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Afina-wow [57]
3 years ago
11

A disoriented physics professor drives 3.22 km north, then 4.75 km west, then 1.90 km south. find the magnitude and direction of

the resultant displacement, using the method of components.
Physics
1 answer:
Andre45 [30]3 years ago
7 0

It can be noted that the directions west and north are perpendicular to each other. hence we can use the right angle formula for calculating total distance in these directions. Also South is just negative north. This technique is called euclidean distance norm. It is the basis of Cartesian coordinate system.

Total distance = 3.22 N + 4.75 W +1.90  S

= 1.32 N + 4.75 W

D = \sqrt{N^2 + W^2}

D = 4.93 Km

direction = atan(N/W) = 15.53 deg north of west

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If the mass of a material is 111 grams and the volume of the material is 23 cm3, what would the density of the material be? g/cm
Butoxors [25]
Density can be kg/m^3 or g/cm3
In g/cm3 density =mass /volume =111g/23cm3
=4.826g/cm3.

In kg/m3,density=mass/volume. converting mass in grams to kg, 1000g=1kg,111g=0.111kg.
cm3 to m3, 1cm3=10^-6m3, 23cm3=0.000023m3
density=0.111kg/0.000023m3 or 2.3*10^-5=4,826.1kg/m3.

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3 years ago
Which statements describe the results of the effect of angle of insolation on absorption experiment? Check all that apply.
lilavasa [31]

Answer:

At 6 minutes, the soil temperature at a 45° angle of insolation is higher than at 0°.

At 15 minutes, the soil temperature at a 90° angle of insolation is higher than at 45°

Explanation:

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3 years ago
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According to Newton’s first law of motion, what will an object in motion do when no external force acts on it?
KonstantinChe [14]

By definition, we have to:

Newton's first law states that any object will remain in a state of rest or with a uniform rectilinear motion unless an external force acts on it.

Therefore, according to the first law of Newton, if the object is already in motion and has no force acting on it then, it will remain with a uniform rectilinear motion.

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The roller coaster car has a mass of 700 kg, including its passenger. If it is released from rest at the top of the hill A, dete
sweet [91]

Answer:

h = 18.75 m

Now when it will reach at point B then its normal force is just equal to ZERO

N_B = 0

F_n = 1.72 \times 10^4

Explanation:

Since we need to cross both the loops so least speed at the bottom must be

v = \sqrt{5 R g}

also by energy conservation this is gained by initial potential energy

mgh = \frac{1}{2}mv^2

v = \sqrt{2gh}

so we will have

\sqrt{2gh} = \sqrt{5Rg}

now we have

h = \frac{5R}{2}

here we have

R = 7.5 m

so we have

h = \frac{5(7.5)}{2}

h = 18.75 m

Now when it will reach at point B then its normal force is just equal to ZERO

N_B = 0

now when it reach point C then the speed will be

mgh - mg(2R_c) = \frac{1}{2]mv_c^2

v_c^2 = 2g(h - 2R_c)

v_c = 13.1 m/s

now normal force at point C is given as

F_n = \frac{mv_c^2}{R_c} - mg

F_n = \frac{700\times 13.1^2}{5} - (700 \times 9.8)

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