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Softa [21]
3 years ago
12

Two blocks are arranged at the ends of a mass-less string as shown in the figure. The system starts from rest. When the 3.26 kg

mass has fallen through 0.386 m, its downward speed is 1.28 m/s. What is the frictional force between the 4.52 kg mass and the table? (round your answer to the 4th decimal place)
Physics
2 answers:
Wewaii [24]3 years ago
8 0

Answer:  002 10.0points Two blocks are arranged at the ends of a massless string as shown in . 96 kg mass has fallen through 0 . 415 m, its downward speed is 1 . 25 m / s.

Explanation: I hoped that helped!!

True [87]3 years ago
6 0

Answer:

2.5 kg

Explanation:

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Alex17521 [72]

may i ask your answer soloutions... and i believe that the correct answer is mass


6 0
3 years ago
What is the tool used for classifying organisms called? A. Dichotomy B. Dichotomous key C.Dichromic key D. Periodic table
ivolga24 [154]
It is B. Dichotomous key.
4 0
3 years ago
a concrete slab 20 m long and weighing 400,000 N is supported by one pillar. A 19,600 N car is parked 8 meters from one end, whe
Elden [556K]

let the distance of pillar is "r" from one end of the slab

So here net torque must be balance with respect to pillar to be in balanced state

So here we will have

Mg(r - L/2) = mg(L/2 - 8)

here we know that

mg = 19600 N

Mg = 400,000 N

L = 20 m

from above equation we have

400,000(r - 10) = 19,600 (10 - 8)

r - 10 = 0.098

r = 10.098 m

so pillar is at distance 10.098 m from one end of the slab

3 0
3 years ago
A toy rocket is launched with an initial velocity of 10.0 m/s in the horizontal direction from the roof of a 38.0-m-tall buildin
Bas_tet [7]

Answer:

Explanation:

Consider downward displacement first .

downward displacement h = 38 m .

downward acceleration = g

downward initial velocity u = 0

h = ut + 1/2 g t ²

38 = 0 + 9.8 t ²

t = 1.97 s

During this period , there will be horizontal displacement with initial velocity u = 10 m /s and acceleration a = 1.6 m /s²

s = ut + 1/2 a t²

= 10 x 1.97 + .5 x 1.6 x 1.97²

= 19.7 + 3.10

= 22.8 m

3 0
3 years ago
Imagine you use Nitrogen as your gas. If you have the cold side as cold as you can without liquefying it (78 K), and run the hot
alina1380 [7]

Answer:

The efficiency of a Stirling engine is 74%

Explanation:

Given:

Temperature of gas when it is cold T_{1} = 78 K

Temperature of gas when it is hot T_{2} = 300 K

The efficiency of a stirling engine,

  \eta =1 - \frac{T_{1} }{T_{2} }

  \eta = 1- \frac{78}{300}

  \eta = 1-0.26

  \eta = 0.74

∴ \eta = 74 \%

Therefore, the efficiency of a Stirling engine is 74%

5 0
3 years ago
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