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leonid [27]
3 years ago
7

The gold has a density of 19300 kg/m3 calculate the mass of one gold bar 1= 2.54cm

Physics
1 answer:
icang [17]3 years ago
5 0
Fair enough, but you'll have to tell us the volume of the bar first.
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Tiles are 6 mm long how many can you fit into a space 54 cm long
vlabodo [156]
It fits 90 times because 54÷6= 9 so if 54÷0.6 it will equal 90

4 0
3 years ago
What is generally TRUE about diagnosing psychological disorders?
Allisa [31]

The statement that says "Psychological disorders can be very difficult to diagnose" is true about diagnosing psychological disorders.

<h2>What are psychological disorders?</h2>

Psychological disorders are those mental, behavioral, emotional and thinking conditions that interfere with the normal performance of the individual in society.

  • Mental disorders are psychiatric conditions that are expressed in a syndrome, verifiable from different diagnostic criteria.

  • The steps to obtain a diagnosis include a medical history, physical examination, and possibly laboratory tests and a psychological evaluation.

Therefore, we can conclude that a psychological disorder is an alteration in the mental balance of a person that requires specialized attention adapted to the characteristics of the dysfunction.

Learn more about psychological disorders here: brainly.com/question/6367767

3 0
3 years ago
If 1 foot is 30.28 centimeters.how many cm is 130 feet
uranmaximum [27]

Answer:3,936.4

Explanation:

If 1 foot is 30.28 centimeters multiply 30.28 by 130

6 0
3 years ago
A projectile is launched from ground level at angle u and speed v0 into a headwind that causes a constant horizontal acceleratio
ale4655 [162]

Answer:

Explanation:

Given

Launch angle =u

Initial Speed is v_0

Horizontal acceleration is a_x=a

At maximum height velocity is zero therefore

v_f=v_i-gt

0=v_0\sin u-gt

t=\frac{v_0\sin u}{g}

Total time of flight T=2t=\frac{2v_0\sin u}{g}

During this time horizontal range is

R=v_o\cos u\cdot 2t-\frac{a(2t)^2}{2}

R=\frac{2v_0^2\sin u\cos u}{g}-\frac{2av_0^2\sin ^u}{g^2}

For maximum range \frac{\mathrm{d} R}{\mathrm{d} u}=0

\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2\cos 2u}{g}-\frac{4av_0^2\sin u\cos u}{g^2}

\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2}{g}\left [ \cos 2u-\frac{a}{g}\sin 2u\right ]=0

\tan 2u=\frac{g}{a}

u=\frac{1}{2}tan ^{-1}\frac{g}{a}

(b)If a =10% g

a=0.1g

thus u=\frac{1}{2}tan^{-1}\frac{g}{0.1g}

u=42.14^{\circ}

7 0
3 years ago
Is this object showing acceleration for the first 2 seconds? explain your answer.
iVinArrow [24]
Yes. The line is increasing. The flat line at the top of the graph is where there is not acceleration and the decreasing line is deceleration.
3 0
3 years ago
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