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3 years ago

What is the mass of modelling clay?

Physics

2 answers:

Maurinko [17]3 years ago

6 0

the mass of modelling clay is 500g

iren2701 [21]3 years ago

6 0

The modeling clay mass white 500g

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The pilot of an airplane flying at an elevation of 5000 feet sights two trees that are 300 feet apart. If the angle of depressio

OLEGan [10]

Answer:

32°

Explanation:

From the diagrammatic representation of the question,

Point A is the pilot point of view from the airplane

Point C is the foot of the first tree

Point D is the foot of the second tree

Line AB is the height of the airplane from the ground level.

Line CD is the distance between the two trees

Considering ΔABC,

tan 33° $= \frac{5000}{BC}$

BC = 5000 / tan 33°

$BC = \frac{5000}{0.6494}$

BC = 7,699.41 feet

BD = BC + CD

BD = 7,699.41 + 300

BD = 7,999.41 feet

Considering ΔABD,

tan θ $= \frac{AB}{BD}$

tan θ $= \frac{5000}{7,999.41}$

tan θ = 0.6250

θ = tan⁻¹ 0.6250

θ = 32°

7 0

3 years ago

B. The coefficient of friction between the tires and the road is 0.850 and the mass of the car is

Mrac [35]

Answer:

156.26N

Explanation:

The data needed are incomplete. Let the acceleration of the body be 3.5m/s²

Other given parameters

Mass = 1.35×10^1 = 13.5kg

coefficient of friction between the tires and the road = 0.850

Acceleration due to gravity = 9.8m/s²

According to Newton's second law:

Fnet = ma

Fnet = Fapp - Ff

Fapp is the applied force

Ff is the frictional force = umg

The equation becomes:

Fapp - Ff = ma

Fapp-umg = ma

Fapp - 0.85(13.5)(9.8) = 13.5(3.5)

Fapp - 109.0125 = 47.25

Fapp = 47.25+109.0125

Fapp = 156.2625N

Hence the applied force that caused the acceleration is 156.26N

Note that the acceleration of the car was assumed. Any value of acceleration can be used for the calculation.

8 0

3 years ago

Consider the model above. It represents the electrical force. As r increases, the attractive force decreases. How would this mod

aivan3 [116]

Answer:

As we keep on increasing the radius the value of the gravitation force of attraction decreases and as we decrease the radius the gravitation force increases.

Explanation:

Like the coulombs law of electrostatics, the law of gravitation also depends inversely on the square of the value of r. Therefore, as we keep on increasing the value of r the value of the gravitation force decreases and as we decrease the value of the r the value of gravitation force increases.

Gravitation Force= $\frac{Gm_{1}m_{2} }{r^{2}}$