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2 years ago

NEWTONS SECOND LAW LAB REPORT

Physics

1 answer:

hram777 [196]2 years ago

8 0

Answer:

Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

Explanation:

here this may help.

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A device that does work with one movement and changes the size or direction of a force is a(n)

tresset_1 [31]

A device that does work with one movement and changes the size or direction of a force is a simple machine.

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3 years ago

Read 2 more answers

I need help in my physics class and show me how it’s done

Korolek [52]

If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula

$A_x = |A|cos(\alpha)\\\\A_y = |A|sin(\alpha)$

Where | A | is the magnitude of the vector and $\alpha$ is the angle that it forms with the x axis in the opposite direction to the hands of the clock.

In this problem we know the value of Ax and Ay and we need the angle $\alpha$ .

Vector A is in the 4th quadrant

So:

$A_x = 6\\\\A_y = -6.5$

So:

$|A| = \sqrt{6^2 + (-6.5)^2}\\\\|A| = 8.846$

So:

$Ay = -6.5 = 8.846cos(\alpha)\\\\sin(\alpha) = \frac{-6.5}{8.846}\\\\sin(\alpha) = -0.7348\\\\\alpha = sin^{- 1}(- 0.7348)$

$\alpha$ = -47.28 ° +360° = 313 °

$\alpha$ = 313 °

Option 4.

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3 years ago

A ship sets sail from Rotterdam, The Netherlands, intending to head due north at 6.5 m/s relative to the water. However, the loc

sattari [20]

Answer:

Explanation:

velocity of ship with respect to water = 6.5 m/s due north

$\overrightarrow{v}_{s,w}=6.5 \widehat{j}$

velocity of water with respect to earth = 1.5 m/s at 40° north of east

$\overrightarrow{v}_{w,e}=1.5\left ( Cos40\widehat{i} +Sin40\widehat{j}\right)$

velocity of ship with respect to water = velocity of ship with respect to earth - velocity of water with respect to earth

$\overrightarrow{v}_{s,w} = \overrightarrow{v}_{s,e} - \overrightarrow{v}_{w,e}$

$\overrightarrow{v}_{s,e} = 6.5 \widehat{j}- 1.5\left (Cos40\widehat{i} +Sin40\widehat{j} \right )$

$\overrightarrow{v}_{s,e} = - 1.15 \widehat{i}+5.54\widehat{j}$

The magnitude of the velocity of ship relative to earth is $\sqrt{1.15^{2}+5.54^{2}}$ = 5.66 m/s

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3 years ago

A 78.5-kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above w

Dima020 [189]

Answer:

A) V_air = 1.295 L

B) Volume is not reasonable

Explanation:

A) Let;

m be total mass of the man

m_p be the mass of the man that pulled out of the water because of the buoyant force that pulled out of the lung

m_3 be the mass above the water with the empty lung

m_5 be the mass above the water with full lung

F_b be the buoyant force due to the air in the lung

V_a be the volume of air inside man's lungs

w_p be the weight that the buoyant force opposes as a result of the air.

Now, we are given;

m = 78.5 kg

m_3 = 3.2% × 78.5 = 2.512 kg

m_5 = 4.85% × 78.5 = 3.80725 kg

Now, m_p = m_5 - m_3

m_p = 3.80725 - 2.512

m_p = 1.29525 kg

From archimedes principle, we have the formula for buoyant force as;

F_b = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

F_b = w_p = 1.29525 × 9.81

F_b = 12.7064 N

As earlier said,

F_b = (ρ_water × V_air × g)

Thus;

V_air = F_b/(ρ_water × × g)

V_air = 12.7064/(1000 × 9.81)

V_air = 1.295 × 10^(-3) m³

We want to convert to litres;

1 m³ = 1000 L

Thus;

V_air = 1.295 × 10^(-3) × 1000

V_air = 1.295 L

B) From research, the average lung capacity of an adult human being is 6 litres of air.

Thus, the calculated lung volume is not reasonable

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2 years ago

Why doesn't every planet have a moon?

GalinKa [24]

Answer:

Up first are Mercury and Venus. Neither of them has a moon. Because Mercury is so close to the Sun and its gravity, it wouldn't be able to hold on to its own moon. Any moon would most likely crash into Mercury or maybe go into orbit around the Sun and eventually get pulled into it.

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2 years ago