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mash [69]
3 years ago
9

A car is traveling 20 m/s and slows down at a uniform rate. It stops in 6 seconds. How far has it traveled in this interval?

Physics
1 answer:
Scilla [17]3 years ago
6 0

Answer:

Car travel a distance of 60.06 m in 6 sec

Explanation:

We have given initial velocity v = 20 m/sec

Time = 6 sec

As the car stops finally so final velocity v = 0

From the first equation of motion

v = u+at (as the car velocity is slows down means it is a case of deceleration)

So v = u-at

0=20-a\times 6

a=3.33m/sec^2

Now from second equation of motion s=ut-\frac{1}{2}at^2=20\times 6-\frac{1}{2}\times 3.33\times 6^2=60.06m

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Why is a shadow formed
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Answer:

Shadows are made by blocking light. Light rays travel from a source in straight lines. If an opaque (solid) object gets in the way, it stops light rays from traveling through it. The size and shape of a shadow depend on the position and size of the light source compared to the object.

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To visit your favorite ice cream shop, you must travel 490 m west on Main Street and then 920 m south on Division Street. Suppos
topjm [15]

Answer:

a) The magnitude of your average velocity during the 121 s is 8.61 m/s.

b) The direction of the average velocity is 61.9° south of west.

c) Your average speed during the trip is 11.7 m/s

Explanation:

Hi there!

a) The average velocity (a.v) is calculated as the displacement divided by the time it took to do such a displacement.

The displacement is calculated as the distance between the initial position and the final position:

Displacement = Δ(x,y) = final position - initial position

Let's consider that your initial position is the origin of our frame of reference and let's also consider that west and south are positive directions (+x and +y respectively). Then the displacement vector will be:

Δ(x,y) = final positon - initial position

Δ(x,y) = (490, 920) m - (0, 0) m = (490, 920) m

The average velocity will be:

a.v = Δ(x,y) / t

a.v = (490, 920) m / 121 s

a.v = (4.05, 7.60) m/s

The magnitude of the average velocity is calculated as follows:

 

The magnitude of your average velocity during the 121 s is 8.61 m/s.

b) To find the direction of the average velocity, we have to use trigonometric rules of right triangles. Notice that the x and y-components of the average velocity (vx and vy) together with the average velocity vector (v), with magnitude 8.61 m/s, form a triangle (see figure).

Also, notice that v is the hypotenuse of the triangle and that vx is the side adjacent to the angle θ while vy is the side opposite to θ.

Using trigonometry, we can calculate the value of the angle θ:

cos θ = adjacent side / hypotenuse

cos θ = vx / v

cos θ = 4.05 m/s / 8.61 m/s

θ = 61.9°

The direction of the average velocity is 61.9° south of west.

c) The average speed (a.s) is calculated as the traveled distance (d) divided by the time it took to cover that distance (t). In total, you traveled (490 m + 920 m) 1410 m in 121 s, then the average speed will be:

a.s = d/t

a.s = 1410 m / 121 s

a.s = 11.7 m/s

Your average speed during the trip is 11.7 m/s

5 0
3 years ago
A subway train starts from rest at a station and accelerates at a rate of 1.68 m/s2 for 14.2 s. It runs at constant speed for 68
liubo4ka [24]

Answer:

total distance = 1868.478 m

Explanation:

given data

accelerate = 1.68 m/s²

time = 14.2 s

constant time = 68 s

speed = 3.70 m/s²

to find out

total distance

solution

we know train start at rest so final velocity will be after 14 .2 s is

velocity final = acceleration × time      ..............1

final velocity = 1.68 × 14.2

final velocity = 23.856 m/s²

and for stop train we need time that is

final velocity = u + at

23.856 = 0 + 3.70(t)

t = 6.44 s

and

distance = ut + 1/2 × at²     ...........2

here u is initial velocity and t is time for 14.2 sec

distance 1 = 0 + 1/2 × 1.68 (14.2)²

distance 1 = 169.37 m

and

distance for 68 sec

distance 2= final velocity × time

distance 2= 23.856 × 68

distance 2 = 1622.208 m

and

distance for 6.44 sec

distance 3 = ut + 1/2 × at²

distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²

distance 3 = 76.90 m

so

total distance = distance 1 + distance 2 + distance 3

total distance = 169.37 + 1622.208 + 76.90

total distance = 1868.478 m

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Answer:

Trial 1 is the largest, trial 3 is the smallest

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M₂ = 6·10²² kg

d₂ = 7 000 km = 7·10⁶ м

<em>Trial  3</em>

M₃ = 3·10²² kg

d₃ = 7 000 km = 7·10⁶ м

___________

F - ?

Gravitational force:

F₁ = G·m·M₁ / d₁² = m·6.67·10⁻¹¹·6·10²² / (3.5·10⁶)² = 0.37·m  (N)

F₂ = G·m·M₂ / d₂² = m·6.67·10⁻¹¹·6·10²² / (7·10⁶)² = 0.08·m  (N)

F₃ = G·m·M₃ / d₃² = m·6.67·10⁻¹¹·3·10²² / (7·10⁶)² = 0.04·m  (N)

Trial 1 is the largest, trial 3 is the smallest

5 0
1 year ago
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