Answer:
Seven
Explanation:
The rules for significant digits are:
- Non-zero digits are always significant.
- Zeros between significant digits are also significant.
- Trailing zeros are significant only after a decimal point.
Here, the 2, 4, 9, and 2 are significant because they are non-zero digits.
The first two 0s are significant because they are between significant digits.
The last 0 is significant because it is a trailing zero after a decimal point.
Therefore, all seven digits are significant.
Answer:
Decreases the time period of revolution
Explanation:
The time period of Cygnus X-1 orbiting a massive star is 5.6 days.
The orbital velocity of a planet is given by the formula,
v = √[GM/(R + h)]
In the case of rotational motion, v = (R +h)ω
ω = √[GM/(R + h)] /(R +h)
Where 'ω' is the angular velocity of the planet
The time period of rotational motion is,
T = 2π/ω
By substitution,
<em>T = 2π(R +h)√[(R + h)/GM] </em>
Hence, from the above equation, if the mass of the star is greater, the gravitational force between them is greater. This would reduce the time period of revolution of the planet.
We will solve this question using the second law of motion which states that force is directly equal to the product of mass and acceleration.

Where,
- F is force
- m is mass
- a is acceleration
In our case,
- F = ?
- m = 2500 kg
- a = 20m/s

<em>Thus, The force of 50000 Newton is required to accelerate a car of 2500 kg...~</em>
Answer:
P = 10 kPa
Explanation:
Given that,
The mass of a small table, m = 4 kg
The area of each leg = 0.001 m²
We need to find the pressure exerted by the table on the floor. Pressure is equal to the force per unit area. So

So, the required pressure is 10 kPa.