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ladessa [460]
3 years ago
8

The greater the mass of an object

Physics
1 answer:
lesya [120]3 years ago
7 0

Answer:

The greater the velocity, the greater the Force needs to be, and the greater the fiction is

Explanation:

I don't know what you are working on so here are a few responses

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1
Darya [45]
  • Mass=0.25kg
  • Force=95N

\\ \bull\sf\dashrightarrow F=ma

\\ \bull\sf\dashrightarrow a=\dfrac{F}{m}

\\ \bull\sf\dashrightarrow a=\dfrac{95}{0.25}

\\ \bull\sf\dashrightarrow a=380m/s^2

3 0
2 years ago
You want to determine whether the mass of an object attached to a parachute affects the time it takes to fall to the ground. In
hammer [34]

Answer:

The Answer is below!!

Explanation:

The larger the area of the parachute, the more air needs to be pushed out of the way, and so the slower it descends. the independent variable is the shape of the parachutes' canopies. The dependent variable is the drop speeds of the parachutes. How long does it take for the parachutes to reach the ground? Measure this using a stopwatch.

Hope I Helped!!

:)

5 0
2 years ago
Two friends leave a movie theater and take different busses to the same ice cream shop. One bus takes a longer route driving on
lisov135 [29]

Answer: short displacement has shorter road and long displacement has longer displacement . now think your self Which one is right

6 0
2 years ago
Why cant soy sauce freeze
juin [17]
The salt content in soy
8 0
2 years ago
Read 2 more answers
The vapor pressure of ethanol at 293 K is 5.95 kPa and at 336.5 K it is 53.3 kPa. Calculate the enthalpy of vaporization of etha
denis-greek [22]

Answer:

H=41.3kJmol^{-1}

Explanation:

The equation relating the the enthalphy, pressure and temperature is expressed as

ln(\frac{P_{2}}{P_{1}} )=\frac{H}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}} ) \\

Where P is the pressure, H is the enthalphy, and T is the temperature.

since the given values are

T_{1}=293k, \\T_{2}=336.5k\\P_{1}=5.95kPA\\p_{2}=53.3kPA\\and R=8.314J.K^{-1}mol_{-1}

if we insert values, we arrive at

ln(\frac{53.3}{5.95} )=\frac{H}{8.314}(\frac{1}{293}-\frac{1}{336.5} )\\2.19=\frac{H}{8.314}(0.00044)\\H=(2.19*8.314)/0.00044\\H=41,268.8Jmol^{-1}\\H=41.3kJmol^{-1}

4 0
2 years ago
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