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3 years ago

A car drives 16 miles south and then 12 miles west. What is the magnitude of the car’s displacement?

Physics

1 answer:

Lorico [155]3 years ago

7 0

20miles (16^2+12^2)^1/2=20

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A frictionless toy car is placed on a ramp, which is inclined at an unknown angle with respect to the horizontal. Starting from

NikAS [45]

The final speed of the toy car at the end of the given time period is 3.58 m/s.

The given parameters;

distance traveled by the car, s = 1.2 m
time of motion of the car, t = 0.67 s
initial velocity of the car, u = 0

The acceleration of the car is calculated as;

$s = ut + \frac{1}{2} at^2\\\\1.2 = 0 + 0.5\times a\times (0.67)^2\\\\1.2 = 0.225a\\\\a = \frac{1.2}{0.225} \\\\a = 5.33 \ m/s^2$

The final velocity of the toy car is calculated as;

$v_f^2 = u^2 + 2as\\\\v_f^2 = 0 + 2\times 5.33 \times 1.2\\\\v_f^2 = 12.792\\\\v_f = \sqrt{12.792} \\\\v_f = 3.58 \ m/s$

Thus, the final speed of the toy car at the end of the given time period is 3.58 m/s.

Learn more here: brainly.com/question/20352766

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2 years ago

Some people will realize that they are dreaming while the dream is happening this is called

algol [13]

Answer:

C. Lucid Dreaming

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3 years ago

PLS HELP WILL MARK BRAINLIEST

MariettaO [177]

Answer:

2.835 Watts

Explanation:

P = I²R

P = 1.5² × 1.26

P = 2.835 Watts

3 0

3 years ago

Read 2 more answers

The change in pitch of a train's horn as it passes while you are standing still can be explained by

alexgriva [62]

The change in pitch of a train's horn as it passes while you are
standing still can be described by the Doppler effect, but that
doesn't explain it.

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3 years ago

In 1996, astronomers discovered an icy object beyond pluto that was given the designation 1996 tl 66. it has a semimajor axis of

antoniya [11.8K]

Answer : 2446 years.

Explanation :

Length of semi major axis is, $a=84\ au= 1.496\times 10^{11}\ m$

According to Kepler's third law, square of time period of an orbit is directly proportional to the cube of the semi major axis.

i.e $T^2=\dfrac{4\pi^2}{GM}a^3$

where G is gravitational constant

M is mass of sun, $M=1.98\times 10^{30}\ Kg$

So, $T^2=\dfrac{4\times (3.14)^2}{6.6\times 10^{-11}Nm^2/Kg\times 1.98\times 10^{30}\Kg}$

$T^2=3\times 10^{-19}\times(84\times 1.496\times 10^{11})^3$

$T^2=3\times 10^{-19}\times 1984415.6\times 10^{33}$

$T^2=59532469.8\times 10^{14}\ s$

$T=7715.7\times 10^7\ s$

since, $1\ sec=3.17\times 10^{-8}\ years$

So, orbital period is approximately 2446 years.

7 0

3 years ago