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3 years ago

11

What is the difference between hypothesis, theory and scientific law?

1 answer:

patriot [66]3 years ago

5 0

Hypothesis is what you think will happen , a theory is a system of ideas to explain something. scientific law means a statement that is based on repeated experimental observations.

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A helicopter carries a 1000-kg-car suspended from a rope below it. The helicopter flies horizontally at a constant speed of 25 m

kolbaska11 [484]

Answer:

(a) Tension, T = 28.653 kN

(b) Wind resistance force, $26.925\ kN$

Solution:

As per the question:

Mass of the car, m = 1000 kg

Speed of the helicopter, v = 25 m/s

Angle made by the rope, $theta = 20^{\circ}$

Now,

(a) To calculate the tension, T in the car:

Tension along the direction of motion, $T_{h} = Tcos20^{\circ}$

Tension along the vertical direction, $T_{v} = Tsin20^{\circ}$

Now, let the force due to the wind directed in the opposite direction of the motion be $F_{W}$ and it balances the horizontal component of the tension, T.

The vertical component is balance by the weight of the car, i.e., mg that acts vertically downwards.

Now,

$T_{v} = mg$

$Tsin20^{\circ} = 1000\times 9.8$

T = 28653 N = 28.653 kN

(b) The force of the wind resistance:

$F_{W} = T_{h}$

$F_{W} = 2cos20^{\circ} = 26925\ N = 26.925\ kN$

(c) Now,

If the angle made by the rope with the vertical is $0^{\circ}$ :

$mg = Tsin(90^{\circ} - 0^{\circ})$

$Tsin90^{\circ} = mg = 9800\ N$

The tension in the rope will be equal to the weight the car.

Wind resistance force, $F_{W} = Tcos90^{\circ} = 0\ N$

If the angle made by the rope with the vertical is $90^{\circ}$ :

$mg = Tsin(90^{\circ} - 90^{\circ})$

T = 0 N

Wind resistance force, $F_{W} = Tsin0^{\circ}$

$Tsin0^{\circ} = mg$

$F_{W} = \infty$

There will be no tension in the rope and wind resistance will be infinite.

3 0

3 years ago

The oscillating electric field in a plane electromagnetic wave is given by <img src="https://tex.z-dn.net/?f=%7B50%5Csin%20%28%5

Scorpion4ik [409]

Here, we are given with:

${:\implies \quad \longrightarrow \begin{cases}\sf E_{0}= 50\: V/m\\ \sf \nu = 2\times 10^{7}Hz\end{cases}}$

(a) So now, we can thus obtain

${:\implies \quad \sf \omega =2\pi \nu}$

${:\implies \quad \sf \omega =2\pi (2\times 10^{7})}$

${:\implies \quad \boxed{\bf{\omega = 4\pi \times 10^{7}\: s^{-1}}}}$

Now, finding $\lambda$

${:\implies \quad \sf \lambda =\dfrac{c}{\omega}=\dfrac{3\times 10^{8}}{4\pi \times 10^{7}}}$

${:\implies \quad \boxed{\bf{\lambda \approx 2.39\:m}}}$

Now, finding k

${:\implies \quad \sf k=\dfrac{2\pi}{\lambda}=\dfrac{200\pi}{239}}$

${:\implies \quad \boxed{\bf{k\approx 2.63\:m^{-1}}}}$

Thus, expression for the electric field is:

${:\implies \quad \boxed{\bf{E=50\sin \bigg(4\pi \times 10^{7}t-\dfrac{263x}{100}\bigg)}}}$

(b) Now, here

${:\implies \quad \sf B_{0}=\dfrac{E_{0}}{c}=\dfrac{50}{3\times 10^{8}}}$

${:\implies \quad \boxed{\bf{B_{0}\approx 16.67×10^{-7}\:T}}}$

Thus, expression for the magnetic field:

${:\implies \quad \boxed{\bf{B_{y}=16.67\times 10^{-7}\sin \bigg(4\pi \times 10^{7}t-\dfrac{263x}{100}\bigg)\:T}}}$

(c) The electromagnetic wave propagates along Z-axis

4 0

2 years ago

On the average, about what percentage of the solar energy that strikes the outer atmosphere eventually reaches the earth's surfa

almond37 [142]

I think the answer to your question is twenty-four percent.

5 0

3 years ago

Which of the following represents an image that is located in front of a lens?

Lostsunrise [7]

-di represents an image in front of a lens

5 0

3 years ago

Read 2 more answers

The velocity of a moving body increases from 10m/s to 15 m/s in 5 sec . calculate its acceleration

Gnoma [55]

Answer:

v=u+at

15=10+a x 5

15-10= 5a

5= 5a

a=1 m/sec^2

4 0

3 years ago