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3 years ago

How much potential energy does a 1kg mess have 10m off the ground?

Physics

1 answer:

777dan777 [17]3 years ago

3 0

Potential Energy = mass x gravitational acceleration x height
potential Energy = 1 x 9.8 x 10 = 98 joules

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The safe load, L, of a wooden beam supported at both ends varies jointly as the width, w, the square of the depth, d, and invers

soldier1979 [14.2K]

Answer:

L' = 555.95 lb

Explanation:

Analyzing the given conditions in the question, we get

The safe load, L is directly proportional to width (w) and square of depth (d²)

also,

L is inversely proportional length (l) i.e L = k/l

combining the above conditions, we get an equation as:

L = k(wd²/l)

now, for the first case we have been given

w = 3 in

d = 6 in

l = 11 ft

L = 1213 lbs

thus,

1213 lb = k ((3 × 6²)/11)

k = 123.54 lbs/(ft.in³)

Now,

Using the calculated value of k to calculate the value of L in the second case

in the second case, we have

w = 6 in

d =3 in

l = 12 ft

Final Safe load L' = 123.54 × (6 × 3²/12)

L' = 555.95 lb

6 0

3 years ago

How close would you have to bring 1 C of positive chargeand 1 C of negative charge for them to exert forces of 1 N onone another

Harman [31]

Answer:

94,800 m

Explanation:

F = kq1 q2/r^2

1 = 9 x 10^9 x 1 / r^2

4 0

2 years ago

Mrs. Smith can walk 1.4 m/s. If it takes her 8.5 seconds to get to the teacher lounge, how far is the teacher lounge from her ro

AlexFokin [52]

Answer:

6 meters away

Explanation:

6*1.4= 8.4 which is pretty close

5 0

3 years ago

The proper order of the cycle of addiction is

Viktor [21]

Answer:

The answer would be drug use, addiction, dependence, tolerance, and withdrawal.

4 0

3 years ago

Read 2 more answers

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

spayn [35]

Answer:

$v_0 = 3.53~{\rm m/s}$

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

$x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}$

For the y-direction gives

$v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t$

Combining both equation yields the y_component of the final velocity

$v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}$

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

$\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}$

6 0

3 years ago