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2 years ago

3- A car was traveled 110 meters in 5.5 seconds. What was the car's velocity?*

Physics

2 answers:

Tatiana [17]2 years ago

7 0

<u>Answer</u>:

20 m/s

<u>Explanation</u>:

* velocity = distance / time *

velocity = 110 / 5.5

velocity = 20 m/s

xxMikexx [17]2 years ago

7 0

Answer:

V= 20m/s

Explanation:

velocity= distance/time

V= d/s

= 110m/5.5 seconds

= 20m/s

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Explanation:

the light ray leaving a medium in contrast to the entering or incident ray.

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2 years ago

Two forces F1 and F2 act on an object at a point P in the indicated directions. The magnitude of F1 is 15 lb and the magnitude o

adoni [48]

Answer

given,

F₁ = 15 lb

F₂ = 8 lb

θ₁ = 45°

θ₂ = 25°

Assuming the question's diagram is attached below.

now,

computing the horizontal component of the forces.

F_h = F₁ cos θ₁ - F₂ cos θ₂

F_h = 15 cos 45° - 8 cos 25°

F_h = 3.36 lb

now, vertical component of the forces

F_v = F₁ sin θ₁ + F₂ sin θ₂

F_v = 15 sin 45° + 8 sin 25°

F_v = 13.98 lb

resultant force would be equal to

$F = \sqrt{F_h^2+F_v^2}$

$F = \sqrt{3.36^2+13.98^2}$

F = 14.38 lb

the magnitude of resultant force is equal to 14.38 lb

direction of forces

$\theta =tan^{-1}(\dfrac{F_v}{F_h})$

$\theta =tan^{-1}(\dfrac{13.98}{3.36})$

θ = 76.48°

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2 years ago

How far from Earth must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances

Tatiana [17]

Answer:

258774.9441 m

Explanation:

x = Distance of probe from Earth

y = Distance of probe from Sun

Distance between Earth and Sun = $x+y=149.6\times 10^6\ m$

G = Gravitational constant

$M_s$ = Mass of Sun = $1.989\times 10^{30}$

$M_e$ = Mass of Earth = $5.972\times 10^{24}\ kg$

According to the question

$\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y$

$x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m$

The probe should be 258774.9441 m from Earth

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2 years ago

Define accleration due to gravity

kaheart [24]

Therefore as you move around the U.S. the acceleration due to gravity (g) varies from about 9.79 to 9.81 meters per second squared. The Earth’s average is 9.80 m/s2 which is generally reported as the acceleration of gravity on Earth.

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2 years ago

Read 2 more answers

A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of ma

ahrayia [7]

Incomplete question as we have not told to find what.So the complete question is here

A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of mass m, moves with velocity (-60 m/s)i and a second piece, also of mass m, moves with velocity (-60 m/s)j. The third piece has mass 3m.Just after the explosion, what are the (a) magnitude and (b) direction of the velocity of the third piece?

Answer:

$V_{3}=(20i+20j)m/s$

Explanation:

Given data

The vessel at rest

Piece one,of mass m,moves with velocity=(-60 m/s)i

Piece two,of mass m,moves with velocity=(-60 m/s)j

Piece three,of mass 3m

As the linear momentum is conserved in this system,Because the system is closed and no external force acting on it

So momentum is given as

$p_{initial}=p_{final}$

As the vessel at rest so the initial momentum is zero

$m_{1}V_{1}+m_{2}V_{2}+m_{3}V_{3}=0\\m_{3}V_{3}=-m_{1}V_{1}-m_{2}V_{2}\\V_{3}=\frac{-m_{1}V_{1}-m_{2}V_{2}}{m_{3}} \\V_{3}=\frac{-m_{1}(-60m/s)i-m_{2}(-60m/s)j}{3m}\\V_{3}=(20i+20j)m/s$

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3 years ago