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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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The air in a room has a pressure of 1 atm, a dry-bulb temperature of 24°C, and a wet-bulb temperature of 17°C. Using the psychro

Licemer1 [7]

Answer:

Given that

Dry-bulb temperature(T) =24°C

Wet-bulb temperature(Tw) = 17°C

Pressure ,P = 1 atm

As we know that psychrometric chart are drawn at constant pressure.

From the diagram

ω= specific humidity

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From chart at point P

Specific humidity,ω = 0.00922 kg/kg

The enthalpy ( h)

h=47.59 KJ/kg

The relative humidity, RH

RH= 49.58 %

Specific volume ,

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3 years ago

What likely happens to the hart rate as the cardiac mucle gets stronger

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3 years ago

Which of the following factors affects the pressure of an enclosed gas?

melamori03 [73]

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3 years ago

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Determine the velocity that a car should have while traveling around a frictionless curve of radius 100m and that is banked 20 d

alex41 [277]

Answer:

$v=18.89\frac{m}{s}$

Explanation:

From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:

$\sum F_x:Nsin(20^\circ)=ma_c(1)\\\sum F_y:Ncos(20^\circ)=mg(2)$

Solving N from (2) and replacing in (1):

$N=\frac{mg}{cos(20^\circ)}\\(\frac{mg}{cos(20^\circ)})sin(20^\circ)=ma_c\\gtan(20^\circ)=a_c$

The centripetal acceleration is given by:

$a_c=\frac{v^2}{r}$

Replacing and solving for v:

$gtan(20^\circ)=\frac{v^2}{r}\\v=\sqrt{grtan(20^\circ)}\\v=\sqrt{9.8\frac{m}{s^2}(100m)tan(20^\circ)}\\v=18.89\frac{m}{s}$

4 0

3 years ago

Two people are standing on a 1.75-m-long platform, one at each end. The platform floats parallel to the ground on a cushion of a

Fiesta28 [93]

Answer:

0.05312 m

Explanation:

Given:

Length of platform L = 1.75 m

mass of ball m_b = 5.76 kg

mass of (people + platform) m_p = 184 kg

Initial Velocity of ball V_i,b = 0

Initial Velocity of ball (people + platform) V_i,p = 0

Find:

How far does the platform recoils to rest

Solution:

Using the conservation of momentum on ust before and after the ball was thrown P_i = P_f :

Where, P_i = 0 (initially at rest)

P_f = m_p*V_f,p + m_b * V_f,b

0 = m_p*V_f,p + m_b * V_f,b

V_f,p = - (m_b /m_p) * V_f,b

V_f,p = - (5.76 / 184)*V_f,b

V_f,p = - 0.0313*V_f,b ....1

The time the ball is in air:

t = L / (V_f,b - V_f,p) ...2

The distance that the platform moves d:

d = V_f,p *t ....3

Substitute 2 into 3

d = V_f,p*L /(V_f,b - V_f,p) .... 4

Solve 1 and 4 simultaneously :

d = - m_b*L / (m_b + m_p)

d = - 5.76*1.75 / (5.76 + 184)

d = -0.05312 m

The platform moves 0.05312 m to the opposite to which the ball is thrown.

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3 years ago