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Nookie1986 [14]
3 years ago
7

An object is launched with an initial speed of 30 m/s at an angle of 60° above the horizontal . What is the maximum height reach

ed by the object
Physics
1 answer:
matrenka [14]3 years ago
6 0

Answer:

H = 34.43 m

Explanation:

Given that,

Initial speed of the object, u = 30 m/s

The angle of projection, \theta=60^{\circ}

We need to find the maximum height reached by the object. Let it is H. Using the formula for maximum height reached by the projectile.

H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(30)^2\times \sin^2(60)}{2\times 9.8}\\\\H=34.43\ m

So, the maximum height reached by the object is 34.43 m.

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The Moon s escape speed will be smaller than Earth's.

  • What is escape speed:

The minimum speed that is required for an object to free itself from the gravitational force exerted by a massive object.

The formula of escape speed is

  • v = \sqrt{\frac{2GM}{R} }

where

v is escape velocity

G is universal gravitational constant

M is mass of the body to be escaped from

r is distance from the center of the mass

we can say that,

Escape speed depends on the gravity of the object trying to hold the spacecraft from escaping.

we know that,

The Moon's surface gravity is about 1/6th as powerful or about 1.6 meters per second per second.

since, v ∝ g

The Moon s escape speed will be smaller than Earth's.

Learn more about escape speed here:

<u>brainly.com/question/15318861</u>

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2 years ago
What causes the phenomena?
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Most pneumonia occurs when a breakdown in your body's natural defenses allows germs to invade and multiply within your lungs.

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3 years ago
I NEED HELP ASAP! BRAINIEST TO THE CORRECT ANSWER. HELP ME NOW!
sergij07 [2.7K]

Answer:

<h3>a)</h3>

\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}

  • V = 12 V
  • P = 24 W

\implies \mathsf{24=\frac{12^2}{R} }

\implies \mathsf{24R=12^2 }

\implies \mathsf{24R=144 }

<u>=> R= 6 Ohms(Ω)</u>

<h3>b)</h3>

\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}

  • Power (P)= 100 W

<em>these lights operate at the usual 240 volts direct from the main electricity supply. Therefore,</em>

  • V = 240 V

\implies \mathsf{100=\frac{240^2}{R} }

<em>R and 100 can interchange places</em>

\implies \mathsf{R=\frac{240^2}{100} }

\implies \mathsf{R=\frac{57600}{100} }

<u>=> R = 576 Ω</u>

<u></u>

By Ohm's Law:

\boxed{\mathsf{Voltage(V)=Current(I) \times Resistance(R)}}

=> 240 = I × 576

=>

=> I = 0.417 A

<h3 /><h3>c)</h3>

I don't know it's resistance,... so sorry

<h3>d)</h3>

The brightness of the bulb in series is <em><u>less than</u></em> when they're placed individually.

For bulbs in series their resistance gets added to form the equivalent resistance of the two bulbs.

Their resistances are nothing but mere numbers and the sum of two numbers(positive of course) is greater than the numbers.

So, the effective resistance of some bulbs in series <u>is more</u> than the individual resistance.

And

<em>Brightness, i. e., Power</em>

\boxed{\mathfrak{Power \propto  \frac{1}{Resistance} }}

If resistance increases, Power decreases.

Here, the effective resistance was for sure larger, therefore resistance was increasing, hence power decreased taking brightness along with it.

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3 years ago
Can someone sent some gravity questions
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What do you mean ???
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2 years ago
Read 2 more answers
Please Help Me I will try and give extra points if I can and if you have the rest of the answers to this then please help me out
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First, you must know that the statement "<span>An object at rest tends to stay at rest. An object in motion tends to stay in motion unless acted upon by an outside force." is true, because it is the first Law of Newton or inercy law.


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After the ball leaves the pitcher's hand, it is subject only to the gravitational attraction of the Earth. That is why the pitcher has to give the appropiate impulse in order to the ball reaches the point that he and the catcher want.

What about if you threw it straight into the air?

It is the same thing. The only force would be the gravitational attraction of the Earth.

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In outer space, at the beginning the baseball would be very far from of other massive objects to feel their gravitational field, so there would not be any forces to slow it down. Although eventulally, after many light-years, it would enter the gravitational field of a galaxy or other massive body and it would attract it.
</span>
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