Which is the larger value 5 * 10^5 or 3* 10^7

A book rests on a table, exerting a downward force on the table. the reaction to this force is:

lapo4ka [179]

The upward force the table exerts on the ground!
Equal and opposite forces.

4 0

3 years ago

Why do we feel the force of gravity from the Earth but we don’t feel the force of gravity from the moon?

Zigmanuir [339]

Answer:

The gravitational force on the moon is less than on Earth because the strength of gravity is determined by an object's mass. The bigger the object, the bigger the gravitational force. Gravity is pretty much everywhere. We just feel it in different ways depending on our state of motion.

Explanation:

Hope this helped!!

8 0

2 years ago

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An idea is being proposed. The steps that lead to the idea are listed below.

gavmur [86]

there were different outcomes each time.

8 0

3 years ago

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A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

raketka [301]

Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

Then –

$\qquad$ $\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

6 0

2 years ago

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Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 × 10^4 Pa.

Triss [41]

Answer:

9.8 × 10⁴Pa

Explanation:

Given:

Velocity V₁ = 12m/s

Pressure P₁ = 3 × 10⁴ Pa

From continuity equation we have

ρA₁V₁ = ρA₂V₂

A₁V₁ = A₂V₂

making V₂ the subject of the equation;

$V_{2} = \frac{A_{1}V_{1}}{A_{2}}$

the pipe is widened to twice its original radius,

r₂ = 2r₁

then the cross-sectional area A₂ = 4A₁

⇒ $V_{2}= \frac{A_{1}V_{1}}{4A_{1}}$

$V_{2}= \frac{V_{1}}{4}$

This implies that the water speed will drop by a factor of $\frac{1}{4}$ because of the increase the pipe cross-sectional area.

The Bernoulli Equation;

Energy per unit volume before = Energy per unit volume after

p₁ + $\frac{1}{2}$ ρV₁² + ρgh₁ = p₂ + $\frac{1}{2}$ ρV₂² + ρgh₂

Total pressure is constant and $P_{T}$ = P = $\frac{1}{2}$ ρV₂²ρV²

p₁ + $\frac{1}{2}$ ρV₁² = p₂ + $\frac{1}{2}$ ρV₂²

Making p₂ the subject of the equation above;

p₂ = p₁ + $\frac{1}{2}$ ρV₁² - $\frac{1}{2}$ ρV₂²

But $V_{2} = \frac{V_{1}}{4}$ so,

p₂ = p₁ + $\frac{1}{2}$ ρV₁² - $\frac{1}{2}$ ρ $\frac{V_{1}^{2}}{4^{2}}$

p₂ = 3.0 x 10⁴ + ( $\frac{1}{2}$ × 1000 × 12²) - ( $\frac{1}{2}$ × 1000 × 12²/4² )

P₂ = 3.0 x 10⁴ + 7.2 × 10⁴ - 4.05 x 10³

P₂ = 9.79 × 10⁴Pa

P₂ = 9.8 × 10⁴Pa

4 0

2 years ago