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san4es73 [151]
3 years ago
13

At exactly 3:14PM, the velocity of a dog running in a park points toward a group of flowers. Which of the following best describ

e the forces acting on the dog at 3:14PM?
There may be many forces acting on the dog, but the forces pointing toward the flowers are larger than the forces pointing away from the flowers.
There may be many forces acting on the dog, but the forces pointing toward the flowers are smaller than the forces pointing away from the flowers.
There may be many forces acting on the dog, but the forces pointing toward the flowers are equal to the forces pointing away from the flowers.
Options A and B are both possible.
Options A and C are both possible.
Options A, B, and C are all possible.
Physics
1 answer:
dmitriy555 [2]3 years ago
4 0

Answer:

Options A, B, and C are all possible.

Explanation:

We know that the instantaneous velocity of the dog at 3:14PM is possitive to toward the flowers. But what about the acceleration to toward the flowers?

If the dog is decreasing speed at 3:14PM, it means that acceleration is negative toward the flowers, hence (since F=ma) the net force points away from the flowers.

If the dog is increasing speed at 3:14PM, it means that acceleration is positive toward the flowers, hence (since F=ma) the net force points toward the flowers.

If the dog is not increasing nor decreasing speed at 3:14PM, it means that acceleration is 0, hence (since F=ma) the net force is null and it does not point neighter to toward the flowers  nor away from the flowers. This happens when the forces acting on the dog are equal to both sides.

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Ksenya-84 [330]

Answer:

Part a) \frac{5}{9}\ \frac{rev}{sec}

Part b) \frac{9}{5}\ \frac{sec}{rev}

Explanation:

Part a) what is its frequency, in rev/s

we have that

An old-fashioned LP record rotates at 33 1/3 RPM

so

33\frac{1}{3}\ \frac{rev}{min}

Convert mixed number to an improper fraction

33\frac{1}{3}\ \frac{rev}{min}=\frac{33*3+1}{3}=\frac{100}{3}\ \frac{rev}{min}

Remember that

1\ min=60\ sec

Convert rev/min to rev/sec

\frac{100}{3}\ \frac{rev}{min}=\frac{100}{3}(\frac{1}{60})=\frac{100}{180}\ \frac{rev}{sec}

Simplify

\frac{5}{9}\ \frac{rev}{sec}

Part b) what is it period, in seconds

we know that

The period is the reciprocal of the frequency

therefore

the frequency is

\frac{9}{5}\ \frac{sec}{rev}

4 0
3 years ago
If vx=9.80 units and vy=-6.40 units, determine the magnitude and direction of v
dexar [7]
The resultant vector can be determined by the component vectors. The component vectors are vector lying along the x and y-axes. The equation for the resultant vector, v is:

v = √(vx² + vy²)
v = √[(9.80)² + (-6.40)²]
v = √137 or 11.7 units
5 0
3 years ago
What is the intensity of 60dB sound?​
polet [3.4K]

Answer:

The intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².

Explanation:

Given;

intensity of the sound level, dB = 60 dB

The intensity of the sound in W/m² is calculated as;

dB = 10 Log[\frac{I}{I_o} ]\\\\

where;

I₀ is threshold of hearing = 1 x 10⁻¹² W/m²

I is intensity of the sound in W/m²

Substitute the given values and for I;

dB = 10 Log[\frac{I}{I_o} ]\\\\60 = 10 Log[\frac{I}{I_o} ]\\\\6 =  Log[\frac{I}{I_o} ]\\\\10^6 = \frac{I}{I_o} \\\\I = 10^6 \ \times \ I_o\\\\I = 10^6 \ \times \ 1^{-12} \ W/m^2 \\\\I = 1\ \times \ 10^{-6} \ W/m^2

Therefore, the intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².

5 0
3 years ago
A compass taken to Earth’s moon does not point in a specific direction on the moon. What is the most likely reason for the diffe
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Most likely, C. the Moon does not have a liquid core (this is what would create a magnetic field) is correct.

I hope this enough to help you!
8 0
3 years ago
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Anastaziya [24]

M = molar mass of the helium gas = 4.0 g/mol

m = mass of the gas given = 18.0 g

n = number of moles of the gas

number of moles of the gas is given as

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V = Volume of balloon = ?

T = temperature = 297 K

R = universal gas constant = 8.314

Using the ideal gas equation

P V = n R T

(202650) V = (4.5) (8.314) (297)

V = 0.055 m³

4 0
3 years ago
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