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3 years ago

Language is most objective when based upon

Physics

2 answers:

Sergeeva-Olga [200]3 years ago

6 0

The answer is D. Evidence
Hope this helps!:)

amid [387]3 years ago

4 0

Answer: (D) Evidence

Explanation:

The language is objective when it is based upon the evidences as the objective language is the type of language which is accurate and fair. The objective language avoid all the kinds of bias and the exaggeration.

The language is more objective when it is based on the evidences as the real evidences justify the given facts true that makes the situation more efficient and true. The evidences support all the true proofs about the given situation and facts.

Therefore, Option (D) is correct.

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A gymnast spring vertical upward from a trampoline as in figure. The gymnast leaves the trampoline at height of 1.20m and reache

vaieri [72.5K]

Answer:

The initial velocity of the gymnast is 8.5 m/s.

Explanation:

We can use the kinematic equation

$v_f^2= v_o^2+2ad$

to figure out the initial velocity $v_o$ of the gymnast.

Now, when the gymnast reaches the maximum height, the distance he has traveled is $d = 4.8m- 1.2m = 3.6 m$ , and his velocity is zero; therefore $v_f =0$ .

Thus, we have

$0 = v_0^2+2(-10m/s^2)(3.6m)$

$v_0^2=72m^2/s^2$

$\boxed{v_0= 8.5m/s}$

4 0

3 years ago

A 30 kg box sits on the floor it requires 275N of force to get it moving once it is moving it only takes 225N of force.what are

gavmur [86]

1) The coefficient of static friction is 0.935

2) The coefficient of kinetic friction is 0.765

Explanation:

When a force is applied on a box sitting on the floor, the force that must be applied in order to make the box moving is equal to the maximum force of static friction between the floor and the box, which is:

$F = \mu_s mg$

where

$\mu_s$ is the coefficient of static friction

m is the mass of the box

g is the acceleration of gravity

Here we have:

m = 30 kg

$g=9.8 m/s^2$

F = 275 N

Therefore, the coefficient of static friction is

$\mu_s = \frac{F}{mg}=\frac{275}{(30)(9.8)}=0.935$

Once the box is in motion, the force that must be applied in order to make the box moving at constant velocity is equal to the force of kinetic friction between the floor and the box, which is:

$F = \mu_k mg$