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3 years ago

The change in elevation from one contour line to the next is called the

Physics

2 answers:

Ad libitum [116K]3 years ago

8 0

Yeah it’s counter interval

Snezhnost [94]3 years ago

3 0

The answer is counter interval

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Environment includes _______.

insens350 [35]

D should be correct.

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3 years ago

Why is the work output of a machine never equal to the work input? A. Some work input is used to overcome friction. B. Input dis

SVETLANKA909090 [29]

The answer is A. <span>Some work input is used to overcome friction. </span>

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3 years ago

Read 2 more answers

Froghopper insects have a typical mass of around 12.5 mg and can jump to a height of 42.3 cm. The takeoff velocity is achieved a

Maksim231197 [3]

Answer:

2065.005 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s$

Now H-h = 42.3 - 0.2 = 42.1 cm = 0.421 m

The final velocity will be the initial velocity

$v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.421\\\Rightarrow a=-\frac{0.421\times -9.81}{0.002}\\\Rightarrow a=2065.005\ m/s^2$

Acceleration of the frog is 2065.005 m/s²

6 0

3 years ago

"A bridge, constructed of 11 beams of equal length L and negligible mass, supports an object of mass M.

fiasKO [112]

F_P + F_Q = M g
F_P = M g - F_Q
Torque, or moment of force:
∑ M_P = 0
∑ M_P = M g L - F_Q · 3 L
0 = M g L - 3 F_Q L / : L
0 = M g - 3 F_Q
3 F_Q = M g
F_Q = M g /3
Finally:
F_P = M g - M g/3
F_P = 4 M g / 3

7 0

3 years ago

Read 2 more answers

You stand on a frictional platform that is rotating at 1.1 rev/s. Your arms are outstretched, and you hold a heavy weight in eac

bezimeni [28]

Answer:

a) The resulting angular speed of platform is 1.38 rev/sec

b) The change in kinetic energy of the system is 53 J.

Explanation:

This question is incomplete. The complete question will be:

You stand on a frictional platform that is rotating at 1.1 rev/s. Your arms are outstretched, and you hold a heavy weight in each hand. The moment of inertia of you, the extended weights, and the platform is 8.8 kg · m2. When you pull the weights in toward your body, the moment of inertia decreases to 7.0 k g .m 2

a) What is the resulting angular speed of the platform? Answer in units of r e v / s .

b)What is the change in kinetic energy of the system? Answer in units of J.

<h3>ANSWER:</h3>

we know that:

Angular Momentum = L = Iω

From conservation of momentum:

Lo = Lf

(Io) (ωo) = (If) (ωf)

ωf = (Io) (ωo)/(If)

ωf = (8.8 kg.m²)(1.1 rev/s)/(7.0 kg.m²)

ωf = (1.38 rev/sec)(2π rad/ 1 rev) = 8.67 rad/sec

ωo = (1.1 rev/sec)(2π rad/ 1 rev) = 6.91 rad/sec

The kinetic energy for rotational motion is given as:

K.E = (1/2)Iω²

Thus, the change in kinetic energy will be:

ΔK.E = (K.E)f - (K.E)o

ΔK.E = (1/2)Ifωf² - (1/2)Ioωo²

ΔK.E = (1/2)(Ifωf² - Ioωo²)

ΔK.E = (1/2)[(7 kg.m²)(8.67 rad/sec)² - (8.8 kg.m²)(6.91 rad/sec)²

5 0

3 years ago