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3 years ago

6. A car travels a distance of 10 miles in 2 minutes towards 695 north east. 3 points

Physics

1 answer:

damaskus [11]3 years ago

5 0

Answer:

C. 5 miles/ minute north east

Explanation:

magnitude = 10 mi / 2 min = 5 mi/min

Velocity is a vector

A vector always has both a magnitude and direction

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How to tell if a circuit is connected in.a series.or parallel?

stepladder [879]

Series circuits split the voltage of resistors, so if you see several diodes connected <em>in series </em>or all next to each other, just a complete loop, it will be in series.

Parallel circuits split the current of resistors, so if you see several diodes connected along different branches or pathways, it will be in parallel.

7 0

3 years ago

Suppose Earth orbited a star whose mass was double the mass of the sun. If the radius of Earth’s orbit remained the same as it i

Vaselesa [24]

Answer:

Two times as much

Explanation:

The equation for gravitational force is: Fg = GMm/r^2 with G being the universal gravitational constant.

So to make things easier we'll set r equal to 1 since it's a constant as well as G.

Then we're left with Fg=Mm with M being the mass of the sun and m being the mass of the earth.

So if m is constant and supposedly equals 1 then Fg=M so Fg is proportional to M therefore if M doubles then Fg doubles.

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2 years ago

A Red-Tailed Hawk leaves its nest and flies 100.0 meters across the sky in 20 seconds, when it spots a garter snake on the groun

igomit [66]

Speed = distance / time.

Speed of him leaving the nest:
S = 100 / 20sec

5 m/s.

Catching the snake:
S2 = 50 / 5sec

10 m/s.

Average of 5& 10 = 7.5

7.5 m/s has to be the answer.

3 0

2 years ago

· A car moves at 12 m/s and coasts up a hill with a

Zigmanuir [339]

Explanation:

Given that,

Initial speed of a car, u = 12 m/s

Aceleration, a = -1.6 m/s²

(a) The displacement of car after 6 seconds is :

$d=ut+\dfrac{1}{2}at^2\\\\d=12\times 6+\dfrac{1}{2}\times (-1.6)\times 6^2\\\\d=43.2\ m$

(b) The displacement of the car after 9 seconds is :

$d=ut+\dfrac{1}{2}at^2\\\\d=12\times 9+\dfrac{1}{2}\times (-1.6)\times 9^2\\\\d=43.2\ m$

Hence, this is the required solution.

7 0

3 years ago

The mass of a star is 1.210×1031 kg and it performs one rotation in 20.30 days. Find its new period (in days) if the diameter su

balandron [24]

The new period will be 2.486 days.

<h3>What is the period?</h3>

The period is found as the ratio of the angular displacement and the angular velocity. Its unit is the second and is denoted by t. The value of time needed to complete the rotation is the total period.

Given data;

Mass of a star,m= 1.210×10³¹ kg

The time period for one rotation of the star, T = 20.30 days

D' = 0.350 D

R' = 0.350 R

From the law of conservation of angular momentum;

$\rm I \omega = I' \omega' \\\\ \frac{2}{5} MR^2 \times \frac{2 \pi }{T}=\frac{2}{5} MR'^2 \\\\ R^2 \times \frac{1}{T}= R'^2 \times \frac{1}{T} \\\\ T' = \frac{R'^2T}{R^2} \\\\ T' = \frac{(0.350 R)^2 \times 26.1 }{R^2} \\\\T' = 0.1225 \times 20.30 \\\\ T'= 2.486 \ days$

Hence, the new period will be 2.486 days.

To learn more about the period, refer to the link;

brainly.com/question/569003

#SPJ1

8 0

1 year ago