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scoundrel [369]
3 years ago
8

On June 9, 1983, the lower part of the Variegated Glacier in Alaska was observed to be moving at a rate of 64 m per day. What is

this speed in kilometers per hour?
Physics
1 answer:
myrzilka [38]3 years ago
3 0

(64 meter/day) x (1 kilometer / 1,000 meters) x (1 day / 24 hours) =

       (64 x 1 x 1) / (1,000 x 24)         (kilometer/hour)  = 

                     1/375  km/hour

                (0.002667 km/hour, rounded)

                 2-2/3 meters/hour

                 (8-ft 9-inches) per hour
 
 
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A bulb converts 9J of energy in each second from an input of 100J of energy. Calculate efficiency of this light bulb
77julia77 [94]
Efficiency is calculated through dividing the actual mechanical advantage by the hypothetical mechanical advantage:

- the actual mechanical advantage is 9J because that's how much work the light bulb doing

- the hypo. mechanical advantage is 100J. Ideally, in a perfect world, the light bulb can convert 100J input into 100J output, but do to resistance and other factors it is not possible.

\frac{9}{100}  = .09
change the decimal to a percentage:

.09 = 9\%
the light bulb had 9% efficiency
8 0
4 years ago
wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the
lys-0071 [83]

Answer:

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

BP = 2 m

A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

E_{A}=\frac{KQ_{A}}{AP^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{6^{2}}=1000 N/C (rightwards)

The electric field due to charge QB at P is EB leftwards

E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

5 0
3 years ago
Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has a magnitude of 3.7m/s2.​ We want
lbvjy [14]

Explanation:

It is given that, Onur drops a basketball from a height of 10 m on Mars, where the acceleration due to gravity has a magnitude of 3.7 m/s².

The second equation of kinematics gives the relationship between the height reached and time taken by it.

Here, the ball is droped under the action of gravity. The value of acceleration due to gravity on Mars is positive.

We want to know how many seconds the basketball is in the air before it hits the ground. So, the formula is :

\Delta x=v_ot+\dfrac{1}{2}at^2

t is time taken by the ball to hit the ground

v_o is initial speed of the ball

So, the correct option is (A).

8 0
3 years ago
What is the answer of A and B question 2
jok3333 [9.3K]
They have the writing
5 0
3 years ago
Read 2 more answers
Imagine you leave the dock on a fishing boat that travels east for 30 kilometers when the captain realizes he left the fishing r
telo118 [61]

Answer:

Average speed is 60 km/hr whereas average velocity is 0 km/ hr.

Explanation:

The average speed of the boat is 60 km/ hr while on the other hand, the average velocity of the boat is 0 km/ hr because average speed is the total distance covered by the boat in total time and average velocity is the displacement covered by the boat in total time. The total distance covered by the boat is 60 km and the total time is 60 minutes which is equals to one hour so the answer is 60 km/hr whereas the displacement is the shortest distance between initial and final position which is 0 in this case so 0 divided by 60 minutes or one hour is also 0.

3 0
3 years ago
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