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Nookie1986 [14]
3 years ago
15

A gorilla drags a log across the jungle floor. The log begins at rest and accelerates at until the gorilla is moving at a fast l

ope. The coefficient of kinetic friction between the log and the mud and plants of the jungle is . Assume that friction and the gorillas pulling force remain constant the whole time. What frictional force acts on the log
Physics
1 answer:
morpeh [17]3 years ago
3 0

Answer:

  fr = 245 N,  T = fr

Explanation:

We can solve this exercise using the equilibrium conditions, let's start by creating a reference system

where we call T the drag force of the gorilla, m the mass of the trunk and μ the coefficient of kinetic friction

Y axis  

           N- W =

           N = W = mg

X axis

          T - fr = 0

          T = fr

the expression for the friction force is

          fr = μ N

          fr = μ mg

we substitute

           T = μ m g

When analyzing these expressions we see that when the trunk reaches the maximum speed that the gorilla carries, the friction force is equal to the gorilla's tension

            fr = T

for a specific calculation we must assign values ​​to:

      μ = 0.25

      m = 100 kg

let's calculate

         T = 0.25 100 9.8

         T = 245 N

therefore for this case

         fr = 245 N

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A chef places an open sack of flour on a kitchen scale. The scale reading of
Novay_Z [31]

Answer:

Explanation:

Given

Initial reading on scale =40 N

So, we can conclude that weight of the sack is 40 N

After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)

This net downward force is the resultant of earth graviational pull and the applied upward force.

So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.

This situation can be diagramatically represented by figure given below  

4 0
3 years ago
After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. T
oksian1 [2.3K]

Answer:

required distance is 233.35 m

Explanation:

Given the data in the question;

Sound intensity I = 1.62 × 10⁻⁶ W/m²

distance r = 165 m

at what distance from the explosion is the sound intensity half this value?

we know that;

Sound intensity I is proportional to 1/(distance)²

i.e

I ∝ 1/r²

Now, let r² be the distance where sound intensity is half, i.e I₂ = I₁/2

Hence,

I₂/I₁ = r₁²/r₂²

1/2 = (165)²/ r₂²

r₂² = 2 × (165)²

r₂² = 2 × 27225

r₂² = 54450

r₂ = √54450

r₂ = 233.35 m

Therefore, required distance is 233.35 m

6 0
3 years ago
Starting at t = 0 s , a horizontal net force F⃗ =( 0.285 N/s )ti^+(-0.460 N/s2 )t2j^ is applied to a box that has an initial mom
Irina-Kira [14]

Answer:

Explanation:

We know that Impulse = force x time

impulse = change in momentum

change in momentum = force x time

Force F = .285 t -.46t²

Since force is variable

change in momentum = ∫ F dt  where F is force

= ∫ .285ti - .46t²j dt

= .285 t² / 2i - .46 t³ / 3 j

When t = 1.9

change in momentum = .285 x 1.9² /2 i  -  .46 x 1.9³ / 3 j

= .514i - 1.05 j

final momentum

= - 3.1 i + 3.9j +.514i - 1.05j

= - 2.586 i + 2.85j

x component = - 2.586

y component = 2.85

7 0
3 years ago
Machmer Hall is 400 m North and 180 m West of Witless.
yan [13]

Answer:

The distance from Witless to Machmer is 438.63 m.

Explanation:

Given that,

Machmer Hall is 400 m North and 180 m West of Witless.

We need to calculate the distance

Using Pythagorean theorem

D = \sqrt{(d_{m})^2+(d_{w})^2}

Where, d_{m} =distance of Machmer Hall

d_{w} =distance of Witless

Put the value into the formula

D = \sqrt{(400)^2+(180)^2}

D=438.63\ m

Hence, The distance from Witless to Machmer is 438.63 m.

5 0
3 years ago
A coin released at rest from the top of a tower hits the ground after falling 5.7 s. What is the speed of the coin as it hits th
Bad White [126]

Given parameters:

Initial velocity of Coin = 0m/s

Time taken before coin hits ground  = 5.7s

Unknown:

Final velocity of the coin  = ?

Velocity is displacement with time. To solve this problem, we have to apply one of the equations of motion.

The fitting one of them here is shown below;

             V = U + gt

where;

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

t is the time taken

Here we use positive value of acceleration due to gravity because the coin is falling with the effect of acceleration and not against it.

Now input the parameters and solve;

               V  = 0 + 9.81 x 5.7

               V = 55.917m/s

Therefore, the final velocity is 55.917m/s.

8 0
3 years ago
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