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d1i1m1o1n [39]
3 years ago
6

Gina is driving her car to work, but she’s stopped at a red light. When the light turns green, she presses the gas pedal and acc

elerates (speeds up) to move the car. Which of these statements is true the moment Gina presses the gas pedal? Assume the road is flat and level.
-The net force on the car is zero in the horizontal direction.
-The net force on the car is zero in the horizontal and the vertical directions.
-The net force on the car is greater than zero in the horizontal direction.
-The net force on the car is greater than zero in the vertical direction.
-The net force on the car is less than zero in the vertical direction.
Physics
2 answers:
rusak2 [61]3 years ago
8 0
The net force of the car is greater than zero in the horizontal direction. If it were not greater thn zero, then the vehicle would remain stationary.
BaLLatris [955]3 years ago
7 0

Answer:

The net force on the car is greater than zero in the horizontal direction.

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A particle with mass 1.09 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.985
e-lub [12.9K]

Answer:

a) f = 0.598\,hz, b) v_{max} = 3.701\,\frac{m}{s}, c) k = 15.385\,\frac{N}{m}, d) U = 1.081\,J, e) K = 6.382\,J, f) v\approx 3.422\,\frac{m}{s}

Explanation:

a) The frequency of oscillation is:

f = \frac{76}{127\,hz}

f = 0.598\,hz

b) The angular frequency is:

\omega = 2\pi \cdot f

\omega = 2\pi \cdot (0.598\,hz)

\omega = 3.757\,\frac{rad}{s}

Lastly, the speed at the equilibrium position is:

v_{max} = \omega \cdot A

v_{max} = (3.757\,\frac{rad}{s} )\cdot (0.985\,m)

v_{max} = 3.701\,\frac{m}{s}

c) The spring constant is:

\omega = \sqrt{\frac{k}{m}}

k = \omega^{2}\cdot m

k = (3.757\,\frac{rad}{s} )^{2}\cdot (1.09\,kg)

k = 15.385\,\frac{N}{m}

d) The potential energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

U = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.375\,m)^{2}

U = 1.081\,J

e) The maximum potential energy is:

U_{max} = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.985\,m)^{2}

U_{max} = 7.463\,J

The kinetic energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

K = U_{max} - U

K = 7.463\,J - 1.081\,J

K = 6.382\,J

f) The speed when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

K = \frac{1}{2}\cdot m \cdot v^{2}

v = \sqrt{\frac{2\cdot K}{m} }

v = \sqrt{\frac{2\cdot (6.382\,J)}{1.09\,kg} }

v\approx 3.422\,\frac{m}{s}

4 0
3 years ago
Derive a relation between time period anf frequency of a wave
Elza [17]

Answer:

The relation between frequency and time period is given by:

f = 1/T

Explanation:

In a wave motion, the particle move about the mean position with the passage of time. The particles rise to reach the highest point which is crest, and similarly falls to reach the lowest point which is trough. The cycle keeps on repeating.

The time period of the wave can be defined as the time taken to complete one such cycle. Time period is given by:

T = 2π/ω

Frequency can be defined as the number of cycles completed in unit time, which can be taken as the inverse of time period. frequency is given by

f = ω/2π

or

f = 1/T

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The magnitude of a component of a vector must be
den301095 [7]
Less than or equal to the magnitude of the vector
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Which best esplains how the body maintains homeostasis.
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a. all systems work together to stabilize the body

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Convert 5g/cm^3 into kg/m^3​
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Answer:

0.01135624

Explanation:

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