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Deffense [45]
3 years ago
8

A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. the hoist weighs 430 n. the ropes,

fastened at different heights, make angles of 50° and 38° with the horizontal. find the tension in each rope and the magnitude of each tension. (let t2 and t3, represent the tension vectors corresponding to the ropes of length 2 m and 3 m respectively. round all numerical values to two decimal places.)

Physics
1 answer:
ICE Princess25 [194]3 years ago
7 0
Refer to the diagram shown below.

For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂                (1)

For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392        (2)

Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N

Answer:
T₂ = 339.06 N
T₃ = 276.57 N

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EastWind [94]

Answer:

50N

Explanation:

Force (N) = mass (kg) × acceleration (m/s²)

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7 0
3 years ago
A baseball pitcher throws the ball towards the batter at 90 mph. His bat connects with the ball for a line drive, after which th
forsale [732]

Answer:

F=-18412.9N, where the minus indicates the direction is opposite to that of the throw.

Explanation:

a)

Since MKS stands for meter-kilogram-second and we know that:

1\ hour = 3600\ seconds

1\ mile = 1600\ meters

1000g = 1kg

We can write that:

\frac{1\ hour}{3600\ seconds}=1

\frac{1600\ meters}{1\ mile}=1

\frac{1kg}{1000g}=1

These are conversion factors, equal to 1, so multiplying our results by them won't change their value, only their units.

So we have that:

90 mph=90 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=40m/s

110 mph=110 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=48.89m/s

145 g=145 g(\frac{1kg}{1000g})=0.145kg

b)

Newton's 2nd Law tells us that F=ma, and the definition of acceleration is a=\frac{\Delta v}{\Delta t}, so we have:

F=m\frac{\Delta v}{\Delta t}=m\frac{v_f-v_i}{t}

Taking the throw direction as the positive one, for our values we have:

F=m\frac{v_f-v_i}{t}=(0.145kg)\frac{(-48.89m/s)-(+40m/s)}{0.0007s}=-18412.9N

4 0
3 years ago
Pls send the answer..pls...​
Sophie [7]

Answer:

The fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C

Explanation:

Given;

initial temperature of the liquid, t₁ = 76.3  +/-  0.4⁰C

final temperature of the liquid, t₂ = 67.7  +/-  0.3⁰C

The change in temperature of the liquid is calculated as;

Δt = t₂  -  t₁

Δt = (67.7 - 76.3)  +/-  (0.3 - 0.4)

Δt = (-8.6)  +/-  (-0.1)

Δt = 8.6 +/- 0.1 ⁰C

Therefore, the fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C

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drek231 [11]
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