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Digiron [165]
3 years ago
6

PLEASE PLEASE PLEASE HELP PLEASE

Physics
1 answer:
RideAnS [48]3 years ago
4 0

Answer:

Because there is too smooth

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A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 24.3 m/s . A 1.0-kg stone is thrown
defon

Answer:

-969.06

-286.74

698.7

-115.6, 12.9

-139.9, 12.9

Explanation:

Given that

Speed v, wrt y = -24.3 m/s

Speed v, wrt x = 12.9 m/s

time t, = 11.8 s

a

Using the formula

H(t) = ut - 1/2gt², where u = v wrt y

H(t) = -24.3 * 11.8 - 1/2 * 9.8 * 11.8²

H(t) = -286.74 - 682.28

H(t) = -969.06 m

b

H = ut, where u = v wrt y

H = -24.3 * 11.8

H = -286.74 m

H(1) = -969.06 - -286.74 = -682 m

c

Horizontal displacement, x = vt. Where v = v wrt x

x = 12.9 * 11.8

x = 152.22 m

d = √(H1² + x²)

d = √682² + 152²

d = 465124 + 23104

d = √488228

d = 698.7 m

d

Vertical component =

-gt - 0 =

-9.8 * 11.8 = -115.6

Horizontal component =

v wrt x - 0

12.9 - 0 = 12.9

e

Vertical component =

-gt - v wrt y =

-9.8 * 11.8 - 24.3 = -139.9

Horizontal component =

v wrt x - 0 =

12.9 - 0 = 0

8 0
3 years ago
What is the current flowing through this circuit? Look at the picture!
castortr0y [4]

Answer:

Explanation:

It doesnt show

6 0
3 years ago
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A single covalent bond is stronger than a single hydrogen bond so why does a group of polar molecules tend to have a higher boil
Naily [24]

Answer:

this question makes no sense

Explanation:

like how do you get this question

5 0
3 years ago
Read 2 more answers
A star that appears to pulsate is called a
Kay [80]
A star that appears to pulsate is called a Variable Star.
8 0
3 years ago
An uncrewed mission to the nearest star, Proxima Centauri, is launched from the Earth's surface as a projectile with an initial
Anna [14]

Answer:

42.96 km/s

Explanation:

From the conservation of Energy

(PE+KE)_i=(PE+KE)_f\\\Rightarrow -\frac{GmM}{R}+\frac{1}{2}mv_i^2=0+\frac{1}{2}mv_f^2

Mass gets cancelled

-\frac{GM}{R}+\frac{1}{2}v_i^2=0+\frac{1}{2}v_f^2\\\Rightarrow -2\frac{GM}{R}+v_i^2=v_f^2\\\Rightarrow -v_e^2+v_i^2=v_f^2\\\Rightarrow v_f=\sqrt{v_i^2-v_e^2}

v_e=\sqrt{\frac{2Gm}{R}} = Escape velocity of Earth = 11.2 km/s

v_i = Velocity of projectile = 44.4 km/s

v_f=\sqrt{44.4^2-11.2^2}\\\Rightarrow v_f=42.96\ km/s

The velocity of the spacecraft when it is more than halfway to the star is 42.96 km/s

6 0
3 years ago
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