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3 years ago

Two objects exert a gravitational force on 8N on one another what would that force be if the mass of both objects were doubled?

1 answer:

3 0

Gravitational force of attraction between two objects can be calculated from the following formula:
$F = \frac{m_1 \cdot m_2}{r^{2}}$
Where m with subscript stands for mass of the object and r is the distance between them.
When we double the mass of those two objects, distance between them stays the same, while in the numerator we have:
$2\cdot m_1 \cdot 2 \cdot m_2 = 4\cdot m_1 \cdot m_2$
When numerator in the second case 4 times greater than the 'original' numerator and denominator stays the same, force becomes 4 times greater.

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IBM has a computer it calls the Blue Gene/L that can do 136.8 teracalculations per second. How many calculations can it do in a

Norma-Jean [14]

There are 1,000,000 micro seconds in one second so multiple 136.8 by 1000000 and you'll get 136,800,000 Tera calculations per second.

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3 years ago

Read 2 more answers

Peter designed a road with a curve of radius 30 m that is banked so that a 950 kg car traveling at 40.0 km/h can round it even i

spayn [35]

Answer:

$v = 15.56 m/s$

$v = 56 km/h$

Explanation:

When coefficient of friction is approximately zero then we have

$F_ncos\theta = mg$

$F_n sin\theta = \frac{mv^2}{R}$

$tan\theta = \frac{v^2}{Rg}$

here we know that

$v = 40 km/h = 11.11 m/s$

R = 30 m

$tan\theta = \frac{11.11^2}{30\times 9.81}$

$\theta = 22.75 degree$

now when friction coefficient is 0.30 then we have

$F_n cos\theta = mg + F_f sin\theta$

$F_f cos\theta + F_n sin\theta = \frac{mv^2}{R}$

now we have

$v = \sqrt{Rg(\frac{\mu + tan\theta}{1 - \mu tan\theta})}$

$v = \sqrt{30(9.81)(\frac{0.30 + tan22.75}{1 - (0.30) tan22.75})}$

$v = 15.56 m/s$

$v = 56 km/h$

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3 years ago

Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o

seraphim [82]

Answer:

$\theta = 15^o \: or\: 75^o$

Explanation:

As we know that the formula of range is given as

$R = \frac{v^2sin2\theta}{g}$

now we know that

maximum value of the range of the projectile is given as

$R_{max} = \frac{v^2}{g}$

now we need to find such angles for which the range is half the maximum value

so we will have

$\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}$

$sin(2\theta) = \frac{1}{2}$

$2\theta = 30 or 150$

$\theta = 15^o \: or\: 75^o$

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3 years ago

Why do we see the sun before it actually rises in the morning?

Lera25 [3.4K]

Answer:

It appears to rise and set because of the Earth's rotation on its axis. It makes one complete turn every 24 hours. It turns toward the east. As the Earth rotates toward the east, it looks like the sun is moving west.

Explanation:

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3 years ago

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3 years ago