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Misha Larkins [42]

2 years ago

8

A cart is rolling down a ramp. if the angle of the ramp is increased to make the ramp steeper the cart will

1 answer:

Westkost [7]2 years ago

8 0

The speed of the carts descent will increase? Not quite sure what you’re asking.

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The rainbow is a result of a. Diffraction of the light.

suter [353]

Rainbows are caused by the dispersion of light, which itself consists of a combination of refraction and reflection of light around little droplets of water.

Choice C

5 0

2 years ago

Using a 683 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t

n200080 [17]

Answer:

10.2 m

Explanation:

The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:

$y=\frac{\lambda (m+\frac{1}{2})D}{d}$

where

y is the position of the m-th minimum

m is the order of the minimum

D is the distance of the screen from the slit

d is the width of the slit

$\lambda$ is the wavelength of the light used

In this problem we have:

$\lambda=683 nm = 683\cdot 10^{-9} m$ is the wavelength of the light

$d=1.1 mm = 0.0011 m$ is the width of the slit

m = 13 is the order of the minimum

$y=8.57 cm = 0.0857 m$ is the distance of the 13th dark fringe from the central maximum

Solving for D, we find the distance of the screen from the slit:

$D=\frac{yd}{\lambda(m+\frac{1}{2})}=\frac{(0.0857)(0.0011)}{(683\cdot 10^{-9})(13+\frac{1}{2})}=10.2 m$

6 0

3 years ago

Bro i need help omg.

sergey [27]

Answer:

25.

Explanation:

8 0

2 years ago

Read 2 more answers

Calculate the kinetic energy of a 750 kg compact car moving 50 m/s.

k0ka [10]

935,500 joules because when we use the KE formula KE=1/2mv^2;
KE=1/2(750)(50)^2
KE=375(2500)
KE=935,500 Joules
Hope it helps

6 0

2 years ago

A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What

AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, $\alpha=3.3 rad/s^2$

Diameter of the wheel, d=21 cm

Radius of wheel, $r=\frac{d}{2}=\frac{21}{2}$ cm

Radius of wheel, $r=\frac{21\times 10^{-2}}{2} m$

1m=100 cm

Magnitude of total linear acceleration, a= $1.7 m/s^2$

We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration, $a_t=\alpha r$

$a_t=3.3\times \frac{21\times 10^{-2}}{2}$

$a_t=34.65\times 10^{-2}m/s^2$

Radial acceleration, $a_r=\frac{v^2}{r}$

We know that

$a=\sqrt{a^2_t+a^2_r}$

Using the formula

$1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}$

Squaring on both sides

we get

$2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}$

$\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}$

$v^4=r^2\times 2.7699$

$v^4=(10.5\times 10^{-2})^2\times 2.7699$

$v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}$

$v=0.418 m/s$

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0

2 years ago