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3 years ago

Which of the following is a field (action-at-a-distance) force?

Physics

1 answer:

Kobotan [32]3 years ago

3 0

Answer:

$F_{grav}$ is a field force, as gravity does not physically apply force and does not require proximity.

Explanation:

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On his way to class, a student on a skateboard is accelerating on a downhill stretch. Which of the following statements is true?

sasho [114]

Answer:

B is correct

Explanation:

Because of gravity

3 0

3 years ago

there are 1.6 km in a mile. the distance between two cities is 248 miles. How many kilometers apart are the two cities?

podryga [215]

1.6 X 248=399.117
hope this helps

3 0

3 years ago

A 70.9-kg boy and a 43.2-kg girl, both wearing skates face each other at rest on a skating rink. The boy pushes the girl, sendin

Lelechka [254]

Answer:

Explanation:

Given

mass of boy $m_b=70.9\ kg$

mass of girl $m_g=43.2\ kg$

speed of girl after push $v_g=4.64\ m/s$

Suppose speed of boy after push is $v_b$

initially momentum of system is zero so final momentum is also zero because momentum is conserved

$P_i=P=f$

$0=m_b\cdot v_b+m_g\cdot v_g$

$v_b=-\frac{m_g}{m_b}\times v_g$

$v_b=-\frac{43.2}{70.9}\times 4.64$

$v_b=-2.82\ m/s$

i.e. velocity of boy is 2.82 m/s towards west

8 0

3 years ago

Light of wavelength 400 nm is incident on a single slit of width 15 microns. If a screen is placed 2.5 m from the slit. How far

olganol [36]

Answer:

0.0667 m

Explanation:

λ = wavelength of light = 400 nm = 400 x 10⁻⁹ m

D = screen distance = 2.5 m

d = slit width = 15 x 10⁻⁶ m

n = order = 1

θ = angle = ?

Using the equation

d Sinθ = n λ

(15 x 10⁻⁶) Sinθ = (1) (400 x 10⁻⁹)

Sinθ = 26.67 x 10⁻³

y = position of first minimum

Using the equation for small angles

tanθ = Sinθ = y/D

26.67 x 10⁻³ = y/2.5

y = 0.0667 m

5 0

3 years ago

You are trying to hear your friend give directions to new store in town. But from your distance (1 point) of 15 m you only hear

Marat540 [252]

Answer:

option D

Explanation:

given,

Intensity of sound = 20 dB

distance = 15 m

intensity of sound is increased to = 50 dB

distance between the sound level = ?

Using relation

$L_2 = L_1 - |20(log \dfrac{r_2}{r_1})|$

L₁ = 20 dB L₂ = 50 dB r₁ = 15 m r₂ = ?

$log (\dfrac{r_2}{r_1}) = \dfrac{L_1 -L_2}{20}$

$\dfrac{r_2}{r_1}= 10^{\dfrac{|L_1 -L_2|}{20}}$

$r_2 =r_1 10^{\dfrac{|L_1 -L_2|}{20}}$

$r_2 =15 \times 10^{\dfrac{|20-50|}{20}}$

$r_2 =15 \times 10^{-1.5}$

r₂ = 0.47 m

r₂ = 47 cm

hence, the correct answer is option D

7 0

3 years ago