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Crank
2 years ago
7

A 160.-kilogram space vehicle is traveling along a

Physics
1 answer:
mario62 [17]2 years ago
4 0

If the object is moving in a straight line at a constant speed, then that's
the definition of zero acceleration.  It can only happen when the sum of
all forces (the 'net' force) on the object is zero.

And it doesn't matter what the object's mass is.  That argument is true
for specks of dust, battleships, rocks, stars, rock-stars, planets, and
everything in between.

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A rock is thrown upward with an initial velocity of 16 ft/s from an initial height of 5 ft. write a quadratic function equation
Andrei [34K]
During upward projection the final velocity is zero, and the gravitational acceleration is -10 m/s² (against the gravity).
Therefore; using the equation;
S = 1/2gt² + ut
Where s is the height h, g is gravitational acceleration, and t is the time and u is the initial velocity u, is 16 ft/s.
Thus; h= 1/2(-10)t² + 16t
We get; h = -5t² + 16t
Therefore; the quadratic equation is 5t² - 16t + h =0
5 0
3 years ago
What is required foe electricity to flow
lukranit [14]
Conductivity is required for the electric current to flow.
5 0
3 years ago
A uniform 1.6-kg rod that is 0.89 m long is suspended at rest from the ceiling by two springs, one at each end. Both springs han
artcher [175]

Answer:

8.27°

Explanation:

To angle difference will be determined by the difference in the displacement of the springs, produced by the weight of the center of mass of the rod.

d=y_1-y_2=\frac{F_1}{k_1}-\frac{F_2}{k_2}=\frac{0.5mg}{31N/m}-\frac{0.5mg}{63N/m}\\\\d=0.5(1.6kg)(9.8m/s^2)[\frac{1}{31N/m}-\frac{1}{63N/m}]=0.128m

by a simple trigonometric relation you obtain that the angle:

sin\theta=\frac{d}{l}=\frac{0.128m}{0.89m}=0.144\\\\\theta=sin^{-1}(0.144)=8.27\°

hence, the angle between the rod and the horizontal is 8.27°

4 0
2 years ago
What force can stop a 1400kg truck accelerating at 7.5m/s2? URGENT PLZZ
Archy [21]

Answer:

10500 N

Explanation:

F = m*a=1400*7.5N=10500N

6 0
2 years ago
Read 2 more answers
A charge of 1.0 × 10-6 μC is located inside a sphere, 1.25 cm from its center. What is the electric flux through the sphere due
Korvikt [17]

Answer:

\Phi_E=0.11\frac{N\cdot m^2}{C}

Explanation:

According to Gauss's Law, the electric flux of a charged sphere is the electric field multiplied by the area of ​​the spherical surface:

\Phi_E=EA\\\Phi_E=E(4\pi R^2)\\\Phi_E=\frac{q}{4\pi \epsilon_0 R^2}(4\pi R^2)\\\Phi_E=\frac{q}{\epsilon_0}

This is identical to the electric flux of a point charge located in the center of the sphere.

\Phi_E=\frac{1*10^{-12}C}{8.85*10^{-12} \frac{C^2}{N\cdot m^2}}\\\Phi_E=0.11\frac{N\cdot m^2}{C}

8 0
3 years ago
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