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Eduardwww [97]
3 years ago
15

the time required for one cycle, a complete motion that returns to its starting point, it called the_____ medium frequency perio

d periodic motion
Physics
1 answer:
telo118 [61]3 years ago
4 0

Answer:

The correct answer to the following question will be "Period".

Explanation:

The Period seems to be the time deemed necessary for such a perfect cycle of vibration to transfer a particular moment. Because as the amplitude of the wave raises, the wavelength falls.

It is denoted by "T" and its formula will be:

⇒  T  = \frac{1}{F}

Where, T = Period

            F = Frequency

The other given choices are not related to the given circumstances. So that the above would be the right answer.

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A 75N force is applied to a 15kg box. If the box
icang [17]

Answer:

The coefficient  of kinetic friction is 0.51.

Explanation:

Given that,

Applied force, F = 75 N

Mass of a box, m = 15 kg

The acceleration of the box is 3 m/s²

We need to find the coefficient  of kinetic friction. It is given by the formula as follows :

F=\mu_k F_N

Where,

F_N=mg, is the normal force

\mu_k=\dfrac{F}{mg}\\\\\mu_k=\dfrac{75}{15\times 9.8}\\\\\mu_k=0.51

So, the coefficient  of kinetic friction is 0.51.

6 0
2 years ago
A car goes from point A to point B, five miles away and then returns to point A. The car is going 15 mph.
Sergeeva-Olga [200]

the velocity would be 0 mph

The answer is B.

3 0
3 years ago
How does increasing the distance between charged objects affect the electric force between them?
bazaltina [42]
The best option would be C.
8 0
3 years ago
You drive on Interstate 10 from San Antonio to Houston, half the time at 54 km/h and the other half at 118 km/h. On the way back
Irina-Kira [14]

Answer:

Average speed: 86 km/h

Explanation:

Driving from San Antonio to Houston:

1st. half time: 54km/h

2nd. half time: 118 km/h

Average speed = [tex] \frac{54 \frac{km}{h}+ 118 \frac{km}{h}  }{2}=86 \frac{km}{h} [\tex]

Driving way back:

1st. half time: 54km/h

2nd. half time: 118 km/h

Average speed = [tex] \frac{54 \frac{km}{h}+ 118 \frac{km}{h}  }{2}=86 \frac{km}{h} [\tex]

As in both routes we have the same average speed, then the average speed for the whole trip is 86 km/h

7 0
3 years ago
Read 2 more answers
An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F
ololo11 [35]

Answer

given,

mass of jogger  = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed  = 1.3 m/s  in  61 ° north of east.

jogger one

P_1 = m_1 v_1 \hat{i}

P_1 = 67 \times 2.3\hat{i}

P_1 = 154.1 \hat{i}

P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}

P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}

P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}

P_2 = 44.12\hat{i} +79.59\hat{j}

now

P = P₁ + P₂

P = 198.22 \hat{i} +79.59 \hat{j}

magnitude

P = \sqrt{198.22^2 + 79.59^2}

P =213.60 kg.m/s

\theta = tan^{-1}\dfrac{79.59}{198.22}

\theta = 21.87

the angle is \theta = 21.87 north of east

7 0
3 years ago
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