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3 years ago

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When you take your 1400 kg car out for a spin, you go around a corner of radius 55 m with a speed of 18 m/s. the coefficient of

static friction between the car and the road is 0.88. assuming the car doesn't skid, what is the force exerted on it by static friction?

1 answer:

Lorico [155]3 years ago

5 0

Are there ant answer choices

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Two spherical objects are separated by a distance of 2.59 × 10-3 m. The objects are initially electrically neutral and are very

kodGreya [7K]

Answer:

Number of electrons, n = 12 electron

Explanation:

Given that,

The distance between charged spheres, $d=2.59\times 10^{-3}\ m$

The object experiences an electrostatic force that has a magnitude of, $F=4.94\times 10^{-21}\ N$

The electric force between spheres is given by :

$F=\dfrac{kq^2}{d^2}$

$q=\sqrt{\dfrac{Fd^2}{k}}$

$q=\sqrt{\dfrac{4.94\times 10^{-21}\times (2.59\times 10^{-3})^2}{9\times 10^9}}$

$q=1.91\times 10^{-18}\ C$

Let there are n number of electrons. Using quantization of electric charge we get :

$q=ne$

$n=\dfrac{q}{e}$

$n=\dfrac{1.91\times 10^{-18}}{1.6\times 10^{-19}}$

n = 11.93 electrons

or

n = 12 electrons

Hence, 12 electrons produce the charge on one of the objects.

4 0

3 years ago

A mosquito of mass 0.15 mg is found to be flying at a speed of 50 cm/s with an uncertainty of 0.5 mm/s. (a) How precisely may it

amid [387]

Answer:

a) Δx = t 0.05 + 0.5

, Δx = 0.5 cm, b) Do not present any problem

Explanation:

The kinematic equation for constant speed is

v = x / t

x = v t

a) the uncertainty can be calculated with

Δx = dx /dv Δv + dx /dt Δt

Δx = t Δv + v Δt

Speed is

v = (50.00 ± 0.05) cm / s

The most common uncertainty for the time of Δt = 0.01 s

We replace

Δx = t 0.05 + 50 0.01

Δx = t 0.05 + 0.5

We must know the time to have an explicit value, if we assume that the measure was t = 1s

Δx = 0.5 cm

b)

Do not present any problem since its value is not very small, we must take as soon as the quantum effects and the velocity are not so high that we must take into account the relativistic effects

5 0

3 years ago

Which term describes the ability to process many things simultaneously?

stellarik [79]

Answer:

Parallel processing

Explanation:

It is a parallel processing because it involves the brain processing many part of problem simultaneously.

The brain natural mode of information involves performing many task .

It is also ability of the brain to simultaneously process incoming stimuli or signals of different quality. It is a part of brain's vision and divide what is seen into four major parts which are color, motion, shape, and depth.

4 0

3 years ago

Read 2 more answers

The age of the universe is thought to be about 14 billion years. Assuming two significant figures, write this in powers of ten i

ira [324]

A ) 14 billion years = 14,000,000,000
$1.4 * 10^{10}$
b ) 1 year = 365 · 24 · 3,600 = 31,536 · 10³
$1.4 * 10 ^{10}*3.1536 * 10^{7} =4.4 * 10 ^{17}$

6 0

4 years ago

Read 2 more answers

Dennis throws a volleyball up in the air. It reaches its maximum height 1.1\, \text s1.1s1, point, 1, start text, s, end text la

rewona [7]

Answer:

If max height = 1.1 meters, then initial velocity is 3.28 m/s

If max height is 1.1 feet, then the initial velocity is 5.93 ft/s

Explanation:

Recall the formulas for vertical motion under the acceleration of gravity;

for the vertical velocity of the object we have

$v=v_0-g \,t$

for the object's vertical displacement we have

$y-y_0=v_0\,t - \frac{g}{2} \,t^2$

If the maximum height reached by the object is given in meters, we use the value for g in $m/s^2$ which is: $9.8\,\,m/s^2$

If the maximum height of the object is given in feet, we use the value for g in $ft/s^2$ which is : $32\,\,ft/s^2$

Now, when the ball reaches its maximum height, the ball's velocity is zero, so that allows us to solve for the time (t) the process of reaching the max height takes:

$v=v_0-g \,t\\0=v_0-g \,t\\g\,\,t=v_0\\t=\frac{v_0}{g}$

and now we use this to express the maximum height in the second equation we typed:

$y-y_0=v_0\,t - \frac{g}{2} \,t^2\\max\,height=v_0\,(\frac{v_0}{g}) - \frac{g}{2} \,(\frac{v_0}{g})^2\\max\,height= \frac{v_0^2}{2\,g}$

Then if the max height is 1.1 meters, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,9.8}\\(9.8)\,(1.1)=v_0^2\\v_0=10.78\\v_0=\sqrt{10.78} \\v_0=3.28\,\,m/s$

If the max height is 1.1 feet, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,32}\\(32)\,(1.1)=v_0^2\\v_0=35.2\\v_0=\sqrt{35.2} \\v_0=5.93\,\,ft/s$

5 0

3 years ago

Read 2 more answers