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kolezko [41]
3 years ago
13

HELLO ........... OVER HERE............. HELP NEEDED .............. THANK YOU

Physics
1 answer:
Hatshy [7]3 years ago
5 0

Explanation:

_z_________z__z_z______

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Let's examine over the next few questions an NFL kick as described in the 3rd Law video. The announcer claimed that the ball is
hichkok12 [17]

Answer:

8451.62109367671 Newtons

Explanation:

1\ kg=1\times 9.8066500286389=9.8066500286389\ N

1 \lb=2.2046226218488\ kg

It is known that

1\ lbs=\dfrac{9.8066500286389}{2.2046226218488}=4.4482216282508\ N

So,

1900\ lbs=1900\times 4.4482216282508\\\Rightarrow 1900\ lbs=8451.6210936765\ N

The force in Newtons is 8451.6210936765 Newtons

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4 years ago
Another word for iron oxide?​
NARA [144]

Answer:

Hydrated iron(III) oxide, or ferric oxide

Explanation:

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2 years ago
Pls place the words in the right place I'll give brainless and the points!
Sphinxa [80]

Explanation:

you measure temperature in degrees celsius using a thermometer. Thermal energy is measured in joules. A larger volume of water will take longer to heat up but will store more energy than the smaller object. However, a smaller object will lose it's heat faster than a larger object. A cup of tea has less thermal energy than a swimming pool.

5 0
3 years ago
2. A rocket blasts off vertically from rest on the launch pad with a constant upward acceleration of 2.5 . At 30.0 s after blast
harina [27]

Answer:

a)The highest point reached by the rocket is 1412 m

b)The rocket crashes after 54.7 s

Explanation:

Hi there!

The equations of height and velocity of the rocket are the following:

h = h0 + v0 · t + 1/2 · a · t² (while the engines work).

h = h0 + v0 · t + 1/2 · g · t² (when the rocket is in free fall).

v = v0 + a · t (while the engines work).

v = v0 + g · t (when the rocket is in free fall).

Where:

h = height of the rocket at a time t.

h0 = initial height of the rocket.

v0 = initial velocity.

t = time.

a = acceleration due to the engines.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the rocket at a time t.

First, let's find the velocity and height reached by the rocket until the engines fail:

h = h0 + v0 · t + 1/2 · a · t²

Let's set the origin of the frame of reference at the launching point so that h0 = 0. Since the rocket starts from rest, v0 = 0. So after 30.0 s the height of the rocket will be:

h = 1/2 · a · t²

h = 1/2 · 2.5 m/s² · (30.0 s)²

h = 1125 m

Now let's find the velocity of the rocket at t = 30.0 s:

v = v0 + a · t (v0 = 0)

v = 2.5 m/s² · 30.0 s

v = 75 m/s

After 30.0s the rocket will continue to ascend with a velocity of 75 m/s. This velocity will be gradually reduced due to the acceleration of gravity. When the velocity is zero, the rocket will start to fall. At that time, the rocket is at its maximum height. So, let's find the time at which the velocity of the rocket is zero:

v = v0 + g · t

0 = 75 m/s - 9.8 m/s² · t (v0 = 75 m/s because the rocket begins its free-fall motion with that velocity).

-75 m/s / -9.8 m/s² = t

t = 7.7 s

Now, let's find the height of the rocket 7.7 s after the engines fail:

h = h0 + v0 · t + 1/2 · g · t²

The rocket begins its free fall at a height of 1125 m and with a velocity 75 m/s, then, h0 = 1125 m and v0 = 75 m/s:

h = 1125 m + 75 m/s · 7.7 s - 1/2 · 9.8 m/s² · (7.7 s)²

h = 1412 m

The highest point reached by the rocket is 1412 m

b) Now, let's calculate how much time it takes the rocket to reach a height of zero (i.e. to crash) from a height of 1412 m.

h = h0 + v0 · t + 1/2 · g · t² (v0 = 0 because at the maximum height the velocity is zero)

0 = 1412 m - 1/2 · 9.8 m/s² · t²

-1412 m / -4.9 m/s² = t²

t = 17 s

The rocket goes up for 30.0 s with an acceleration of 2.5 m/s².

Then, it goes up for 7.7 s with an acceleration of -9.8 m/s².

Finally, the rocket falls for 17 s with an acceleration of -9.8 m/s²

The rocket crashes after (30.0 s + 7.7 s + 17 s) 54.7 s

6 0
3 years ago
You charge a capacitor and then discharges it through a resistor. You notice that, after four time constants, the voltage across
ollegr [7]

Answer: half of it's value?

Explanation:

5 0
3 years ago
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