Q: What is the distance of 2 m?

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To convert 2 km 90 m to meters, we first convert 2 km to meters by multiplying 2 by 1000 (1 km = 1000 m), which equals 2000 m. Then, we add the 90 m to get a total of 2090 meters. Therefore, 2 km 90 m is equal to 2090 meters.

3 1/2

distance/time2, or d/t2

distance equals two meters

Multiply the distance in km by 1000 to get the distance in m.

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long distance swimming can meAN 2 things 1 you swim for along distance e.e 1500 m (100m=4lengths) 2 you race in a 50 m pool instead of 25mpool

Just divide the distance by the time. The answer will be in m/sec.

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2

The top to bottom distance of the disturbance would be 2 meters since amplitude represents half of this full distance.

distance/time2, or d/t2

distance equals two meters

None. M, M and M are the same so they are coincident and, consequently, the distance around them is zero.

The masses, and the distance. The formula for gravitional attraction is: F = G m(1) m(2) / r2, where G is a constant, m(1) and m(2) are the two masses, and r is the distance.

Given a force of -500 N, which implies braking, the stopping distance of the car can be calculated using the equation ( d = v^2 / 2a ), where ( d ) is the stopping distance, ( v ) is the initial velocity (20 m/s), and ( a ) is the acceleration produced by the force. Using Newton's second law, we have ( a = F / m = -500 / 1000 = -0.5 , \text{m/s}^2 ). Substituting ( v = 20 , \text{m/s} ) and ( a = -0.5 , \text{m/s}^2 ) into the stopping distance equation, we get ( d = 20^2 / (2 \times 0.5) = 400 , \text{m} ). Hence, the stopping distance for the car will be 400 meters.

The total distance covered by the jogger is the sum of the distances covered in the first 3 minutes and the last 2 minutes. Using the formula distance = speed × time, the total distance is 5 m/s × 3 min + 4 m/s × 2 min = 15 m + 8 m = 23 m. The total time taken is 3 min + 2 min = 5 min. Therefore, the average speed is total distance / total time = 23 m / 5 min = 4.6 m/s.

Multiply the distance in km by 1000 to get the distance in m.