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Pie
3 years ago
10

A 66-kg fellow stands on a digital scale in an elevator that accelerates upwards from rest to 4.5 m/s in 2.00 s. show answer Inc

orrect Answer Calculate the scale reading, in newtons, during this process (The scale exerts an upward force on him equal to its reading.)
Physics
1 answer:
Nina [5.8K]3 years ago
7 0

Answer:

Explanation:

mass of the fellow ( m ) = 66kg

acceleration of fellow a

v = u + at

4.5 = 0 + a x 2

a = 4.5 /2

= 2.25 m / s²

Net force acting on fellow in upward direction by the surface of  elevator

R - mg where R is reaction force of the surface of the elevator

Applying Newton's law of motion

R - mg = ma

R = m (g +a )

= 66 x ( 9.8 + 2 )

= 778.8 N

This will be the scale reading .

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Honeybees accumulate charge as they fly, and they transfer charge to the flowers they visit. Honeybees are able to sense electri
Vilka [71]

Answer:

ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

Explanation:

For this exercise let's use the electric field expression

         E = k q / r²

where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee

let's calculate the field for each charge

 

Q = 24 pC = 24 10⁻¹² C

         E₁ = 9 10⁹ 24 10⁻¹² / 0.20²

         E₁ = 5.4 N / C

Q = 32 pC = 32 10⁻¹² C

         E₂ = 9 10⁹ 32 10⁻¹² / 0.2²

         E₂ = 7.2 N / C

let's find the difference between these two fields

         ΔE = E₂ -E₁

         ΔE = 7.2 - 5.4

         ΔE = 1.8 N / C

the minimum detection field is

         E_minimum = 0.77 N / C

        ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

8 0
3 years ago
What is a net force on an object that has a mass of 20.0 kg, an applied force of 100 n moving on a surface with a friction coeff
sergiy2304 [10]

The net force on the object as described is; 58.84N

Two forces acting on the object are;

  • The <em>applied force and the frictional force.</em>

In essence; the frictional force can be evaluated as;

  • Frictional force; = coefficient × Weight of object.

  • Frictional force = 0.21 × 20 × 9.8.

  • Frictional force = 41.16N

  • The Net force = Applied force - frictional force

  • Net force = 100 - 41.16N

Net Force = 58.84 N.

Read more:

brainly.com/question/94428

5 0
2 years ago
Which of the following statements would be classified as an aphorism?
Veseljchak [2.6K]
A bird in the hand is worth two in the bush
or
actions speak louder than words
or
her bark is worse than her bite
i hope this helps these are examples of aphorism
5 0
3 years ago
A block of weight w sits on a frictionless inclined plane, which makes an angle theta with respect to the horizontal, as shown.
zavuch27 [327]

Answer:

a.

Explanation:

6 0
3 years ago
If it takes a planet 2.8 × 108 s to orbit a star with a mass of 6.2 × 1030 kg, what is the average distance between the planet a
Levart [38]

Answer:

9.36*10^11 m

Explanation

Orbital velocity  v=√{(G*M)/R},

G = gravitational constant =6.67*10^-11 m³ kg⁻¹ s⁻²,

M = mass of the star

R =distance from the planet to the star.

v=ωR, with ω as the angular velocity and R the radius

ωR=√{(G*M)/R},

ω=2π/T,

T = orbital period of the planet

To get R we write the formula by making R the subject of the equation

(2π/T)*R=√{(G*M)/R}

{(2π/T)*R}²=[√{(G*M)/R}]²,

(4π²/T²)*R²=(G*M)/R,

(4π²/T²)*R³=G*M,

R³=(G*M*T²)/4π²,

R=∛{(G*M*T²)/4π²},

Substitute values

R=9.36*10^11 m

5 0
2 years ago
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