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3 years ago

For all simple machines, when the output force is greater than the input force,

1 answer:

7 0

For all simple machines, when the output force is greater than the input force, <span>the input force is exerted over a larger distance than the output force.

In short, Your Answer would be Option D

Hope this helps!</span>

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____________ is the number of complete movements of a wave per second.

Citrus2011 [14]

hertz: A unit for measuring frequency, equal to one cycle per second. If a sound wave has a frequency of 20,000 Hz,

8 0

2 years ago

Read 2 more answers

Some believe that the positions of the planets at the time

laila [671]

Answer:

$1.4007\times 10^{-8}\ N$

$1.50075\times 10^{-6}\ N$

$0.000000667\ N$

Explanation:

$m_1$ = Mass of baby = 3 kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Distance between objects

Gravitational force of attraction is given by

$F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 70\times 3}{1^2}\\\Rightarrow F=1.4007\times 10^{-8}\ N$

The force between baby and obstetrician is $1.4007\times 10^{-8}\ N$

$F=\dfrac{6.67\times 10^{-11}\times 3\times 2.7\times 10^{27}}{(6\times 10^{11})^2}\\\Rightarrow F=1.50075\times 10^{-6}\ N$

The force between the baby and Jupiter is $1.50075\times 10^{-6}\ N$

$F=\dfrac{6.67\times 10^{-11}\times 3\times 2.7\times 10^{27}}{(9\times 10^{11})^2}\\\Rightarrow F=0.000000667\ N$

The force between the baby and Jupiter is $0.000000667\ N$

7 0

3 years ago

An observer is approaching at stationary source at 17.0 m/s. Assuming the speed of sound is 343 m/s, what is the frequency heard

UNO [17]

Answer:

the frequency heard by the observer is equal to 2677 Hz

Explanation:

given,

velocity of the observer = 17 m/s

speed of the sound = 343 m/s

velocity of the source = 0 m/s

frequency emitted from the source = 2550 Hz

$f = f_0(\dfrac{v-v_0}{v-v_s})$

$f = 2550\times (\dfrac{343+17}{343-0})$

velocity of observer is negative as it is approaching the source. f = 2676.38 Hz ≈ 2677 Hz

hence, the frequency heard by the observer is equal to 2677 Hz

7 0

3 years ago

an always be used to calculate the electric field. relates the electric field at points on a closed surface to the net charge en

Temka [501]

Complete Question:

Gauss's law:

Group of answer choices

A. can always be used to calculate the electric field.

B. relates the electric field throughout space to the charges distributed through that space.

C. only applies to point charges.

D. relates the electric field at points on a closed surface to the net charge enclosed by that surface.

E. relates the surface charge density to the electric field.

Answer:

D. relates the electric field at points on a closed surface to the net charge enclosed by that surface.

Explanation:

Gauss's law states that the total (net) flux of an electric field at points on a closed surface is directly proportional to the electric charge enclosed by that surface.

This ultimately implies that, Gauss's law relates the electric field at points on a closed surface to the net charge enclosed by that surface.

This electromagnetism law was formulated in 1835 by famous scientists known as Carl Friedrich Gauss.

Mathematically, Gauss's law is given by this formula;

ϕ = (Q/ϵ0)

Where;

ϕ is the electric flux.

Q represents the total charge in an enclosed surface.

ε0 is the electric constant.

8 0

2 years ago

insens350 [35]

Answer:

774.8 secs

Explanation:

distance(d)= speed(v)* time(t)

calculate speed:

refractive index = speed of light (c)/ speed of light in medium (v)

1.56 = 3*10^8*v

v=192307692.3 m/s

d = v *t

t = d/v

on substituting values:

t = 774.8 secs

4 0

2 years ago