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Rina8888 [55]
3 years ago
8

Several students in the group made comments to the Coach about the back pack. Which student has the correct analysis of the forc

es and motion of the back pack?
Physics
1 answer:
kolbaska11 [484]3 years ago
7 0

Answer:

20%

Explanation:

Efficiency of an engine is a ratio in percentage of the work done to the energy intake of the engine.

η=(q₁-q₂)/q₁×100%

η is efficiency, q₁ is the energy intake, q₂ is energy wasted.

the information provided is:

energy intake = 400 j

work done = 80 j = (400-320) =q₁-q₂

energy wasted =320 j

η= 80/400×100%

=20%

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Answer:

Fluorine

Explanation:

The more number of electrons on the outer shell makes the atom more reactive

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3 years ago
A car accelerates from rest at a constant rate of 2 m/s^2 for 5 s. what is the speed of the car at the end of that time? g
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Hope this helps you! This is my step-by-step work. Lemme know if you have any questions!

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3 years ago
Atoms and molecules are constantly in motion. The more kinetic energy atoms or molecules have, the faster they move about. In wh
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3 years ago
A 0.25 ball is thrown with the speed of v=4.5m/s at a resting stick that can rotate around one end at the axis O. The ball hits
malfutka [58]

Answer:

\mathbf{^\to Q =0.84375 \ (- \hat z)  \ kgm^2s^{-1}}

Explanation:

mass of the ball = 0.25  kg

its speed v = 4.5 m/s  

r = 0.75 m

let the speed be in the horizontal direction and the distance r be in the vertical direction we have :

^ {\to}v = 4.5 \ x  \ \  m/s

^ {\to}r = 0.75 \  y \ \  m/s

Let the momentum about the center of intersection be P;

SO;

^ \to} P = ^{\to} mv

^ \to} P = 0.25* 4.5 \ x  \ \  m/s

^ \to} P = 1.125 \ x  \ \  m/s

Let the angular momentum be Q;

^\to Q = ^ \to r*  ^ \to P

^\to Q = (0.75*1.125) kgm^2s^{-1} *(\hat y * \hat x)

\mathbf{^\to Q =0.84375 \ (- \hat z)  \ kgm^2s^{-1}}

6 0
4 years ago
A coin 15.0 mm in diameter is placed 15.0 cm from a spherical mirror. The coin's image is 5.0 mm in diameter and is erect. Is th
s344n2d4d5 [400]

Answer:

15 cm

Explanation:

h_{o} = Diameter of the coin = 15 mm

h_{i} = Diameter of the image of coin =  5 mm

d_{o} = distance of the coin from mirror = 15 cm

d_{i} = distance of the image of coin from mirror = ?

Using the equation

\frac{d_{i}}{d_{o}} = \frac{- h_{i}}{h_{o}}

\frac{d_{i}}{15} = \frac{- (5)}{15}

d_{i} = - 5 cm

R = radius of curvature

Using the mirror equation

\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{2}{R}

\frac{1}{15} + \frac{1}{- 5} = \frac{2}{R}

R = - 15 cm

6 0
3 years ago
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