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tensa zangetsu [6.8K]
3 years ago
12

when an object falls, eventually the force of air resistance (drag) = the force of gravity. This is called ____​

Physics
1 answer:
jarptica [38.1K]3 years ago
4 0
Terminal velocity dkdkmfocmdpdmfpfmfl
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How many joules of energy are required to accelerate one kilogram of mass from rest to a velocity of 0.866c?
Arlecino [84]

Answer:

the amount of energy needed is 1.8 x 10¹⁷ J.

Explanation:

Given;

mass of the object, m₀ = 1 kg

velocity of the object, v = 0.866 c

By physics convection, c is the speed of light = 3 x 10⁸ m/s

The energy needed is calculated as follows;

E = Mc²

As the object approaches the speed of light, the change in the mass of the object is given by Einstein's relativity formula;

M = \frac{M_0}{\sqrt{1- \frac{v^2}{c^2} } } \\\\  M = \frac{1}{\sqrt{1- \frac{(0.866c)^2}{c^2} } }\\\\  M = \frac{1}{\sqrt{1- \frac{0.74996c^2}{c^2} } }\\\\  M = \frac{1}{\sqrt{0.25} } \\\\ M = 2 \ kg

The energy required is calculated as;

E = 2 x (3 x 10⁸)²

E = 1.8 x 10¹⁷ J

Therefore, the amount of energy needed is 1.8 x 10¹⁷ J.

3 0
3 years ago
In root pass welding, that electrode we are going to use?​
postnew [5]

Answer:

The root pass is made with a 5/32” (4.0mm) diameter electrode. A cellulosic electrode (E-XX10) is being used. The root pass is welded with reverse (DC+) polarity.

Explanation:

5 0
3 years ago
Tara's cell phone plan costs $39.00 a month, which includes 100 text messages. After she uses all of her text messages, it will
vekshin1
I think it is $585.hope it is right
5 0
3 years ago
A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.5 m>s. Two seconds later
ivanzaharov [21]

Answer:

4.28 s

Explanation:

after two seconds (2 s) His friends is

d = 3.5 m/s x 2 s = 7 meter ahead.

in this state, a bicylist start from initial velocity vo = 0 m/s and accelerat 2.4 m/s²

then, when bicylist reach his friend

t friend = t bicyclist = t

d bicylist = d friend + d

-------

d friend = 3.5 . t

d bicylist = vo . t + ½ a t²

d friend + d = vo . t + ½ a t²

3.5 t + 7 = 0 . t + ½ . 2.4 . t²

3.5 t + 7 = 1.2 t²

0 = 1.2 t² - 3.5 t - 7

t = -1.363 and t = 4.28

take the positive one

6 0
3 years ago
Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance
PSYCHO15rus [73]

Answer:

No she cannot.

Explanation:

Let v_h be the horizontal component of the ball velocity when it's kicked, assume no air resistance, this is a constant. Also let v_v be the vertical component of the ball velocity, which is affected by gravity after it's kicked.

The time it takes to travel 95m accross the field is

t = 95 / v_h or v_h = 95/t

t is also the time it takes to travel up, and the fall down to the ground, which ultimately stops the motion. So the vertical displacement after time t is 0

s = v_vt + gt^2/2= 0

where g = -9.8m/s2 in the opposite direction with v_v

v_vt - 4.9t^2 = 0

v_vt = 4.9t^2

v_v = 4.9t

Since the total velocity that the goal keeper can give the ball is 30m/s

v = v_v^2 + v_h^2 = 30^2 = 900

(4.9t)^2 + \left(\frac{95}{t})^2 = 900

24.01t^2 + \frac{9025}{t^2} = 900

Let substitute x = t^2 > 0

24.01 x + \frac{9025}{x} = 900

We can multiply both sides by x

24.01 x^2 + 9025 = 900x

24.01x^2 - 900x + 9025 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{900\pm \sqrt{(-900)^2 - 4*(24.01)*(9025)}}{2*(24.01)}

As (-900)^2 - 4*24.01*9025 = -56761 < 0

The solution for this quadratic equation is indefinite

So it's not possible for the goal keeper to do this.

6 0
3 years ago
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