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alexgriva [62]
3 years ago
15

A 205-kg crate is pushed horizontally with a force of 720 n. if the coefficient of friction is 0.20, calculate the acceleration

of the crate. m/s2
Physics
1 answer:
Marat540 [252]3 years ago
6 0
Google it...............
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A thin cylindrical shell and a solid cylinder have the same mass and radius. The two are released side by side and roll down, wi
Vitek1552 [10]

Answer:

Explanation:

Let the velocity be v

Total energy at the bottom

= rotational + linear kinetic energy

= 1/2 Iω² + 1/2 mv² ( I moment of inertia of shell  = mr² )

= 1/2 mr²ω² + 1/2 mv² ( v = ω r )

= 1/2 mv² +1/2 mv²

= mv²

mv² = mgh ( conservation of energy )

v² = gh

v = √gh

= √9.8 x 1.8

= 4.2 m /s

8 0
3 years ago
What is the equation for potential energy?
Mrrafil [7]
You can calculate potential energy by:
U = m.g.h

Where, U = potential energy
m = mass
g = acceleration due to gravity
h = height

Hope this helps!
7 0
3 years ago
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When did galileo discover projectile motion?
Tcecarenko [31]
Between 1589-1592 when he discovered projecctile motion
4 0
2 years ago
Can someone please help me ASAP???!!!
olganol [36]

Answer: The answers are:

Explanation:

1. C

2. A

3. C

4. B

5. D

8 0
1 year ago
The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371
arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

q=m\times c\times (T_2-T_1)

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = 4.18J/g.K

T_1 = initial temperature of water = 293.0 K

T_2 = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

q=250g\times 4.18J/g.K\times (371.2-293.0)K

q=81719J

Now we have to calculate the enthalpy of combustion of octane.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

\Delta H=\frac{-81719J}{0.015mole}

\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0
3 years ago
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