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strojnjashka [21]
4 years ago
5

The nine ring wraiths want to fly from barad-dur to rivendell. rivendell is directly north of barad-dur. the dark tower reports

that the wind is coming from the west at 58 miles per hour. in order to travel in a straight line, the ring wraiths decide to head northwest. at what speed should they fly (omit units)?
Physics
2 answers:
Katarina [22]4 years ago
5 0
The west constituent of their sequence needs to cancel out 58 mph crosswind. Subsequently a northwest direction is a 45-degree angle up to even with the destination. That is the third point out of the triangle and the right angle is at the destination. The top side is the west constituent of their flight the vertical side is their resultant travel and the hypotenuse is their definite distance flown. Since the 58 mph crosswind was negated by flying northwest, the distance from the beginning to the destination must be the same distance as the west component of their travel. The hypotenuse is square root of twice the side since it has 2 identical sides.

c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02 
 
Alternative solution:

c = sqrt (2) * 58 = 1.414 * 58 = 82.02

Therefore, they have to fly 82.02 mph
agasfer [191]4 years ago
5 0

Answer:

V = 82.02 mph

Explanation:

Speed of the wind, v = 58 mph

We need to find the speed with which it fly. It will form a right angled triangle. It can be calculated as :

V=\sqrt{58^2+58^2}

V = 82.02 mph

So, the speed of the wind is 82.02 mph. Hence, this is the required solution.

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To move a heavy object, like a refrigerator what could be used to help decrease the frictional force?
Olenka [21]

Answer:

A. Remove everything in the refrigerator to lighten the load.

B. Put a lubricant between the surface of the object and the floor

C. Use round objects, like pencils , to decrease the friction and push the refrigerator over the pencils more easily

Explanation:

Force of friction is a resistance force which acts between two surfaces which are in relative motion. Friction is both boon and bane. Due to friction, we are able to sit, walk etc but also, due to friction there is dissipation of energy. Friction can be reduced by applying lubricants, reducing contact area, reducing the load.

F = μN where N is the normal force which depends on the mass.

Thus, by reducing the load, force of friction can be reduced. Round objects like wheels can also be used. By this the contact area reduces.

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3 years ago
What is power?
hammer [34]
We can define power as the rate of doing work, it is the work done in unit time. The SI unit of power is Watt (W) which is joules per second (J/s). Sometimes the power of motor vehicles and other machines are given in terms of Horsepower (hp) which is approximately equal to 745.7 watts.


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3 years ago
Read 2 more answers
How many joules of work are done on an object when a force of 10 N pushes it 5 m?
zhenek [66]

Answer:

option C

Explanation:

given,                            

Force on the object = 10 N

distance of push = 5 m

Work done = ?              

we know,              

work done is equal to Force into displacement.

W = F . s            

W = 10 x 5              

W = 50 J                

Work done by the object when 10 N force is applied is equal to 50 J

Hence, the correct answer is option C

5 0
4 years ago
Please need help fast
iVinArrow [24]

(a) See graph in attachment

The appropriate graph to draw in this part is a graph of velocity vs time.

In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.

Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at

t = 10 s

v = 15 m/s

(b) 1.5 m/s^2

The average acceleration of the horse can be calculated as:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval

In this problem,

u = 0

v = 15 m/s

t = 10 s

Substituting,

a=\frac{15-0}{10}=1.5 m/s^2

(c) 75 m

For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance travelled

u is the initial velocity

t is the time interval

a is the acceleration

In this problem,

u = 0

t = 10 s

a=1.5 m/s^2

Substituting,

s=0+\frac{1}{2}(1.5)(10)^2=75 m

(d) See attached graphs

In a uniformly accelerated motion:

- The distance travelled (x) follows the equation mentioned in part c,

x=ut+\frac{1}{2}at^2

So, we see that this has the form of a parabola: therefore, the graph x vs t will represents a parabola.

- The acceleration is constant during the motion, and its value is

a=1.5 m/s^2 (calculated in part b)

therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.

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