High tide weird question

An airplane takes off from Dallas Texas to fly to new york city traveling ne for 2,760 km the Same plane returns that day to Dal

Kipish [7]

<h3>Answer</h3>

1104 km/hour

<h3>Explanation</h3>

Distance between Dallas Texas to New York = 2760 km

Time the plane took from Dallas to New York = 2 hours

Time the plane took from New York back to Dallas = 2.5 hours

Formula to use

<h3>distance = speed x time </h3>

Speed the plane took from Dallas to New York

2760 = 2 x speed

speed = 2760 / 2

= 1380 km/hour

Speed the plane took from New York to Dallas (ROUND TRIP)

2760 = 2.5 x speed

speed = 2760 / 2.5

= 1104 km/hour

3 0

3 years ago

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Two bodies are falling with negligible air resistance, side by side, above a horizontal plane. If one of the bodies is given an

tankabanditka [31]

Answer:4-strikes the plane at same time as the other body

Explanation:

Given

If both bodies is falling on a horizontal plane and second body is given an acceleration in horizontal direction then it does not change the time to reach the Horizontal Plate as there is no change in vertical direction.

Horizontal acceleration will give only horizontal range and horizontal velocity.

8 0

3 years ago

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A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago

Which of the following choices concerning the net magnetic flux through any enclosed surface is true according to Gauss' law for

11Alexandr11 [23.1K]

Answer:

e. The net magnetic flux in this case would be equal to zero.

Explanation:

As per Gauss law of magnetism we need to find the net magnetic flux through a closed loop

here we know that net magnetic flux is the scalar product of magnetic field vector and area vector

so here we have

$\int B. dA$ = net magnetic flux

since we know that magnetic field always forms closed loop so if we find the integral over a closed loop

then in that case the value of the close integral must be zero

so correct answer would be

e. The net magnetic flux in this case would be equal to zero.

6 0

3 years ago

A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a rope that goes over an ideal pulley. If the masses are

irga5000 [103]

Answer:

The time taken will be 0.553 seconds.

Explanation:

We should start off by finding the force exerted by the rope on the 3kg weight in this case.

Weight of 5kg mass = 5 * 9.81 = 49.05 N

Weight of 3kg mass = 3 * 9.81 = 29.43 N

The force acting upward on the 3kg mass will equal the weight of the 5kg mass. Thus the resultant force acting on the 3kg mass is:

Total force = 49.05 - 29.43 = 19.62 N (upwards)

We can now find the acceleration:

F = m * a

19.62 = 3 * a

a = 6.54 m/s^2

We now use the following equation of motion to get the time taken to travel 1 meter:

$s=u*t+\frac{1}{2} (a*t^2)$

$1=0*t+\frac{1}{2} (6.54*t^2)$

t = 0.553 seconds

4 0

4 years ago