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brilliants [131]
3 years ago
10

Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin

wedge. One edge of the wedge is vertical, and the tip makes an angle of 33. The block that hangs vertically weighs 2.85 kg, and the block on the incline weighs 2.94 kg. If the two blocks do not move, what is magnitude of the force of friction on the second second block
Physics
1 answer:
likoan [24]3 years ago
6 0

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma    (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0     and T - mgcos57 - F = m(0) = 0

Mg - T = 0    (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F  

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

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Answer:

2m/s

Explanation:

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When do two different position-time graphs have<br> matching velocity-time graphs?
Ivan

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when they have the same slope

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A bus took 8 hours to travel 639 km. For the first 5 hours, it
vladimir1956 [14]

Answer:

93 km/h

Explanation:

Given that a bus took 8 hours to travel 639 km. For the first 5 hours, it travelled at an average speed of 72 km/h

Let the first 5 hours journey distance = F

From the formula of speed,

Speed = distance/time

Substitute speed and time

72 = F/5

F = 72 × 5 = 360 km

The remaining distance will be:

639 - 360 = 279km

The remaining time will be:

8 - 5 = 3 hours

Speed = 279/3

Speed = 93 km/h

Therefore, the average speed for the remaining time of the journey is equal to 93 km/h

8 0
3 years ago
A 0.15 g honeybee acquires a charge of 22 pC while flying. The electric field near the surface of the earth is typically 100 N/C
Rus_ich [418]

Answer:

1.50\ *10^{-6} }

Explanation:

Given

e=100 N/C

M=0.15 g

q=\ 22\  pC\\=\ 22\ *10^{-2}

The  ratio of the electric force on the bee to the bee's weight can be determined by the following formula

\frac{fe}{M*9.81}

\frac{22*10^{-12\ *\ 100} }{0.15*\ 10^{-3} *\ 9.81}

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4 0
3 years ago
The brain mass of a human fetus during a particular trimester can be accurately estimated from the circumference of the head by
Paladinen [302]

Answer:

a. If c = 20 cm, then the mass of the brain is m = 5 g.

b. At c = 20 cm, the brain's mass is increasing at a rate of 15.75 g/cm.

Explanation:

From the equation

m\left(c\right) = \frac{c^3}{100}-\frac{1500}{c}

we have

a. for c = 20 cm

m\left(20\right)=\frac{20^3}{100}-\frac{1500}{20}=5,

then the mass is m(20) = 5 g.

b. In order to find the rate of change, first we derivate

\frac{dm}{dc}=\frac{3c^2}{100}+\frac{1500}{c^2}.

Evaluated at c = 20 cm, we have

\frac{dm}{dc}|_{c=20}=\frac{3\times 20^2}{100}+\frac{1500}{20^2}=15.75.

So, at c = 20 cm, the mass of the brain is increasing at a rate of 15.75 g/cm.

3 0
3 years ago
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