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ioda
3 years ago
3

Identify the scenarios below as to whether they would increase, decrease, or not change

Physics
1 answer:
tekilochka [14]3 years ago
7 0

Answer:1. Increase 2. Decrease 3. Decrease

Explanation:

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Joe places two solid objects in contact with each other.
Tasya [4]

Answer:

B. temperature of the objects

5 0
3 years ago
The voltage across the terminals of a 9.0 v battery is 8.5 v when the battery is connected to a 60 ω load. part a what is the ba
snow_lady [41]
Refer to the diagram shown below.

i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω

Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5         (10

Also,
R₂*i = 9.5         (2)

Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A

From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)

Answer: 0.08 Ω

3 0
3 years ago
Read 2 more answers
Vectors have which two properties?
stiks02 [169]

Answer:

D. magnitude and direction

Explanation:

3 0
4 years ago
It takes a minimum distance of 41.14 m to stop a car moving at 11.0 m/s by applying the brakes (without locking the wheels). Ass
padilas [110]

Answer:

Find the time it took for the car to stop at 11.0m/s

V = deltax/t

t = 41.14/11.0 = 3.74s

Now find at what rate it was decelerating, so find the acceleration during that interval of time.

vf = vi + at

-11.0m/s = a3.74s

a = -2.94m/s^2

The acceleration is negative because is pulling the car towards its opposite direction to make it stop.

Now find how much time it would take for the car to stop at 28.0m/s but with the same acceleration, the car is the same so its acceleration to stop the car will remain the same.

vf = vi + at

0 = 28.0 - 2.94t

t = 9.52

Once the time is obtained, you can find the final position, xf, by plugging the time acceleration and velocity values.

xf = 0 + (28m/s)(9.52s) + 1/2(-2.94)(9.52s)^2

xf = 266.6m - 133.23m = 133m

8 0
4 years ago
300N effort is applied to lift the load of 900N by using a first class lever. If the distance between the load and fulcrum is 20
Klio2033 [76]

Answer:

Effort=300N

Load=900N

Load distance=20m

Now,

Ed=(L*Ld)/E

Ed=(900*20)/300

Ed=1800/300

Ed=6m

Explanation:

7 0
2 years ago
Read 2 more answers
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