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Ne4ueva [31]
2 years ago
10

In a doorknob, the knob is connected to a shaft. When the knob turns, the shaft turns,which moves the door latch. The radius of

the shaft is 8 mm. The radius of the doorknob is 56 mm. What is the mechanical advantage of this wheel and axle?
Physics
1 answer:
HACTEHA [7]2 years ago
4 0

Answer:

`7

Explanation:

i know

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How many times did john glenn orbit the earth
galben [10]

Answer:

2 times

Explanation:

1st time on Feb. 20, 1962 at age 41

2nd time on Oct. 29, 1998 at age 77

8 0
3 years ago
How does the balloon react to the cloth item?
almond37 [142]

Answer: When rubbing a balloon with a wool cloth, it puts negative charges on the balloon. Negative charges attract to positive charges. If a balloon is not rubbed with the wool cloth, it has an equal amount of negative to positive charges, so it will attract to a rubbed balloon.

8 0
3 years ago
A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed
fgiga [73]

Answer:

Explanation:

Time taken to accelerate to 28 m /s

= 28 / 2 = 14 s

a ) Total length of time in motion

= 14 + 41 + 5

= 60 s .

b )

Distance covered while accelerating

s = ut + 1/2 at²

= 0 + .5 x 2 x 14²

= 196 m .

Distance covered while moving in uniform motion

= 28 x 41

= 1148 m

distance covered while decelerating

v = u - at

0 = 28 - a x 5

a = 5.6 m / s²

v² = u² - 2 a s

0 = 28² - 2 x 5.6 x s

s = 28² / 2 x 5.6

= 70 m .

Total distance covered

= 196 + 1148 + 70

= 1414 m

total time taken = 60 s

average velocity

= 1414 / 60

= 23.56 m /s .

8 0
3 years ago
2. What do you understand by balanced and unbalanced force​
labwork [276]

Answer:

forces that are equal in size and opposite in direction. Balanced forces do not result in any change in motion. unbalanced. forces: forces applied to an object in opposite directions that are not equal in size. Unbalanced forces result in a change in motion.

\\

hope helpful ~

8 0
2 years ago
Find the speed vfinal of the joined cars after the collision. mastering physics
Tanya [424]
<span>Px = 0 Py = 2mV second, Px = mVcosφ Py = –mVsinφ add the components Rx = mVcosφ Ry = 2mV – mVsinφ Magnitude of R = âš(Rx² + Ry²) = âš((mVcosφ)² + (2mV – mVsinφ)²) and speed is R/3m = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²) simplifying Vf = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²) Vf = (1/3)âš((Vcosφ)² + (2V – Vsinφ)²) Vf = (V/3)âš((cosφ)² + (2 – sinφ)²) Vf = (V/3)âš((cos²φ) + (4 – 2sinφ + sin²φ)) Vf = (V/3)âš(cos²φ) + (4 – 2sinφ + sin²φ)) using the identity sin²(Ď)+cos²(Ď) = 1 Vf = (V/3)âš1 + 4 – 2sinφ) Vf = (V/3)âš(5 – 2sinφ)</span>
6 0
3 years ago
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