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Citrus2011 [14]

2 years ago

11

Marcy pulls a backpack on wheels down the 100 m hall. The 60N force is applied at an angle of 30° above the horizontal. How much

work done on the weights? show full working out

1 answer:

shutvik [7]2 years ago

6 0

Work= Fcos∆×S

W=60N×cos 30⁰×100

W=60×0.866×100=5196.1J

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Two UFPD are patrolling the campus on foot. To cover more ground, they split up and begin walking in different directions. Offic

miss Akunina [59]

Answer:

0.256 hours

Explanation:

<u>Vectors in the plane </u>

We know Office A is walking at 5 mph directly south. Let $X_A$ be its distance. In t hours he has walked

$X_A=5t\ \text{miles}$

Office B is walking at 6 mph directly west. In t hours his distance is

$X_B=6t\ \text{miles}$

Since both directions are 90 degrees apart, the distance between them is the hypotenuse of a triangle which sides are the distances of each office

$D=\sqrt{X_A^2+X_B^2}$

$D=\sqrt{(5t)^2+(6t)^2}$

$D=\sqrt{61}t$

This distance is known to be 2 miles, so

$\sqrt{61}t=2$

$t =\frac{2}{\sqrt{61}}=0.256\ hours$

t is approximately 15 minutes

3 0

2 years ago

mojhsa [17]

The marble's volume is equal to 1.2 $cm^{3}$
Density= mass/volume
Mass= 3.05 Grams
Volume= 1.2 cm^3
3.05/1.2= 2.92 <span>g/mL</span>

7 0

2 years ago

the face of a cube towards A is brightly and shiny and the face towards V is full black.State with reason the adjustments that s

MariettaO [177]

Explanation:

increase the distance of cube from black and dull substance

5 0

2 years ago

a stomp rocket takes 1.5 seconds to reach its maximum height what was the initial velocity and what was the maximum height ?

Zinaida [17]

1) Vf = Vo - gt; Vf = 0 => Vo = gt = 9.8m/s^2 * 1.5s = 14.7 m/s

2) d = Vo*t - gt^2 /2 = 14.7m/s*1.5 - 9.8m/s^2 * (1.5s)^2 / 2 = 11.02 m

5 0

2 years ago

Read 2 more answers

A Person whose weight is 5.20 x 10^2 N is being pulledup

Gnoma [55]

Answer:

Explanation:

Given

Weight of Person $W=5.20\times 10^{2} N$

Cave is $h=35.1 m$ deep

Breaking stress $T=569 N$

Net Force on Person

$F_{net}=569-520=49 N$

$a_{net}=\frac{F_{net}}{\frac{W}{g}}$

$a_{net}=\frac{49}{\frac{520}{9.8}}$

$a_{net}=0.923 m/s^2$

The shortest time such that the person can be taken out of cave

$h=ut+\frac{1}{2}at^2$

where

h=distance moved

t=time

a=acceleration

$35.1=0+\frac{1}{2}(0.923)(t)^2$

$t^2=76.05$

$t=\sqrt{76.05}$

$t=8.72\ s$

6 0

2 years ago