All categories

3 years ago

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

2 answers:

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

ioda3 years ago

5 0

I took the test and got 25 N correct

You might be interested in

Two speakers separated by a distance of 4.40 m emit sound. The speakers have opposite phase. A person listens from a location 3.

Hoochie [10]

Answer:

f = 147.21 Hz

Explanation:

In order to have a destructive interference, as the source emit in opposite phases, the path difference between the distance to the person, measured in a straight line from the speakers, must be equal to an integer number of wavelengths.

We need to know the distance from the listener to the other speaker, located 4.4 m from the one which is directly in front of him, which we can find using Pythagorean theorem, as follows:

l₂ = √(3)²+(4.4)² = 5.33 m

The difference in path will be, then:

d = l₂-l₁ = 5.33 m - 3.00 m = 2.33 m

For the lowest frequency that gives destructive interference, the wavelength will be highest possible, which happens when the distance is just one wavelength.

⇒ d = λ = 2.33 m

In any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

v = λ*f Κ ⇒ f = v/λ

Taking the speed of sound as 343 m/s, and solving for f, we get:

f= 343 m/s / 2.33 m = 147.21 Hz

3 0

3 years ago

What fills the void between stars and galaxies.

Lana71 [14]

Answer: The voids between stars in our galaxy can be filled with tenuous clouds of gas and other molecules. ... That material gets "ripped away" from the galaxies by the force of gravity, and often enough it collides with other material.

HOPE IT HELPED:) HAVE A NICE DAY

6 0

3 years ago

A form of erosion in which particles of sand or dust rub across the surface of rocks.

gayaneshka [121]

The answer is B) Abrasion.

7 0

3 years ago

Read 2 more answers

A student wants to determine the impulse delivered to the lab cart when it runs into the wall. The student measures the mass of

forsale [732]

Impulse = Force * times and also Impulse = change in momentum.

Given that the mass does not change, change if momentum = mass * (final velocity - initial velocity)

Given that you know mass and initial velocity (which is the velicity before the cart hits the wall) you need the final velocity (which is the velocity after the cart hits the wall).

Answer: the velocity of the cart after it hits the wall.

6 0

3 years ago

(15pts) A hungry 12.0 kg fish is coasting from west to east at 75 cm/s when it suddenly swallows a 1 kg fish swimming towards it

faust18 [17]

Answer:

The speed of the big fish after swallowing the small fish is 0.38 m/s.

Explanation:

Consider west to east direction as positive and the opposite direction as negative.

Given:

Mass of big fish (m₁) = 12.0 kg

Initial velocity of big fish (u₁) = 75 cm/s = 0.75 m/s

Mass of small fish (m₂) = 1 kg

Initial velocity of small fish (u₂) = -4 m/s (Direction is opposite to u₁)

After swallowing the small fish, both the fishes move together with same velocity. Let the velocity be 'v'.

So, as there are no effects of drag or any other forces, the given scenario can be considered as a case of inelastic collision where the objects move together with same velocity after collision.

The momentum is conserved in inelastic collision. Therefore,

Initial momentum of the fishes = Final momentum of the fishes

$m_1u_1+m_2u_2=(m_1+m_2)v\\\\v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}$

Now, plug in the given values and solve for 'v'. This gives,

$v=\frac{12.0\times 0.75+1\times (-4)}{12.0+1}\\\\v=\frac{9-4}{13}\\\\v=\frac{5}{13}=0.38\ m/s$

Therefore, the speed of the big fish after swallowing the small fish is 0.38 m/s

3 0

3 years ago