Lus

A worker pushes a large rock to the north while another worker helps by pushing it to the east. If they both exert equal force,

alexandr1967 [171]

let the magnitude of force applied by each worker be "F"

consider east-west direction along X-axis and north-south direction along Y-axis

In unit vector form, force vector by worker pushing in east direction is given as

$\underset{A}{\rightarrow}$ = F $\hat{i}$ + 0 $\hat{j}$

In unit vector form, force vector by worker pushing in north direction is given as

$\underset{B}{\rightarrow}$ = 0 $\hat{i}$ + F $\hat{j}$

resultant force is given as the vector sum of two vector forces as

$\underset{R}{\rightarrow}$ = $\underset{A}{\rightarrow}$ + $\underset{B}{\rightarrow}$

$\underset{R}{\rightarrow}$ = (F $\hat{i}$ + 0 $\hat{j}$ ) + (0 $\hat{i}$ + F $\hat{j}$ )

$\underset{R}{\rightarrow}$ = F $\hat{i}$ + F $\hat{j}$

direction of the force is hence given as

θ = tan⁻¹(F/F)

θ = tan⁻¹(1)

θ = 45 degree north of east

hence the direction is north-east

3 0

3 years ago

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of t

muminat

<h2><em><u>⇒</u></em>Answer:</h2>

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Step-by-Step Solution:

Solution 35PE

This question discusses about the increased range. So, we shall assume that the angle of jumping will be as the horizontal range is maximum at this angle.

Step 1 of 3<

/p>

The legs have an extension of 0.600 m in the crouch position.

So, m

The person is at rest initially, so the initial velocity will be zero.

The acceleration is m/s2

Acceleration m/s2

Let the final velocity be .

Step 2 of 3<

/p>

Substitute the above given values in the kinematic equation ,

m/s

Therefore, the final velocity or jumping speed is m/s

Explanation:

3 0

3 years ago

Read 2 more answers

A fisherman is fishing from a bridge and is using a "42.0-N test line." In other words, the line will sustain a maximum force of

lara31 [8.8K]

Answer:

(a) 42 N

(b)36.7 N

Explanation:

Nomenclature

F= force test line (N)

W : fish weight (N)

Problem development

(a) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled in at constant speed

We apply Newton's first law of equlibrio because the system moves at constant speed:

∑Fy =0

F-W= 0

42N -W =0

W = 42N

(b) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled with an acceleration whose magnitude is 1.41 m/s²

We apply Newton's second law because the system moves at constant acceleration:

m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity

∑Fy =m*a

m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity

F-W= ( W/9.8 )*a

42-W= ( W/9.8 )*1.41

42= W+0.1439W

42=1.1439W

W= 42/1.1439

W= 36.7 N

8 0

3 years ago

A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom w

nirvana33 [79]

Answer:

$|D_{depth} |=19.697m$

Explanation:

To find Depth D of lake we must need to find the time taken to hit the water.So we use equation of simple motion as:

Δx=vit+(1/2)at²

$x_{f}-x_{i}=v_{i}t+(1/2)at^{2}\\ -5.0m=(o)t+(1/2)(-9.8m/s^{2} )t^{2}\\ -4.9t^{2}=-5.0\\ t^{2}=5/4.9\\t=\sqrt{1.02} \\t=1.01s$

As we have find the time taken now we need to find the final velocity vf from below equation as

$v_{f}=v_{i}+at\\v_{f}=0+(-9.8m/s^{2} )(1.01s) \\v_{f}=-9.898m/s$

So the depth of lake is given by:

first we need to find total time as

t=3.0-1.01 =1.99 s

$|D_{depth} |=|vt|\\|D_{depth} |=|(-9.898m/s)(1.99s)|\\|D_{depth} |=19.697m$

6 0

3 years ago

A 25.0 kg pickle is accelerated from rest through a distance of 6.0m in 4.0s across a level floor . If the friction force betwee

SIZIF [17.4K]

Add the KE increase and the work done against friction.

The final velocity is twice the average, or 3.0 m/s
The final KE is (1/2)*25*3^2 = 112.5 J

The friction work done is 6*3.8 = 22.8 J
hope this is correct

8 0

3 years ago