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yulyashka [42]
2 years ago
6

Does anyone know how to solve this??

Physics
1 answer:
Scrat [10]2 years ago
7 0

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wtjfavyw

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The average speed of a nitrogen molecule in air is about 6.70×102 m/s, and its mass is 4.68×10-26 kg.
Otrada [13]

Answer:

a)   a = 3.06 10¹⁵ m / s , b)    F= 1.43  10⁻¹⁰ N, c)    F_total = 14.32 10⁻²⁶ N

Explanation:

This exercise will average solve using the moment relationship.

a ) let's use the relationship between momentum and momentum

          I = ∫ F dt = Δp

          F t = m v_{f} - m v₀

          F = m (v_{f} -v₀o) / t

 in the exercise indicates that the speed module is the same, but in the opposite direction

          F = m (-2v) / t

if we use Newton's second law

          F = m a

we substitute

            - 2 mv / t = m a

            a = - 2 v / t

let's calculate

            a = - 2 4.59 10²/3 10⁻¹³

            a = 3.06 10¹⁵ m / s

b)      F= m a

        F= 4.68 10⁻²⁶ 3.06 10¹⁵

        F= 1.43  10⁻¹⁰ N

c) if we hit the wall for 1015 each exerts a force F

            F_total = n F

            F_total = n m a

            F_total = 10¹⁵  4.68 10⁻²⁶ 3.06 10¹⁵

            F_total = 14.32 10⁻²⁶ N

8 0
3 years ago
Hank and Harry are two ice skaters whiling away time by playing 'tug of war' between practice sessions. They hold on to opposite
Alex73 [517]

Answer:

the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1

Explanation:

Given the data in the question;

Hank and Harry are two ice skaters, since both are on top of ice, we assume that friction is negligible.

We know that from Newton's Second Law;

Force = mass × Acceleration

F = ma

Since they hold on to opposite ends of the same rope. They have the same magnitude of force |F|, which is the same as the tension in the rope.

Now,

Mass_{Hank × Acceleration_{Hank = Mass_{Henry × Acceleration_{Henry

so

Mass_{Hank /  Mass_{Henry = Acceleration_{Henry / Acceleration_{Hank

given that; magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration,

Mass_{Hank /  Mass_{Henry = 1 / 1.26

Mass_{Hank /  Mass_{Henry = 0.7937 or [ 0.7937 : 1 ]

Therefore, the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1 ]

8 0
2 years ago
What single property was the most important in Jesseca’s material?
aleksandrvk [35]

Answer:

Jesseca wanted to create a material that reflected most of the light that fell on it.

Explanation:

Plato Answer

7 0
2 years ago
Read 2 more answers
A major-league pitcher can throw a ball in excess of 40.1 m/s. If a ball is thrown horizontally at this speed, how much will it
mote1985 [20]

Answer:

The ball will drop 0.881 m by the time it reaches the catcher.

Explanation:

The position of the ball at time "t" is described by the position vector "r":

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

When the ball reaches the catcher, the position vector will be "r final" (see attached figure).

The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.

Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):

x = x0 + v0x · t

17.0 m = 0 m + 40.1 m/s · t

t = 17.0 m/ 40. 1 m/s = 0.424 s

With this time, we can calculate the y-component of the vector "r final", the drop of the ball:

y = y0 + v0y · t + 1/2 · g · t²

Initially, there is no vertical velocity, then, v0y = 0.

y = 1/2 · g · t²

y = -1/2 · 9.8 m/s² · (0.424 s)²

y = -0.881 m

The ball will drop 0.881 m by the time it reaches the catcher.

8 0
2 years ago
Person is lifting a 250 N dumbbell. The weight is 30 cm from the pivot point of the elbow. What force must be exerted five from
qwelly [4]
Refer to the diagram shown below.

The force, F, is applied at 5 cm from the elbow.

For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N

Answer: 1500 N

4 0
3 years ago
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