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Delicious77 [7]
3 years ago
13

What should you do when a crate is too heavy to be lifted by the pulley?

Physics
1 answer:
kenny6666 [7]3 years ago
7 0
Use a wedge, I would say. It would depend on the answers.
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Suppose an event is measured to be at a = (0,-2, 3, 5) in one reference frame. Find the components of this event in another refe
LenaWriter [7]

Answer:

The components of the moving frame is (8.07c, -2, 3, 9.493)

Solution:

As per the question:

Velocity of moving frame w.r.t original frame v_{m} 0.85c

Point 'a' of an event in one reference frame corresponds to the (x, y, z, t) coordinates of the plane

a = (0, - 2, 3, 5)

Now, according the the question, the coordinates of moving frame, say (X, Y, Z, t'):

New coordinates are given by:

X = \frac{x - v_{m}t}{\sqrt{1 - \frac{v_{m}^{2}}{c^{2}}}}

X = \frac{0 - 0.85c\times 5}{\sqrt{1 - \frac{(0.85c)^{2}}{c^{2}}}}

X = 8.07 c

Now,

Y = y = - 2

Z = z = 3

Now,

t' = \frac{t - \frac{vx}{c}^{2}}{\sqrt{1 - (\frac{v}{c})^{2}}}

t' = \frac{5 - 0}{\sqrt{1 - (\frac{0.85c}{c})^{2}}} = 9.493 s

4 0
3 years ago
Elements with positive valences usually ______ electrons
Leno4ka [110]
The answer is donate, therefore elements with positive valences usually donate electrons
7 0
2 years ago
What is the mass of an object if a force of 2N and gives it an acceleration of 2 m/s2​
tino4ka555 [31]

Answer:

1kg

Explanation:

Use the formula

F=ma

m=F/a

m=2N/2m/s^2

m=1kg

6 0
2 years ago
Write two main group of animals​
Nesterboy [21]

Answer:

Animals are classifield as two main groups in the animal kingdom : Vertebrates and invertebrates .

Explanation:

keep it up....

7 0
3 years ago
The block in the figure below has a mass of 5.1 kg and it rests on an incline of angle . You pull on the rope with a force F = 3
viktelen [127]

42.9°

Explanation:

Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:

x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;

\Rightarrow mg\sin{\theta} = F

Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at \theta. Solving for the angle, we get

\sin{\theta} = \dfrac{F}{mg}

or

\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)

\;\;\;=  \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]

\;\;\;=42.9°

6 0
2 years ago
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