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lara31 [8.8K]
3 years ago
14

Calculate the mass (in SI units) of (a) a 160 lb human being; (b) a 1.9 lb cockatoo. Calculate the weight (in English units) of

(c) a 2300 kg rhinoceros; (d) a 22 g song sparrow.
Physics
1 answer:
kondaur [170]3 years ago
5 0

Explanation:

1. Force applied on an object is given by :

F = W = mg

(a) A 160 lb human being, F = 160 lb

g = acceleration due to gravity, g = 32 ft/s²

m=\dfrac{F}{g}

m=\dfrac{160\ lb}{32\ ft/s^2}

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

m=\dfrac{F}{g}

m=\dfrac{1.9\ lb}{32\ ft/s^2}

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

W=2300\ kg\times 32\ ft/s^2=73600\ lb

(b) A 22 g song sparrow, m = 22 g = 0.022 kg

W=0.022\ kg\times 32\ ft/s^2=0.704\ lb

Hence, this is the required solution.

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Olegator [25]

Answer:

I believe the answer is B.

Explanation:

Newton's First Law of Gravity states, "The greater the weight (or mass) of an object, the more inertia it has. Heavy objects are harder to move than light ones because they have more inertia. "

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3 years ago
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Imagine that you have a 6.50 l gas tank and a 4.50 l gas tank. you need to fill one tank with oxygen and the other with acetylen
Margarita [4]
V₁(O2) = 6.50<span>  L                
</span>p₁(O2) = 155 atm          
V₂(acetylene) = <span>4.50 L                   
</span>p₂(acetylene) =?                             

According to Boyle–Mariotte law (At constant temperature and unchanged amount of gas, the product of pressure and volume is constant) we can compare two gases that have ideal behavior and the law can be usefully expressed as:

V₁/p₁ = V₂/p₂

6.5/155 = 4.5/p₂

0.042 x p₂ = 4.5

p₂ = 107.3 atm



7 0
3 years ago
Shanti is riding on a train that is moving at a speed of 90 km/h. He is carrying a power cord for his phone that is 1.2 m long.
faltersainse [42]

Answer:equal to 1.2 m

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3 years ago
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Sammy squirrel is steering his boat at a heading of 327 degree at 18mph. The current is flowing at 4mph at a heading of 60 degre
Tanya [424]

Answer:

  • 59.97 º at 18.23 mph

Explanation:

To find Sammy's course you have to add the two velocities (vectors), 18 mph 327º and 4 mph 60º.

To add the two vectors analytically you decompose each vector into their vertical and horizontal components.

<u>1. 18 mph 327º</u>

  • Horizontal component: 18 mph × cos (327º) = 15.10 mph

  • Vertical component: 18 mph × sin (327º) = - 9.80 mph

  • Vector notation:

       15.10\hat i-9.80\hat j

<u>2. 4 mph 60º</u>

  • Horizontal component: 4 mph × cos (60º) = 2.00 mph

  • Vertical component: 4 mph × sin (60º) = 3.46 mph

  • Vector notation:

       2.00\hat i+3.46\hat j

<u>3. Addition:</u>

You add the corresponding components:

15.10\hat i-9.80\hat j+2.00\hat i+3.46\hat j\\ \\ 17.10\hat i-6.34\hat j

To find the magnitude use Pythagorean theorem:

  • \sqrt{17.1^2+6.34^2}= 18.23

<u>4. Direction:</u>

Use the tangent ratio:

  • tan(\alpha )=opposite/adjacent=3.46/2.00=1.73

Find the inverse:

  • arctan (1.73) ≈ 59.97º
5 0
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Why might it be necessary to ignore some of the data points just before and just after the collision?
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We have no idea. We need to examine the experimental set-up. You've given us no information, except that there may have been some sort of collision.
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