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svlad2 [7]
3 years ago
15

Beginning at the NW corner of the intersection of Pine & 675, thence north 950 feet, thence west 380 feet, thence south 950

feet, thence east 380 feet. Is this an acceptable metes and bounds description?
Physics
1 answer:
Serjik [45]3 years ago
6 0

Answer:

this description is valid for mediadle displacement, bone is an acceptable description

Explanation:

The description of a person's position must be done with a position vector. These vectors must have magnitude, a given direction and a starting point.

In the description this has a starting point corner NO of pine and 675.

Each displacement occurs with respect to the previous one, indicating the magnitude of the displacement and its direction.

After analyzing  this description is valid for mediadle displacement, bone is an acceptable description

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A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in
german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

\Sigma F = -f = m\cdot a (Eq. 1)

Where:

f - Kinetic friction force, measured in newtons.

m - Mass of the box, measured in kilograms.

a - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight (W = m\cdot g) and uniform accelerated motion (v = v_{o}+a\cdot t), we expand the previous expression:

-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)

And the magnitude of the friction force exerted on the box is calculated by this formula:

f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right) (Eq. 1b)

Where:

W - Weight, measured in newtons.

g - Gravitational acceleration, measured in meters per square second.

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

t - Time, measured in seconds.

If we know that W = 52.4\,N, g = 9.807\,\frac{m}{s^{2}}, v_{o} = 1.37\,\frac{m}{s}, v = 0\,\frac{m}{s} and t = 2.8\,s, the magnitud of the kinetic friction force exerted on the box is:

f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s}  }{2.8\,s} \right)

f = 2.614\,N

The magnitude of the friction force exerted on the box is 2.614 newtons.

5 0
3 years ago
All of the noble gases, Group 18, have eight valence electrons in its outer shell (excluding helium which only has two). Which o
ikadub [295]

Answer:

Zero or +2

Explanation:

The noble gases already have a avplete outermost shell. They are the least reactive elements of earth?

Their normal oxidation number is zero but some have been shown to be reactive.

6 0
3 years ago
Read 2 more answers
A 6000 kg roller coaster goes around a loop of radius 30 m at 6 m/s. What is the centripetal acceleration?
Margarita [4]

Answer:

The answer to the question is 7200

7 0
3 years ago
the total positive charge is QQQ = 1.62×10−6 CC , what is the magnitude of the electric field caused by this charge at point P,
balu736 [363]

Answer:

6.1 × 10^9 Nm-1

Explanation:

The electric field is given by

E= Kq/d^2

Where;

K= Coulombs constant = 9.0 × 10^9 C

q = magnitude of charge = 1.62×10−6 C

d = distance of separation = 1.53 mm = 1.55 × 10^-3 m

E= 9.0 × 10^9 × 1.62×10−6/(1.55 × 10^-3 )^2

E= 14.58 × 10^3/2.4 × 10^-6

E= 6.1 × 10^9 Nm-1

8 0
3 years ago
Snow tends to sublime outside during cold, clear winter days when the atmospheric pressure is low. During this proces
olga nikolaevna [1]
A I think because sublimation is when something skips melting and goes directly to vapor
3 0
3 years ago
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