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2 years ago

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A small remote controlled car with mass 1.60 kg moves at a constant speed of12.0 m/s in a vertical circle with a radius of 5.0 m

.What is the magnitude ofthe normal force exerted on the car by the walls of the circle at pointA?

1 answer:

kenny6666 [7]2 years ago

5 0

Answer:

61.76 N.

Explanation:

Given the mass of the car, m = 1.60 kg.

The speed of the car, v = 12.0 m/s.

The radius of the circle, r = 5 m.

As car is moving in circular motion, so net force ( normal force + weight of the car) is equal to centripetal force enables the car to reamins in circular path.

Let N is the normal force.

So, $N - mg = F_c$

$N-mg=\frac{mv^2}{r}$

Now substitute the given values, we get

$N-1.60kg\times9.8m/s^2=\frac{1.60kg\times(12.0m/s)^2}{5.0m}$

$N=15.68+46.08$

N = 61.76 N.

Thus, the magnitude ofthe normal force exerted on the car by the walls is 61.76 N.

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Answer:

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Explanation:

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In Niels Bohr's 1913 model of the hydrogen atom, the single electron is in a circular orbit of radius 5.29×10⁻¹¹m and its speed

Svet_ta [14]

The magnitude of the magnetic moment due to the electron's motion is $87.87 * 10^{-37}$ .

<h3>What is magnetic moment?</h3>

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Calculations:

radius= $5.29 * 10^{-11} m\\$

velocity= $2.9* 10^{6} m/s$

Working formula, M=N/A

$I=\frac{charge flow }{time taken} =\frac{e}{time taken\\}$

$T= \frac{2xr}{v} =\frac{2xx * 5.29 * 10^{-11} }{2.9* 10^{6} }$

= $15.16 * 10^{-5} s$

$I= \frac{1.6 * 10^{-19} }{15.16 * 10^{-5} }$ $= 0.10 * 10^{-14}$

= $1 * 10^{-15} C$

M= $1x (1* 10^{-15} * (5.29 * 10^{-11} )^{2}$

= $87.87 * 10^{-37}$

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8 months ago

A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.

Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

1)

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

$y=(u sin \theta)t-\frac{1}{2}gt^2$ (1)

where:

$u sin \theta$ is the initial vertical velocity of the shell, with $u=156 ft/s$ and $\theta=49.0^{\circ}$

$g=32 ft/s^2$ is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

$x=(ucos \theta)t$

where $u cos \theta$ is the initial horizontal velocity of the shell.

We can re-write this last equation as

$t=\frac{x}{u cos \theta}$ (1b)

And substituting into (1),

$y=xtan\theta -\frac{1}{2}gt^2$ (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

$y=x tan \alpha$ (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$

Which has 2 solutions:

x = 0 (origin)

and

$tan \theta - tan \alpha=\frac{gx}{2u^2 cos^2 \theta}=0\\x=(tan \theta - tan \alpha) \frac{2u^2 cos^2\theta}{g}=1322 ft$

So, the distance d down the hill at which the shell strikes the hill is

$d=\frac{x}{cos \alpha}=\frac{1322}{cos(-41.0^{\circ})}=1752 ft=534 m$

2)

In order to find how long the mortar shell remain in the air, we can use the equation:

$t=\frac{x}{u cos \theta}$

where:

x = 1322 ft is the final position of the shell when it strikes the hill

$u=156 ft/s$ is the initial velocity of the shell

$\theta=49.0^{\circ}$ is the angle of projection of the shell

Substituting these values into the equation, we find the time of flight of the shell:

$t=\frac{1322}{(156)(cos 49^{\circ})}=12.9 s$

3)

In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.

The horizontal component of the velocity is constant and it is

$v_x = u cos \theta =(156)(cos 49^{\circ})=102.3 ft/s$

Instead, the vertical component of the velocity is given by

$v_y=usin \theta -gt$

And substituting at t = 12.9 s (time at which the shell strikes the hill),

$v_y=(156)(cos 49^{\circ})-(32)(12.9)=-310.4ft/s$

Therefore, the final speed of the shell is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(102.3)^2+(-310.4)^2}=326.8 ft/s=99.6 m/s$

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2 years ago

The time constant for RC circuit with the values of R1 and C1 is 5ms. What will be the time constant for a new RC circuit with t

lions [1.4K]

Answer:

d. 25 ms

Explanation:

In a RC circuit we call time constant to the product of the resistance times the capacitance, which represents the time when the charge reaches to the 63% of the final value, as follows:

$\tau_{1} = R_{1} *C_{1} = 5 ms (1)$

If we have a new circuit with new values for R and C, the time constant will be defined in the same way, as follows:

$\tau_{2} =10* R_{1} *0.5*C_{1} = 5*(R_{1}* C_{1}) = 5* \tau_{1} = 5* 5 ms = 25 ms (2)$

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2 years ago