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3 years ago

What is most often the basis for racist behavior in the United States?

Physics

2 answers:

olganol [36]3 years ago

4 0

Unfair laws will be the answer cause that is mostly what we see

leonid [27]3 years ago

3 0

Legally or socially authorized rights and powers given to white Americans but denied to all other races is most often the basis for racist behavior in the United States. The White people in the United States were the rich and affluent people who were given some important rights that made them different from the Black or other immigrant people. Racism is actually a theory that believes in the superiority of one particular race over the others, which often occurs in prejudice and discrimination towards people on the basis of their race and ethnicity. Thus, in America the basis of racist behavior was sectioned by the State.

You might be interested in

Match the correct term with the statement that is true about it:

icang [17]

ANSWER:
1. Cardiovascular Fitness
2. Muscular Fitness
3. Flexibility
4. Motivation
5. Money
6.
7. Family Behaviors
8. Understanding

is there another option for number 6? i could only find 7 word options

6 0

4 years ago

Read 2 more answers

1. How much force would you have to apply to a 15kg object in order to accelerate it! a 2 m/s?

mihalych1998 [28]

30N

Explanation:

Given parameters:

Mass of object = 15kg

Acceleration = 2m/s

Unknown:

Force = ?

Solution:

Force is given as the product of mass and acceleration:

F = m x a

m is the mass

a is the acceleration

Inputting the parameters:

F = 15 x 2 = 30N

The unit of force is newtons, N .

Learn more:

Force brainly.com/question/10470406

#learnwithBrainly

3 0

3 years ago

Two people are pulling a boat through the water. Each exerts a force of 600 N directed at a 30.0 angle relative to the forward m

VashaNatasha [74]

It sounds as though the two people are standing in front of the boat on opposite sides of it, so that they both make an angle of 30.0° with the axis of the boat, as in the attached free body diagram (ignoring the force of buoyancy and the weight of the boat).

By Newton's second law, the net vertical force is

∑ F = P₁ sin(60.0°) + P₂ sin(120.0°) - R = 0

where upward is positive and downward is negative, and the right side is 0 because the boat moves with constant velocity and thus zero acceleration.

We're told that P₁ = P₂ = 600 N, and we know sin(60°) = sin(120°), so the above reduces to

R = 2 P sin(60.0°) = 2 (600 N) sin(60.0°) ≈ 1040 N

7 0

3 years ago

A student plucks a fixed-end string, creating a standing wave with 6.00 nodes (including any nodes at the ends). The string is t

Vesna [10]

1) 2.5 wavelengths

2) 0.208 m

3) 1731 Hz

Explanation:

Standing waves are waves that do not propagate, but instead the particles of the medium just oscillate around a fixed position. Examples of standing waves are the waves produced on a string with fixed ends.

The points of a standing wave in which the amplitude of the oscillation is always zero are called nodes.

The two fixed ends of the string are two nodes. In this problem, we have a total of 6 nodes along the string: this means that there are 4 additional nodes apart from the two ends of the string.

Therefore, this also means that the string oscillate in 5 different segments.

One wavelength is equal to 2 segments of the oscillation: therefore, since here there are 5 segments, this means that the number of wavelengths that we have in this string is

$n=\frac{5}{2}=2.5$

The wavelength of a wave is the distance between two consecutive crests (or throughs) of the wave.

The wavelength of a standing wave can be also measured as the distance between the nth-node and the (n+2)-th node: so, basically, the wavelength in a standing wave is twice the distance between two nodes:

$\lambda = 2 d$

where

$\lambda$ is the wavelength

d is the distance between two nodes

Here the length of the string is

L = 0.520 m

And since it oscillates in 5 segments, the distance between two nodes is

$d=\frac{L}{5}=\frac{0.520}{5}=0.104 m$

And therefore, the wavelength is

$\lambda=2d=2(0.104)=0.208 m$

The frequency of a wave is the number of complete oscillations of the wave per second.

The frequency of a wave is related to its speed and wavelength by the wave equation:

$v=f\lambda$

where

v is the speed

f is the frequency

$\lambda$ is the wavelength

In this problem:

v = 360 m/s is the speed of the wave

$\lambda=0.208 m$ is the wavelength

Therefore, the frequency is

$f=\frac{v}{\lambda}=\frac{360}{0.208}=1731 Hz$

3 0

3 years ago

Kruka [31]

Answer:

The weight of the object X is approximately 3.262 N (Acting downwards)

The weight of the object Y is approximately 8.733 N (Acting downwards)

Explanation:

The question can be answered based on the principle of equilibrium of forces

The given parameters are;

The weight of Z = 12 N (Acting downwards)

The weight of the pulleys = Negligible

From the diagram;

The tension in the in the string attached to object Z = The weight of object Z = 12 N

The tension in the in the string attached to object X = The weight of the object X

The tension in the in the string attached to object Y = The weight of the object Y

Given that the forces are in equilibrium, we have;

The sum of vertical forces acting at a point, $\Sigma F_y$ = 0

Therefore;

$T_{1y} + T_{2y} + T_{3y} = 0$

$T_{1y} = -( T_{2y} + T_{3y} )$

Where;

$T_{1y}$ = The weight of object Z = 12 N

$T_{1y}$ = 12 N

$T_{2y}$ = The vertical component of tension, T₂ = T₂ × sin(24°)

∴ $T_{2y}$ = T₂ × sin(156°)

Similarly;

$T_{3y}$ = T₃ × sin(50°)

From $T_{1y} = -( T_{2y} + T_{3y} )$ , and $T_{1y}$ = 12 N, we have;

12 N = -(T₂ × sin(156°) + T₃ × sin(50°))...(1)

Given that the forces are in equilibrium, we also have that the sum of vertical forces acting at a point, ∑Fₓ = 0

Therefore at point B, we have;

T₁ₓ + T₂ₓ + T₃ₓ = 0

The tension force, T₁, only has a vertical component, therefore;

∴ T₁ₓ = 0

∴ T₂ₓ + T₃ₓ = 0

T₂ₓ = -T₃ₓ

T₂ₓ = T₂ × cos(156°)

T₃ₓ = T₃ × cos(50°)

From T₂ₓ = -T₃ₓ, we have;

T₂ × cos(156°) = - T₃ × cos(50°)...(2)

Making T₃ the subject of equation (1) and (2) gives;

Making T₃ the subject of equation in equation (1), we get;

12 = -(T₂ × sin(156°) + T₃ × sin(50°))

∴ T₃ = (-12 - T₂ × sin(156°))/(sin(50°))

Making T₃ the subject of equation in equation (2), we get;

T₂ × cos(156°) = - T₃ × cos(50°)

∴ T₃ = T₂ × cos(156°)/(-cos(50°))

Equating both values of T₃ gives;

(-12 - T₂ × sin(156°))/(sin(50°)) = T₂ × cos(156°)/(-cos(50°))

-12/(sin(50°)) = T₂ × cos(156°)/(-cos(50°)) + T₂ × sin(156°)/(sin(50°))

∴ T₂ = -12/(sin(50°))/((cos(156°)/(-cos(50°)) + sin(156°)/(sin(50°))) ≈ -8.02429905283

∴ T₂ ≈ -8.02 N

From T₃ = T₂ × cos(156°)/(-cos(50°)), we have;

T₃ = -8.02× cos(156°)/(-cos(50°)) = -11.3982199717

∴ T₃ ≈ -11.4 N

The weight of the object X = -T₂ × sin(156°)

∴ The weight of the object X ≈ -(-8.02 × sin(156°)) = 3.262 N

The weight of the object X ≈ 3.262 N (Acting downwards)

The weight of the object Y = -(T₃ × sin(50°))

∴ The weight of the object Y = -(-11.4 × sin(50°)) ≈ 8.733 N

The weight of the object Y ≈ 8.733 N (Acting downwards)

4 0

3 years ago