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Kamila [148]
3 years ago
12

CNA Secondary and Preparatory School

Physics
1 answer:
ale4655 [162]3 years ago
3 0

Answer:

2:13 2) cm' c6462.

Explanation:

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When an object threw to the free space to make an angle of 25 degree at an initial speed of 15 m/sec, the ball takes time to rea
alekssr [168]

Answer:

The horizontal distance traveled by the projectile is 15.23 m.

Explanation:

Given;

angle of projection, θ = 25⁰

initial velocity of the projectile, u = 15 m/s

time of flight, t = 1.12 s

The the travelling path of the object is calculated as the range of the projectile

R = u_x t\\\\R = (15\ Cos \ 25^0) \times 1.12\\\\R = 13.595 \times 1.12\\\\R = 15.23 \ m

Therefore, the horizontal distance traveled by the projectile is 15.23 m.

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3 years ago
A small hot-air balloon is filled with 1.01×106 l of air (d = 1.20 g/l). as the air in the balloon is heated, it expands to 1.10
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<span>Volume of air in the balloon 1.01 x 10^6 L
 Density of air is 1.20 g/l
 Mass = Density X Volume

 So mass of the air in the Balloon= ( 1.01 x 10^6) X 1.20 = 1.212 x 10^6 g
 As the air is heated, the volume of air in the balloon expands to 1.10x 10^6 L Density= Mass/ voume So the Density of heated air = 1.212 x 10^6/ 1.10x 10^6 = 1.101 g/l The answer is 1.101 g/l.</span>
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3 years ago
What is an exothermic chemical reaction? A. It is a reaction that converts matter to heat. B. It is a reaction that requires hea
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it is D exo means realease

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3 years ago
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A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250
Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block m=0.5 kg

inclination \theta =30

coefficient of static friction \mu =0.35

coefficient of kinetic friction \mu _k=0.25

distance traveled d=77.3 cm

spring constant k=35 N/m

work done by gravity+work done by friction=Energy stored in Spring

mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}

mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}

0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}

x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}

x=0.234 m

x=23.4 cm

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3 years ago
True or False:
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