PLEASE HELP I WILL GOVE BRAINLIEST TO FIRST CORRECT ANSWER!!!!!

A car drives to the right. There is a large amount or air resistance and the cars engine provides the cars forward motion. Which

scZoUnD [109]

Answer:

the answer to some may be Force A

Explanation:

7 0

3 years ago

Read 2 more answers

In __________ waves, the motion of the particles in a medium is across the direction of the wave (perpendicular).

Alex_Xolod [135]

In longitudinal waves, the motion of the particles in a medium is across the direction of the wave.

4 0

2 years ago

A coil of wire with 60 turns lies on the plane of the page, and a uniform magnetic field points into the page with a strength of

baherus [9]

Answer:

0.85V

Explanation:

The emf is calculated by using the Lenz's Law

$\epsilon=-\frac{\Delta\Phi_B}{\Delta t}$

But for this case we have that the magnetic field is constant. Hence we have

$\epsilon=-\frac{B(A_{2}-A_{1})}{t_2-t_1}=-\frac{(1.10T)(0.270m^{2}-0.89m^{2})}{0.8s}\\\\\epsilon=0.85V$

where we have taken that the intial time is t1=0

I hope this is useful for you

regards

4 0

2 years ago

A mass weighing 4 lb stretches a spring 2 in. Suppose that the mass is given an additional 6-in displacement in the positive dir

givi [52]

Answer:

$\frac{1}{8} y'' + 2y' + 24y=0$

Explanation:

The standard form of the 2nd order differential equation governing the motion of mass-spring system is given by

$my'' + \zeta y' + ky=0$

Where m is the mass, ζ is the damping constant, and k is the spring constant.

The spring constant k can be found by

$w - kL=0$

$mg - kL=0$

$4 - k\frac{1}{6}=0$

$k = 4\times 6 =24$

The damping constant can be found by

$F = -\zeta y'$

$6 = 3\zeta$

$\zeta = \frac{6}{3} = 2$

Finally, the mass m can be found by

$w = 4$

$mg=4$

$m = \frac{4}{g}$

Where g is approximately 32 ft/s²

$m = \frac{4}{32} = \frac{1}{8}$

Therefore, the required differential equation is

$my'' + \zeta y' + ky=0$

$\frac{1}{8} y'' + 2y' + 24y=0$

The initial position is

$y(0) = \frac{1}{2}$

The initial velocity is

$y'(0) = 0$

6 0

2 years ago

a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the f

34kurt

Answer:

$\frac{h_{liquid} }{ h_{body} }$ = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

B = ρ_liquid g V_liquid

let's write the translational equilibrium condition

B - W = 0

let's use the definition of density

ρ_body = m / V_body

m = ρ_body V_body

W = ρ_body V_body g

we substitute

ρ_liquid g V_liquid = ρ_body g V_body

$\frac{\rho_{body} }{\rho_{liquid} } } = \frac{V_{liquid} }{V_{body} } }$

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

V = A h_bogy

Thus

$\frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }$

we substitute

5/9 = $\frac{h_{liquid} }{ h_{body} }$

8 0

2 years ago