If (1.001)1259 = 3.52 and (1.001)2062 = 7.85, then (1.001)3321 =
27.64
This problem requires us to determine the value of a power, specifically \( (1.001)^{3321} \), by leveraging two other given power values that share the same base. To approach this, we will use a fundamental property of exponents.
A crucial rule in algebra for exponents states that when you multiply two exponential terms that have the same base, you can add their exponents. This rule is often written as:
In our particular question, the base of the exponential terms is \( 1.001 \).
We are provided with the following values:
Our objective is to find the value of \( (1.001)^{3321} \). Let's examine if there's a connection between the target exponent \( 3321 \) and the exponents given in the problem, \( 1259 \) and \( 2062 \). If we add these two exponents:
This sum matches the exponent we need to calculate, which confirms we can use the multiplication rule for exponents.
Since \( 3321 \) is the sum of \( 1259 \) and \( 2062 \), we can express \( (1.001)^{3321} \) as a product of two powers:
Now, substitute the numerical values given in the problem into this equation:
To get the final answer, we multiply the two decimal numbers:
The calculated value is \( 27.632 \).
When comparing our result of \( 27.632 \) with the provided options, we look for the closest value. It's common in these types of problems for intermediate values or final answers to be rounded slightly. The closest option to \( 27.632 \) is \( 27.64 \).
Therefore, \( (1.001)^{3321} \) is approximately \( 27.64 \).
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A sphere of volume V is made of a material with lower density than water. While on Earth, it floats on water with its volume f1V (f1 < 1) submerged. On the other hand, on a spaceship accelerating with acceleration a < g (g is the acceleration due to gravity on Earth) in outer space, its submerged volume in water is f2V. Then:
A railway wagon (open at the top) of mass M1 is moving with speed v1 along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M2 and speed becomes v2. Taking the rain to be falling vertically and the water stationery inside the wagon, the relation between the two speeds v1 and v2 is :
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1. Distance between the longitudes becomes zero on North Pole and South Pole.
2. Distance between the longitudes is maximum on the Equator.
3. Number of longitudes is more than number of latitudes.
Which of the statements given above is/are correct?
One block of 2⋅0 kg mass is placed on top of another block of 3⋅0 kg mass. The coefficient of static friction between the two blocks is 0⋅2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s2, the maximum value of the frictional force is :