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3 years ago

Why do astronauts feel weightless in space?

Physics

1 answer:

JulsSmile [24]3 years ago

6 0

Weight = mass × gravitational field strength
At space gravitational field strength is very low so its"weightless".

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A pair of closely spaced parallel conducting plates, charged with equal and opposite electric charges, produces a uniform electr

Sedbober [7]

Answer:

Plate B.

Explanation:

If the direction of the electric field is from plate A to plate B, then this means that plate A is positively charged and plate B is negatively charged. If we are to move an electron between the plates, then we should place the electron on plate B, so the negatively charged electron can be attracted by the positive charges on plate A.

3 0

3 years ago

Read 2 more answers

A body of mass 400 kg is suspended at a lower end of a light vertical chain and is being pulled up vertically. Initially the bod

yKpoI14uk [10]

Answer:31.62 m/s

Explanation:

Given

mass of body $m=400 kg$

Pull on chain is $F_1=6000g N=60 kN$

Pull get smaller at the rate of $F_2=360g N/m$

Net Upward Force $F=6000 g-360 g\times 10=24 kN$

net acceleration $a=\frac{F}{m}$

$a=\frac{24\times 1000}{m}$

$a=\frac{24000}{400}$

$a=60 m/s^2$

but g is acting downward

$a_{net}=a-g=60-10=50 m/s^2$

using $v^2-u^2=2 as$

here initial velocity is zero

$v^2=2\times 50\times 10$

$v=31.62 m/s$

7 0

4 years ago

Pure water has a pH of 7. Pure water _______. A. is a neutral substance B. could be either an acid or a base C. is a base D. is

malfutka [58]

A. is a neutral substance.

Only neutral substances (like pure water) can have a pH of 7.

5 0

3 years ago

Read 2 more answers

A 4 kg block is launched up a 30° ramp with an initial speed of 5 m/s. The coefficient of kinetic friction between the block and

zlopas [31]

Answer:

The speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.

Explanation:

Given;

mass of block, m = 4 kg

coefficient of kinetic friction, μk = 0.25

angle of inclination, θ = 30°

initial speed of the block, u = 5 m/s

From Newton's second law of motion;

F = ma

a = F/m

Net horizontal force;

∑F = mgsinθ + μkmgcosθ

$a = \frac{F_{NET}}{m} = \frac{mgsin \theta + \mu_kmgcos \theta}{m} \\\\a = gsin \theta + \mu_kgcos \theta\\\\a = 9.8sin (30) + 0.25*9.8cos(30)\\\\a = 4.9 + 2.1217\\\\a = 7.022 \ m/s^2$

At the top of the ramp, energy is conserved;

Kinetic energy = potential energy

¹/₂mv² = mgh

¹/₂ v² = gh

¹/₂ x 5² = 9.8h

12.5 = 9.8h

h = 12.5/9.8

h = 1.28 m

Height of the ramp is 1.28 m

Now, calculate the speed of the block (in m/s) when it has returned to the bottom of the ramp;

v² = u² + 2ah

v² = 5² + 2 x 7.022 x 1.28

v² = 25 + 17.976

v² = 42.976

v = √42.976

v = 6.56 m/s

Therefore, the speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.

4 0

4 years ago

Which is an example of a scientist using a mathematical model to describe the weather?

mr Goodwill [35]

Answer:

The scientist finds an equation that predicts how long the rainstrom will last.

Explanation:

I just answered it :)

6 0

3 years ago