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4 years ago

5

how does the force of attraction between large masses compare with the force of attraction between small masses at the same dist

ance?

1 answer:

Veseljchak [2.6K]4 years ago

4 0

The force and amount of attracton is great for larger masses

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The cue ball is deflected so that it makes an angle of 30.0° with its original direction of travel.

Alexus [3.1K]

Answer:

(a) θ= 43.89°

(b) $v_{1} = 1.88\sqrt{3} \frac{m}{s}$

$v_{2} = 6.79 \frac{m}{s}$

Explanation:

Ball 1:

$u_{1} = 7.52\frac{m}{s}$

Ball 2:

$u_{2} = 0\frac{m}{s}$

As the collision is elastic, it means that kinetic energy and momentum are conserved. Following that, we apply the law of conservation of energy and momentum:

$\frac{1}{2} m_{1}u_{1}^{2} = \frac{1}{2} m_{1}v_{1}^{2} + \frac{1}{2} m_{2}v_{2}^{2}$

and

$m_{1}u_{1} = m_{1}v_{1} + m_{2}v_{2}$

Where u is the velocity before the collision, and v are the velocities after the collision. Both previous equations can be simplified as:

$u_{1}^{2} = v_{1}^{2} + v_{2}^{2}$

and

$u_{1} = v_{1} + v_{2}$

This because the two balls have the same mass. We know that the cue ball is deflected and makes an angle of 30°. From conservation in x-direction, we get::

$56.55 = v_{1} ^{2} + v_{2} ^{2}$

and

$u_{1} = v_{1}cos(30) + v_{2}cos(\theta)$

Solving we get:

$(\frac{2u_{1}-\sqrt{3}v_{1} }{2cos(\theta)})^{2} = 56.55-v_{1} ^{2}$

From conservation in y-direction, we get:

$0 = v_{1}sin(30) - v_{2}sin(\theta)$

From this, we solve this equation system and get the answers. Remember to add 30° to the angle obtained.

8 0

3 years ago

Se lanza un cohete en un ángulo de 53° sobre la horizontal con una rapidez inicial de 100 m/s. El cohete se mueve por

likoan [24]

Answer:

Explanation:

v = u + at

v₃ = 100 +30.0(3.00) = 190 m/s

s = vt + ½at²

y₃ = (100sin53)(3.00) + ½(30sin53)(3.00²) = 347.4 m

x₃ = (100cos53)(3.00) + ½(30cos53)(3.00²) = 261.8 m

a) v² = u² + uas

s = (v² - u²) / 2a

ymax = 347.4 + (0² - (190sin53)²) / (2(-9.80)) = 1,522 m

b) t₁ = 3.00 s

t₂ = (190sin53) / 9.80 = 15.5 s

t₃ = √(2(1522) / 9.80) = 17.6 s

t = 3.00 + 15.5 + 17.6 = 36.1 s

c) xmax = 261.8 + (190cos53)( 15.5 + 17.6) = 4,047 m

3 0

3 years ago

A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 7.5 oz . Aluminum has a density of 2.70 g/cm3. W

Tems11 [23]

Answer:

The thickness of the foil is 0.017 mm.

Explanation:

Given that,

Weight = 7.5 oz = 212.6175 gm

Density = 2.70 g/cm³

Area of aluminium = 50 ft² = 46451.52 cm²

We need to calculate the thickness of the foil

Using formula of density

$\rho=\dfrac{m}{V}$

$\rho=\dfrac{m}{A\times t}$

$t=\dfrac{m}{A\times \rho}$

Where, A = area

t = thickness

m = mass

Put the value into the formula

$t=\dfrac{212.6175}{46451.52\times2.70}$

$t=0.00170\ cm$

$t=0.017\ mm$

Hence, The thickness of the foil is 0.017 mm.

3 0

4 years ago

What is a plane mirror? state the characteristics of the image formed by a plane mirror

Nadusha1986 [10]

A plane mirror always forms a virtual image. the image and the object are the same distance from a flat mirror, the image size is the same as the object, and the image is upright!

3 0

3 years ago

If the boy and his bike actually had a mass of 40 kg instead of 50 kg how much force would it take for him to accelerate at 0.8

klio [65]

Answer:

F = 32 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = forces applied [N] (units of Newtons)

m = mass = 40 [kg]

a = acceleration = 0.8 [m/s²]

Now replacing:

$F=m*a\\F = 40*0.8\\F = 32 [N]$

8 0

3 years ago