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netineya [11]
3 years ago
11

A circular dartboard has a radius of 2 meters and a red circle in the center. Assume you hit the target at a random point. For w

hat radius of the red center region does P(hitting red) = 0.6?
Physics
1 answer:
Sati [7]3 years ago
3 0

Answer:

1.549 m

Explanation:

Given:

The radius of the circular board, r = 2 m

The probability of hitting the red is given as 0.6

Now, this probability of hitting the red can be conclude as

0.6 = (Area of red)/ (Total area of the board)

Total area of the board = πr² = π × 2²

let the radius of the red area be R

thus, area of red circle, = πR²

on substituting the value of the area, we have

0.6 = (πR²)/ (π × 2²)

or

R² = 2.4

or

R = 1.549 m

Thus, the radius of the red circle is 1.549 m

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Kelly is riding a bicycle, moves with an initial velocity of 5 m/s. Ten seconds later, she is moving at 15 m/s. What is her acce
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Answer:

her acceleration is 1 m/sec

Explanation:

The following information is given in the question

The initial velocity is 5 m/s

After 10 seconds, she would be moved at 15 m/s

We need to find the acceleration

As we know that

Acceleration = Change in speed ÷ time

Acceleration = (15 - 5) ÷ (10)

= 1 m/sec

Hence, her acceleration is 1 m/sec

The same would be considered  

3 0
3 years ago
A 1,600 kg train car rolling freely on level track at 16 m/s bumps into a 1.0 × 103 kg train car moving at 10.0 m/s in the same
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This item is solved through the concept of the conservation of momentum which states that the momentum before and after collision should be equal. 
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The value of x is 13.69 m/s. Thus, their final speed is approximately letter D. 14 m/s. 
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4 years ago
Please answer as fast as possible
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Explanation:

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where h is height,p is density, g is 10m/s²

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3 0
3 years ago
A 10-g bullet moving horizontally with a speed of 1.8 km/s strikes and passes through a 5.0-kg block initially at rest on a hori
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Answer:

6.4 J

Explanation:

m = mass of the bullet = 10 g = 0.010 kg

v = initial velocity of bullet before collision = 1.8 km/s = 1800 m/s

v'  = final velocity of the bullet after collision = 1 km/s = 1000 m/s

M = mass of the block = 5 kg

V = initial velocity of block before collision = 0 m/s

V'  = final velocity of the block after collision = ?

Using conservation of momentum

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(0.010) (1800) + (5) (0) = (0.010) (1000) + (5) V'

V' = 1.6 m/s

Kinetic energy of the block after the collision is given as

KE = (0.5) M V'²

KE = (0.5) (5) (1.6)²

KE = 6.4 J

4 0
4 years ago
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