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Vedmedyk [2.9K]
3 years ago
5

A horse pulls a cart along a flat road. Consider the following four forces that arise in this situation. (1) the force of the ho

rse pulling on the cart (3) the force of the horse pushing on the road (2) the force of the cart pulling on the horse (4) the force of the road pushing on the horse Which two forces form an "action-reaction" pair that obeys Newton's third law?
Physics
1 answer:
Paraphin [41]3 years ago
8 0

Answer:

(1) - (2)

and

(3) - (4)

Explanation:

Hi!

The Newton's third law states that for every action there's and equal and opposite reaction.

Lets us consider the first force:

<em>(1) the force of the horse pulling on the cart</em>

Therefore, its counterpart, will be a force of the cart pulling on the horse, which is actually:

<em> (2) the force of the cart pulling on the horse</em>

<em />

Analogously, the third force:

<em>(3) the force of the horse pushing on the road</em>

and the fourth force:

<em>(4) the force of the road pushing on the horse </em>

<em />

From a action reaction pair, since the former force acts on the road by the horse, and the latter on the horse by the road

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A car is traveling at an average speed of 70 m/s. How many km would the car travel in 6.5 hrs. ?
docker41 [41]

Answer:

<h2>38,769.23 miles</h2>

Explanation:

given:

A car is traveling at an average speed of 70 m/s.

find:

How many km would the car travel in 6.5 hrs. ?

solution:

distance = velocity over time

let velocity = 70 m/s

           time = 6.5 hrs.

convert velocity 70 m/s into m/h for consistency of units.

<u>70 mi. </u> x   <u>3600 sec.</u>  =  252,000 mi/hour

  sec.          1 hr.

now plugin values into the formula d = v/t

d = <u>252,000 miles/hour </u>

              6.5 hours

d = 38,769.23 miles

therefore, the distance travelled in 6.5 hours with a speed of 70 m/s is 38,769.23 miles

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A temperature of 20°C is equivalent to approximately?
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The answer is A.68 degrees

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A jet transport has a weight of 2.25 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel
Rudik [331]

Answer:

Explanation:

Given that,

Weight of jet

W = 2.25 × 10^6 N

It is at rest on the run way.

Two rear wheels are 16m behind the front wheel

Center of gravity of plane 10.6m behind the front wheel

A. Normal force entered on the ground by front wheel.

Taking moment about the the about the real wheel.

Check attachment for better understanding

So,

Clock wise moment = anti-clockwise moment

W × 5.4 = N × 16

2.25 × 10^6 × 5.4 = 16•N

N = 2.25 × 10^6 × 5.4 / 16

N = 7.594 × 10^5 N

B. Normal force on each of the rear two wheels.

Using the second principle of equilibrium body.

Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces

ΣFy = 0

Nr + Nr + N — W = 0

2•Nr = W—N

2•Nr = 2.25 × 10^6 — 7.594 × 10^5

2•Nr = 1.491 × 10^6

Nr = 1.491 × 10^6 / 2

Nr = 7.453 × 10^5 N

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3 years ago
A lamp is rated for 240v,2.5A.what is the cost of using the lamp for 3hrs if 1kWh of electricity cost #12​
marshall27 [118]

Answer:

≈ 22¢

Explanation:

240 / 1000 = 0.240 kV

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6 0
3 years ago
U = 3, 9 , v = 4, 2 (a) find the projection of u onto v. (b) find the vector component of u orthogonal to v.
ExtremeBDS [4]

Answer:

(a) At U = 3, 9 , v = 4, 2, the projection of u onto v is w1=<2,8>

(b)At U = 3, 9 , v = 4, 2,  the vector component of u orthogonal to v is w2 = <4,-1>

Explanation:

A

The projection of u onto v is given by:

w1= projvu = (u⋅v||v||2)v

Given that  u= <6,7> and v=<1,4>, we can find the projection of u onto v as shown below:

w1= projvu = (u⋅v||v||2v=(<6,7>⋅<1,4><1,4>⋅<1,)

=(6⋅1+7⋅41⋅1+4⋅4)<1,4>

=3417<1,4>

=<2,8>

Part B

The vector component of u orthogonal to v is given by:

Using the given vectors and the projection found in part (a), we can find the vector component of u orthogonal to v as shown below:

w2=u−projvu

=<6,7≻<2,8>

=<(6−2),(7−8)>

=<4,−1>

To learn more about vector component, click brainly.com/question/17016695

#SPJ4

5 0
2 years ago
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