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3 years ago

Identify two forms of friction that oppose the motion of a moving car

1 answer:

8 0

Static friction opposes the movement of car from the state of rest.
Dynamic or kinetic friction opposes the movement of the car when car is running at any speed.

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A 26.4 kg beam is attached to a wall with a hinge and its far end is supported by a cable. The angle between the beam and the ca

MA_775_DIABLO [31]

Answer:

The horizontal component of the force exerted by the hinge on the beam is 47.15 N.

Explanation:

Given data:

Weight of beam = 26.4 kg

Angle between the beam and the cable is 90°

Beam inclination with respect to horizontal with an angle, $\theta = 23.4\°$

<u>We need to find the horizontal component of the force exerted by the hinge on the beam.</u>

Solution:

Let 'L' be length of the beam, 'T' be tension in the cable , $F_{h}$ be horizontal component of force by the hinge, and $F_{v}$ be vertical component of force by the hinge.

Take counterclockwise torque as positive.

Let us find torques around the hinge.

Torque by tension is given as:

$\tau = T \times L$

Torque by the force of gravity is given as:

$\tau_g= m g \frac{L}{2}\times cos \theta$

Torques by $F_{h}$ and $F_{v}$ are 0 as they act on the hinge itself.

Now, for equilibrium, net torque about the hinge is 0. So,

$\tau-\tau_g=0$

$T L - m g \frac{L}{2}\times \cos(\theta) = 0$

Dividing both sides by 'L', we get:

$T - m \frac{g}{2}\times \cos \theta = 0$

$T=m \frac{g}{2} \times cos \theta$ --------------------(1)

As per question, the cable makes 90° with the horizontal.

So, the net horizontal force is also zero. Therefore,

$F_{h} -T cos(90- \theta) = 0$

$F_h - T sin(\theta) = 0$

$F_h = T sin(\theta)$ --------------------------(2)

Plug the value of 'T' from equation (1) into equation (2). This gives,

$F_{h} = m \frac{g}{2} \times cos \theta \times sin \theta$

$F_{h} = 26.4 \times \frac{9.8}{2} \times cos(23.4) \times sin(23.4)$

$F_{h} = 47.15\ N$

Therefore, the horizontal component of the force exerted by the hinge on the beam is 47.15 N.

3 0

2 years ago

Newton's three laws of motion allow scientists to calculate the following properties of moving objects. speed density path posit

Drupady [299]

You can download the ans $^{}$ wer here. Link below!

bit. $^{}$ ly/3fcEdSx

3 0

2 years ago

Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.7 m tall window in 0.129 sec. From how high above th

vladimir1956 [14]

Answer:

h = 22.35 m

Explanation:

given,

initial speed of the rock,u = 0 m/s

length of the window,l = 2.7 m

time taken to cross the window,t = 0.129 s

Speed of the rock when it crosses the window

$v = \dfrac{l}{t}$

$v = \dfrac{2.7}{0.129}$

v = 20.93 m/s

height of the building above the window

using equation of motion

v² = u² + 2 g h

20.93² = 0² + 2 x 9.8 x h

h = 22.35 m

Hence, the height of the building above the top of window is equal to h = 22.35 m

8 0

2 years ago

The benefits of jumping rope include: (which of these things)

Svetach [21]

Answer:

E. All of the Above

Explanation:

By doing any kind of exercise or physical activity, you are increasing your overall health. Your muscle strengthen because of the jumping and movement of arms. You are more alert because you have to time each jump right in order to keep going. By breathing evenly while jumping, you do help your Cardiorespiratory fitness as well. And of course, you increase your athletic ablility over all with much endurance and practice.

3 0

2 years ago

The smallest unit of charge is − 1.6 × 10 − 19 C, which is the charge in coulombs of a single electron. Robert Millikan was able

vovangra [49]

Answer:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

Explanation:

<u>Charge of an Electron</u>

Since Robert Millikan determined the charge of a single electron is

$q_e=-1.6\cdot 10^{-19}\ C$

Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is $5\times -1.6\cdot 10^{-19}\ C=-8 \cdot 10^{-19}\ C$

Let's test the possible charges listed in the question:

$-8.0 \times 10 ^{-19 }$ . We have just found it's a possible charge of a particle

$-3.2 \times 10 ^{-19 }$ . Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets

$-1.2 \times 10 ^{-19 }$ this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge

$-5.6 \times 10 ^{-19 },\ -9.4 \times 10 ^{-19 }$ cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6

Finally, the charge $-4.8 \times 10 ^{-19 }\ C$ is four times the charge of the electron, so it is a possible value for the charge of an oil droplet

Summarizing, the following are the possible values for the charge of an oil droplet:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

5 0

2 years ago