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2 years ago

Which statement represents a difference between horizontal and vertical relationships?

1 answer:

6 0

D.) Vertical relationships involve unequal status, while horizontal relationships represent equal status.
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State Newtons first law of motion.

Fofino [41]

Answer:

An object at rest will stay at rest and an object in motion will stay in motion unless it is acted upon by an unbalanced/external force.

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3 years ago

What is the electric field at the point x=6.5 m? positive e-fields point to the right.?

Mademuasel [1]

6,5 6,4 6,3 6,2 6
if you know,1

5 0

2 years ago

A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police

Nastasia [14]

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. $x=x_{0}+vt$

2. $x=x_{0}+v_{0}t+\frac{at^{2}}{2}$

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

$vt = v_{0}t+\frac{at^2}{2}\\\\ 16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}$

We simplify the fraction present and rearrange for our formula so that it equals 0:

$0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0$

In the very last step we factored a common factor t. There is two possible solutions to the equation at $t=0$ and:

$0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\ 0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s$

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at $t=0 s$ (when the SUV passed the police car) and $t=15.56s$ (when the police car catches up to the SUV)

8 0

2 years ago

g We saw in class that in a pendulum the string does no work. We also saw that the normal force does no work on an object slidin

Ymorist [56]

Answer:

From question (a) and (b) the pendulum motion is perpendicular to the force so the normal force will do no work and the tension in the string of the pendulum will not work

$i.e Normal \ Force(N) = mg cos \theta$

And $\theta = 90$ so

$N = 0$

An example will be a where a stone is attached to the end of a string and is made to move in a circular motion while keeping the other end of the string in a fixed position

A dog walking along a surface which has friction, here the frictional force would acting in the direction of the motion and this would do positive work

Explanation:

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2 years ago

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A tennis ball forgotten on the court after match is:

maw [93]

C is what i always have so ima go with C.

6 0

2 years ago

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