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zmey [24]
3 years ago
8

The attraction between the Sun's gravity and Earth's gravity holds Earth in is ?

Physics
1 answer:
soldi70 [24.7K]3 years ago
5 0
The attraction between the suns gravity and earths gravity holds earth in it's orbit
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It's velocity changes, but its speed remains the same.<br> The<br> true or false
statuscvo [17]

Answer:

True

Explanation:

Velocity is a vector quantity, which means that it carries both magnitude and direction. Hence when direction of a particle changes, although magnitude (speed) may remain same, it's velocity changes due to direction change. For ex. A particle is m... A particle is moving along x axis with speed 1m/s, it's velocity will be represented as 1i (i represents unit vector along x)

But if it now starts moving along y axis, it's velocity is 1j (j represents unit vector along y axis). Hence velocity changes with direction.

brainllest pls .

7 0
3 years ago
If you had a mass of 44 g and a volume of 40.5ml what is the density
klio [65]
Here, We know, Density = Mass / Volume
Here, mass = 44 g
volume = 40.5 ml = 40.5 cm³

Substitute their values, 
d = 44 / 40.5
d = 1.086 g/cm³

In short, Your Answer would be  1.086 g/cm³

Hope this helps!
4 0
3 years ago
The x-coordinates of two objects moving along the x-axis are given as a function of time (t). x1= (4m/s)t x2= -(161m) + (48m/s)t
Kitty [74]
The two displacement functions are
x₁ = 4t
x₂ = -161 + 48t - 4t²
where
x₁, x₂ are in meters
t is time, s

The distance between the two objects is
x = x₁ - x₂
   =  4t + 161 - 48t + 4t²
x = 4t² - 44t + 161

Write this equation in the standard form for a parabola.
x = 4[t² - 11t] + 161
  = 4[ (t - 5.5)² - 5.5² ] + 161
 x = 4(t-5)² + 40

Ths is a parabola that faces up and has its vertex (lowest point) at (5, 40).
Therefore the closest approach of the two objects is 40 m.
The graph of x versus t confirms the result.

Answer: The distance of the closest approach is 40 m.

5 0
3 years ago
A 0.015 kg marble sliding to the right at 22.5 cm/s on a frictionless surface makes an elastic head-on collision with a 0.015 kg
slava [35]

Answer:

Explanation:

mass of first marble m_1=m=0.015\ kg

Initial velocity of the first marble u_1=22.5\ m/s

considering right side as positive

Mass of second marble

m_2=m=0.015\ kg\\u_2=-18\ cm/s

After collision first marble moves to the left with a velocity of 18 cm/s

i.e. v_1=-18\ cm/s

considering v_2 be the velocity of second marble after collision

The Coefficient of restitution is 1 for an elastic collision

e=\frac{v_2-v_1}{u_1-u_2}

Putting values

1=\frac{v_2-(-18)}{22.5-(-18)}\\22.5+18=v_2+18\\v_2=22.5\ m/s

So, the velocity of the second marble is 22.5 m/s to the right after the collision

(b)Initial kinetic energy =0.5\times 0.015\times (22.5\times 10^{-2})^2+0.5\times 0.015\times (18\times 10^{-2})^2=6.22\times 10^{-4}\ J

Final kinetic energy=

0.5\times 0.015\times (18\times 10^{-2})^2+0.5\times 0.015\times (22.5\times 10^{-2})^2=6.22\times 10^{-4}\ J

6 0
3 years ago
The current required to stimulate the heart during ventricular fibrillation is about 110mA (1000mA = 1A). Assuming that a conduc
algol [13]

Answer:

Explanation:

Given

current required I=110 mA

Internal Resistance R=300 \Omega

According to ohm law

Current flows in a conductor is directly Proportional to the voltage applied.

V\propto I

V=IR

V=110\times 10^{-3}\times 300

V=33 V  

5 0
3 years ago
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