A net horizontal force of 2000 n is applied to an 800-kg car at rest. the car's speed after 5 s will be

A 5 meter long ladder leans against a wall. The bottom of the ladder slides away from the wall at the constant rate of 1 3 m/s.

Oksana_A [137]

Answer:9.75 m/s

Explanation:

Given

Length of ladder $(L)=5 m$

Foot the ladder is moving away with speed of $\frac{\mathrm{d} x}{\mathrm{d} t}=13 m/s$

From diagram

$x^2+y^2=L^2$ ------1

at $x=3$

$y^2=25-9=16$

$y=4 m$

Now differentiating equation 1 w.r.t time

$2x\frac{\mathrm{d} x}{\mathrm{d} t}+2y\frac{\mathrm{d} y}{\mathrm{d} t}=0$

$x\frac{\mathrm{d} x}{\mathrm{d} t}=-y\frac{\mathrm{d} y}{\mathrm{d} t}$

$3\times 13=-4\times \frac{\mathrm{d} y}{\mathrm{d} t}$

$\frac{\mathrm{d} y}{\mathrm{d} t}=-\frac{3\times 13}{4}=-9.75 m/s$

negative indicates distance is decreasing with time

5 0

2 years ago

A horizontal beam of light of intensity 25 W/m2 is sent through two polarizing sheets. The polarizing direction of the first mak

Zina [86]

Answer:

option (B)

Explanation:

Intensity of unpolarised light, I = 25 W/m^2

When it passes from first polarisr, the intensity of light becomes

$I'=\frac{I_{0}}{2}=\frac{25}{2}=12.5 W/m^{2}$

Let the intensity of light as it passes from second polariser is I''.

According to the law of Malus

$I'' = I' Cos^{2}\theta$

Where, θ be the angle between the axis first polariser and the second polariser.

$I'' = 12.5\times Cos^{2}15$

I'' = 11.66 W/m^2

I'' = 11.7 W/m^2

7 0

3 years ago

What is the equivalent resistance between points A and C if R1=1430, R2=1350, R3=1100, R4=1350, and R5=1150.

Marianna [84]

R1 + R4 = 1430 + 1350 = 2780 = R14 series combination of R1 & R4

R2 + R5 = 1350 + 1150 = 2500 = R25

The circuit has been reduced to 3 resistors in parallel

R314 = 2780 * 1100 / (2780 + 1100) = 788 this is the resistance of the parallel combination of R14 and R3

R31425 = 2500 * 788 / (2500 + 788) = 599 which is the equivalent of the circuit - you can also use the formula for 3 resistors in parallel but this seems simpler

7 0

3 years ago

Calculate the wavelengths of the first five emission lines of the Balmer series for hydrogen

In-s [12.5K]

Answer:

Explanation:

The Balmer series in a hydrogen atom relates the possible electron transitions down to the n = 2 position to the wavelength of the emission that scientists observe. In quantum physics, when electrons transition between different energy levels around the atom (described by the principal quantum number, n) they either release or absorb a photon. The Balmer series describes the transitions from higher energy levels to the second energy level and the wavelengths of the emitted photons. You can calculate this using the Rydberg formula.

5 0

2 years ago

A. Use the graph and the element made in question 2 to determine the mass of the star.

Minchanka [31]

Black balls with blue waffles

4 0

2 years ago