago
How to solve it? Three capacitors with capacities of 600 pF, 300 pF, 200 pF are connected in series. The 60 V voltage is applied
adell [148]

Answer:

1. Voltage across 600 pF is 10 V.

2. Voltage across 300 pF is 20 V.

3. Voltage across 200 pF is 30 V.

Explanation:

We'll begin by calculating the total capacitance of capacitor. This can be obtained as follow:

Capicitance 1 (C₁) = 600 pF

Capicitance 2 (C₂) = 300 pF

Capicitance 3 (C₃) = 200 pF

Total capacitance (Cₜ) =?

1/Cₜ = 1/C₁ + 1/C₂ + 1/C₃

1/Cₜ = 1/600 + 1/300 + 1/200

1/Cₜ = 1 + 2 + 3 / 600

1/Cₜ = 6/600

1/Cₜ = 1/100

Cₜ = 100 pF

Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:

1 pF = 1×10¯¹² F

Therefore,

100 pF = 100 pF × 1×10¯¹² F / 1 pF

100 pF = 1×10¯¹⁰ F

Thus, 100 pF is equivalent to 1×10¯¹⁰ F.

Next, we shall determine the charge. This can be obtained as follow:

Voltage (V) = 60 V

Capicitance (C) = 1×10¯¹⁰ F

Charge (Q) =?

Q = CV

Q = 60 × 1×10¯¹⁰ F

Q = 6×10¯⁹ C

1. Determination of the voltage across 600 pF.

Capicitance 1 (C₁) = 600 pF = 6×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 1 (V₁) =?

Q = C₁V₁

6×10¯⁹ = 6×10¯¹⁰ × V₁

Divide both side by 6×10¯¹⁰

V₁ = 6×10¯⁹ / 6×10¯¹⁰

V₁ = 10 V

2. Determination of the voltage across 300 pF.

Capicitance 2 (C₂) = 300 pF = 3×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 2 (V₂) =?

Q = C₂V₂

6×10¯⁹ = 3×10¯¹⁰ × V₂

Divide both side by 3×10¯¹⁰

V₂ = 6×10¯⁹ / 3×10¯¹⁰

V₂ = 20 V

3. Determination of the voltage across 200 pF.

Capicitance 3 (C₃) = 200 pF = 2×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 3 (V₃) =?

Q = C₃V₃

6×10¯⁹ = 2×10¯¹⁰ × V₃

Divide both side by 2×10¯¹⁰

V₃ = 6×10¯⁹ / 2×10¯¹⁰

V₃ = 30 V

7 0
2 years ago
A boat leaves the dock at t = 0.00 s and, starting from rest, maintains a constant acceleration of (0.461 m/s2)i relative to the
liberstina [14]

Answer:

At t=4.82 s, the boat is moving at 3.464 m/s.

At t=4.82 s, the boat is 13.112 m from the dock.

Explanation:

The speed of the boat in j'th direction remains constant for all times (vj=2.16 m/s), however, the speed in i'th direction is changing due to the constant acceleration (0.461 m/s^2)i.

In order to find the velocity of the boat a t=4.82 s, first we need to compute the velocity of the boat relative to the water in the i direction (vi_b) at t=4.82 s:

vi_b = a*t = (0.461 m/s^2)*(4.82 s) = 2.222 m/s

Now, we add this velocity to the velocity of the water in the i direction:

vi = vi_b + vi_w = 2.222 m/s + 0.486 m/s = 2.708 m/s

Therefore, the speed of the boat at t = 4.82 s is: v = (vi, vj) = (2.708, 2.16) m/s. Finally, to find its speed, we just calculate the magnitude of v and obtain that the speed is: 3.464 m/s.

For the second question, first we will find the distance that the boat moved in the i'th direction and then in the j'th direction.

The speed in the i'th direction, for all times, is given by:

(0.485 + 0.461*t) and in order to find the distance advanced in the i'th direction (di) during 4.82 s, we need to integrate this velocity:

di = 0.485*t + (0.461*t^2)/2 (evaluated from t=0 to t =4.82) = 0.485*(4.82) + (0.461*(4.82)^2)/2 = 2.337 + 5.634 = 7.971 m

The speed in j'th direction, for all times, is given by:

2.16 and in order to find the distance advanced in the j'th direction (dj) during 4.82 s, we need to integrate this velocity:

dj = 2.16*t (evaluated from t=0 to t =4.82) = (2.16)*(4.82) = 10.411 m

Using Pythagoras' Theorem, we find that the the boat is at 13.112 m from the dock at t = 4.82 s.

4 0
2 years ago
The a992 steel rod bc has a diameter of 50 mm and is used as a strut to support the beam. determine the maximum intensity w of t
EastWind [94]

Answer:

w = 11.211 KN/m

Explanation:

Given:

diameter, d = 50 mm

F.S = 2

L = 3

Due to symmetry, we have:

Ay = By = \frac{w * 6}{2} = 3w

P_c_r = 3w * F.S = 3w * 2.0 = 6w

I = \frac{\pi}{64} (0.05)^4 = 3.067*10^-^7

To find the maximum intensity, w, let's take the Pcr formula, we have:

P_c_r = \frac{\pi^2 E I}{(KL)^2}

Let's take k = 1

E = 200*10^9

Substituting figures, we have:

6w = \frac{\pi^2 * 200*10^9 * 3.067*10^-^7}{(1 * 3)^2}

Solving for w, we have:

w = \frac{67266.84}{6}

w = 11211.14 N/m = 11.211 KN/m

Since Area, A= pi * (0.05)²

\sigma _c_r = \frac{w}{A}

\sigma _c_r = \frac{11.211}{\pi (0.05)^2} = 1.4 MPA < \sigma y. This means it is safe

The maximum intensity w = 11.211KN/m

3 0
3 years ago
Other questions:
  • Which activity best reflects a muscle movement working against gravity? a lifting the torso during a sit up b lowering the body
    13·2 answers
  • Please Please Please Can Help Me On This Question!!!!! I Give Thanks!!!!
    5·1 answer
  • Coulomb's law predicts that charged bodies may attract or repel each other. true or false
    5·1 answer
  • A police car is at rest parallel to the highway and measures the speed of cars. It sends the signal with a frequency of 1200 Hz,
    15·1 answer
  • Which statement describes how a covalent molecule would be formed? A) an atom of an element with atomic number 12 reacts with on
    15·2 answers
  • What is the frequency of radio waves with wavelength of 20m?
    15·2 answers
  • Which of the following best defines the science of separating color wavelengths?
    5·2 answers
  • In a certain ideal heat engine, 10.00 kJ of heat is withdrawn from the hot source at 273 K and 3.00 kJ of work is generated. Wha
    6·2 answers
  • A car driving at 20 m/s accelerates continuously 2m/s2​ ​ for 3 seconds. What is its final velocity
    11·1 answer
  • The current through two identical light bulbs connected in series is 0.25 A. The voltage across both bulbs is 110 V. The resista
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!