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2 years ago

6

What kind of model is shown below? Qin = W + Qout A. Physical model B. Mathematical model C. Computer model D. Experimental mode

l

1 answer:

amm18122 years ago

7 0

Answer: mathematical model

Explanation: just took the test

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An object moves from point A to point C along the rectangle shown in the figure below.

Naily [24]

Answer:

Hello friend where is the figure of the question

4 0

3 years ago

BaLLatris [955]

part 1

mass = ρ x V

mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg

PE (potential energy)= mgh

PE = 6608.2 kg x 9.81 x 403

PE = 2.61 x 10⁷ J

part 2

megaton of TNT (Mt) =4.2 x 10¹⁵ J

convert PE to Mt:

2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt

4 0

3 years ago

Can someone help me ASAP

Phantasy [73]

Well I don't know. Let's actually LOOK at the picture and see if that helps.

A, B, C, and D all have the same TOTAL length, but A has the most waves crammed into that same total length.

By golly, that means the length of <u><em>each</em></u> wave in A must be shorter than each wave in B, C, or D.

The correct choice is <em> A </em>. Looking at the picture did the trick !

7 0

3 years ago

Light shined through a single slit will produce a diffraction pattern. Green light (565 nm) is shined on a slit with width 0.210

kondor19780726 [428]

Answer:(a)9.685 mm

(b)4.184 mm

Explanation:

Given

Wavelength of light $(\lambda )=565nm \approx 565\times 10^{-9}m$

Width of slit(b)=0.210

(a)Width of central maximum located 1.80m from slit

$=\frac{2\lambda L}{b}$

$=\frac{2\times 565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}$

=9.685 mm

(b)Width of the first order bright fringe

$Y_1=\frac{\lambda \times L}{b}$

$Y_1=\frac{565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}$

$Y_1=4.84mm$

5 0

3 years ago

A person hits a tennis ball with a mass of 0.058 kg against a wall.

horrorfan [7]

Explanation:

Mass of the ball, m = 0.058 kg

Initial speed of the ball, u = 11 m/s

Final speed of the ball, v = -11 m/s (negative as it rebounds)

Time, t = 2.1 s

(a) Let F is the average force exerted on the wall. It is given by :

$F=\dfrac{m(v-u)}{t}$

$F=\dfrac{0.058\times (11-(-11))}{2.1}$

F = 0.607 N

(b) Area of wall, $A=3\ m^2$

Let P is the average pressure on that area. It is given by :

$P=\dfrac{F}{A}$

$P=\dfrac{0.607\ N}{3\ m^2}$

P = 0.202 Pa

Hence, this is the required solution.

8 0

3 years ago