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3 years ago

Is the relationship between velocity and centripetal force a direct, linear or nonlinear square relationship?

Physics

1 answer:

Svet_ta [14]3 years ago

5 0

Answer:

non linear square relationship

Explanation:

formula for centripetal force is given as

a = mv^2/r

here a ic centripetal acceleration , m is mass of body moving in circle of radius r and v is velocity of body . If m ,and r are constant we have

a = constant × v^2

a α v^2

hence non linear square relationship

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48 grams 12cm^3, what would the density of the material be

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$D = \frac{m}{V}$

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If you put that into the equation, you will get your density:

$D = \frac{m}{V} = \frac{48g}{12cm^{3} } =4g/ cm^{3}$

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3 years ago

Read 2 more answers

How do I go about this?

Anna71 [15]

Hi there!

(a)

Recall that:
$W = F \cdot d = Fdcos\theta$

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

$W =248(56)cos(30) = 12027.36 J$

To the nearest multiple of ten:
$W_A = \boxed{12030 J}$

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
$\boxed{W_g = 0 J}$

(d)

Recall that the force of kinetic friction is given by:
$F_{f} =\mu_k mg$

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
$W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J$

In multiples of ten:
$\boxed{W_f = -3070 J}$

(e)
Simply add up the above values of work to find the net work.

$W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J$

Nearest multiple of ten:
$\boxed{W_{net} = 8950 J}}$

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
$F_{net} = F_{Ax} - F_f$

$W = F_{net} \cdot d = (F_{Ax} - F_f)$

$W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J$

Nearest multiple of ten:
$\boxed{W_{net} = 8950 J}$

5 0

2 years ago

Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s

Alenkasestr [34]

Answer: 1.289 m

Explanation:

The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:

<u>x-component: </u>

$x=V_{o}cos\theta t$ (1)

Where:

$x$ is the horizontal distance traveled by the venom

$V_{o}=3.10 m/s$ is the venom's initial speed

$\theta=47\°$ is the angle

$t$ is the time since the venom is spitted until it hits the ground

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0.44 m$ is the initial height of the venom

$y=0$ is the final height of the venom (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Let's begin with (2) to find the time it takes the complete path:

$0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}$ (3)

Rewritting (3):

$-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 m=0$ (4)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula: