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MaRussiya [10]
3 years ago
9

The slowest animal ever discovered was a crab found in the Red Sea. It traveled with an average speed of 5.70km/yr. how long wou

ld it take this crab to travel 100km?
Physics
2 answers:
Vaselesa [24]3 years ago
8 0
If it takes 1 year for this crab to travel 5,70km

Then, it will takes approximatly 1000/5,70 = 175 years to travel 1000 km 
svlad2 [7]3 years ago
7 0

Answer:

Time taken by the crab, t = 17.54 years.

Explanation:

It is given that,

Speed of a crab, v = 5.70 km/yr

Distance travelled by the crab, d = 100 km

We need to find the time taken by the crab to cover this much distance. Time taken is given by :

t=\dfrac{d}{v}

t=\dfrac{100\ km}{5.70\ km/yr}

t = 17.54 years

So, the time taken by the crab is 17.54 years. Hence, this is the required solution.

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krok68 [10]

I have found this tool to hep you solve your question.

https://quizlet.com/76366884/ch15ch16physical-science-waves-sound-flash-cards/

8 0
3 years ago
An object with a mass of m = 122 kg is suspended by a rope from the end of a uniform boom with a mass of M = 74.9 kg and a lengt
ehidna [41]

The the tension in the horizontal rope is 7,019.4 N.

<h3>What is tension?</h3>

Tension is described as the pulling force transmitted axially by the means of a string, a cable, chain, or similar object, or by each end of a rod.

Tension can also be described as the action-reaction pair of forces acting at each end of said elements.

The unit of tension is Newton (same unit as force, since tension is also a force).

<h3>Tension in the horizontal rope</h3>

The tension in the horizontal rope is calculated as follows;

Apply the principle of torque;

T(L/2) cosθ = Mg (L/2) sinθ + mg L sinθ

T = (M + 2m)g sinθ/cosθ

T =  (M + 2m)g tanθ

<em>let θ = 66⁰ (this value should be given in the question)</em>

<em />

T =  (M + 2m)g tanθ

T = (74.9    +   2 x 122)(9.8)(tan 66)

T = 7,019.4 N

Thus, the the tension in the horizontal rope is 7,019.4 N.

Learn more about tension here: brainly.com/question/918617

#SPJ1

3 0
2 years ago
A football is kicked into the air from an initial height of 4 feet. The height, in feet, of the football above the ground is giv
kakasveta [241]

Answer: 0.5 seconds or 2.625 seconds

Explanation:

At t = 0, The ball is 4 ft above the ground.

The height of the football varies with time in the following way:

s(t) = -16 t² + 50 t + 4

we need to find the time in which the height would of the football would be 25 ft:

⇒25 = -16 t² + 50 t + 4

we need to solve the quadratic equation:

⇒ 16 t² - 50 t + 21 = 0

t = \frac{50 \pm \sqrt{50^2-4\times 16\times 21}}{2\times16}

⇒ t = 0.5 s or 2.625 s

Therefore, at t = 0.5 s or 2.625 s, the football would be 25 ft above the ground.

3 0
3 years ago
Read 2 more answers
A rifle is aimed horizontally at a target 50.0 m away. The bullet hits the target 2.90 cm below the aim point. . . Whats the bul
Marina CMI [18]
The problem ask to calculate the bullet's flight time and the bullet's speed as it left the barrel. So base on the problem, the answer would be that the flight time is 0.076 seconds and the speed of the bullet is 657.9 m/s. I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications. 
5 0
3 years ago
A 740-kg boulder is raised from a quarry 119 m deep by a long uniform chain having a mass of 550 kg . This chain is of uniform s
Naddika [18.5K]

Answer:

A) the maximum acceleration the boulder can have and still get out of the quarry

B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

A)

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is:   w_{c} =m_{c} g   and maximum tension is T=2.50 m_{c} g=1.41*10^4N

total mass and weight is :

M =m_{c}+ m_{b} =740kg+550kg=1290 kg

w_{M} =1.2650*10^4N

∑F_{y} =ma_{y}

T-M_{g} =Ma_{y}

a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg

=0.645m/s^2

B)

maximum acceleration

a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0

using y-y_{0} =v_{oy} t+1/2(a_{y} )t^2

to solve for t

t=\sqrt{2(y-y_{0} )/a_{y} }

t=\sqrt{2(119m)/0.645m/s^2} =19.20s

6 0
3 years ago
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