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Nookie1986 [14]

2 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

1 answer:

morpeh [17]2 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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2 years ago

This graph indicates that the kinetic energy of an object is increasing. What is the most reasonable explanation for this observ

soldi70 [24.7K]

B) The object's velocity doubled.

Explanation:

The graph is missing: find it in attachment.

The kinetic energy of an object is the energy possessed by the object due to its motion. It is calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the velocity of the object

We notice that:

The kinetic energy is directly proportional to the mass
The kinetic energy is proportional to the square of the velocity

In the graph, one of the two quantities (either mass or speed) is represented on the x-axis, while the quantity on the y-axis is the kinetic energy.

First of all, we notice that the relationship is not linear: this means that the quantity on the x-axis cannot be the mass, so it must be the velocity.

Moreover, we notice that when the quantity on the x-axis increases from 1 to 2 (so, it doubles), the kinetic energy increases by a factor of 4. This means that the object's velocity has doubled, therefore

B) The object's velocity doubled.

Learn more about kinetic energy:

brainly.com/question/6536722

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5 0

3 years ago

A stone is thrown vertically upwards with an initial velocity of 20m/sec. Find the maximum height ot reaches and the time taken

MAXImum [283]

Answer:

The height reached is 20m, The time taken to reach 20m is 2 seconds

Explanation:

Observing the equations of motion we can see that the following equation will be most helpful for this question.

$v^{2} = u^{2} + 2as$

We are given initial velocity, u

We know that the stone will stop at its maximum height, so final velocity, v

Acceleration, a

And we are looking for the displacement (height reached), s

Substitute the values we are given into the equation

$0^{2} = 20^{2} + 2(10)s$

Rearrange for s

$0^{2} -20^{2} =20s$

$-400=20s$

$\frac{-400}{20} =s$

s = -20 (The negative is just showing direction, it can be ignored for now)

The height reached is 20m

Use a different equation to find the time taken

$s = vt - \frac{1}{2} at^{2}$

Substitute in the values we have

$-20=(0)t - \frac{1}{2} (10)t^{2}$

Rearrange for t

$-20 =0 -5 t^{2}$

$\frac{-20}{-5} =t^{2}$

$4 = t^{2}$

t = 2s

The time taken to reach 20m is 2 seconds

4 0

2 years ago

A constant force of 12N is applied for 3.0s to a body initially at rest. The final velocity of the body is 6.0ms–1. What is the

sp2606 [1]

From the question,
$u = 0m {s}^{ - 1}$
$v = 6m {s}^{ - 1}$

$t = 3s$
$F=12N$

Using Impulse, the product of the constant force, F and time t equals the product of the mass of the body and change in velocity.

$Ft =m(v-u)$

$12(3.0)=m(6.0- \: 0)$
This implies that

$36.0 = 6m$
$m = \frac{36.0}{6.0}$
$\therefore \: m = 6.0kg$

You can also use the equation of linear motion,
$v = u + at$
$6 = 0 + a(3)$
$6 = 3a$
$a = \frac{6}{3}$

$a = 2 {ms}^{ - 2}$
But
$F=ma$
$12 = m(2)$
$12 = 2m$
$\frac{12}{2} = m$
$\therefore \: m = 6kg$

4 0

3 years ago

It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the sp

Ede4ka [16]

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

$W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m$

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

$W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J$

So, the new work is more than 130 J.

6 0

2 years ago