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mojhsa [17]
3 years ago
13

Need help with this THANKS TO WHO HELPS <3

Physics
2 answers:
nekit [7.7K]3 years ago
8 0

Answer:

because it's north

Explanation:

u if yyggggggggggggggg hgfth

kirza4 [7]3 years ago
6 0

The North Star is to the north, if you keep going north the will eventually find help.

Good Luck!

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A 450.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tensi
bogdanovich [222]

Answer:

1) W_{object} = 400 N

2) x = 0.28 m from cable A.

Explanation:

1 ) Let's use the first Newton to find the , because bar is in equilibrium.

\sum F_{Tot} = 0

In this case we just have y-direction forces.

\sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0

Now, let's solve the equation for W(object).

W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N

2 ) To find the position of the heaviest weight we need to use the torque definition.

\sum \tau = 0

The total torque is evaluated in the axes of the object.

Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.

First let's find the positions from each force to the P point.

L = 1.50 m  ; total length of the bar.

D_{AP} = x  ; distance between Tension A and P point.

D_{BP} = L-x ; distance between Tension B and P point.

D_{W_{bar}P} = \frac{L}{2}-x ; distance between weight of the bar (middle of the bar) and P point.

Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.

\sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0

Finally we just need to solve it for x.

x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}}

x = 0.28 m

So the distance is x = 0.28 m from cable A.

Hope it helps!

Have a nice day! :)

8 0
4 years ago
What do lenses do?
Dovator [93]
A convex lens makes light rays converge (come together) at the focal point or focus. The distance from the center of the lens to the focal point is the focal length of the lens. Convex lenses are used in things like telescopes and binoculars to bring distant light rays to a focus in your eyes.
3 0
4 years ago
Read 2 more answers
Mary and Jane are standing at each end of an ice board 1.0 m in length. 2 points
madam [21]

Answer:

0.053 m ang answer my phsiycal

3 0
3 years ago
Three identical very dense masses of 7500 kg each are placed on the x axis. One mass is at x1 = -100 cm , one is at the origin,
sukhopar [10]

Answer:

0.00354 (N)

Explanation:

Convert to metric system:

x_1 = -100 cm = 1 m

x_2 = 420 cm = 4.2 m

Formula for gravitational force:

F_g = G\frac{mM}{s^2}

where s is the distance between 2 bodies masses m and M

Substitute the number to the formula above and since the 2 forces are acting in opposite direction, the total net gravitational force on the mass of origin be:

F_g = F_{g1} - F_{g2}

F_g = G\frac{m_1M}{x_1^2} - G\frac{m_2M}{x_2^2}

F_g = GM(\frac{m_1}{x_1^2} - \frac{m_2}{x_2^2})

F_g = 6.67*10^{-11} * 7500 (\frac{7500}{1^2} - \frac{7500}{4.2^2})

F_g = 5*10^{-7}(7500 - 425.17)

F_g = 5*10^{-7} * 7074.83

F_g = 0.00354 (N)

5 0
3 years ago
A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra
skelet666 [1.2K]

Answer:

So the acceleration of the child will be 8.05m/sec^2

Explanation:

We have given angular speed of the child \omega =1.25rad/sec

Radius r = 4.65 m

Angular acceleration \alpha =0.745rad/sec^2

We know that linear velocity is given by v=\omega r=1.25\times 4.65=5.815m/sec

We know that radial acceleration is given by a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2

Tangential acceleration is given by

a_t=\alpha r=0.745\times 4.65=3.464m/sec^

So total acceleration will be a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2

7 0
3 years ago
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