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Deffense [45]
3 years ago
13

A football is kicked 18m/s at 20 degrees. what is the acceleration rate in the horizontal direction

Physics
1 answer:
fomenos3 years ago
6 0

If we ignore air resistance, as we always do,
there is no horizontal acceleration.

Gravity acts only in the vertical direction, and
no other forces act on the ball after the punt.

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Define orbital velocity
sveta [45]

the Orbital Velocity is the velocity sufficient to cause a natural or artificial satellite to remain in orbit. Inertia of the moving body tends to make it move on in a straight line, while gravitational force tends to pull it down. The orbital path, elliptical or circular, representing a balance between gravity and inertia, and it follows a rue that states that the more massive the body at the centre of attraction is, the higher is the orbital velocity for a particular altitude or distance.

8 0
3 years ago
An apple falls from a tree and hits the ground 9.98 m below. with what speed will it hit the ground
cupoosta [38]
The answer to this question is <span>13,537</span>
8 0
3 years ago
A motor-driven winch pulls a 50.0 kg student 5.00 m up the rope at a constant speed of 1.25 m/s. how much power does the motor u
nadya68 [22]
Power is the rate work done given by dividing work done by unit time. It is measured in watts equivalent to J/s.
In this case the force by the student is mg = 490 N (taking g as 9.8m/s²)
Work done is given by force × distance,
Therefore, Power =(force × distance)/ time, but velocity/speed =distance/time
Thus, Power = force × speed/velocity
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7 0
3 years ago
Determine the normal boiling point of a substance whose vapor pressure is 55.1 mm hg at 35°c and has a δhvap of 32.1 kj/mol.
Novosadov [1.4K]

Answer:

389.78681 K

Explanation:

P_1 = Initial pressure = 55.1 mmHg

P_2 = Final pressure = 1 atm = 760 mmHg

T_2 = Boiling point

T_1 = Initial temperature = 35°C

\Delta H_{vap} = Heat of vaporization = 32.1 kJ/mol

From the Clausius-Claperyon equation

ln\dfrac{P_2}{P_1}=(-\dfrac{\Delta H_{vap}}{R})(\dfrac{1}{T_2}-\dfrac{1}{T_1})\\\Rightarrow \dfrac{1}{T_2}=-ln\dfrac{P_2}{P_1}\dfrac{R}{\Delta H_{vap}}+\dfrac{1}{T_1}\\\Rightarrow \dfrac{1}{T_2}=-ln\dfrac{760}{55.1}\dfrac{8.314}{32.1\times 10^{3}}+\dfrac{1}{273.15+35}\\\Rightarrow T_2=\left(-ln\left(\frac{760}{55.1}\right)\frac{8.314}{32.1\times \:10^3}+\frac{1}{273.15+35}\right)^{-1}\\\Rightarrow T_2=389.78681\ K

The normal boiling point of the substance is 389.78681 K

3 0
3 years ago
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MaRussiya [10]
Storm!! Hope I could help
8 0
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