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KATRIN_1 [288]
2 years ago
15

9) For a horizontally launched projectile, decreasing the height of the

Physics
1 answer:
Volgvan2 years ago
4 0

Answer:

Increasing the launch height increases the downward distance, giving the horizontal component of the velocity greater time to act upon the projectile and hence increasing the range.

Explanation:

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Determine the field strength, E, experienced by a test charge, q, if a charge of 7.0 × 10-5 coulombs is placed on q and a force
kkurt [141]
Formula for feild strength= F/q
q=7.0^10-5 coulombs
F=5.2 N
E=5.2 / 7.0^10-5
E=
7 0
3 years ago
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Newton’s third law in football?
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Newtons third law of motion applies to football as when the ball is thrown, force is applied to the football thrown forward and also force back to the player. Therefore it states that every action has a reaction. The player does not get a huge force back from throwing the football due to the player having a bigger mass than the football being thrown.

In short, when the football is throw, there is a reaction with the ball and the player accordingly with the force put into the ball.
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3 years ago
If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol
xenn [34]

Answer:

Average density of Sun is 1.3927 \frac{g}{cm}.

Given:

Radius of Sun = 7.001 ×10^{5} km = 7.001 ×10^{10} cm

Mass of Sun = 2 × 10^{30} kg = 2 × 10^{33} g

To find:

Average density of Sun = ?

Formula used:

Density of Sun = \frac{Mass of Sun}{Volume of Sun}

Solution:

Density of Sun is given by,

Density of Sun = \frac{Mass of Sun}{Volume of Sun}

Volume of Sun = \frac{4}{3} \pi r^{3}

Volume of Sun = \frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}

Volume of Sun = 1.436 × 10^{33} cm^{3}

Density of Sun = \frac{ 2\times 10^{33} }{1.436 \times 10^{33} }

Density of Sun = 1.3927 \frac{g}{cm}

Thus, Average density of Sun is 1.3927 \frac{g}{cm}.

4 0
3 years ago
Point masses m1 m2 are placed at opposite ends
tiny-mole [99]

(a) x = \frac{m_2L}{m_1+m_2}

<u>Explanation:</u>

Given:

Moment of Inertia of m₁ about the axis, I₁ = m₁x²

Moment of Inertia of m₂ about the axis. I₂ = m₂ (L - x)²

Kinetic energy is rotational.

Total kinetic energy is E = \frac{1}{2} I_1w_0^2 + \frac{1}{2}I_2w_0^2 = \frac{1}{2} w_0^2(m_1x^2 + m_2(L-x)^2)

Work done is change in kinetic energy.

To minimize E, differentiate wrt x and equate to zero.

m_1x - m_2(L-x) = 0\\\\x = \frac{m_2L}{m_1+m_2}

Alternatively, work done is minimum when the axis passes through the center of mass.

Center of mass is at \frac{m_2L}{m_1 + m_2}

7 0
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Why does the food processor produce sound when its switched on
dybincka [34]
Because thats the motor working hard to process food
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