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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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FrozenT [24]

Answer:

1.25 m/s

Explanation:

Given,

Mass of first ball=0.3 kg

Its speed before collision=2.5 m/s

Its speed after collision=2 m/s

Mass of second ball=0.6 kg

Momentum of 1st ball=mass of the ball*velocity

=0.3kg*2.5m/s

=0.75 kg m/s

Momentum of 2nd ball=mass of the ball*velocity

=0.6 kg*velocity of 2nd ball

Since the first ball undergoes head on collision with the second ball,

momentum of first ball=momentum of second ball

0.75 kg m/s=0.6 kg*velocity of 2nd ball

Velocity of 2nd ball=0.75 kg m/s ÷ 0.6 kg

=1.25 m/s

4 0
2 years ago
Q6) A speed skater moving to the left across frictionless ice at 8.0 m/s hits a 5.0-m-wide patch of rough ice. She slows steadil
77julia77 [94]

Answer:

-2.8 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity The S. I unit of acceleration is m/s²

Using the equation of motion,

v² = u² + 2as................... Equation 1

Where v = Final velocity, u = initial velocity, a = acceleration, s = distance,

Given: v = 6.0 m/s, u = 8.0 m/s, s = 5.0 m.

Substituting into equation 1

6² = 8²+2(a)5

36 = 64 + 10a

10a = 36-64

10a = -28

10a/10 = -28/10

a = -2.8 m/s²

Note: a is negative because because the skater decelerate on the rough ice

Hence the magnitude of her acceleration is  = -2.8 m/s²

6 0
3 years ago
NEED ANSWER ASAP WILL MARK BRAINLIEST
lukranit [14]
Speed=30 m/s - 1.5 m/s = 28.5 m/s forward
7 0
3 years ago
A record turntable rotates through 5.0 rad in 2.8 s as it is accelerated uniformly from rest. What is the angular velocity at th
Usimov [2.4K]

Answer:

\omega_f = 3.584\ rad/s

Explanation:

given,

turntable rotate to, θ = 5 rad

time, t = 2.8 s

initial angular speed  = 0 rad/s

final angular speed = ?

now, using equation of rotational motion

\theta = \omega_i t + \dfrac{1}{2}\alpha t^2

5 = 0+ \dfrac{1}{2}\alpha\times 2.8^2

\alpha= \dfrac{10}{2.8^2}

       α = 1.28 rad/s²

now, calculation of angular velocity

\omega_f = \omega_i + \alpha t

\omega_f =0 +1.28\times 2.8

\omega_f = 3.584\ rad/s

hence, the angular velocity at the end is equal to 3.584 rad/s

4 0
3 years ago
An object 16.8 cm tall is placed in front of a converging lens. A real image, 46 cm tall, is formed on the other side of the len
worty [1.4K]

Answer:

2.74

Explanation:

Magnification = image distance/object distance

Mag = v/u

Given

v = 46cm

u = 16.8

Magnification = 46/16.8

Magnification = 2.74

Hence the magnification is 2.74

7 0
3 years ago
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