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2 years ago

10

Explain some reasons why Muslims are hated in America. Be sure to distinguish between fact and falsehood. In other words, what i

s being factually (or falsely) said about the Muslim community

1 answer:

Drupady [299]2 years ago

6 0

People believed they caused 911 not sure if that’s a answer choice but that’s my answer

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A burning candle provides :

mixer [17]

This would be B

Hope this helped

8 0

2 years ago

A beam of electrons passes through a single slit, and a beam of protons passes through a second, but identical, slit. The electr

ryzh [129]

Answer:

(b) The electrons, because they have the smaller momentum and, hence, the larger de Broglie wavelength

Explanation:

de Broglie wavelength λ = h / m v

Since both electrons and protons have same velocity , momentum mv will be less for electrons because mass of electron is less .

for electron , momentum is less so . Therefore de Broglie wavelength λ will be more for electrons .

Amount of diffraction that is angle of diffraction is proportional to λ

Therefore electrons having greater de Broglie wavelength will show greater diffraction.

7 0

2 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

8 0

2 years ago

1. Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of wood is sitting on a frictionless

Maslowich

Since momentum is a vector quantity, take any direction as positive and other as negative. Answer won't change.

4 0

2 years ago

Read 2 more answers

Suppose a car approaches a hill and has an initial speed of

kvv77 [185]

Answer:

a) 1.73*10^5 J

b) 3645 N

Explanation:

106 km/h = 106 * 1000/3600 = 29.4 m/s

If KE = PE, then

mgh = 1/2mv²

gh = 1/2v²

h = v²/2g

h = 29.4² / 2 * 9.81

h = 864.36 / 19.62

h = 44.06 m

Loss of energy = mgΔh

E = 780 * 9.81 * (44.06 - 21.5)

E = 7651.8 * 22.56

E = 172624.6 J

Thus, the amount if energy lost is 1.73*10^5 J

Work done = Force * distance

Force = work done / distance

Force = 172624.6 / (21.5/sin27°)

Force = 172624.6 / 47.36

Force = 3645 N

5 0

2 years ago