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3 years ago

What is the wavelength of an earthquake wave if it has a speed of 13 km/s and a frequency of 14 Hz?

2 answers:

8 0

Wavelength = velocity/frequency

wavelength = v/f
v= 13km/s = change this to m/s = 13000m/s
f= 14Hz

wavelength = 13000m/s÷14Hz =928.7 m

mojhsa [17]3 years ago

6 0

Hey!

Wavelength= speed/frequency

= 13/14

= 0.92 m

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A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a

Zielflug [23.3K]

Answer: $F_{v} =\mu_{k} mg$

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to start crate moving is $F_{s} =\mu_{s} mg$ .

but once crate starts moving the force of friction is reduced $F_{v} =\mu_{k} mg$ .

Hence to keep crate moving at constant velocity we have to reduce the force pushing crate ie $F_{v} =\mu_{k} mg$ .

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as forces are balanced.

Magnitude of the force

$F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9 N$

4 0

3 years ago

One of your delivery trucks traveled 1,200 miles on 55 gallons of gas. How many miles per gallon did the truck get? (Round off y

Aleksandr-060686 [28]

The word "Per" means divide

"miles per gallon" is the same as "miles / gallon"

The truck went 1,200 miles
on 55 gallons

1,200 ÷ 55 = 21.81

7 0

3 years ago

Read 2 more answers

The ratio of the speed of light in a medium to the speed of light in a vacuum

Liula [17]

Answer:D.Refractive Indez

Explanation:

It is usually expressed the other way: the ratio of the speed of light in a vacuum to the speed of light in a medium. In that case, it is called the "index of refraction".

7 0

3 years ago

Read 2 more answers

A force acts on a 9.90 kg mobile object that moves from an initial position of to a final position of in 5.40 s. Find (a) the wo

horrorfan [7]

Given that,

Mass of object = 9.90 kg

Time =5.40 s

Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.

We need to calculate the displacement

Using formula of displacement

$s=r_{2}-r_{1}$

Where, $r_{1}$ = initial position

$r_{2}$ = final position

Put the value into the formula

$s= (-4.10i + 3.30j + 5.40k)-(2.70i - 2.90j + 5.50k)$

$s= -6.80i+6.20j-0.1k$

(a). We need to calculate the work done on the object

Using formula of work done

$W=F\cdot s$

Put the value into the formula

$W=(2.00i + 9.00j + 5.30k)\cdot (-6.80i+6.20j-0.1k)$

$W=-13.6+55.8-0.53$

$W=41.67\ J$

(b). We need to calculate the average power due to the force during that interval

Using formula of power

$P=\dfrac{W}{t}$

Where, P = power

W = work

t = time

Put the value into the formula

$P=\dfrac{41.67}{5.40}$

$P=7.71\ Watt$

(c). We need to calculate the angle between vectors

Using formula of angle

$\theta=\cos^{-1}(\dfrac{r_{1}r_{2}}{|r_{1}||r_{2}|})$

Put the value into the formula

$\theta=\cos^{-1}\dfrac{(-4.10i + 3.30j + 5.40k)\cdot(2.70i - 2.90j + 5.50k)}{7.54\times6.778})$

$\theta=79.7^{\circ}$

Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.

(b). The average power due to the force during that interval is 7.71 Watt.

(c). The angle between vectors is 79.7°

7 0

4 years ago

In a 400-m relay race the anchorman (the person who runs the last 100 m) for team A can run 100 m in 9.8 s. His rival, the ancho

melisa1 [442]

Answer:

largest lead = 3 m

Explanation:

Basically, this problem is about what is the largest possible distance anchorman for team B can have over the anchorman for team A when the final leg started that anchorman for team A won the race. This show that anchorman for team A must have higher velocity than anchorman for team B to won the race as at the starting of final leg team B runner leads the team A runner.

So, first we need to calculate the velocities of both the anchorman

given data:

Distance = d = 100 m

Time arrival for A = 9.8 s

Time arrival for B = 10.1 s

Velocity of anchorman A = D / Time arrival for A

=100/ 9.8 = 10.2 m/s

Velocity of anchorman B = D / Time arrival for B

=100/10.1 = 9.9 m/s

As speed of anchorman A is greater than anchorman B. So, anchorman A complete the race first than anchorman B. So, anchorman B covered lower distance than anchorman A. So to calculate the covered distance during time 9.8 s for B runner, we use

d = vt

= 9.9 x 9.8 = 97 m

So, during the same time interval, anchorman A covered 100 m distance which is greater than anchorman B distance which is 97 m.

largest lead = 100 - 97 = 3 m

So if his lead no more than 3 m anchorman A win the race.

5 0

3 years ago