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2 years ago

What centripetal force is needed to keep a 7kg mass moving in a circle of 4 meters radius at 15m/s

Physics

1 answer:

liq [111]2 years ago

7 0

Answer:

We conclude that the centripetal force needed to keep a 7kg mass moving in a circle of 4 meters radius at 15m/s is 393.75 N.

Explanation:

Given

Mass m = 7 kg

Radius r = 4 m

Velocity v = 15m/s

To determine

We need to determine the centripetal force needed to keep a 7kg mass moving in a circle of 4 meters radius at 15m/s.

We know a centripetal force acts on a body to keep it moving along a curved path.

We can determine the centripetal force using the formula

$F_c=\frac{mv^2}{r}$

where

$m$ is the mass

$v$ is the velocity

$r$ is the radius

$F_c$ is the centripetal force

substitute m = 7, r = 4, and v = 15 in the formula

$F_c=\frac{mv^2}{r}$

$F_c=\frac{7\left(15\right)^2}{4}$

$F_c=\frac{1575}{4}$

$F_c=393.75$ N

Therefore, we conclude that the centripetal force needed to keep a 7kg mass moving in a circle of 4 meters radius at 15m/s is 393.75 N.

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Is it possible for a population with a high birth rate to decrease in size.

matrenka [14]

Sure, if the mortality (death) rate is even higher than the birth rate.

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2 years ago

if you drop a softball from just above your knee, the kinetic energy of the ball just before it hits the ground is about 1 jule,

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False.

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This corresponds to the total mechanical energy of the ball at the moment it is dropped, because there is no kinetic energy (the ball starts from rest). Then the ball is dropped, and just before it hits the ground, all this energy is converted into kinetic energy: but energy cannot be created, so its final kinetic energy cannot be greater than 0.6 J.

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2 years ago

Let’s look at a radio-controlled model car. Suppose that at time t1=2.0st1=2.0s the car has components of velocity vx=1.0m/svx=1

Nonamiya [84]

Answer:

$a_x=6\ \text{m/s}^2$ and $a_y=0\ \text{m/s}^2$

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Explanation:

$t_1=2\ \text{s}$

$v_x=1\ \text{m/s}$

$v_y=3\ \text{m/s}$

$t_2=2.5\ \text{s}$

$v_x=4\ \text{m/s}$

$v_y=3\ \text{m/s}$

Average acceleration in the different axes

$a_x=\dfrac{\Delta v_x}{\Delta t}\\\Rightarrow a_x=\dfrac{4-1}{2.5-2}\\\Rightarrow a_x=6\ \text{m/s}^2$

$a_y=\dfrac{\Delta v_y}{\Delta t}\\\Rightarrow a_y=\dfrac{3-3}{2.5-2}\\\Rightarrow a_y=0\ \text{m/s}^2$

The components of the acceleration is $a_x=6\ \text{m/s}^2$ and $a_y=0\ \text{m/s}^2$

The magnitude of acceleration

$a=\sqrt{a_x^2+a_y^2}\\\Rightarrow a=\sqrt{6^2+0^2}\\\Rightarrow a=6\ \text{m/s}^2$

Direction

$\theta=\tan^{-1}\dfrac{a_y}{a_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{0}{6}\\\Rightarrow \theta=0^{\circ}$

The magnitude of accleration is $6\ \text{m/s}^2$ and the direction is $0^{\circ}$ .

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2 years ago

A barometer accidentally contains 6.5 inches of water on top of the mercury column (so there is also water vapor instead of a va

viktelen [127]

Answer:

(a). 14.4 lbf/in^2.

(b). 27.8 in, AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.

Explanation:

So, from the question above we are given the following parameters which are going to help us in solving this particular Question;

=> The "barometer accidentally contains 6.5 inches of water on top of the mercury column (so there is also water vapor instead of a vacuum at the top of the barometer)"

=> "On a day when the temperature is 70oF, the mercury column height is 28.35 inches (corrected for thermal expansion)."

With these knowledge, let us delve right into the solution;

(a). The barometric pressure = water vapor pressure + acceleration due to gravity (ft/s^2) × water density(slug/ft^3) × {ft/12 in}^3 × [ height of mercury column + specific gravity of mercury × height of water column].

The barometric pressure= 0.363 + {(62.146) ÷ (12^3) × 390.6425}. = 14.4 lbf/in^2.

(b). { (13.55 × length of mercury) + 6.5 } × (62.15÷ 12^3) = 14.4 - 0.603.

Length of mercury = 27.8 in.

AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.

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3 years ago

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salantis [7]

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so
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The frictional force can be rewritten as
$F_f = \mu m g$
where $m=50 kg$ , $g=9.81 m/s^2$ . Re-arranging, we can solve this equation to find $\mu$ , the coefficient of dynamic friction:
$\mu = \frac{F}{mg}= \frac{99 N}{(50 kg)(9.81 m/s^2)} =0.20$

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2 years ago