652-06-01-01-00_files/i0190000.jpg What has rubbing done to the charges on the objects shown in diagrams A and B?

Sammy squirrel is steering his boat at a heading of 327 degree at 18mph. The current is flowing at 4mph at a heading of 60 degre

Tanya [424]

Answer:

59.97 º at 18.23 mph

Explanation:

To find Sammy's course you have to add the two velocities (vectors), 18 mph 327º and 4 mph 60º.

To add the two vectors analytically you decompose each vector into their vertical and horizontal components.

1. 18 mph 327º

Horizontal component: 18 mph × cos (327º) = 15.10 mph

Vertical component: 18 mph × sin (327º) = - 9.80 mph

Vector notation:

$15.10\hat i-9.80\hat j$

2. 4 mph 60º

Horizontal component: 4 mph × cos (60º) = 2.00 mph

Vertical component: 4 mph × sin (60º) = 3.46 mph

Vector notation:

$2.00\hat i+3.46\hat j$

3. Addition:

You add the corresponding components:

$15.10\hat i-9.80\hat j+2.00\hat i+3.46\hat j\\ \\ 17.10\hat i-6.34\hat j$

To find the magnitude use Pythagorean theorem:

$\sqrt{17.1^2+6.34^2}= 18.23$

4. Direction:

Use the tangent ratio:

$tan(\alpha )=opposite/adjacent=3.46/2.00=1.73$

Find the inverse:

arctan (1.73) ≈ 59.97º

5 0

3 years ago

On the surface of the earth the weight of an object is 200 lb. Determine the height of the

siniylev [52]

Answer:

The height of the object is 5007.4 miles.

Explanation:

Given that,

Weight of object = 200 lb

We need to calculate the value of $Gmm_{e}$

Using formula of gravitational force

$F=\dfrac{Gmm_{e}}{r^2}$

Put the value into the formula

$200=\dfrac{Gmm_{e}}{(3958.756)^2}$

$200\times(3958.756)^2=Gmm_{e}$

$Gmm_{e}=3.134\times10^{9}$

We need to calculate the height of the object

Using formula of gravitational force

$F=\dfrac{Gmm_{e}}{r^2}$

Put the value into the formula

$125=\dfrac{200\times(3958.756)^2}{r^2}$

$r^2=\dfrac{200\times(3958.756)^2}{125}$

$r^2=25074798.5$

$r=\sqrt{25074798.5}$

$r=5007.4\ miles$

Hence. The height of the object is 5007.4 miles.

7 0

3 years ago

Sarah weighs 500 newtons. She climbs a set of stairs 6 meters high. How much work does she do? 83 J Work equals force times dist

jek_recluse [69]

Answer:

Answer: Work equals force times distance. 3,000 J

Explanation:

Work Done By A Force

When some force $\vec F$ is applied and a displacement $\vec r$ is achieved, the work done by the force is given by

$W=\vec F \cdot \vec r$

Note that the work is a scalar magnitude as the result of the dot-product of two vectors. If the force and the displacement are parallel, then the vectors can be replaced as its magnitudes F,x and the work is

$W=F.x$

The dot product becomes a simple arithmetic product, i.e force times distance.

Sara weighs 500 Nw and she climbs up a 6 meter set of stairs. She needs to lift her weight up, so the force is the weight and the distance is the height of the stairs, thus

$W=500\ Nw.\ 6\ m=3,000\ Joule$

Answer: Work equals force times distance. 3,000 J

7 0

2 years ago

A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m>s when it is at point P in Fig. E10.35. (a) At this instant, wha

zaharov [31]

Answer:

(A) L = 115.3kgm²/s

(B) dL/dt = 94.1kgm²/s²

Explanation:

The magnitude of the angular momentum of the rock is given by the foemula

L = mvrSinθ

We have been given θ = 36.9°, m = 2.0kg, v = 12.0m/s and r = 8.0m.

Therefore L = 2.00 × 12 × 8.0 × Sin 36.9° =

115.3 kgm²/s

(B) The magnitude of the rate of angular change in momentum is given by

dL /dt = d(mvrSinθ)/dt = mgrSinθ = 2.00 × 9.8 × 8.0× Sin36.9 = 94.1kgm²/s²

7 0

2 years ago

You stop for a cappuccino at a coffee shop and notice that the tiny white bubbles of steamed milk remain on the surface of the c

Rom4ik [11]

Answer:

C. less dense than the coffee but more dense than the air above thecoffee.

Explanation:

3 0

3 years ago