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3 years ago

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A spider spins a web with silk threads of density 1300 kg/m3 and diameter 3.0 μm . a typical tension in the radial threads of su

ch a web is 7.0 mn. suppose a fly hits this web. part a which will reach the spider first: the very slight sound of the impact or the disturbance traveling along the radial thread of the web?

1 answer:

Tema [17]3 years ago

5 0

Answer:

Explanation:

The velocity of a wave in a string is equal to:

v = √(T / (m/L))

where T is the tension and m/L is the mass per length.

To find the mass per length, we need to find the cross-sectional area of the thread.

A = πr² = π/4 d²

A = π (3.0×10⁻⁶ m)²

A = 2.83×10⁻¹¹ m²

So the mass per length is:

m/L = ρA

m/L = (1300 kg/m³) (2.83×10⁻¹¹ m²)

m/L = 3.68×10⁻⁸ kg/m

So the wave velocity is:

v = √(T / (m/L))

v = √(7.0×10⁻³ N / (3.68×10⁻⁸ kg/m))

v ≈ 440 m/s

The speed of sound in air at sea level is around 340 m/s. So the spider will feel the vibration in the thread before it hears the sound.

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Plz solve this. plz plz plz plz simple machine

nalin [4]

Answer:

Explanation:

i. CW moment = 10 N (10 cm) + 30 N (30 cm) - 60 N (40 cm) = - 1400 N-cm

ii. ACW momenet = 60 N (40 cm) - 10 N (10 cm) + 30 N (30 cm) = 1400 N-cm

iii. No. The lever is not balanced in the situation. Because the moment is ± 1400 N-cm. if balance, the moment must be Zero.

iv. the location of 10N by keeping the other loads unchanged to balance the lever is 150 cm

take moment from Δ (support)

60(40) = 10(x) + 30(30)

2400 = 10x + 900

10x = 2400 - 900

10x = 1500

x = 1500/10

x = 150 cm

therefore, the location of 10N by keeping the other loads unchanged to balance the lever is 150 cm

5 0

3 years ago

A 20-kg child is coasting at 3.3 m/s over flat ground in a 4.0-kg wagon. The child drops a 1.0-kg ball out the back of the wagon

Shalnov [3]

Answer:

The final velocity is $u_f = 3.44 \ m/s$

Explanation:

From the question we are told that

The mass of the child is $m_1 = 20 \ kg$

The initial speed of the child is $u_1 = 3.3 \ m/s$

The mass of the wagon is $m_w = 4.0 \ kg$

The initial speed of the wagon is $u_w = 3.3 \ m/s$

The mass of the ball is $m_2 = 1.0 \ kg$

The initial speed off the ball is $u_2 = 3.3 \ m/s$

Generally the initial speed of the system (i.e the child , wagon , ball) is

$u_1 = u_w = u_2 = u =3.3 \ m/s$

Generally from the law of linear momentum conservation

$p_i = p_f$

Here $p_i$ is the momentum of the system before the ball is dropped which is mathematically represented as

$p_i = ( m_1 + m_2 + m_3 ) * u$

=> $p_i = ( 20 + 4 + 1 ) * 3.3$

=> $p_i = 82.5 \ kg \cdot m/s$

and

$p_f$ is the momentum of the system after the ball is dropped which is mathematically represented as

$p_f = ( m_1 + m_w ) * u_f$

=> $p_i = ( 20 + 4 ) * u_f$

So

$82.5 = 24 * u_f$

=> $u_f = 3.44 \ m/s$

4 0

3 years ago

A tennis player tosses a tennis ball straight up and then catches it after 2.00 s at the same height as the point of release. (a

Readme [11.4K]

Answer:

Part a)

a = -9.81 m/s/s

Part b)

v = 0

Part c)

v = 9.81 m/s

Part d)

$H = 4.905 m$

Explanation:

Part a)

During the motion of ball it will have only gravitational force on the ball

so here the acceleration of the ball is only due to gravity

so it is given as

$a = g = 9.81 m/s^2$

Part b)

As we know that ball is moving against the gravity

so here the velocity of ball will keep on decreasing as the ball moves upwards

so at the highest point of the motion of the ball the speed of ball reduce to zero

$v_f = 0$

Part c)

We know that the total time taken by the ball to come back to the initial position is T = 2 s

so in this time displacement of the ball will be zero

$\Delta y = 0 = v_y t + \frac{1}{2} at^2$

$0 = v_y (2) - \frac{1}{2}(9.81)(2^2)$

$v_y = 9.81 m/s$

Part d)

at the maximum height position we know that the final speed will be zero

so we will have

$v_f^2 - v_i^2 = 2 a d$

here we have

$0 - (9.81^2) = 2(-9.81)H$

$H = 4.905 m$

4 0

3 years ago

The largest and the smallest balls used in the experiment are with diameter 9.52 mm, and 2.38 mm respectively. For a glycerin wi

svlad2 [7]

Answer

given,

largest diameter of balls = 9.52 mm = 0.00476 m

radius = 0.00476

smallest diameter of ball = 2.38 mm = 0.00238 m

radius = 0.00119

viscosity = 1.5 Pa.s

density of the ball = 1.42 g/cm

$F = 6\pi \eta r V_t$

$F = \dfrac{mv}{t}$

$F = \dfrac{\dfrac{4}{3}\pi\ r^3\times (\rho-\sigma) \times 0.99 V_T}{t}$

$6\pi \eta r V_t = \dfrac{\dfrac{4}{3}\pi\ r^3\times rho \times 0.99 V_T}{t}$

$t = \dfrac{\dfrac{4}{3}\pi\ r^3\times rho \times 0.99 V_T}{6\pi \eta r V_t}$

$t= \dfrac{0.22 r^2 (\rho-\sigma) }{\eta}$

for small balls

$t= \dfrac{0.22\times 0.00119^2 (1460-1300)}{1.5}$

t = 0.033 ms

for larger ball

$t= \dfrac{0.22\times 0.00476^2 (1460-1300)}{1.5}$

t = 0.531 ms

6 0

4 years ago

The pressure exerted by a phonograph needle on a record is surprisingly large, due to the very small width of the needle. If the

Alex

Answer:

102597.6 Pa

Explanation:

mass, m = 1.25 g

Force, F = m x g = 1.25 x 9.8 x 10^-3 = 0.01225 N

radius, r = 0.195 mm = 0.195 x 10^-3 m

Area, A = πr² = 3.14 x 0.195 x 0.195 x 10^-6 m^2

A = 1.19 x 10^-7 m^2

Pressure is defined as the thrust acting per unit area.

P = Force / Area

P = 0.01225 N / (1.10 x 10^-7)

P = 102597.6 Pa

Thus, the pressure exerted is 102597.6 Pa.

6 0

3 years ago