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Alex17521 [72]
2 years ago
15

If four vertical cables are holding up a rectangular, 100- Newton sign, how much tension is on each cable?

Physics
2 answers:
trasher [3.6K]2 years ago
6 0
We may balance the forces in order to calculate the tension.
We know that there are no horizontal forces being applied. Next, the vertical forces are:
Weight (W) which is acting downwards
Tension in each wire (T) which is acting upwards
Because the sign is stationary, the downward force must equal the upward force. So we may write:
Weight = 4 * tension
W = 4T

100 = 4T
T = 25 N

The tension in each wire is 25 Newtons
ioda2 years ago
5 0

I took the test and got 25 N correct

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During a tornado in 2008 the Peachtree Plaza Westin Hotel in downtown Atlanta suffered damage. Suppose a piece of glass dropped
Darina [25.2K]

Answer:

Time = t = 6.62 s

Explanation:

Given data:

Height = h = 215 m

Initial velocity = v_{i} = 0 m/s

gravitational acceleration = g = 9.8 m/s²

Time = t = ?

According to second equation of motion

                            h = v_{i}t + \frac{1}{2}gt^{2}

As initial velocity is zero, So the first term of right hand side of above equation equal to zero.

                                 h = \frac{1}{2}gt^{2}

                                        t² = \frac{2h}{g}

                                         t =\sqrt{\frac{2h}{g} }

                                         t = \sqrt{\frac{(2)(215) }{9.8} }

                                         t = 6.62 s

3 0
3 years ago
(15 POINTS) Victor drew a diagram to show the life cycle of a low-mass star.
Evgen [1.6K]
The answer is c so the area can grow
4 0
3 years ago
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During a tennis serve, a racket is given an angular acceleration of magnitude 150 rad/s^2. At the top of the serve, the racket h
QveST [7]

Answer:

270 m/s²

Explanation:

Given:

α = 150 rad/s²

ω = 12.0 rad/s

r = 1.30 m

Find:

a

The acceleration will have two components: a radial component and a tangential component.

The tangential component is:

at = αr

at = (150 rad/s²)(1.30 m)

at = 195 m/s²

The radial component is:

ar = v² / r

ar = ω² r

ar = (12.0 rad/s)² (1.30 m)

ar = 187.2 m/s²

So the magnitude of the total acceleration is:

a² = at² + ar²

a² = (195 m/s²)² + (187.2 m/s²)²

a = 270 m/s²

3 0
3 years ago
A single loop of current is immersed in an externally applied uniform magnetic field of 3 Tesla oriented in the positive y direc
vlabodo [156]

Answer:

mu=12Tm^2

Explanation:

the magnetic moment mu of a single loop is given by:

\mu = I A B

where I is the current, B is the magnetic field and A is the area of the loop. By replacing we obtain:

\mu=(0.5A)(4m*2m)(3T)=12Tm^2

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6 0
3 years ago
When changing a tire sized 195/75 R15 to a 5 percent lower profile, the correct tire size would be _____
valentinak56 [21]

Answer and explanation:

When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.

The total car tire diameter can be calculated as:

D_{tot orig}=\frac{195 \cdot 75}{2540 \cdot 2}+15''=26.5''

The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:

pr=0.95·75=71.25

The problem is that the profiles are normalized and the nearest profile available is 70.

If we take a theorical tire with a profile of 71.25:

D_{tot orig}=D_{new}\\\frac{195 \cdot 75}{2540 \cdot 2}+15''=\frac{X \cdot 71.25}{2540 \cdot 2}+15''\\X=205.26

The theorical tire size should be 205/71 R15.

If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.

The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').

3 0
3 years ago
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