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lisabon 2012 [21]
3 years ago
8

A race car accelerates on a straight road from rest to speed of 180 km/hr in 25s . Assuming uniform acceleration of the car thro

ugh out find the find the distance covered in this time .
Physics
1 answer:
adelina 88 [10]3 years ago
3 0
X-x0=0.5(v0+v)t ergo x=0.5(50)25=625m
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13. An object, travelling along a straight path, covers 35 m distance in 4 seconds. In next 6
WARRIOR [948]

Answer:

(4) 8.5 m/s

Explanation:

You add both the meters together and both the seconds together and then divide them both.

5 0
3 years ago
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Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g
Ne4ueva [31]

Answer:

C

Explanation:

- Let acceleration due to gravity @ massive planet be a = 30 m/s^2

- Let acceleration due to gravity @ earth be g = 30 m/s^2

Solution:

- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:

                                 t = v / a

                                 t = v / 30

- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:

                                 t = v / g

                                 t = v / 9.81

- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C

7 0
2 years ago
consider a solid sphere and a solid disk wiht the same radius and the same mass. explain why the solid disk has a greater moment
andriy [413]

Answer:

Moment of inertia of the solid sphere:

I

s

=

2

5

M

R

2

.

.

.

.

.

.

.

.

.

.

.

(

1

)

Is=25MR2...........(1)

Here, the mass of the sphere is

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2 years ago
What is electron capture
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7 0
3 years ago
Two bicycle tires are set rolling with the same initial speed of 4.0 m/s along a long, straight road, and the distance each trav
umka2103 [35]

Answer:

The coefficient of rolling friction will be "0.011".

Explanation:

The given values are:

Initial speed,

v_i = 4.0 \ m/s

then,

v_f=\frac{4.0}{2}

    =2.0 \ m/s

Distance,

s = 18.2 m

The acceleration of a bicycle will be:

⇒ a=\frac{v_f^2-v_i^2}{2s}

On substituting the given values, we get

⇒    =\frac{(2.0)^2-(4.0)^2}{2\times 18.2}

⇒    =\frac{4-8}{37}

⇒    =\frac{-4}{37}

⇒    =0.108 \ m/s^2

As we know,

⇒  f=ma

and,

⇒  \mu_rmg=ma

⇒       \mu_r=\frac{a}{g}

On substituting the values, we get

⇒       =\frac{0.108}{9.8}

⇒       =0.011

7 0
3 years ago
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