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3 years ago

Calculate the tension (in N) in a vertical strand of spiderweb if a spider of mass 5.00 ✕ 10-5 kg hangs motionless on it.

Physics

1 answer:

Mandarinka [93]3 years ago

3 0

Answer:

4.9 x 10⁻⁴N

Explanation:

Given parameters:

Mass of the vertical strand of spiderweb = 5 x 10⁻⁵kg

Unknown:

Tension in the vertical strand = ?

Solution:

The tension is the vertical strand will be the weight of the strand.

Weight of strand = mg

m is the mass

g is the acceleration due to gravity = 9.8m/s²

Tension in the vertical strand = 5 x 10⁻⁵kg x 9.8m/s²

Tension in the vertical strand = 4.9 x 10⁻⁴N

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A soccer ball takes 20 s to roll 10 m. What is the average speed of the soccer ball.

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Explanation:

velocity is distance divided by time.

the average speed of the ball is 10m/20s

= 0.5 m/s

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2 years ago

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The ratio between the force of sliding friction and the normal force of an object is called?

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2 years ago

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A pool ball moving 1.33 m/s strikes an identical ball at rest. Afterward, the first ball moves 0.750 m/s at a 33.30 angle. What

kramer

Explanation:

We need to apply the conservation law of linear momentum to two dimensions:

Let $p_{1}$ = momentum of the 1st ball

$p_{2}$ = momentum of the 2nd ball

In the x-axis, the conservation law can be written as

$(p_{1} \cos \theta_{1})_{i} + (p_{2} \cos \theta_{2})_{i} = (p_{1} \cos \theta_{1})_{f} + (p_{2} \cos \theta_{2})_{f}$

$(m_{1}v_{1})_{i}= (m_{1}v_{1}\cos \theta_{1})_{f} + (m_{2}v_{2}\cos \theta_{2})_{f}$

Since we are dealing with identical balls, all the m terms cancel out so we are left with

$(v_{1})_{i} = (v_{1})_{f}\cos \theta_{1} + (v_{2})_{f}\cos \theta_{2}$

Putting in the numbers, we get

$1.33 = (0.750) \cos(33.30) + (v_{2})_{f} \cos \theta_{2}$

$= > (v_{2})_{f} \cos \theta_{2} = 0.703$

In the y-axis, there is no initial y-component of the momentum before the collision so we can write

$0 = (v_{1}\sin \theta_{1})_{f} + (v_{2}\sin \theta_{2})_{f}$

$= > (v_{2})_{f} \sin \theta_{2} = (0.750) \sin(33.30) = 0.412$

Taking the ratio of the sine equation to the cosine equation, we get

$\frac{ \sin \theta _{2}}{ \cos \theta_{2} } = \tan \theta_{2} = \frac{0.412}{0.703} = 0.586$

$\theta_{2} = { \tan}^{ - 1} (0.586) = 30.4$

Solving now for $(v_{2})_{f}$ ,

$(v_{2})_{f} = \frac{0.412}{ \sin(30.4) } = 0.815 \: \frac{m}{s}$

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3 years ago

A football is kicked from a tee at 12 m/s at 72° above the horizontal. What is the flight time of the football? __s

Ierofanga [76]

2.3 seconds

Ignoring air resistance, the flight time is merely a function of gravity and vertical velocity. The vertical velocity will be the initial velocity multiplied by the sine of the angle above the horizon. So:

V = sin(72)*12 m/s

V = 0.951056516 * 12 m/s

V = 11.4126782 m/s

Gravitational acceleration is 9.8 m/s, so divide the vertical velocity by gravitational acceleration to get how long it takes for the ball to reach its apex.

11.4126782 m/s / 9.8 m/s^2 = 1.164559 s

And the old saying "What goes up, must come down" really applies here. And conveniently, it's also symmetric, in that the time it takes to fall will match the time it takes to reach its apex. So multiply the time by 2.

1.164559 s * 2 = 2.329117999 s

Rounding the result to 2 significant figures gives 2.3 seconds.

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2 years ago