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meriva
3 years ago
10

in a historical movie, two knights on horseback start from rest at 88.0 m spartans ride directly toward each other to do battle.

Sir George’s acceleration has a magnitude of 0.300 m/s^2, while Sir Alfred’s has a magnitude of 0.200 m/s^2. Relative to Sir George’s starting point, where do the knights collide?
Physics
1 answer:
Tema [17]3 years ago
7 0
The distance formula:d = v i · t + a · t² / 2Sir G.: ( v i = 0 )
d G = 0 + 0.3 · t² / 2Sir A. : d A = 0 + 0.2 t² / 288 m = 0.3 t² / 2   + 0.2 t² / 2    / · 2 ( multiple both sides by 2 )176 = 0.3 t² + 0.2 t²176 = 0.5 t²t² = 176 : 0.5t² = 352t = √352t = 18.76 sd G. = 0.3 · 18.76² / 2 = 0.3 · 352 / 2 = 52.8 mAnswer: The knights collide at 52.8 m relative to Sir George`s starting point.


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An accepted value for the acceleration due to gravity is 9.801 m/s2. In an experiment with pendulums, you calculate that the val
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the difference between the ideal and observed value is 0.381.

hence the error is -\frac{0.381}{9.801} *100

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In developed nations, the water delivered to homes by pipe is generally potable. Please select the best answer from the choices
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Answer:

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Explanation:

From the question we are told that

Mass of pitcher   M= 2.6kg

Force on pitcher f=8.8N

Distance traveled 48cm=>0.48m

Coefficient of friction \mu=0.28

a)Generally frictional force is mathematically given by

F=\mu N

F=0.28*2.6*9.8

F=7.1344N

Generally work done on the pitcher is mathematically given as

W_n_e_t=W_f+W_F

W_f=8.8*0.48=>4.224N\\W_F=7.1344*0.48=>3.424512N

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b)Generally K.E can be given mathematically as

K.E= W_n_e_t

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