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3 years ago

What feature of an object does not affect air resistance?

1 answer:

7 0

The amount of air resistance an object experiences depends on its speed, its cross-sectional area, its shape and the density of the air. Air densities vary with altitude, temperature and humidity. Nonetheless, 1.29 kg/m3 is a very reasonable value. The shape of an object affects the drag coefficient (Cd)

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How a diode can act as a capacitor?

maks197457 [2]

A diod can act as a capacitor the deplition region in diod acts as a capacitor as it also stores energy similar to capacitor in electric field.

3 0

3 years ago

A light ray in air enters and passes through a block of glass. What can be stated with regard to its speed after it emerges from

Alik [6]

Answer:

Speed is same as that before it entered glass.

Explanation:

Given:

A light ray enters and passes through the glass as shown in the diagram.

We have to analyze its speed.

Speed of light in air is $3\times 10^8\ ms^-^1$ and speed of light in glass is $2.25\times 10^8\ ms^-^1$

Whenever a light ray enters a glass block or slab there is bending of light at the interface of the two media.

So speed of light will decrease in glass medium and again it passes to the air.

Speed of light in air will again increase or will be equivalent to the earlier speed when it was entering the glass block.

Finally

Speed is same as that before it entered glass as it in the same medium (air).

6 0

3 years ago

Which of the following is not an element of installment credit?

olasank [31]

Answer:

I believe it's A. An installment credit has an interest rate, repayment terms and fees, and fixed payment.

3 0

3 years ago

1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

Tanzania [10]

Answer with explanation:

We are given that

Mass of ball, $m_1=$ 75 g= $\frac{75}{1000}=0075kg$

1 kg=1000 g

Height, $h_1=1.6 m$

$h_2=0.6 m$

Horizontal velocity, $v_x=2 m/s$

Mass of plate $m_2=400 g=\frac{400}{1000}=0.4 kg$

a.Initial velocity of plate, $u_2=0$

Velocity before impact= $u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s$

Where $g=9.8 m/s^2$

Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

According to law of conservation of momentum

$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

Substitute the values

$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

$0.4v_2=0.075\times 5.6+0.075\times 3.4$

$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

Velocity of plate=1.69 m/s

b.Initial energy= $\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J$

Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

6 0

3 years ago

The greater the mass of an object, the greater its force due to gravity

Serga [27]

What is the question?

6 0

3 years ago