A mouse in a maze scurries 41 cm south and then takes a 90-degree turn and scurries 64 cm west to get a piece of cheese. Find the mouses displacement
A) 105 cm
B) southwest
C) 76 cm southwest
D) zero
2) what is the range of a bullet fired horizontally at a height of 1.5 m with an initial velocity of 120 m/s
A) -9.8 m/s^2
B) about 0.55 s
C) about 66.4 m
D) there is not enough info
3) A temporary structure loses 60 kWh of energy each day. If modifications to the structure are made so that the energy flux decreases by a factor of 2, what will the new energy loss per day be?
a) 60 kWh
b) 120 kWh
c) 30 kWh
d) 15 kWh
4) a 1500 kg cars speed changes from 30 m/s to 15 m/s after the brakes are applied. Calculate the work done onto the car from the brakes.
a) -506,250 J
b) 506,250 J
c) 168,750 J
d) -168,750 J
so im just gonna give yall the answers
Physics A Semester Exam
1. scientific theory
2. 0.240lbs
3. It starts with a problem to solve or a question to answer.
4. 76 cm southwest
5. If the object moves in a straight line in one direction represented as positive, then the magnitude of average velocity will be equal to the average speed.
6.3 m/s2
7.≈ 66.4 m
8. weight, friction
9.spring force
10. 12 N, right
11.Wbrick−Tsin θ
12.It increases.
13.By increasing the normal force exerted by the wood on the screw thread
14.2.5 m
15.−506,250 J
16. 60 kWh
17.The weather depends on so many conditions that it is not possible to account for them all in any model.
18.3.3 m
19.A
20.gravitational force
21.12 kg-m/s
22.3 J
23.A length of 12 in. and a thickness of 2 in.
24.3/4
25.35,900 km
26. 0.0167
27. the same amount of time
28. 22.3 AU
good luck everyone
okay so my test was different and only had 25 questions and this is the answer to that exam
1. scientific hypothesis
2. 0.240
3. it starts with a problem to solve or a question to answer
4. 76 cm southwest
5. if the object moves in a straight line in one direction represented as positive, then the magnitude of average velocity will be equal to the average speed.
6. 3 m/s
7. ≈ 66.4
8. weight, friction
9. spring force
10. 12 N, right
11. it increases
12. by increasing the normal force exerted by the wood on the screw thread
13. 2.5 m
14. -506,250
15. The weather depends on so many conditions that it is not possible to account for them all in any model.
16. A
17. friction force
18. 12 kg-m/s
19. 3 J
20. a length of 12 in. and a thickness of 2 in.
21. 3/4
22. 35,900
23. 0.0167
24. the same amount if time
25. 22.3 AU
This is for physics A year 11 semester exam on connections :)
1) c 76cm southwest
2) c 66.4 m
🍀 is correct on the first two. The other two are:
3) 60 kWh
4) -506,250 J
25) 35,900 km
26) 0.0167
27) the same amount of time
28) 22.3 AU
THANK YOU SO MUCH
Is What’s Poppin answers correct?
Yes sir they are all correct. Thank you Whats Poppin
1) To find the mouse's displacement, we need to calculate the total horizontal and vertical distances traveled by the mouse.
The distance traveled south is 41 cm, which corresponds to a negative displacement in the vertical direction.
The distance traveled west is 64 cm, which corresponds to a negative displacement in the horizontal direction.
To find the total displacement, we can use the Pythagorean theorem.
Displacement = √((-64 cm)^2 + (-41 cm)^2)
Displacement = √(4096 cm^2 + 1681 cm^2)
Displacement = √5777 cm^2
Displacement ≈ 76 cm
So, the mouse's displacement is approximately 76 cm southwest.
Answer: C) 76 cm southwest
2) The range of a projectile fired horizontally depends on the time of flight (the time it takes to reach the ground) and the horizontal velocity.
In this case, the initial velocity is 120 m/s and the vertical height is 1.5 m.
The time of flight can be calculated using the equation:
-1.5 m = (1/2) * (-9.8 m/s^2) * t^2
t^2 = 0.306122...
t ≈ 0.553 s (taking the positive value)
The horizontal range can be calculated using the equation:
Range = Horizontal Velocity * Time of Flight
Range = 120 m/s * 0.553 s
Range ≈ 66.4 m
So, the range of the bullet fired horizontally is approximately 66.4 m.
Answer: C) about 66.4 m
3) If the energy loss per day is initially 60 kWh, and modifications are made so that the energy flux decreases by a factor of 2, we need to divide the initial energy loss by 2.
New energy loss per day = 60 kWh / 2
New energy loss per day = 30 kWh
Therefore, the new energy loss per day will be 30 kWh.
Answer: c) 30 kWh
4) The work done on an object can be calculated using the equation:
Work = Change in Kinetic Energy
Work = (1/2) * mass * (final velocity^2 - initial velocity^2)
In this case, the mass of the car is 1500 kg, initial velocity is 30 m/s and final velocity is 15 m/s.
Work = (1/2) * 1500 kg * ((15 m/s)^2 - (30 m/s)^2)
Work = (1/2) * 1500 kg * (-225 m^2/s^2)
Work = -168,750 J
So, the work done on the car from the brakes is -168,750 J.
Answer: d) -168,750 J
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