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3 years ago

Does heavy force work under conditions of weightlessness in such an artificial earth satellite?

1 answer:

3 0

Answer: That is to say, a satellite is an object upon which the only force is gravity. Once launched into orbit, the only force governing the motion of a satellite is the force of gravity. Newton was the first to theorize that a projectile launched with sufficient speed would actually orbit the earth.

Explanation: Because Earth-orbiting objects follow elliptical paths around Earth and not a straight line, forces cannot, by definition, be balanced. Force is directional. It is a push or a pull in a particular direction.

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1. Two forces F~ 1 and F~ 2 are acting on a block of mass m=1.5 kg. The magnitude of force F~ 1 is 12N and it makes an angle of

VLD [36.1K]

Answer:

Normal force=7.48 N

Explanation:

N+F~1 sinθ-mg=0

=>N=1.5*9.8-12 sin37◦

=>N=14.7-7.22=7.48 N

6 0

3 years ago

A 10-kg rock and 20-kg rock are dropped from the same height and experience no significant air resistance. if it takes the 20-kg

bulgar [2K]

<span>Mass of rock 1 is m1 = 10 kg Mass of rock 2 is m2 = 20 kg 10-kg rock takes T (the same time ) to reach the ground as similar to 20-kg rock that takes time T to reach the ground. If no air resistance is present, the rate of descent depends only on how far the object has fallen, no matter how heavy the object is. This means that two objects will reach the ground at the same time if they are dropped simultaneously from the same height. This statement follows from the law of conservation of energy and has been demonstrated experimentally by dropping a feather and a lead ball in an airless tube. When air resistance plays a role, the shape of the object becomes important. THUS 10-KG & 20-KG ROCK reaches the ground in T-time</span>

3 0

4 years ago

A well is pumped at Q = 300 m3 /hr in a confined aquifer. The aquifer transmissivity is 25 m2 /hr and the storage coefficient is

Effectus [21]

Answer:

(a). The draw-down at a distance 200 m from the well after pumping for 50 hr is 5.383 m.

(b). The draw-down at a distance 200 m from the well after pumping for 50 hr is 6.707 m.

Explanation:

Given that,

Energy $Q=300\ m^3/hr$

Transmissivity $T = 25\ m^2/hr$

Storage coefficient $S=2.5\times10^{-4}$

Distance r= 200 m

We need to calculate the draw-down at a distance 200 m from the well after pumping for 50 hr

Using formula of draw-down

$s= \dfrac{Q}{4\pi T}(-0.5772-ln\dfrac{r^2S}{4tT})$

Put the value into the formula

$s=\dfrac{300}{4\pi\times25}(-0.5772-ln\dfrac{(200)^2\times2.5\times10^{-4}}{4\times25\times50})$

$s=5.383\ m$

We need to calculate the draw-down at a distance 200 m from the well after pumping for 200 hr

Using formula of draw-down

$s= \dfrac{Q}{4\pi T}(-0.5772-ln\dfrac{r^2S}{4tT})$

Put the value into the formula

$s=\dfrac{300}{4\pi\times25}(-0.5772-ln\dfrac{(200)^2\times2.5\times10^{-4}}{4\times25\times200})$

$s=6.707\ m$

Hence, (a). The draw-down at a distance 200 m from the well after pumping for 50 hr is 5.383 m.

(b). The draw-down at a distance 200 m from the well after pumping for 200 hr is 6.707 m.

3 0

3 years ago

3. A 40.0-kg wagon is towed up a hill inclined at 18.5 with respect to the horizontal. The tow rope is parallel to the incline a

inn [45]

Force=tension-fg sin ∅
=140-mg sin 18.5
=140-124.35
=15.62N

a=f/m=15.62/40=0.39
now,
v²=u²+2as
=2×0.39×80
v²=62.4
v=7.8m/s

4 0

3 years ago

Calculate the weight of a 4.3 kg bunny WITH PROOF/WORK

ValentinkaMS [17]

In pounds? Cuz if so 2.2 x 4.3 = 9.46

3 0

2 years ago