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4 years ago

5

A projectile is launched at an angle of 60° above the horizontal. compared to the vertical component of the initial velocity of

the projectile, the vertical component of the projectile's velocity when it has reached its maximum height is

1 answer:

jeka944 years ago

8 0

Equal to zero. Because at maximum height before it comes down, it experiences a momentary stop

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For the velocity-time graph showns, which statement describes what happens to the velocity between approximately 11 and 15 secon

dalvyx [7]

Answer:

C) The lander’s velocity decrease toward the reference.

Explanation:

took the test

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3 years ago

A particle of mass 3m is located 1.00 m from a particle of mass m. (a) Where should you put a third mass M so that the net gravi

Airida [17]

Explanation:

It is given that net gravitational force on M is exactly equal to zero. Hence, distance to M from the bigger mass is 3m. Therefore, expression for net force will be as follows.

$F_{net} = F_{1} + F_{2} = 0$

So,

$\frac{-G(3m)(M)}{x^{2}} + \frac{G(m)(M)}{(1 - x)^{2}} = 0$

The first term is negative as the third mass is located between the other two masses. This means that 3 m will be pulling it leftwards (negative x direction) and m will be pulling it rightwards (positive x direction).

$\frac{G(m)(M)}{(1 - x)^{2}} = \frac{G(3m)(M)}{(x)^{2}}$

On dividing both sides of the equation by G.m.M, we get the following.

$\frac{1}{(1 - x)^{2}} = \frac{3}{x^{2}}$

$x^{2} = 3 - 6x + 3x^{2}$

0 = $3 - 6x + 2x^{2}$

Using the formula, $\frac{-b \pm \sqrt{(b)^{2} - 4ac}}{2a}$ the value of x comes out to be equal to +2.37 (not usabale) and -0.634 (usable).

Hence, we can conclude that the third mass will be located 0.634 meters away from the 3 m mass.

7 0

3 years ago

_mg(oh)2 + _HBr _mgBr2 + _HOH

s344n2d4d5 [400]

A) that's not physics, that's chemistry B) assuming you want it balanced, 1,2,1,2

3 0

3 years ago

Three ropes A, B and C are tied together in one single knot K. (See figure.)

noname [10]

The tension in the rope B is determined as 10.9 N.

<h3>Vertical angle of cable B</h3>

tanθ = (6 - 4)/(5 - 0)

tan θ = (2)/(5)

tan θ = 0.4

θ = arc tan(0.4) = 21.8 ⁰

<h3>Angle between B and C</h3>

θ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰

Apply cosine rule to determine the tension in rope B;

A² = B² + C² - 2BC(cos A)

B = C

A² = B² + B² - (2B²)(cos A)

A² = 2B² - 2B²(cos 43.6)

A² = 0.55B²

B² = A²/0.55

B² = 65.3/0.55

B² = 118.73

B = √(118.73)

B = 10.9 N

Thus, the tension in the rope B is determined as 10.9 N.

Learn more about tension here: brainly.com/question/24994188

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3 0

2 years ago

JUST PLZ HELP!!! Why does the lightbulb in the right electrical circuit turn on but not the one on the left?

makkiz [27]

Because,

In left image pin is not touch to the wire.

In right image pin is touch to the wire.

Hope it helps you.....

Plz...Plz...Plz...Plz…Plz…

Mark be Brainliest.....

Please.....

And..

Please thanks me.....

Plz.....Plz.....

8 0

3 years ago

Read 2 more answers