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notsponge [240]

3 years ago

14

How does the vertical component of a projectile’s motion compare with the motion of vertical free fall when air resistance is ne

gligible?

1 answer:

likoan [24]3 years ago

5 0

Answer:

Explanation:

The vertical component of velocity remains same as the free fall. The vertical motion of the projectile is same as the free fall motion.

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What are the benefits of scientists using the metric system worldwide

emmainna [20.7K]

Both systems have benefits and disadvantages.

Benefits of the Metric System

<span>Easier for anything involving conversions.Celsius is based on the the freezing and boiling points of water.Is the most popular system of measurement worldwide. This gives interchangeability internationally.Other systems of measurement, like U.S. standard units, are tied directly to metric units so they can be converted back and forth.<span>You can sell to the government since governments (including the U.S. government) require government contractors to use the metric system for most things.</span></span>

8 0

3 years ago

Read 2 more answers

An iron anchor of density 7810.00 kg/m^3 appears 252 N lighter in water than in air. (a) What is the volume of the anchor?(b) Ho

RSB [31]

Answer:

(a)The volume of the iron anchor = 25.2 m³

<em>(b)The mass of iron = 196812 kg</em>

Explanation:

<em>Density:</em> This is the ratio of the mass of a body to its volume. The S.I unit of density is <em>kg/m³</em>

(a)

From Archimedes principle, the iron will displace a volume of water that is equal to its volume. And the weight of water displaced is equal to loss in weight or upthrust.

Therefore, Upthrust = lost in weight = weight of water displace.

Therefore, Volume of the iron = mass of water displace/density of water

Vi = m₁/D₁........................... Equation 1

Where Vi = volume of iron, m₁ = mass of water displaced, D₁ = Density of water.

<em>Given: upthrust = 252 N, ∴ m₁ = 252/10 = 25.2 kg.</em>

<em>Constant: D₁ = 1.0 kg/m³</em>

<em>Substituting these values into equation 1,</em>

Vi = 25.2/1.0

Vi = 25.2 m³

Therefore the volume of the iron anchor = 25.2 m³

(b)

<em>Density of the iron = mass of the iron/volume of the iron.</em>

<em>Therefore, Mass of the iron = Density of the iron × volume of the iron...........................................equation 2</em>

<em>Given: Density of the iron = 7810 kg/m³, Volume of the iron = 25.2 m³</em>

<em>Substituting these values into equation 2,</em>

<em>Mass of the iron = 7810×25.2</em>

<em>Mass of the iron = 196812 kg.</em>

<em>Therefore the mass of iron = 196812 kg</em>

8 0

3 years ago

Consider a 30-cm-diameter hemispherical enclosure. The dome is maintained at 600 K, and heat is supplied from the dome at a rate

SIZIF [17.4K]

Answer:

$\epsilon_2=0.098$

Explanation:

Diameter $d=30cm=0.3m$

Temperature $T=600k$

Rate of supply $r=65W$

Emissivity of base surface $\in_b =0.55$

Temperature at base $T_b=400k$

Generally the equation for Area of base surface is mathematically given by

$A_b=\frac{\pi}{4}d^2$

$A_b=\frac{\pi}{4}0.3^2$

$A_b=0.0707m^2$

Generally the equation for Area of Hemispherical dome is mathematically given by

$A_h=\frac{\pi}{2}d^2$

$A_h=\frac{\pi}{2}0.3^2$

$A_h=0.1414m^2$

Since base is a flat surface

$F_{11}+F_{12}=1$

$F_{11}=0$

Therefore

$F_{12}=1$

$A_b=0.0707m^2$

Generally the equation for Net rate of radiation heat transfer between two surfaces is mathematically given by

$Q_{21}=-Q_{12}$

$Q_{21}=\frac{\sigma(T_1^4-T_2^4)}{\frac{1-\epsilon}{A_b\epsilon_1} +\frac{1}{A_bF_{12}} +\frac{1-\epsilon_2}{A_h*\epsilon_2} }$

Where

$\sigma=5.67*10^{-8}$

Therefore

$65=\frac{(5.67*10^{-8}(400^4-600^4))}{\frac{1-0.55}{0.0707*0.55}+\frac{1}{0.0707}+\frac{1-\epsilon_2}{0.1414*\epsilon_2}}$

$\epsilon_2=0.098$

$\epsilon_2 \approx 0.1$

Therefore the emissivity of the dome is

$\epsilon_2=0.098$

3 0

3 years ago

A 1325 kg car and a 2050 kg pickup truck approach a curve on a highway that has a radius of 255 m. At what angle should the high

Scorpion4ik [409]

Answer:

the banking angle of the road is 24.2⁰

Explanation:

Given;

speed of the vehicles considered, v = 75 mi/h

Speed in m/s ⇒ 1 mi/h --------> 0.44704 m/s

75 mi/h --------> ?

= 75 x 0.44704 m/s = 33.528 m/s

radius of the curve, r = 255 m

The banking angle of the road is calculated as;

$\theta = tan^{-1} (\frac{v^2}{rg} )\\\\\theta = tan^{-1} (\frac{33.528^2}{255\times 9.8} )\\\\\theta = tan^{-1}(0.44983)\\\\\theta =24.2^0$

Therefore, the banking angle of the road is 24.2⁰

8 0

2 years ago

Suppose that you hear a clap of thunder 16.2 s (seconds) after seeing the associated lightning stroke. The speed of sound waves

spayn [35]

Since the speed of sound is very high, we consider the time period of thunder and lightening negligible.

Now, d = vt

thus, d = 343 x 16.2

thus, d = 5556.6 metres

Thus, we are 5556.6 metres away from lightening ( approximately)

5 0

3 years ago