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Flura [38]
2 years ago
13

Brainliest!!! Write: Forces are all around us. Imagine that your teacher has asked you to teach a lesson to your peers about for

ces. Explain, in detail, how you experience forces in your everyday life.
Physics
1 answer:
Mademuasel [1]2 years ago
8 0
A force is a push or pull upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects. When the interaction ceases, the two objects no longer experience the force. Forces only exist as a result of an interaction.
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What is the Acceleration of a car traveling 25 m/s and in 5 seconds<br> traveling 65 m/s? *
WARRIOR [948]

Answer:

8m/s²

Explanation:

a = change in v/t

a = (65-25)/5

a = 8

7 0
2 years ago
Suppose that the acceleration of a model rocket is proportional to the difference between 100 ft/sec and the rocket's velocity.
sp2606 [1]

Answer:

Explanation:

initial velocity, u = 0

final velocity, v = 80 ft/s

acceleration, a = 150 ft/s²

Let the time taken is t.

v = u + at

80 = 0 + 150 x t

t = 0.53 second

3 0
3 years ago
A ship sails east from a harbor for 16 nautical miles. it then sails in the direction of n 55 e for 18 nautical miles. how far i
Katena32 [7]
<span> Use the Law of Cosines, where you have a triangle with included angle of 145 degrees and sides of 16 and 18. You are then solving the equation: </span>

<span>d^2 = 16^2 + 18^2 - 2(16)(18)cos(145)        </span>
8 0
2 years ago
Read 2 more answers
For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be requ
slava [35]

Explanation:

First we will convert the given mass from lb to kg as follows.

        157 lb = 157 lb \times \frac{1 kg}{2.2046 lb}

                   = 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

         180 \frac{mg}{kg} \times 71.215 kg

         = 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

      0.65% (w/w) = \frac{0.65 g}{100 g}

                           = \frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}

                           = \frac{0.65 g}{100 ml}

Therefore, calculate the volume which contains the amount of caffeine as follows.

   12818.7 mg = 12.8187 g = \frac{12.8187 g}{\frac{0.65 g}{100 ml}}

                       = 1972 ml

Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.

5 0
3 years ago
a car traveling at a velocity of 2 m/s undergoes an acceleration of 4.5 m/s^2 over a distance of 340 m. How fast will it be goin
ra1l [238]
Vi = 2m/s
a= 4.5 m/s 
d= 340 m
vf= ?

use this equation ...  vf^2=vi<span>^2+2ad

you should get vf = 55.3
hope this helps </span>
3 0
3 years ago
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