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4 years ago

If gravity puts a force of 3N an object, how much force would it put on an second object that was twice as heavy

Physics

2 answers:

MrMuchimi4 years ago

8 0

I am thinking the answer might be 6N considering the second object would be twice as heavy.

Harman [31]4 years ago

8 0

What we call "heaviness" is almost always the object's weight, and
that weight is exactly the force of gravity pulling the object and the Earth
together.

So an object that's twice as heavy as 3 Newtons is an object whose weight
is twice as much as 3 N = 6 newtons .

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Diana observed that the plants in her garden were not growing well due to poor soil conditions. She tested the soil and used the

n200080 [17]

The appropriate answer is c. silty clay loam. This is the most likely soil that was present in the garden before sand was added to balance it. This type of soil contains an even mix of silt and clay. This type of soil does not drain well and tends to hold water. This would not be suitable for most garden variety plants. Adding sand to the soil ensures better drainage and removes moisture that would rot roots or create conditions for fungi to develop.

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3 years ago

A horizontal force of 100 N is used to pull a crate, which weighs 500 N, at constant velocity across a horizontal floor. What is

Ilya [14]

Answer:

0.2

Explanation:

Horizontal force=100N

Weight of crate=500 N

We have to find the coefficient of kinetic friction.

Normal ,N=Weight=500N

Horizontal force, $F_x=\mu_kN$

Where $F_x$ =Horizontal force

N=Normal force

$\mu_k$ =Coefficient of kinetic friction

Substitute the values in the formula

$100=\mu_k(500)$

$\mu_k=\frac{100}{500}=0.2$

Hence, the coefficient of kinetic friction =0.2

8 0

3 years ago

Derive the following equations of motion

xz_007 [3.2K]

Answer:

___________________________________

<h3>a. Let us assume a body has initial velocity 'u' and it is subjected to a uniform acceleration 'a' so that the final velocity 'v' after a time interval 't'. Now, By the definition of acceleration, we have:</h3>

$a = \frac{v - u}{t} \\ or \: at = v - u \\ v = u + at \:$

It is first equation of motion.

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<h3>b. Let us assume a body moving with an initial velocity 'u'. Let it's final body 'v' after a time interval 't' and the distance travelled by the body becomes 's' then we already have,</h3>

$v = u + at...........(i) \\ s = \frac{u + v}{2} \times t.........(ii)$

Putting the value of v from the equation (i) in equation (ii), we have,

$s= \frac{u + (u + at)}{2} \times t \: \: \\ or \: s = \frac{(2u + at)t}{2} \\ or \: s = \frac{2ut + a {t}^{2} }{2} \\ s = ut + \frac{1}{2} a {t}^{2}$

It is third equation of motion.

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<h3>c. Let us assume a body moving with an initial velocity 'u'. Let it's final velocity be 'v' after a time and the distance travelled by the body be 's'. We already have,</h3>

$v = u + at.....(i) \\ s = \frac{u + v}{2} \times t......(ii) \\$

$v = u + at \\ or \: at = v - u \\ t = \frac{v - u}{a}$

Putting the value of t from (i) in the equation (ii)

$s = \frac{u + v}{2} \times \frac{v - u}{a} \\ or \: s = \frac{ {v}^{2} - {u}^{2} }{2a} \\ or \: 2as = {v}^{2} - {u}^{2} \\ {v}^{2} = {u}^{2} + 2as$

It is forth equation of motion.

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Hope this helps...

Good luck on your assignment..

3 0

3 years ago

Serena drops a 5-kg book down the stairwell. What is the book's final velocity vf after it falls

elena55 [62]

Answer:

39.2m/s

Explanation:

Given parameters:

Mass of book = 5kg

Time taken for fall = 4s

Unknown:

Final velocity of the book = ?

Solution:

Serena dropped the book from rest therefore, the initial velocity of the book is 0.

Let us find the appropriate motion equation to solve this problem;

V = U + gt

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity = 9.8m/s²

t is the time taken

Insert the given parameters and solve;

V = 0 + 9.8 x 4 = 39.2m/s

3 0

3 years ago

What is the wavelength in nm of a light whose first order bright band forms a diffraction angle of 30 degrees, and the diffracti

Artist 52 [7]

Answer:

The wavelength is 3500 nm.

Explanation:

d= $\frac{1}{700 lines per mm} = 0.007mm = 7000 nm$

n= 1

θ= 30°

λ= unknown

Solution:

d sinθ = nλ

λ = $\frac{7000 nm sin 30}{1}$

λ = 3500 nm

8 0

4 years ago