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3 years ago

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A certain light truck can go around a flat curve having a radius of 150 m with a maximum spee dof 32.0 m/s. With what maximum sp

eed can it go around a curve having a radius of 75.0 m?

1 answer:

Dvinal [7]3 years ago

7 0

Answer

Maximum speed at 75 m radius will be 22.625 m /sec

Explanation:

We have given radius of the curve r = 150 m

Maximum speed $v_{max}=32m/sec$

Coefficient of friction $\mu =\frac{v_{max}^2}{rg}=\frac{32^2}{150\times 9.8}=0.6965$

Now new radius r = 75 m

So maximum speed at new radius $v_{max}=\sqrt{\mu rg}=\sqrt{0.6965\times 75\times 9.8}=22.625m/sec$

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A 4,000-kg car traveling at 20m/s hits a wall with a force of 80,000 N and comes to a stop. What was the impact time?

Nookie1986 [14]

Rate of change of momentum = impact force
(m*v-m*u)/t = F
4000*20/t = 80000 (note: v is zero as it stopped)
<span>soo, t = 1 sec</span>

7 0

3 years ago

monochromatic light from a distant source is incident on a slit 0.75 mm wide. on screen 2 m away, the distance from the central

hjlf

Displacement from the center line for minimum intensity is 1.35 mm , width of the slit is 0.75 so Wavelength of the light is 506.25.

<h3>How to find Wavelength of the light?</h3>

When a wave is bent by an obstruction whose dimensions are similar to the wavelength, diffraction is observed. We can disregard the effects of extremes because the Fraunhofer diffraction is the most straightforward scenario and the obstacle is a long, narrow slit.

This is a straightforward situation in which we can apply the

Fraunhofer single slit diffraction equation:

y = mλD/a

Where:

y = Displacement from the center line for minimum intensity = 1.35 mm

λ = wavelength of the light.

D = distance

a = width of the slit = 0.75

m = order number = 1

Solving for λ

λ = y + a/ mD

Changing the information that the issue has provided:

λ = 1.35 * 10^-3 + 0.75 * 10^-3 / 1*2

=5.0625 *10^-7 = 506.25

so

Wavelength of the light 506.25.

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1 year ago

Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a dow

Inessa [10]

The magnitude of the current in wire 3 is 2.4 A and in a direction pointing in the downward direction.

The force per unit length between two parallel thin current-carrying $I_1$ and $I_2$ wires at distance ' r ' is given by $f=\frac{u_0I_1I_2}{2\pi r}$ ....(1) .

If the current is flowing in both wires in the same direction, and the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.

A schematic of the information provided in the question can be seen in the image attached below.

From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3

$F_2_1=F_2_3$

Using equation (1) , we get

$\frac{u_0I_2I_1}{0.2} =\frac{u_0I_2I_3}{0.32} \\\\\frac{I_1}{0.2} =\frac{I_3}{0.32} \\\\\frac{1.50}{0.2} =\frac{I_3}{0.32} \\\\0.48=0.2I_3\\\\I_3=2.4A$

I₃ = 2.4 A and the current is pointing in the downward direction

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1 year ago

A 125 g pendulum bob hung on a string of length 35 cm has the same period as when the bob is hung from a spring and caused to os

Bingel [31]

Answer:

k = 3.5 N/m

Explanation:

It is given that the time period the bob in pendulum is the same as its time period in spring mass system:

$Time\ Period\ of\ Pendulum = Time\ Period\ of\ Spring-Mass\ System\\2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{m}{k}$

$\frac{l}{g} = \frac{m}{k}\\\\ k = g\frac{m}{l}$

where,

k = spring constant = ?

g = acceleration due to gravity = 9.81 m/s²

m = mass of bob = 125 g = 0.125 kg

l = length of pendulum = 35 cm = 0.35 m

Therefore,

$k = (9.81\ m/s^2)(\frac{0.125\ kg}{0.35\ m})\\\\$

<u>k = 3.5 N/m</u>

4 0

3 years ago

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m

mixas84 [53]

Answer:

r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

U = $k \frac{q_1q_2}{r_{12}}$

for the proton at x = -1 m

U₁ = $- k \frac{e^2 }{r+1}$

for the electron at x = 1 m

U₂ = $k \frac{e^2 }{r-1}$

starting point.

Em₀ = K + U₁ + U₂

Em₀ = $\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}$

final point

Em_f = $k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})$

energy is conserved

Em₀ = Em_f

\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})

\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²( $\frac{2}{(r_2+1)(r_2-1)}$ )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ $- \frac{1}{20+1} + \frac{1}{20-1}$ ) = 9 109 (1.6 10-19) ²( $\frac{2}{r_2^2 -1}$ )

2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( $\frac{1}{r_2^2 -1}$ )

2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷ $\frac{1}{r_2^2 -1}$

$\frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}$

r₂² -1 = (4.443 10⁸)⁻¹

r2 = $\sqrt{1 + 2.25 10^{-9}}$

r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0

3 years ago