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3 years ago

In a uniform circular motion map, what is always true? Check all that apply.

Physics

2 answers:

Katen [24]3 years ago

7 0

Answer:

Acceleration vectors point toward the center of the circle.

Velocity vectors are the same length.

Explanation:

grandymaker [24]3 years ago

6 0

Answer:

Velocity vectors are always perpendicular to the circle.

Acceleration vectors point toward the center of the circle.

Velocity vectors are the same length.

Explanation:

(THESE ARE NOT MY WORDS BTW)

1) Acceleration and velocity are vectorial quantities, which means they have magnitude and direction.

2) In a circular motion velocity direction changes all the time, which means that it is accelerated.

3) In a uniform circular motiion, the velocity changes in a constant value. This is the rate of change of velocity, which is the magnitude of the acceleration, is constant (uniform).

4) The velocity is perpendicular to the path, i.e. the circle. You can see it if you think that if the object stopped changing the direction, then the object would follow a straight path (as per inertia principle). That is why this velocity is called tangential velocity (to differentiate it of the angular velocity).

This is what the option C says "Velocity vectors are always perpendicular to the circle". Then this is true.

5) The constant change of direction in a circular path, means that the object is been pushed, accelerated, toward the center of a circle. This is, all the time the object in motion tries to follow the perpendicular path but a push (a force) directed to the center of the circle changes its direction. Such force accelerates the object toward the center of the circle. So, the acceleration vectors point toward the center of the circle, which is what the option D says. So, this is also true.

6) Since the motion is uniform, the magnitude or length of the velocity vectors are always the same, are constant. So, the option E. is also true.

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F=ma

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If you are in an elevator that speed up then the apparent weight is

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Answer:

Fnet - Fg

Explanation:

When an object is in an elevator, its weight varies with respect to the direction of movement of the elevator and the elevators acceleration.

The weight, W, of an object can be expressed as;

W = mg

where m is the object's mass, and g is the acceleration due gravity.

If the object is in an elevator that speed up, an apparent weight would be felt since both mass and elevator are moving against gravitational pull of the earth.

So that,

$W_{net}$ = mg + ma

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2 years ago

What are three key aerodynamics principles?

mafiozo [28]

Answer:

Explanation:

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3 years ago

The law of universal gravitation states that any two objects in the universe, without exception,

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A. attract each other.

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3 years ago

Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina

Alecsey [184]

(a) 23.4

The fiber-to-matrix load ratio is given by

$\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}$

where

$E_f = 131 GPa$ is the fiber elasticity module

$E_m = 2.4 GPa$ is the matrix elasticity module

$V_f=0.3$ is the fraction of volume of the fiber

$V_m=0.7$ is the fraction of volume of the matrix

Substituting,

$\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4$ (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

We can rewrite (1) as

$F_m = \frac{F_f}{23.4}$

And inserting this into (2):

$F=F_f + \frac{F_f}{23.4}$

Solving the equation, we find the actual load carried by the fiber phase:

$F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N$

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

Using

F = 46500 N

$F_f = 44594 N$

We can immediately find the actual load carried by the matrix phase:

$F_m = F-F_f = 46,500 N - 44,594 N=1,906 N$

(d) 437 MPa

The cross-sectional area of the fiber phase is

$A_f = A V_f$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_f=0.3$ , we have

$A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2$

And the magnitude of the stress on the fiber phase is

$\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa$

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

$A_m = A V_m$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_m=0.7$ , we have

$A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2$

And the magnitude of the stress on the matrix phase is

$\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa$

(f) $3.34\cdot 10^{-3}$

The longitudinal modulus of elasticity is

$E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa$

While the total stress experienced by the composite is

$\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa$

So, the strain experienced by the composite is

$\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}$

3 0

3 years ago