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gtnhenbr [62]
3 years ago
6

A car traveling at 90 m/s can stop in a distance of 110 m. What is the magnitude of the cars acceleration as it slows down?

Physics
1 answer:
LenKa [72]3 years ago
8 0

Please find attached photograph for your answer. Hope it helps! Please do comment

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Vesnalui [34]

more deceleration.

in vertical motion downwards => terminal velocity ... raindrops etc

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To measure one wavelength, you would measure the distance between?
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A and B, one wavelength is crest too crest


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What is the expected wavelength (in cm) of 10.5 ghz microwaves in free space?
Sindrei [870]

Wavelength = (speed) / (frequency)

Wavelength = (300 thousand km per second) / (10.5 billion per second)

Wavelength = (300 / 10.5) (thousand km per second) / (billion per second)

Wavelength = (28.57) (million meters / second) / (thousand million / second)

Wavelength = (28.57) (meters / second) / (thousand / second)

Wavelength = (28.57) (meters / thousand)

<em>Wavelength = (28.57) (millimeters) </em>

5 0
3 years ago
1. A small package is dropped from the Golden Gate Bridge. What is the velocity of the package
Sidana [21]

The velocity of the package  after it has fallen for 3.0 s is 29.4 m/s

From the question,

A small package is dropped from the Golden Gate Bridge.

This means the initial velocity of the package is 0 m/s.

We are to calculate the velocity of the package  after it has fallen for 3.0 s.

From one of the equations of kinematics for objects falling freely,

We have that,

v = u + gt

Where

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

and t is time

To calculate the velocity of the package  after it has fallen for 3.0 s

That means, we will determine the value of v, at time t = 3.0 s

The parameters are

u = 0 m/s

g = 9.8 m/s²

t = 3.0 s

Putting these values into the equation

v = u + gt

We get

v = 0 + (9.8×3.0)

v = 0 + 29.4

v = 29.4 m/s

Hence, the velocity of the package  after it has fallen for 3.0 s is 29.4 m/s

Learn more here: brainly.com/question/13327816

6 0
2 years ago
What is the strength of the electric field 3.0 cm from a small glass bead that has been charged to + 8.0 nc ?
Debora [2.8K]
Thinking the small glass bead as a single point charge, the electric field generated by it is given by
E(r) = k_e  \frac{Q}{r^2}
where
k_e = 8.99 \cdot 10^9 N m^2 C^{-2} is the Coulomb's constant
Q=8.0 nC=8.0 \cdot 10^{-9} C is the charge of the bead
r=3.0 cm=0.03 m is the distance at which we calculate the field.

Using these data, we find:
E=(8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{(8.0 \cdot 10^{-9} C}{0.03 m)^2}=8.0 \cdot 10^4 N/C
6 0
3 years ago
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