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Olin [163]
3 years ago
7

How do I calculate speed, velocity, and acceleration? I need the formulas too

Physics
1 answer:
vodomira [7]3 years ago
5 0
Speed is the secular unit, which means it only measures "how fast it goes."
speed= \frac{distance}{time}
Velocity is a victory unit, which means it measures "both how fast it goes AND the direction."
velocity= displacement/time
Acceleration is the change of velocity over time. 
acceleration =  \frac{change in velocity}{time}
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What is the awnser to Blank occurs when a substance changes from a liquid to a gas
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Evaporation (or another word to use is water vapor.)
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How is thermal energy naturally transferred between objects?
Sladkaya [172]

The second law of thermodynamics establishes restrictions on the flow of thermal energy between two bodies. This law states that the energy does not flow spontaneously from a low temperature object T1, to another object that is at a high temperature T2.

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Finally we can say that the correct option is B: From the hotter object to the cooler object

5 0
2 years ago
A net force of 24 N is acting on a 4.0-kg object. Find the acceleration in m/s2.
Inessa [10]

Hi there!

We can use Newton's Second Law:
\Sigma F = ma

ΣF = Net force (N)
m = mass (kg)
a = acceleration (m/s²)

We can rearrange the equation to solve for the acceleration.

a = \frac{\Sigma F}{m}\\\\a = \frac{24}{4} = \boxed{6 \frac{m}{s^2}}

6 0
2 years ago
Mr. hitch taught us about sedimentary, metamorphic, and igneous rocks. he described how they were formed, what they contain, and
Tems11 [23]
Mr. Hitch taught us about sedimentary, metamorphic, and igneous rocks. He described how they were formed, what they contain, and showed us samples of each. He is a good geologist. 

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3 0
2 years ago
A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet
Gnom [1K]

Answer:

d. 37 °C

Explanation:

m_{m} = mass of lump of metal = 250 g

c_{m} = specific heat of lump of metal  = 0.25 cal/g°C

T_{mi} = Initial temperature of lump of metal = 70 °C

m_{w} = mass of water = 75 g

c_{w} = specific heat of water = 1 cal/g°C

T_{wi} = Initial temperature of water = 20 °C

m_{c} = mass of calorimeter  = 500 g

c_{c} = specific heat of calorimeter = 0.10 cal/g°C

T_{ci} = Initial temperature of calorimeter = 20 °C

T_{f} = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) +  m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C

6 0
3 years ago
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