Can anyone help me with this?

A wildebeest runs with an average speed of 4.0\,\dfrac{\text m}{\text s}4.0 s m 4, point, 0, start fraction, start text, m, en

devlian [24]

Answer:

60m

Explanation:

4 0

3 years ago

How does the time difference between the first p-wave and the first s-wave relate to the distance to the epicenter?

Lady bird [3.3K]

S-waves do, the seismograph will detect P-waves arriving first, and S-waves will follow. The time difference, as recorded on a clock, between when the P-waves and S-waves ... earthquake waves speed up with increasing distance, and the lag time graph

6 0

3 years ago

You've learned about simple circuits and resistors at this point. So, let's talk about how we can power different circuits. Supp

DIA [1.3K]

Answer:

Explanation:

Total resistance of 2 bulb in series

= 100 +100 = 200 ohm

current in each bulb

= 9/200 = .045 A

In case of parallel combination of bulb

total resistance

= 100 x 100 / (100+100)

50 ohm

current in the main circuit

= 9 / 50 A

current in each bulb

= 9 / (2 x 50)

.09

So in the second case current is more , hence heat generated too will be more . So brightness too will be more.

First set ( in series ) will be dimmer.

3 0

3 years ago

A motorcyclist drove 7 km at 57km/h and then another 7 km at 81 km/h. What was the average speed?

alex41 [277]

Answer:

The average speed of the car is 66.9 km/h

Explanation:

Here distance covered with the speed 57 km/h=7 km 

distance covered with the speed of 81 km/h=7 km

Average speed is equal to the ratio of total distance to the total time.

total distance= 7 + 7= 14 km

$time= \frac{distance}{speed}$

time taken to cover the first 7 km= 7/57 h

time taken to cover the second part of the journey = 7/81 h

average speed = $14/(7/57+7/81)=(14 \times 57 \times 81)/945=66.9 km/h$ 

Shortcut: 

When equal distances are covered with different speeds average speed=2 ab/(a+b) where a and b are the variable speeds in the phases. 

7 0

3 years ago

A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ

Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3$

(c) From any of the equations, solve for Length of the string (L);

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m$

(a) the fundamental frequency is calculated as;

$f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o = 65 \ Hz$

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

7 0

3 years ago