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3 years ago

A map suggests that Atlanta is 730 miles in a direction of

Physics

1 answer:

yuradex [85]3 years ago

6 0

Answer:

The displacement from Dallas to Chicago is 788 miles.

Explanation:

Given that,

A map shows that,

Atlanta is 730 miles in a direction of 5.00° north of earth from Dallas.

Same map shows that,

Chicago is 560 miles in a direction of 21.0° west of north from Atlanta.

We need to calculate the displacement from Dallas to Chicago

We need to figure,

In figure,

C = Atlanta

B = Chicago

A = Dallas

Using cosine law,

$c=\sqrt{a^2+b^2-2ab\cosC}$

Where, a = distance between Dallas to Atlanta

b = distance between Atlanta to Chicago

Put the value into the formula

$c=\sqrt{(560)^2+(730)^2-2\times560\times730\cos74}$

$c=788\ miles$

Hence, The displacement from Dallas to Chicago is 788 miles.

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A 5.0 kg cannonball is fired horizontally at 68 m/s from a 15-m-high cliff. A strong tailwind exerts a constant 12 N horizontal

jeka57 [31]

Answer:

3.7 m

Explanation:

ASSUMING this means extra distance beyond where the cannonball would land WITHOUT the wind assistance but in general ignoring air resistance. Hmmmmmm...tricky

The ball drops from vertical rest to ASSUMED horizontal ground 15 m below in a time of

t = √ (2h/g) = √(2(15)/9.8) = 1.75 s

Without the tail wind, the ball travels horizontally

d = vt = 68(1.75) = 119 m

The tailwind exerts a constant acceleration on the ball of

a = F/m = 12/5.0 = 2.4 m/s²

The average horizontal velocity during the flight is

v(avg) = (68 + (68 + 2.4(1.75)) / 2 = 70.1 m/s

so the distance with tailwind is

d = v(avg)t = 70.1(1.75) = 122.675 m

The extra distance is 122.675 - 119 = 3.675 = 3.7 m

8 0

3 years ago

There is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted s

lukranit [14]

Answer:

Option B

$a=19 m/s^{2}$

Explanation:

Given information

Radius of container, r=12cm=12/100=0.12m

Angular velocity= 2 rev/s, converted to rad/s we multiply by 2π

Angular velocity, $\omega=2*2\pi =12.56637061$

We know that speed, $v=r\omega$

Centripetal acceleration, $a=\frac {v^{2}}{r}$ and substituting $v=r\omega$ we obtain that

$a=r\omega^{2}$

Substituting \omega for 12.56637061 and r for 0.12

$a=0.12*(12.56637061)^{2}=18.94964045 m/s^{2}$

Rounded off, $a=19 m/s^{2}$

7 0

4 years ago

The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

$\sin\theta=\dfrac{3}{5}$

The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

$T=658.33\ N$

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0

3 years ago

an object of mass m is traveling at constant speed v in a circular path of radius r. how much work is done by the centripetal fo

vlada-n [284]

The work done is by the centripetal force on mass m during an angular displacement of 2π revolutions mv²2π /r J

Centripetal force - a force acts on an moving object in circular path.

the centripetal force is given by

F= mv²/r (equation1)

Work done is given by

W = Fd (equation 2)

d = 2π

work is done by the centripetal force on mass m during an angular displacement of 2π revolutions is given by:

to calculate work done using equation 1 in 2 we get

W = mv² d/r

W = mv² × 2π /r J

The work done is by the centripetal force on mass m during an angular displacement of 2π revolutions mv²2π /r J

To know more about centripetal force :

brainly.com/question/13031430

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6 0

2 years ago

Which statement provides the most complete description of an object's motion?

vivado [14]

Answer:

A. The bird watcher followed the south trail a distance of five kilometers in 45 minutes.

5 0

3 years ago