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oee [108]
1 year ago
6

A body moving with an acceleration 2 m/s?then what is the change in velocity in 4sec.​

Physics
1 answer:
enot [183]1 year ago
7 0

Answer:

As Per Provided Information

Moving body has 2m/s² acceleration

Time taken by body is 4 second

We are asked to find the 'change in velocity' ( ∆V) by the body.

<u>Formula Used here</u>

\boxed{\bf{\Delta \: V \:  =  acceleration \:  \times time \:}}

<u>Substituting </u><u>the </u><u>given </u><u>value</u>

<u>\sf\longrightarrow\Delta\:V \:  = 2 \times 4 \\  \\  \\ \sf\longrightarrow\Delta\:V \:  =8m {s}^{ - 1}</u>

<u>Therefore</u><u>,</u>

  • <u>Change </u><u>in </u><u>velocity </u><u>is </u><u>8</u><u> </u><u>m/</u><u>s</u>
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If a transmission line in a cold climate collects ice, the increased diameter tends to cause vortex formation in a passing wind.
AleksAgata [21]

Answer:

a) f_1=5.587Hz

b) f_{n+1}-f_n=5.587Hz

Explanation:

The frequency of the n^{th} harmonic of a vibrating string of length <em>L, </em>linear density \mu under a tension <em>T</em> is given by the formula:

f_n=\frac{n}{2L} \sqrt{\frac{T}{\mu}

a) So for the <em>fundamental mode</em> (n=1) we have, substituting our values:

f_1=\frac{1}{2(347m)} \sqrt{\frac{65.4\times10^6N}{4.35kg/m}}=5.587Hz

b) The <em>frequency difference</em> between successive modes is the fundamental frequency, since:

f_{n+1}-f_n=\frac{n+1}{2L} \sqrt{\frac{T}{\mu}}-\frac{n}{2L} \sqrt{\frac{T}{\mu}}=(n+1-n)\frac{1}{2L} \sqrt{\frac{T}{\mu}}=\frac{n}{2L} \sqrt{\frac{T}{\mu}}=f_1=5.587Hz

3 0
2 years ago
What force is necessary to accelerate a 5.0 kg mass from rest to a final velocity of 10.0 m/s in 5.0 s?
vesna_86 [32]

Answer:

10 N

Explanation:

F = ma = m(Δv/t) = 5.0(10.0 - 0)/5.0 = 10 N

4 0
3 years ago
Rita jeptoo of kenya was the first female finisher in the 110th boston marathon. she ran the first 10.0 km in a time of 0.5689 h
stira [4]

Part a

Answer: 17.58 km/h

Average speed=\frac{Total\hspace{1mm}Distance}{Total\hspace{1mm}Time}

Total Distance =10 km

Total time =0.5689 h

\Rightarrow Average speed=\frac{10\hspace{1mm}km}{0.5689\hspace{1mm}h}=17.6 \hspace{1mm}km/h

Part b

Answer: 17.626 km/h

Average speed=\frac{Total\hspace{1mm}Distance}{Total\hspace{1mm}Time}

Total Distance =42.195 km

Total time =2.3939 h

\Rightarrow Average speed=\frac{42.195\hspace{1mm}km}{2.3939\hspace{1mm}h}=17.626\hspace{1mm}km/h

8 0
2 years ago
Consider a block on a spring oscillating on a frictionless surface. The amplitude of the oscillation is 11 cm, and the speed of
IRISSAK [1]

Answer:

The angular frequency of the block is ω = 5.64 rad/s

Explanation:

The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.

Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.

The angular frequency of the oscillation ω is

ω = v/r

ω = 62 cm/s ÷ 11 cm

ω = 5.64 rad/s

So, the angular frequency of the block is ω = 5.64 rad/s

6 0
3 years ago
The left side of the lever was forced down 10 inches in order to raise the rock 7 inches. The ideal mechanical advantage is
Yanka [14]

Answer:

1.42

Explanation:

<em> got it right on my homework </em>

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