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2 years ago

What force is needed to give a 4800.0 kg truck an acceleration of 6.2 m/s2 over a level road?

Physics

1 answer:

tekilochka [14]2 years ago

6 0

Answer:

the force needed to give the truck the acceleration is 29,760 N.

Explanation:

Given;

mass of truck, m = 4800 kg

acceleration of the truck, a = 6.2 m/s²

The force needed to give the truck the acceleration is calculated as;

F = ma

F = 4800 x 6.2

F = 29,760 N

Therefore, the force needed to give the truck the acceleration is 29,760 N.

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2 years ago

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Anon25 [30]

Answer:

D. Principle of original horizontality

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Explanation:

Question 1

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It states that "sedimentary rocks are laid down flatly on top one another in a sedimentary basin".

Sedimentary rocks will only vary vertically, but laterally, they are uniform and internally homogeneous in space. This is why most sedimentary rocks are stratified and laid layers upon layers just like the pages of a book.

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Question 2

Principle of faunal succession succession was proposed by Williams Smith, an English Geologist and a canal worker in the 19th century.

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This principle is hinged on theory of evolution.
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Question 3

The principle of uniformitarianism was one of the disruptive proposition in earth science.

A Scottish name James Hutton while in the country side made this proposition as he observed how landform in his native changed with each episode of season.

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Question 4

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3 years ago

If it takes 100 N to move a box 5 meters, what is the work done on the box?

jeyben [28]

Answer: D. 500 J

=========================================================

Explanation:

To find the amount of work done, we multiply the force by displacement

work = force*displacement

work = (100 N)*(5 m)

work = (100*5) Nm

work = 500 J

In this case, "Nm" refers to "Newton meters" and not "nanometers"

1 newton meter is equal to 1 joule

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2 years ago

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2 years ago

Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400

hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

$v^2=u^2+2as$

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

$v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s$

The launching speed of the beetle is 6.4 m/s.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

$v=u+at$

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

$v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms$

The time taken by the beetle to launch itself upwards is 1.62 ms.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

$v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m$

The beetle can jump to a height of 2.1 m

7 0

2 years ago

What force is needed to give a 4800.0 kg truck an acceleration of 6.2 m/s2 over a level road? ​

What force is needed to give a 4800.0 kg truck an acceleration of 6.2 m/s2 over a level road?