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777dan777 [17]
3 years ago
9

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!!

Physics
2 answers:
Vadim26 [7]3 years ago
8 0
I'm say true because vertical and horizontal work together
drek231 [11]3 years ago
4 0
I would say true. If you are calculating using vectors than it would need both...
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Energy is ____________________ when light is directed upward into the night sky and lost into ____________________.
lisov135 [29]
1: Lost (?)
2: Space (?)
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7 0
4 years ago
14. The United States Tobacco Settlement between the major tobacco companies and 46 states caused the price of cigarettes to jum
AysviL [449]

Answer:

B. The elasticity of demand is -0.126

Explanation:

% Change in Quality demand = -2.65% (negative because of drop)

% Change in price = 21%

Elasticity of demand is given by

\text{Elasticity of demand}=\dfrac{\text{\% Change in Quality demand}}{\text{\% Change in price}}\\\Rightarrow \text{Elasticity of demand}=\dfrac{-2.65}{21}\\\Rightarrow \text{Elasticity of demand}=-0.12619

The Elasticity of demand is -0.12619

8 0
4 years ago
Charlie Brown kicks a football at 24.5 m/s at 35.0. What is the maximum height of the ball?
lozanna [386]

Answer:

d = 10.076 m

Explanation:

We need to obtain the velocity of the ball in the y direction

Vy  = 24.5m/s * sin(35) = 14.053 m/s

To obtain the distance, we use the formula

vf^2 = v0^2 -2*g*d

but vf = 0

d = -vo^2/2g

d = (14.053)^2/2*(9.8) = 10.076 m

5 0
4 years ago
Oberon is a moon of Uranus. It has a mass of 3.01 x 1021 kg and is 761.4 km away from Uranus. If the force of gravity
ZanzabumX [31]

Answer:

acc to formula= F= G m1m2/rsq

mass of Uranus= F×r sq/G×m

M=3.00×10^25×761.4×761.4×10^6/6.67×10^ -11×3.01×10^21

=3×10^25-8×7614×7614/667×301×10^-2-11+21-2

=173918988×10^11/200767

=866×10^11(approx)

4 0
3 years ago
A slender rod 100.00cm long is used as a meter stick. Twoparallel axes which are perpendicular to the rod are considered.The fir
vredina [299]

Answer:

 I /I_{cm}  = 1.48 ,    The correct answer is d

Explanation:

the moment of inertia is given by

        I = ∫ r² dm

For figures with symmetry it is tabulated.   In the case of a thin variation, the moment of inertia with respect to its center of mass is

       I_{cm} = 1/12 M L2

There is a widely used theorem, which is the parallel axis theorem, where the moment of inertia of any parallel axis, is the moment of mass inertia plus the moment of inertia of the body taken as a particle

      I = I_{cm}  + M D²

Let's put these expressions to our case.

As the bar is one meter long its center of mass that this Enel midpoint corresponds to

      I_{cm}  = 1/12 m L²

      I_{cm}  = 1/12 m 1.00²

      I_{cm}  = 8.33 10⁻² m

Let's use the parallel axes theorem for the axis that passes through x = 30 cm.  The distance from the enrode masses to the axis is

       D = x_{cm}  - 0.30

       D = 0.50 - 0.30 = 0.20 m

       I = I_{cm}  + m D²

       I = 8.33 10⁻² m + m 0.2²

       I = (8.33 10⁻² + ​​4 10⁻²) m    

       I = 12.33 10⁻² m

The relationship between these two moments of inertia

       I /I_{cm}  = 12.333 10⁻² / 8.333 10⁻²

       I /I_{cm}  = 1.48

The correct answer is d

3 0
3 years ago
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