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3 years ago

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Two dogs pull on a toy. One pulls with a force of 10N, the other with a force of 15N. In what relative directions can the dogs a

ct to give the toy the smallest acceleration?

1 answer:

skad [1K]3 years ago

5 0

They must act one opposite of the other, so the resultant hits the minimum of 5 N. When they go in the same direction you get the maximum of 15 N.

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A block slides down a plane inclinded at an angle of 37° to the vertical with an acceleration of 6m/s each second. what is the c

Irina18 [472]

Answer:

0.315

Explanation:

resolving weight into components and applying second law of motion we obtain the result

All has been explained in attachment

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3 years ago

Consider the following situation, in which a compass is influenced by not only the force of Earth's magnetic field, but also an

DochEvi [55]

Answer:

1) The angle of deflection will be less than 45° ( C )

2) The angle of deflection will be greater than 45° but less than 90° ( E )

Explanation:

1) Assuming that the force applied has a direction which is perpendicular to the Earth's magnetic field

∴ Fearth > Fapplied hence the angle of deflection will be < 45°

2) when the Fearth < Fapplied

the angle of deflection will be : > 45° but < 90°

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3 years ago

A car runs at a constant speed of 15ms-1 for 30secs, and then accelerates uniformly to a speed of 25ms-1 over a time of 20secs.

Otrada [13]

Answer:

a. Please find the attached velocity time graph of the car's motion created with Microsoft Excel

b. The total distance traveled by the car is 8,725 meters

c. The average speed of the car is 22.9605263 m/s

Explanation:

The given parameters of the motion are;

The initial speed of the car, v₁ = 15 m/s

The time during which the car runs at the initial speed, t₁ = 30 seconds

The new speed the car then accelerates at 'a₁' to, v₂ = 25 m/s

The duration it takes for the car to accelerate to the new speed = 20 seconds

The time during which the car runs at the initial speed = 300 seconds

The time it takes the car to be brought to rest with a deceleration, 'a₂' from the new speed (20 m/s) = 30 seconds

The final speed of the car at rest, v₃ = 0 m/s

The acceleration, a₁ = (v₂ - v₁)/t₁ = (25 - 15)/20 = 1/2 m/s²

The deceleration , a₂ = (v₃ - v₂)/t₁ = (0 - 25)/30 = -5/6 m/s²

a. Please find attached the drawing of the velocity time graph of the motion created with Microsoft Excel

b. The total distance traveled by the car, 'Δx', is given b the area under the velocity time graph as follows;

Area of trapezoid, A₁ = (320 + 300)/2 × 10 = 3,100

Area of rectangle, A₂ = 15 × 350 = 5,250

Area of triangle, A₃ = 1/2×30×25 = 375

The total area under the velocity time graph = A₁ + A₂ + A₃ = 3,100 + 5,250 + 375 = 8,725

The total area under the velocity time graph = The total distance traveled by the car, Δx = 8,725 meters

c. The average speed of the car is given as follows;

$The \ average \ speed \ of \ the \ car, \overline v =\dfrac{\Delta x}{\Delta t} = \dfrac{The \ total \ distance \ traveled by \ the \ car}{The \ total \ time \ the \ car \ travels}$

Where;

Δt = The total time during which the car travels

∴ The average speed of the car = 8,725 m/(380 s) = 22.9605263 m/s

4 0

3 years ago

To understand the decibel scale. The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the d

frez [133]

The question is incomplete. Here is the complete question.

To understand the decibel scale. The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel scale is logarithmic, it changes by an additive constant when the intensity when the intensity as measured in W/m² changes by a multiplicative factor. The number of decibels increase by 10 for a factor of 10 increase in intensity. The general formula for the sound intensity level, in decibels, corresponding to intensity I is

$\beta=10log(\frac{I}{I_{0}} )dB$ ,

where $I_{0}$ is a reference intensity. for sound waves, $I_{0}$ is taken to be $10^{-12} W/m^{2}$ . Note that log refers to the logarithm to the base 10.

Part A: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 10 times the reference intensity, i.e. $I=10I_{0}$ ? Express the sound intensity numerically to the nearest integer.

Part B: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 100 times the reference intensity, i.e. $I=100I_{0}$ ? Express the sound intensity numerically to the nearest integer.

Part C: Calculate the change in decibels ( $\Delta \beta_{2},\Delta \beta_{4}$ and $\Delta \beta_{8}$ ) corresponding to f = 2, f = 4 and f = 8. Give your answer, separated by commas, to the nearest integer -- this will give an accuracy of 20%, which is good enough for sound.

Answer and Explanation: Using the formula for sound intensity level:

A) $I=10I_{0}$

$\beta=10log(\frac{10I_{0}}{I_{0}} )$

$\beta=10log(10 )$

β = 10

<u>The sound Intensity level with intensity 10x is </u><u>10dB</u>.

B) $I=100I_{0}$

$\beta=10log(\frac{100I_{0}}{I_{0}} )$

$\beta=10log(100)$

β = 20

<u>With intensity 100x, level is </u><u>20dB</u>.

C) To calculate the change, take the f to be the factor of increase:

For $\Delta \beta_{2}$ :

$I=2I_{0}$

$\beta=10log(\frac{2I_{0}}{I_{0}} )$

$\beta=10log(2)$

β = 3

For $\Delta \beta_{4}$ :

$I=4I_{0}$

$\beta=10log(\frac{4I_{0}}{I_{0}} )$

$\beta=10log(4)$

β = 6

For $\Delta \beta_{8}$ :

$I=8I_{0}$

$\beta=10log(\frac{8I_{0}}{I_{0}} )$

β = 9

Change is

$\Delta \beta_{2},\Delta \beta_{4}$ , $\Delta \beta_{8}$ = 3,6,9 dB

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