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3 years ago

The speed limmit on an interstate highway is posted at 75mi/h. What is the speed in kilometers per hour? In feet per second?

Physics

1 answer:

jok3333 [9.3K]3 years ago

4 0

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wpazsebu

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You are presently taking a weather observation. The sky is full of wispy cirrus clouds estimated to be about 10 km overhead. If

marshall27 [118]

Answer:

x = 2000 Km

Explanation:

Given

y = 10 km

Slope: 1 : 200

x = ?

We can apply the formula

y / x = 1 / 200 ⇒ x = 200*y = 200*10 Km

⇒ x = 2000 Km

7 0

3 years ago

Science and Technology are interdependent. Advances in one lead to advances in the other. Give an example of this phenomenon

Pepsi [2]

If more microscopic technology is invented, then more things in the world can be discovered. Likewise, if a new element is discovered, then people can use it in their technology.

7 0

3 years ago

5/137 Under the action of its stern and starboard bow thrusters, the cruise ship has the velocity vB = 1 m/s of its mass center

quester [9]

The image is missing, so i have attached it;

Answer:

A) V_rel = [-(2.711)i - (0.2588)j] m/s

B) a_rel = (0.8637i + 0.0642j) m/s²

Explanation:

We are given;

the cruise ship velocity; V_b = 1 m/s

Angular velocity; ω = 1 deg/s = 1° × π/180 rad = 0.01745 rad/s

Angular acceleration;α = -0.5 deg/s² = 0.5 x π/180 rad = -0.008727 rad/s²

Now, let's write Velocity (V_a) at A in terms of the velocity at B(V_b) with r_ba being the position vector from B to A and relative velocity (V_rel)

Thus,

V_a = V_b + (ω•r_ba) + V_rel

Now, V_a = 0. Thus;

0 = V_b + (ω•r_ba) + V_rel

V_rel = -V_b - (ω•r_ba)

From the image and plugging in relevant values, we have;

V_rel = -1[(cos15)i + (sin15)j] - (0.01745k * -100j)

V_rel = - (cos15)i - (sin15)j - 1.745i

Note that; k x j = - i

V_rel = [-(2.711)i - (0.2588)j] m/s

B) Let's write the acceleration at A with respect to B in terms of a_b.

Thus,

a_a = a_b + (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel

a_a and a_b = 0.

Thus;

0 = (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel

a_rel = - (α*r_ba) - (ω(ω•r_ba)) - (2ω*v_rel)

Plugging in the relevant values with their respective position vectors, we have;

a_rel = - (-0.008727k * -100j) - (0.01745k(0.01745k * -100j)) - (2*0.01745k * [-(2.711)i - (0.2588)j])

a_rel = 0.8727i - (0.01745² x 100)j + 0.0946j - 0.009i

Note that; k x j = - i and k x i = j

Thus,simplifying further ;

a_rel = 0.8637i + 0.0642j m/s²

4 0

3 years ago

The ground state energy of an electron in a one-dimensional trap with zero potential energy in the interior and infinite potenti

ElenaW [278]

Answer:

0.5 eV

Explanation:

$E_1$ = Initial potential energy = $2\ eV$

$E_2$ = Final potential energy

$L_1$ = Initial width

$L_2$ = Final width = $2L_1$

Energy of an electron in a one-dimensional trap is given by

$E=\dfrac{n^2h^2}{8mL^2}$

From the equation we get

$E\propto \dfrac{1}{L^2}$

So,

$\dfrac{E_1}{E_2}=\dfrac{L_2^2}{L_1^2}\\\Rightarrow E_2=\dfrac{E_1L_1^2}{L_2^2}\\\Rightarrow E_2=\dfrac{2L_1^2}{4L_1^2}\\\Rightarrow E_2=0.5\ eV$

The ground state energy will be 0.5 eV

6 0

3 years ago

Question 3

Lina20 [59]

Answer: D

Explanation:

7 0

3 years ago