Wavelength multiplied by frequency equals

In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The fron

slega [8]

Answer:

a) v_average = 11 m / s, b) t = 0.0627 s

, c) F = 7.37 10⁵ N

, d) F / W = 35.8

Explanation:

a) truck speed can be found with kinematics

v² = v₀² - 2 a x

The fine speed zeroes them

a = v₀² / 2x

a = 22²/2 0.69

a = 350.72 m / s²

The average speed is

v_average = (v + v₀) / 2

v_average = (22 + 0) / 2

v_average = 11 m / s

b) The average time

v = v₀ - a t

t = v₀ / a

t = 22 / 350.72

t = 0.0627 s

c) The force can be found with Newton's second law

F = m a

F = 2100 350.72

F = 7.37 10⁵ N

.d) the ratio of this force to weight

F / W = 7.37 10⁵ / (2100 9.8)

F / W = 35.8

.e) Several approaches will be made:

- the resistance of air and tires is neglected

- It is despised that the force is not constant in time

- Depreciation of materials deformation during the crash

5 0

3 years ago

What's the difference between Uniform magnetic field and Non-uniform magnetic field?

Galina-37 [17]

Magnetic field is the magnetic effect of electric currents and magnetic materials. The magnetic field at any given point specified by both direction and a magnitude. So uniform magnetic field has equal amount of force or magnetic force in both side of the object while the non uniform magnetic field has one of the object exert more than the other

4 0

3 years ago

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

Since $\displaystyle \omega = \frac{2\pi}{T}$ , it can be concluded that:

$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

For the first mass $m_1$ , if the time period is $T_1$ , then the spring constant would be:

$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

Similarly, for the second mass $m_2$ , if the time period is $T_2$ , then the spring constant would be:

$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

6 0

3 years ago

A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.

SVEN [57.7K]

Answer:

a) Knowns, initial speed $v_{i}=13.0 m/s$ , final speed $v_{f}=0 m/s$ and gravity due it is a constant $g=9.8m/s^{2}$

b) The maximum high reached by the dolphin is $y_{max}=8.62 m$

c) Total time is $t=2.65s$

Explanation:

a) First of all the initial speed is given at the start of the problem, gravity is constant and final speed is known any object thrown straight up reaches its max high at $0m/s$ speed.

b) Second, now that we know final speed we use $v_{f} =v_{i}-gt$ , as we clear for $t=\frac{v_{i}-v_{f} }{g}=\frac{20.0m/s}{9.8m/s^{2} }=1.32 s$ .

Then we use $y=v_{i}t-\frac{1}{2} gt^{2}=(20.0m/s)(1.32s)-\frac{1}{2} (9.8m/s^{2} ) (1.32s)^{2} =8.62m$

c)Third, finally we can use $y=v_{i}t-\frac{1}{2} gt^{2}$ , as we know $y=0m$ when the dolphin fall into the water again and $v_{i} =13.0m/s$ , then we have $0=(13m/s)t-\frac{1}{2} (9.8m/s^{2} )t^{2}$ is a quadratic form $0=t(13.0-4.9t)$ so we have $t_{i}=0s$ and $t_{f}=\frac{13}{4.90} =2.65s$