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3 years ago

8

the police department determined that the force required to drag a 130 N(29 lb) car tire across the pavement at a constant veloc

ity is called 100N (23 lb ) specifications from the trucks manufacturer claim that the effecrive coefficient of kinetic friction between the tires and road for both new car and truck??

1 answer:

ki77a [65]3 years ago

4 0

The force of friction is given by:
f = μR, where μ is the friction coefficient and R is the reaction force, which will be equal to the weight.
100 = μ x 130
μ = 0.77

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A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff

strojnjashka [21]

(a) $6.43\cdot 10^5 J$

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

$E=K+U$

The initial kinetic energy is:

$K=\frac{1}{2}mv^2$

where m = 58.0 kg is the mass of the projectile and $v=140 m/s$ is the initial speed. Substituting,

$K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J$

The initial potential energy is given by

$U=mgh$

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

$U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J$

So, the initial mechanical energy is

$E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J$

(b) $-1.67 \cdot 10^5 J$

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

$K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J$

while the potential energy is

$U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J$

So, the mechanical energy is

$E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J$

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

$W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J$

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

$W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J$

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

$E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J$

And this is only kinetic energy:

$E=K=\frac{1}{2}mv^2$

So, we can solve to find the final speed:

$v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s$

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3 years ago

Select the situation for which the torque is the smallest.

alina1380 [7]

Answer:

e. The torque is the same for all cases.

Explanation:

The formula for torque is:

τ = Fr

where,

τ = Torque

F = Force = Weight (in this case) = mg

r = perpendicular distance between force an axis of rotation

Therefore,

τ = mgr

a)

Here,

m = 200 kg

r = 2.5 m

Therefore,

τ = (200 kg)(9.8 m/s²)(2.5 m)

<u>τ = 4900 N.m</u>

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b)

Here,

m = 20 kg

r = 25 m

Therefore,

τ = (20 kg)(9.8 m/s²)(25 m)

<u>τ = 4900 N.m</u>

<u></u>

c)

Here,

m = 8 kg

r = 62.5 m

Therefore,

τ = (8 kg)(9.8 m/s²)(62.5 m)

<u>τ = 4900 N.m</u>

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Hence, the correct answer will be:

<u>e. The torque is the same for all cases.</u>

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4 years ago

Write a hypothesis for Part II of the lab, which is about the relationship described by F = ma. In the lab, you will use a toy c

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If you increase the mass m of the car, the force F will increase, while acceleration a is kept constant. Because F and m are directly proportional.
If you increase the acceleration a of the car, the force F will increase, while mass m is kept constant. Because F and a are directly proportional.

How can Newton's laws be verified experimentally; is by setting this experiment, and changing one variable while keeping the other constant, then observe the change in F.

Hope this helps.

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3 years ago

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Maru [420]

Maybe show a picture ? I don’t get the question .

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3 years ago

A parallel-plate capacitor has plates with an area of 451 cm2 and an air-filled gap between the plates that is 2.51 mm thick. Th

Nostrana [21]

To solve this problem we will apply the concepts related to Energy defined in the capacitors, as well as the capacitance and load. From these three definitions we will build the solution to the problem by defending the energy with the initial conditions, the energy under new conditions and finally the change in the work done to move from one point to the other.

Energy in a capacitor can be defined as

$E = \frac{1}{2}CV^2 = \frac{1}{2}\frac{Q^2}{C}$

Here,

V = Potential difference across the capacitor plates

Q = Charge stored on the capacitor plates

At the same time capacitance can be defined as,

$C = \epsilon_0 (\frac{A}{d})$

Here,

$\epsilon_0 =$ Vacuum permittivity constant

A = Area

d = Distance

Replacing with our values we have that,

$C = (8.85*10^{-12})(\frac{0.0451}{2.51*10^{-3}})$

$C = 1.5901*10^{-10}F$

PART A) Energy stored in the capacitor is

$E = \frac{1}{2} CV^2$

$E = \frac{1}{2} (1.5901*10^{-10})(575)^2$

$E = 2.628*10^{-5}J$

PART B) We know first that everything that the load can be defined as the product between voltage and capacitance, therefore

$Q = CV$

$Q = (1.59*10^{-10})(575)$

$Q = 9.1425*10^{-8}C$

Now if $d = 10.04*10^{-3}m$ we have that the capacitance is

$C = \epsilon_0 (\frac{A}{d})$

$C = (8.85*10^{-12})(\frac{0.0451}{10.04*10^{-3}})$

$C = 3.9754*10^{-11}F$

Then the energy stored is

$E = \frac{1}{2} \frac{Q^2}{C}$

$E = \frac{1}{2} (\frac{(9.1425*10^{-8})^2}{3.9754*10^{-11}})$

$E = 1.051*10^{-4} J$

PART C) The amount of work or energy required to carry out this process is the difference between the energies obtained, therefore

$W = 1.051*10^{-4} J -2.628*10^{-5}J$

$W = 7.882*10^{-5} J$

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3 years ago