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Bad White [126]
3 years ago
15

TRUE or FALSE: Feeling "weightless" can be the result of accelerating downwards towards Earth.

Physics
1 answer:
goblinko [34]3 years ago
4 0
I believe it would be false because in outer space, everything is weightless due to the weak gravitational force which makes things float in the air. Weightless means no weight to it and if we’re accelerating down that means the gravitational is working but in this case it would probably be false
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You kick a ball with a speed of 14 m/s at an angle of 51°. How far away does the ball land?
In-s [12.5K]
-- The vertical component of the ball's velocity is 14 sin(<span>51°) = 10.88 m/s

-- The acceleration of gravity is 9.8 m/s².

-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
same amount of time before it hits the ground.

-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
==================================

-- The horizontal component of the ball's velocity is  14 cos(</span><span>51°) = 8.81 m/s

-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = <em><u>19.56 meters</u></em>
before it hits the ground.


As usual when we're discussing this stuff, we completely ignore air resistance.
</span>
4 0
3 years ago
Read 2 more answers
In half wave rectifier circuit the diode and load resistance are connected in ...to ac power source​
Fudgin [204]
Working of a Half wave rectifier
The diode is connected in series with the secondary of the transformer and the load resistance RL. The primary of the transformer is being connected to the ac supply mains. The ac voltage across the secondary winding changes polarities after every half cycle of the input wave.
5 0
3 years ago
A car moving with an initial speed v collides with a second stationary car that is one-half as massive. After the collision the
Mashutka [201]

Answer:

4v/3

Explanation:

Assume elastic collision by the law of momentum conservation:

m_1v = m_1v_1 + m_2v_2

where v is the original speed of car 1, v1 is the final speed of car 1 and v2 is final speed of car 2. m1 and m2 are masses of car 1 and car 2, respectively

Substitute m_2 = m_1/2 \& v_1 = v/3

m_1v = \frac{m_1v}{3} + \frac{m_1v_2}{2}

Divide both side by m_1, then multiply by 6 we have

6v = 2v + 3v_2

3v_2 = 4v

v_2 = \frac{4v}{3}

So the final speed of the second car is 4/3 of the first car original speed

5 0
3 years ago
A team of dogs accelerates a 290kg dogsled from 0 to 6.0m/s in 3.0 s. Assume that the acceleration is constant.Part AWhat is the
Mademuasel [1]

Answer:

(a) a=2m/sec^2

(b) 5220 j

(c) 1740 watt

(d) 3446.66 watt

Explanation:

We have given mass m = 290 kg

Initial velocity u = 0 m/sec

Final velocity v = 6 m/sec

Time t = 3 sec

From first equation of motion

v = u+at

So a=\frac{v-u}{t}=\frac{6-0}{3}=2m/sec^2

(a) We know that force is given by

F = ma

So force will be F=290\times 2=580N

(b) From second equation of motion we know that

s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 2\times 3^2=9m

We know that work done is given by

W = F s = 580×9 =5220 j

(c) Time is given as t = 3 sec

We know that power is given as

P=\frac{W}{t}=\frac{5220}{3}=1740Watt

(d) Time t = 1.5 sec

So P=\frac{W}{t}=\frac{5220}{1.5}=3466.66Watt

5 0
4 years ago
The average depth of Indian Ocean is about 3000m.Calculate the fractional compression. ▲v/v, of water at the bottom of the ocean
Klio2033 [76]

Answer:

∴ fractional compression = 1.34 × 10⁻²

Explanation:

given,

depth of Indian ocean = 3000 m

Bulk modulus of the water = 2.2 x 10⁹ N/m²

We know,

P = P₀ + ρgh

P₀ is the atmospheric pressure

P₀ = 10⁵ N/m²

ρ is the density of the water, 1000 Kg/m³

P = 10⁵ + 1000 × 9.8 × 3000 = 2.94 × 10⁷ N/m²

using formula,

B = P/{-∆V/V}

B is bulk modulus and { -∆V/V} is the fractional compression

\dfrac{-\Delta V}{V} = \dfrac{2.94 \times 10^7}{2.2 \times 10^9}

\dfrac{-\Delta V}{V} =1.34 \times 10^{-2}

∴ fractional compression = 1.34 × 10⁻²

7 0
3 years ago
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