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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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And 8 kg bowling ball is rolling along the frictionless alley
VLD [36.1K]
It will stop eventually
8 0
3 years ago
A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri
exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

a)

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

U_i +K_i = U_f + K_f

where

U_i is the initial potential energy, at the top

K_i is the initial kinetic energy, at the top

U_f is the final potential energy, at halfway

K_f is the final kinetic energy, at halfway

The equation can be rewritten as

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m = 2.7 kg is the mass of the block

g=9.8 m/s^2 is the acceleration of gravity

h_i = 0.54 is the initial height

u = 0 is the initial speed

h_f = 0.25 m is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s

b)

The total mechanical energy of the block can be calculated from the initial conditions, and it is

E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

E=E_e = \frac{1}{2}kx^2

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m

c)

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

E=E_e = 14.3 J

while the final energy is the energy at the point of maximum height, where all the energy has been converted into gravitational potetial energy:

E=U_f = mg h_f

where h_f is the maximum height reached. Solving for this quantity, we find

h_f = \frac{E}{mg}=\frac{14.3}{(2.7)(9.8)}=0.54 m

which is the initial height: this is correct, because the total mechanical energy is conserved, so the block must return to its initial position.

Learn more about kinetic and potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

brainly.com/question/6536722

#LearnwithBrainly

5 0
3 years ago
A government agency estimated that air bags have saved over 14,000 lives as of April 2004 in the United States. (They also state
balu736 [363]

To solve this problem it is necessary to apply the concepts related to momentum, momentum and Force. Mathematically the Impulse can be described as

I = F*t

Where,

F= Force

t= time

At the same time the moment can be described as a function of mass and velocity, that is

P = m\Delta v \rightarrow P=m(v_1-v_2)

Where,

m = mass

v = Velocity

From equilibrium the impulse is equal to the momentum, therefore

I = p

Ft = m(v_1-v_2)

PART A) Since the body ends at rest, we have the final speed is zero, so the momentum would be

p=m(v_1-v_2)

p = 75*0.15

p = 1125Kg\cdot m/s

Therefore the magnitude of the person's impulse is 1125Kg.m/s

PART B) From the equation obtained previously we have that the Force would be:

Ft = m(v_1-v_2)

F(0.025)= 1125

F= 45000N

Therefore the magnitude of the average force the airbag exerts on the person is 45000N

6 0
3 years ago
A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3
Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,  

\text { Electric power }(p)=\frac{V^{2}}{R}

\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}

\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}

For calculating number of turns

\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}

Given that,

80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }

\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}

\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .

We need to find the number of turns in the secondary winding \left(N_{S}\right) to run the bulb at 120W \left(P_{2}\right)

Firstly find the secondary voltage in the transformer use, \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}

\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}

V_{2}^{2}=\frac{120^{2} \times 120}{80}

V_{2}^{2}=\frac{1728000}{80}

V_{2}^{2}=21600

V_{2}=\sqrt{21600}

V_{2}=146.9 \mathrm{V}=V_{S}

Now, finding the number of turns in secondary coil. Use, \frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}

\frac{30}{N_{S}}=\frac{65}{146.9}

N_{S}=\frac{30 \times 146.9}{65}

N_{S}=\frac{4407}{65}N_{S}=67.8

The number of turns in the secondary winding are 67.8 turns.

6 0
3 years ago
An X-ray tube emits X-rays with a wavelength of 1.0 x 10-11 m. Calculate the potential that must be applied across the X-ray tub
sergey [27]

Answer:

1.24 x 10 to the 5 ev = 124,000 ev          its B

Explanation:

E = hc/lambda = 1.24 ev-micrometer/1.0x10 to the -5 micrometers = 1.24 x 10 to the 5 ev = 124,000 ev

h = Planck's constant = 6.626 × 10 to the -34 joule·s

c = speed of light = 2.998 × 10 to the 8 m/s

lambda is the given wavelength

E is the desired photon energy

3 0
3 years ago
Read 2 more answers
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