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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

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The longer the lever, the greater the

Simora [160]

Answer: The longer the lever, the greater the force on the load will be.

Explanation:

4 0

3 years ago

Rank the pressures from highest to lowest:

n200080 [17]

Answer:

P₃ > P₁ > P₂

Explanation:

To rank pressure of the given situation

a) we know

Pressure at height h below

P = ρ g h

density of salt water, ρ = 1029 kg/m³

P₁ = 1029 x 10 x 0.2

P₁ = 2058 Pa

b) density of fresh water, ρ = 1000 kg/m³

P₂ = 1000 x 10 x 0.2

P₂ = 2000 Pa

c) density of mercury, ρ = 13593 kg/m³

P₃ = 13593 x 10 x 0.05

P₃ = 6796.5 Pa

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P₃ > P₁ > P₂

3 0

3 years ago

The acceleration of a baseball pitcher's hand as he delivers a pitch is extreme. For a professional player, this acceleration ph

SashulF [63]

Answer:

116 N.

Explanation:

Given,

mass of the ball, m = 0.145 Kg

initial speed = 0 m/s

Final speed = 40 m/s

time = 50 ms = 0.05 s

Force is equal to change in momentum per unit time

$F = \dfrac{m\Delta v}{\Delta t}$

$F = \dfrac{0.145(40-0)}{0.05}$

F = 116 N

Force of the pitcher hand on the ball is equal to 116 N.

4 0

3 years ago

Dmitrij [34]

Take 10m/s^2 for the gravitational acceleration, as we know this is a free fall, we can use the equation: d=1/2*g*t^2

Substitute g=10m/s^2, t=5s, d=125m

7 0

3 years ago

Suppose a child drives a bumper car head on into the side rail, which exerts a force of 3900 N on the car for 0.55 s. Use the in

-Dominant- [34]

Answer:

The impulse is 2145 kg-m/s

The final velocity is -8.34 m/s or 8.34 m/s in he opposite direction.

Explanation:

Force on the rail = 3900 N

Elapsed time of impact = 0.55 s

Impulse is the product of force and the time elapsed on impact

I = Ft

I is the impulse

F is force

t is time

For this case,

Impulse = 3900 x 0.55 = 2145 kg-m/s

If the initial velocity was 2.95 m/s

and mass of car plus driver is 190 kg

neglecting friction, the initial momentum of the car is given as

P = mv1

where P is the momentum

m is the mass of the car and driver

v1 is the initial velocity of the car

initial momentum of the car P = 2.95 x 190 = 560.5 kg-m/s

We know that impulse is equal to the change of momentum, and

change of momentum is initial momentum minus final momentum.

The final momentum = mv2

where v2 is the final momentum of the car.

The problem translates into the equation below

I = mv1 - mv2

imputing values, we have

2145 = 560.5 - 190v2

solving, we have

2145 - 560.5 = -190v2

1584.5 = -190v2

v2 = -1584.5/190 = -8.34 m/s

6 0

4 years ago