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3 years ago

14

How far will you travel if you walk for 10.0 minutes at a speed of 1.50 m/s?

1 answer:

fredd [130]3 years ago

7 0

Answer:

15

Explanation:

distance = speed × time.

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calculate the period of a wave whose frequency is 5 Hertz and whose wavelength is one centimeter give your answer in a decimal f

olga2289 [7]

The period of the wave is the reciprocal of its frequency.

1 / (5 per second) = 0.2 second .

The wavelength is irrelevant to the period. But since you
gave it to us, we can also calculate the speed of the wave.

Wave speed = (frequency) x (wavelength)

= (5 per second) x (1cm) = 5 cm per second

4 0

3 years ago

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.3 c

Mashcka [7]

Answer with Explanation:

We are given that

Diameter of pipe, $d_1=0.573 m$

$v_1=13.5 m/s$

$v_2=5.83 m/s$

Volume flow rate of the petroleum along the pipe= $Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})$

$Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s$

By equation of continuity

$A_1v_1=A_2v_2$

$\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2$

$d^2_2=\frac{v_1}{v_2}d^2_1$

$d_2=\sqrt{\frac{v_1}{v_2}}d_1$

$d_2=0.573\sqrt{\frac{13.5}{5.83}}$

$d_2=0.87 m$

$d_2=0.87\time 100=87 cm$

1 m=100 cm

4 0

3 years ago

Why it is difficult to run fast in sand

balu736 [363]

Running on sand requires 1.6 times more energy spent than running on hard surface, so the force applied by our foot on sand is less.

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3 years ago

Read 2 more answers

A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar

m_a_m_a [10]

Answer:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

#Where $u$ is in meters and $t$ in seconds.

Explanation:

Given that : $m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s$

From $\omega=kL$ we have:

$k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2$

From $F_d(t)=-\gamma u\prime(t)$ we have that:

$\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m$

Now,given that the initial value problem is given by:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

Hence,the position of u at time t is given by

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$ , $u$ in meters, $t$ in seconds.

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3 years ago

In which of the following does personality psychology specialize?

ValentinkaMS [17]

B, Studying how ones' character changes over time... That's the only one that has to deal with personality.

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3 years ago

Read 2 more answers