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3 years ago

Which planet's gravitational pull is closest to that of Earth?

1 answer:

3 0

The acceleration of gravity on the surface of Venus
is 8.87 m/s² ... about 9.6% less than on Earth's surface.
That's the most similar of any body in our solar system.

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A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 86.0 m/s2

Harman [31]

When the fuel of the rocket is consumed, the acceleration would be zero. However, at this phase the rocket would still be going up until all the forces of gravity would dominate and change the direction of the rocket. We need to calculate two distances, one from the ground until the point where the fuel is consumed and from that point to the point where the gravity would change the direction.

Given:
a = 86 m/s^2
t = 1.7 s

Solution:

d = vi (t) + 0.5 (a) (t^2)
d = (0) (1.7) + 0.5 (86) (1.7)^2
d = 124.27 m

vf = vi + at
vf = 0 + 86 (1.7)
vf = 146.2 m/s (velocity when the fuel is consumed completely)

Then, we calculate the time it takes until it reaches the maximum height.
vf = vi + at
0 = 146.2 + (-9.8) (t)
t = 14.92 s

Then, the second distance
d= vi (t) + 0.5 (a) (t^2)
d = 146.2 (14.92) + 0.5 (-9.8) (14.92^2)
d = 1090.53 m

Then, we determine the maximum altitude:
d1 + d2 = 124.27 m + 1090.53 m = 1214.8 m

5 0

3 years ago

A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second, and the t

slega [8]

Answer:

The tension increases to four times its original value.

Explanation:

v = Velocity

r = Radius

m = Mass of stone

The centripetal force is

$F_c=m\dfrac{v^2}{r}$

The tension will balance the centripetal force

$T=m\dfrac{v^2}{r}$

$\\\Rightarrow T\propto v^2$

$\dfrac{T_1}{T_2}=\dfrac{v_1^2}{v_2^2}\\\Rightarrow \dfrac{T_1}{T_2}=\dfrac{v_1}{2^2v_1^2}\\\Rightarrow \dfrac{T_1}{T_2}=\dfrac{1}{4}\\\Rightarrow T_2=4T_1$

The new tension will be 4 times the old tension

5 0

3 years ago

Which spacecraft appears the longest to an outside observer? W X Y Z

Ivenika [448]

Answer:

Explanation:

7 0

3 years ago

Read 2 more answers

A certain gas occupies a volume of 3.7 L at a pressure of 0.91 atm and a temperature of 283 K. It is compressed adiabatically to

Nostrana [21]

Answer:

(a)

P₂ = 7.13 atm

(b)

T₂ = 157.14 K

Explanation:

(a)

V₁ = initial volume = 3.7 L = 3.7 x 10⁻³ m³

V₂ = final volume = 0.85 L = 0.85 x 10⁻³ m³

P₁ = Initial Pressure of the gas = 0.91 atm = 0.91 x 101325 = 92205.75 Pa

P₂ = Final Pressure of the gas = ?

Using the equation

$P_{1} V{_{1}}^{\gamma } = P_{2} V{_{2}}^{\gamma }$

$(92205.75) (3.7\times 10^{-3})^{1.4 } = P_{2} (0.85\times 10^{-3})^{1.4 }$

$P_{2}$ = 722860 Pa

$P_{2}$ = 7.13 atm

(b)

T₁ = initial temperature =283 K

T₂ = Final temperature = ?

using the equation

$P{_{1}}^{1-\gamma } T{_{1}}^{\gamma } = P{_{2}}^{1-\gamma } T{_{2}}^{\gamma }$

$(92205.75)^{1-1.4 } (283)^{1.4 } = (722860)^{1-1.4 } T{_{2}}^{1.4 }$

T₂ = 157.14 K

6 0

3 years ago

A 2,000- kilogram railroad car moving at 5m/sec to the east collides with a 6,000-kilogram railroad car moving at 3m/sec to the

otez555 [7]

So we wan't to know what is the velocity after a collision of two railroad cars, one moving to the east and the other moving to the west if m1=2000kg, v1=5m/s and m2=6000kg, v2=3m/s. We can find the solution using the law of conservation of momentum for plastic collisions that states that the momentum must remain constant before (left side of the equation) and after (right side of the equation) the collision: m1*v1+m2*v2=(m1+m2)*v. So now we simply plug in the numbers and get: 2000kg * 5m/s + 6000kg * 3m/s = (2000kg + 6000kg)*v. Now we can write: 10000 kgm/s + 18000 kgm/s = 8000kg * v. To get v, the velocity of both railroad cars after the collision we simply divide both sides of the equation with 8000 kg: so v=3.5m/s to the west.

Read more on Brainly.com - brainly.com/question/183156#readmore

3 0

3 years ago