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3 years ago

An object traveling at 343 m/s, is also traveling at the _____

1 answer:

4 0

Traveling at 343 m/s is emitted by the foghorn of a tugboat. This would be the speed of sound.

Hope you have a nice day! :D

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Volcanic ash and sulfur dioxide spewed out of Mt. Pinatubo in 1991. These materials can reflect incoming solar radiation. Over t

gogolik [260]

The suspended ash made for some some spectacular sunsets! Sulfuric acid was spread worldwide, increasing acidity of rain. Ash deflected energy from the sun, causing a slight drop on global temps for a few years.

7 0

2 years ago

Read 2 more answers

A horizontal spring with stiffness 0.4 N/m has a relaxed length of 11 cm (0.11 m). A mass of 21 grams (0.021 kg) is attached and

riadik2000 [5.3K]

Answer:

0.6983 m/s

Explanation:

k = spring constant of the spring = 0.4 N/m

L₀ = Initial length = 11 cm = 0.11 m

L = Final length = 27 cm = 0.27 m

x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m

m = mass of the mass attached = 0.021 kg

v = speed of the mass

Using conservation of energy

Kinetic energy of mass = Spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(0.021) v² = (0.4) (0.16)²

v = 0.6983 m/s

5 0

3 years ago

a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien

Law Incorporation [45]

Answer:

The magnitude of the friction force is 8197.60 N

Explanation:

Using the definition of the centripetal force we have:

$\Sigma F=ma_{c}=m\frac{v^{2}}{R}$

Where:

m is the mass of the car
v is the speed
R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

$F_{f}=m\frac{v^{2}}{R}$

$F_{f}=2100\frac{18^{2}}{83}$

$F_{f}=8197.60 \: N$

Therefore the magnitude of the friction force is 8197.60 N

I hope it helps you!

7 0

3 years ago

4.1 A steel spur pinion has a pitch of 5 teeth/in, 20 full-depth teeth, and a 20° pressure angle. The pinion runs at a speed of

Vesnalui [34]

Answer:

15.07 ksi

Explanation:

Given that:

Pitch (P) = 5 teeth/in

Pressure angle ( $\Phi$ ) = 20°

Pinion speed ( $n_p$ ) = 2000 rev/min

Power (H) = 30 hp

Teeth on gear ( $N_G$ ) = 50

Teeth on pinion ( $N_p$ ) = 20

Face width (F) = 1 in

Let us first determine the diameter (d) of the pinion.

Diameter (d) = $\frac{N}{P}$

= $\frac{20}{5}$

= 4 in

From the values of Lewis Form Factor Y for ( $n_p$ ) = 20 ; at 20°

Y = 0.321

To find the velocity (V); we use the formula:

$V = \frac{\pi d n_p}{12}$

$V = \frac{\pi *4*2000}{12}$

V = 2094.40 ft/min

For cut or milled profile; the velocity factor $(K_v)$ can be determined as follows:

$K_v = \frac{2000+V}{2000}$

$K_v = \frac{2000+2094.40}{2000}$

= 2.0472

However, there is need to get the value of the tangential load $(W^t)$ , in order to achieve that, we have the following expression

$W^t=\frac{T}{\frac{d}{2} }$

$W^t = \frac{63025*H}{\frac{n_pd}{2}}$

$W^t = \frac{63025*30}{2000*\frac{4}{2}}$

$W^t = 472.69 lbf$

Finally, the bending stress is calculated via the formula:

$\sigma = \frac{K_vW^tp}{FY}$

$\sigma = \frac{2.0472*472.69*5}{1*0.321}$

$\sigma = 15073.07 psi$

$\sigma =$ 15.07 ksi

∴ The estimate of the bending stress = 15.07 ksi

4 0

3 years ago

Read 2 more answers

Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid ra

BabaBlast [244]

Explanation:

It is given that,

Length of the helicopter, l = 3.1 m

The helicopter rotates, the length of helicopter will become the radius of circular path, r = 3.1 m

Angular speed of the helicopter, $\omega=280\ rev/min=29.32\ rad/s$

(a) The centripetal acceleration in terms of angular velocity is given by :

$a_c=r\times \omega^2$

$a_c=3.1\times (29.32)^2$

$a_c=2664.95\ m/s^2$

(b) Let v is the linear speed of the tip. The relation between the linear and angular speed is given by :

$v=r\times \omega$

$v=3.1\times 29.32$

v = 90.89 m/s

$\dfrac{v_{tip}}{v_{sound}}=\dfrac{90.89}{340}=0.267$

Hence, this is the required solution.

4 0

3 years ago