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Kipish [7]
3 years ago
6

Consider the following arrangement with a frictionless/massless pulley. Determine the force F required to move block A if the co

efficient of static friction between block A and the floor is 0.35 and the coefficient of static friction between block A and block B is 0.25
Physics
1 answer:
Delicious77 [7]3 years ago
6 0

Answer: The free - body diagrams for blocks A and B. frictionless surface by a constant horizontal force F = 100 N. Find the tension in the cord between the 5 kg and 10 kg blocks. The string that attaches it to the block of mass M2 passes over a frictionless pulley of negligible mass. The coefficient of kinetic friction Hk between M.

Explanation: Hope this helped :)

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ElenaW [278]

I would say your answer is A.

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1. The two images below show the area swept out by the same planet during two separate time spans. If the first picture represen
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its A

Explanation:

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86 Km/h convertir a (m/min)
gogolik [260]

Answer: 1433.3 m/min

Explanation:

For 86 Km/h converted to a (m/min), convert kilometers to meters, and hour to minutes

So, 86 Km/h means 86 kilometers per 1 hour

- If 1 kilometer = 1000 metres

86 kilometers = 86 x 1000 = 86,000m

- If 1 hour = 60 minute

1 hour = 60 minutes

In m/min: (86,000m / 60 minute)

= 1433.3 m/min

Thus, 86 Km/h convert to 1433.3 m/min

5 0
3 years ago
A 2498 kg car is moving at 17.1 m/s slams on its brakes and slows to 2.6 m/s. What is the magnitude (absolute value) of the impu
bija089 [108]

Answer:

<em>J=36221 Kg.m/s</em>

Explanation:

<u>Impulse-Momentum Theorem</u>

These two magnitudes are related in the following way. Suppose an object is moving at a certain speed v_1 and changes it to v_2. The impulse is numerically equivalent to the change of linear momentum. Let's recall the momentum is given by

p=mv

The initial and final momentums are, respectively

p_1=mv_1,\ p_2=mv_2

The change of momentum is

\Delta p=p_2-p_1=m(v_2-v_1)

It is numerically equal to the Impulse J

J=\Delta p

J=m(v_2-v_1)

We are given

m=2498\ kg,\ v_1=17.1\ m/s,\ v_2=2.6\ m/s

The impulse the car experiences during that time is

J=2498(2.6-17.1)=2498(-14.5)

J=-36221 Kg.m/s

The magnitude of J is

J=36221 Kg.m/s

8 0
3 years ago
A scientist kept 1,000 grams of a radioactive material in a container. After 66 days, he observed that 125 grams of the radioact
bixtya [17]
The sample appears to have gone through 3 half-lives
1st half life: 1000 to 500 g
2nd half life: 500 to 250 g
3rd half life: 250 to 125 g
The duration of a half-life, therefore, can be inferred to be 66 ÷ (3) = 22 days. 

After a 4th half life, there will be 125÷2= 62.5 g. 
At this point, an additional 22 days will have passed, for a total of 88 days.
Answer is C. 

8 0
3 years ago
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