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adell
5 days ago
14

If you stand next to a wall on a frictionless skateboard and push the wall with a force of 40n, how hard does the wall push on y

ou? if your mass is 80 kg, show that your acceleration is 0.5m/s2.
Physics
1 answer:
Ostrovityanka [2.2K]5 days ago
4 0
The wall exerts an equal force in response to your push, so it's 40N. To illustrate that your acceleration is 0.5 m/s², we apply Newton's second law F=ma, which translates into 40N=80kg*a. Knowing that 1 Newton is equivalent to 1 kg*m/s², we can express it as 40kg*m/s²=80kg*a. By isolating 'a,' we find 40kg*m/s²/80kg = a, simplifying down to (40m/s²)/80 = a, which results in 0.5 m/s² for a.
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Consider a 4-mg raindrop that falls from a cloud at a height of 2 km. When the raindrop reaches the ground, it won't kill you or
inna [2210]

Answer:

The work performed by air resistance totals -0.0782 J

Explanation:

Hello!

According to the principle of conservation of energy, the energy of a raindrop must remain constant.

At the outset, the raindrop possesses only gravitational potential energy:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the raindrop.

g = gravitational acceleration (9.8 m/s²)

h = height.

Let's determine the initial potential energy of the raindrop:

(4 mg should be converted into kg: 4 mg · 1 kg / 1 × 10⁶ mg = 4 × 10⁻⁶ kg)

PE = 4 × 10⁻⁶ kg · 9.8 m/s² · 2000 m

PE = 0.0784 J

As the raindrop descends, some of its potential energy converts into kinetic energy while the rest is lost to the air resistance. Upon reaching the ground, all initial potential energy has been either turned into kinetic energy or spent overcoming air resistance:

initial PE = final KE + Work by air

Where:

KE = kinetic energy.

Work by air = work done by air resistance.

The kinetic energy at ground level is computed as follows:

KE = 1/2 · m · v²

Where:

m = mass

v = velocity

<pThus:

KE = 1/2 · 4 × 10⁻⁶ kg · (10 m/s)²

KE = 2 × 10⁻⁴ J

Now, we can find the work done by air resistance:

initial PE = final KE + Work by air

0.0784 J = 2 × 10⁻⁴ J + Work by air

Work by air = 0.0784 J - 2 × 10⁻⁴ J

Work by air = 0.0782 J

Since work is performed in the opposite direction to movement, this results in a negative value. Therefore, the work done by air resistance is -0.0782 J.

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25 days ago
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A person drops a stone down a well and hears the echo 8.9 s later. if it takes 0.9 s for the echo to travel up the well, approxi
Sav [2230]
The total duration from the stone's drop to the echo being heard is 8.9 seconds. The sound takes 0.9 seconds to reach the listener, meaning it takes the stone 8 seconds to descend to the well's bottom. Using gravitational acceleration, the well's depth calculates to be approximately 313.6 meters.
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4 days ago
A small object slides along the frictionless loop-the-loop with a diameter of 3 m. what minimum speed must it have at the top of
ValentinkaMS [2433]
<span>3.834 m/s. To solve this problem, we must ensure that the centripetal force equals or exceeds the gravitational force acting on the object. The formula for centripetal force is F = mv^2/r while the equation for gravitational force is F = ma. Since the mass (m) cancels out in both equations, we can equate them, leading to a = v^2/r. Now, inserting the given values (where the radius is half the diameter) allows us to find v: 9.8 m/s^2 <= v^2/1.5 m, which simplifies to 14.7 m^2/s^2 <= v^2. Therefore, we find that the minimum velocity required is 3.834057903 m/s <= v. Thus, the necessary speed is 3.834 m/s.</span>
4 0
16 days ago
A long cylindrical rod of diameter 200 mm with thermal conductivity of 0.5 W/m⋅K experiences uniform volumetric heat generation
Ostrovityanka [2208]

Answer:

a, 71.8° C, 51° C

b, 191.8° C

Explanation:

Given the data:

D(i) = 200 mm

D(o) = 400 mm

q' = 24000 W/m³

k(r) = 0.5 W/m.K

k(s) = 4 W/m.K

k(h) = 25 W/m².K

The heat generation formula can be articulated as follows:

q = πr²Lq'

q = π. 0.1². L. 24000

q = 754L W/m

Thermal conduction resistance, R(cond) = 0.0276/L

Thermal conduction resistance, R(conv) = 0.0318/L

Applying the energy balance equation,

Energy In = Energy Out

This equates to q, which is 754L

From the initial analysis, the temperature at the interface between the rod and sleeve is found to be 71.8° C

Additionally, the outer surface temperature records as 51° C

Furthermore, based on the second analysis, the calculated temperature at the center of the rod is determined to be 191.8° C

6 0
21 day ago
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