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klemol [59]
3 years ago
12

I takes you 42 minutes to travel 32 miles to Ann Arbor. What was your speed getting there?

Physics
1 answer:
frez [133]3 years ago
3 0

Answer: 45.71 mi/h

Explanation:

Speed S is defined a:

S=\frac{d}{t}

Where:

d=32 miles is the traveled distance

t=42 min \frac{1 h}{60 min}=0.7 h is the time

Solving:

S=\frac{32 miles}{0.7 h}

S=45.71 mi/h This is the speed

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Write an expression for the potential energy UD of a particle when it is at a distance D from the force center, in terms of B an
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Answer:

E+mc+UD-3

Explanation:

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A student weighing 700 N climbs at constant speed to the top of an 8 m vertical rope in 10 s. The average power expended by the
Goshia [24]

To solve this problem we will use the concepts related to power, defined as the amount of energy applied over a period of time.

The energy in this case is the accumulated in the form of potential energy, over a period of time. Thus we will have that the mathematical expression of the power can be expressed as

P = \frac{E}{t}

Here,

E = Energy

t = time

As the energy is equal to the potential Energy we have tat

P = \frac{mgh}{t}

The weight (mg) of the man is 700N, the height (h) is 8m and the time is 10s, then:

P = \frac{700*8}{10}

P = 560W

Therefore the correct answer is A.

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3 years ago
Clois what is the weight of a body in the earth, if its weig is 5Nin moon? <br>​
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3 years ago
4. An airplane maintains a constant acceleration of 4.0 m/s2 [E] as it speeds up from 16 m/s [E] to 28 m/s [E].
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3 years ago
A 0.325 g wire is stretched between two points 57.7 cm apart. The tension in the wire is 650 N. Find the frequency of first harm
Galina-37 [17]

Answer:

f = 931.1 Hz

Explanation:

Given,

Mass of the wire, m = 0.325 g

Length of the stretch, L = 57.7 cm = 0.577 m

Tension in the wire, T = 650 N

Frequency for the first harmonic = ?

we know,

v =\sqrt{\dfrac{T}{\mu}}

μ is the mass per unit length

μ = 0.325 x 10⁻³/ 0.577

μ = 0.563 x 10⁻³ Kg/m

now,

v =\sqrt{\dfrac{650}{0.563\times 10^{-3}}}

   v = 1074.49 m/s

The wire is fixed at both ends. Nodes occur at fixed ends.

For First harmonic when there is a node at each end and the longest possible wavelength will have condition

          λ=2 L

          λ=2 x 0.577 = 1.154 m

we now,

       v = f λ

      f = \dfrac{v}{\lambda}

      f = \dfrac{1074.49}{1.154}

             f = 931.1 Hz

The frequency for first harmonic is equal to f = 931.1 Hz

7 0
3 years ago
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