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2 years ago

An object traveling at a constant

Physics

1 answer:

mel-nik [20]2 years ago

4 0

Answer:

pi / 2 radians / s

Explanation:

One revolution = 2 pi Radians in 4 seconds

2 pi / 4 = pi/2 radians / s

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a concrete slab 20 m long and weighing 400,000 N is supported by one pillar. If a 19,600 N car is parked 8 meters from one end.

LUCKY_DIMON [66]

let the distance of pillar is "r" from one end of the slab

So here net torque must be balance with respect to pillar to be in balanced state

So here we will have

$Mg(r - L/2) = mg(L/2 - 8)$

here we know that

mg = 19600 N

Mg = 400,000 N

L = 20 m

from above equation we have

$400,000(r - 10) = 19,600 (10 - 8)$

$r - 10 = 0.098$

$r = 10.098 m$

so pillar is at distance 10.098 m from one end of the slab

7 0

3 years ago

Two sacks contain the same number of identical apples and are separated by a distance r. The two sacks exert a gravitational for

stepan [7]

Answer:

Explanation:

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3 0

3 years ago

What is the chemical name for a sand

vovikov84 [41]

It is known as silicon dioxide or silica!

Hope this helps!

7 0

3 years ago

An object is pulled to the left by a force of 50 N. The same amount of force pulls it to the right. The object will ____.

11111nata11111 [884]

<h2><u>Required</u><u> </u><u>Answer</u><u>:</u></h2>

The body will <u>stay at rest </u>(Option D). It is because a force of magnitude 50 N is pulled towards left and another force is pulling it towards right with same magnitude 50 N. So, the direction of force is opposite and magnitude is same i.e. 50 N. So, they will cancel each other and net force is 0. Hence, there would be no acceleration.

Option A - Showing acceleration
Option B - Showing acceleration
Option C - Change of direction due to Net force

Hence, these options are incorrect because they are only possible when net external force is non-zero. Staying at rest i.e. Option D means there is no motion and hence no acceleration, this shows that net force is 0 $.$

<u>━━━━━━━━━━━━━━━━━━━━</u>

3 0

3 years ago

Compare the gravitational acceleration on the following objects compared to the Sun using:

arsen [322]

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765 x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86$

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}$

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}$

Learn more about acceleration due to gravity here: brainly.com/question/88039

3 0

3 years ago