Relative speed depends on the individual speeds and

Two round rods, one steel andthe other copper, are joined end to end. Each rod is 0.750 mlong and 1.50 cmin diameter. The combin

nikitadnepr [17]

Answer: a) Strain on Steel rod = 0.0001078

b) elongation on the steel rod = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain on Copper Rod = 0.000189

Explanation: a) To obtain the strain of the steel rod, we invoke Hooke's law which states that, provided the elastic limit of A material isn't exceeded, the stress it undergoes is directly proportional to its strain.

(Stress, σ) ∝ (Strain, ε)

The constant of proportionality is called Young's modulus, E.

σ = Eε

For steel, Younger Modulus as obtained from literature = 210GPa.

Strain = Stress/Young's Modulus

Stress = (Force or Load applied)/ Cross sectional Area.

Force applied For the steel = 4000N

Cross sectional Area = (π(D^2))/4

D = 1.50cm = 0.015m

A = 0.0001767 m2

σ = 4000/0.0001767 = 22637238.257 N/m2

ε = σ/E = 22637238.257/(210 × (10^9)) = 0.0001078.

b) To solve for elongation.

Strain, ε = (elongation, dl)/(original length, lo)

Elongation, dl = strain × original length

dl = 0.0001078 × 0.75 = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain in Copper

ε = σ/E; σ = 22637238.257 N/m2

Young's modulus of Copper, from literature, = 120GPa

ε = 22637238.257/(120 × (10^9)) = 0.000189

4 0

3 years ago

Why are microwaves beneficial to use for transmitting information

Darya [45]

Microwaves. Microwave radiation can be used to transmit signals such as mobile phone calls. ... Certain microwave radiation wavelengths pass through the Earth's atmosphere and can be used to transmit information to and from satellites in orbit

5 0

4 years ago

A 11kg block slides up a 30° inclined plane at a constant velocity. The coefficient of friction between the block and the plane

astra-53 [7]

Answer:

The magnitude of the applied force is 94.74 N

Explanation:

Mass of the block, m = 11 kg

Angle of inclination of the plane, $\theta = 30^{\circ}$

Friction coefficient, $\mu_{k} = 0.2$

Now,

Normal force that acts on the block is given by:

$F_{N} = mgcos\theta + Fsin\theta$ (1)

Now, to maintain the equilibrium parallel to ramp the forces must be balanced.

Thus

$Fcos\theta = \mu_{k}F_{N}$ (2)

From eqn (1) and (2)

$Fcos\theta = \mu_{k}(mgcos\theta + Fsin\theta)$

$F(cos\theta - \mu_{k}sin\theta) = \mu_{k}mgcos\theta$

$F = \frac{\mu_{k}mgcos\theta}{cos\theta - \mu_{k}sin\theta}$

$F = \frac{0.2\times 11\times 9.8cos30^{\circ}}{cos30^{\circ} - 0.2\times sin30^{\circ}}$

F = 94.74 N

3 0

4 years ago

A ball is thrown vertically downward from the top of a 37.4-m-tall building. The ball passes the top of a window that is 15.4 m

Effectus [21]

Answer:

v= 20.8 m/s

Explanation:

Assuming no other forces acting on the ball, from the instant that is thrown vertically downward, it's only accelerated by gravity, in this same direction, with a constant value of -9.8 m/s2 (assuming the ground level as the zero reference level and the upward direction as positive).
In order to find the final speed 2.00 s after being thrown, we can apply the definition of acceleration, rearranging terms, as follows:

$v_{f} = v_{o} + a*t = v_{o} + g*t (1)$

We have the value of t, but since the ball was thrown, this means that it had an initial non-zero velocity v₀.
Due to we know the value of the vertical displacement also, we can use the following kinematic equation in order to find the initial velocity v₀:

$\Delta y = v_{o} *t + \frac{1}{2} * a* t^{2} (2)$