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2 years ago

Describe two examples of activities that you do where you depend on friction, and would not be able to do them without friction.

Physics

1 answer:

kaheart [24]2 years ago

7 0

Answer: One is when a bike stops. The other is when you rub your hands together to get warm.

Explanation: Force and friction affect our daily lives in numerous amounts of ways. For instance, when a football is kicked, it moves faster later after some time its force decreases due to friction. A common example of friciton is when a bike stops. When the brakes are applied the friction on the pads cause the bike to stop. The other is because you are getting warm with friction.

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1) A rock thrown horizontally from the top of a

9966 [12]

Dx = 20m
V1 = 10m/s
g = 9.8m/s^2
(delta-t) = 2sec
dy = 19.6m

3 0

2 years ago

A 2.0 kg pendulum has an initial total energy of 20 J. Calculate the energy lost as heat if the pendulum is 0.10 m high and is t

Maurinko [17]

The correct answer is (A) 2.0 J

Total energy of the pendulum is the sum of its kinetic and potential energy. At the instant of time, when the pendulum is at a height h and has a speed v, Its energy is given by,

$E=mgh+\frac{1}{2} mv^2$

Substitute 2.0 kg for m, the mass of the pendulum, 9.81 m/s² for g, the acceleration due to gravity, 0.10 m for h and 4.0 m/s for v.

$E=mgh+\frac{1}{2} mv^2\\ =(2.0kg)(9.81m/s^2)(0.10 m)+\frac{1}{2}(2.0kg)(4.0m/s)^2\\ =17.962J$

The pendulum has an initial energy of 20 J. the energy lost is given by,

$\Delta E=(20J)-(17.962J)\\ =2.038J=2.0J$

Thus, the energy lost by the pendulum is (A) 2.0 J

4 0

2 years ago

Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity i

leonid [27]

Answer:

Explanation:

To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.

For the wavelength;

Recall that, the separation between two successive max. and min. intensity points are $\dfrac{\lambda}{2}$

Thus; for both speakers; the wavelength of the sound is:

$\dfrac{\lambda}{2} = (10+30) cm$

$\dfrac{\lambda}{2} = (40) cm$

λ = 80 cm

The relation between the path difference(Δx) and the phase difference(Δ∅) is:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

where;

Δx = 10 cm

λ = 80 cm

Δ∅ = π rad

∴

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{80 \ cm}(10 \ cm) + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{8}+ \Delta \phi_o$

$\pi \ rad = \dfrac{ \pi}{4}+ \Delta \phi_o$

$\Delta \phi_o = \pi -\dfrac{ \pi}{4}$

$\Delta \phi_o = \dfrac{ 4\pi - \pi}{4}$

$\Delta \phi_o = \dfrac{ 3\pi}{4} \ rad$

Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm

Thus, we have:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\Delta \phi = \dfrac{2 \pi}{\lambda}(0 \ cm ) + \dfrac{3 \pi}{4} \ rad$

$\Delta \phi = \dfrac{3 \pi}{4} \ rad$

∴

The amplitude of the sound wave if the two speakers are placed side-by-side is:

$A = 2a \ cos \bigg (\dfrac{\Delta \phi }{2} \bigg)$

$A = 2a \ cos \bigg (\dfrac{\dfrac{3 \pi}{4} }{2} \bigg)$

$A = 2a \ cos \bigg ({\dfrac{3 \pi}{8} } \bigg)$

A = 0.765a

7 0

2 years ago

Can anyone tell me the ans of this last question quick please

tia_tia [17]

Explanation:

Speed: distance/time

Average speed: total distance/total time

Total distance: 400

Total time: 60

Average spead: 400/60= 6.67m/s

8 0

2 years ago

The heating of groundwater forms

KIM [24]

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6 0

3 years ago

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