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4 years ago

List two general reasons why scientists are likely to collaborate

Physics

1 answer:

Sergeu [11.5K]4 years ago

3 0

Two scientists may try to collaborate under Few circumstances :-

Every scientist have their II level of knowledge and creativity on combining two it might result in a better outcome Mistakes of one can be analysed and covered by the other thereby increasing the chances of getting good results.

A rather greatest spectrum of knowledge would be present as the knowledge of two scientists would always be more than one Which is also true in the case of expertise in more than one subject can be there.

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,how do charged objects react???quiet urgent

son4ous [18]

Any charged object can<span> exert the force upon other objects ... i think tell me if im right</span>

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3 years ago

Read 2 more answers

The Sun exerts a gravitational force of 29.9 N on a rock that's located in a river bed here on Earth. How much gravitational for

11Alexandr11 [23.1K]

Answer:

F = 29.9 N

Explanation:

It is given that, The Sun exerts a gravitational force of 29.9 N on a rock that's located in a river bed here on Earth. We need to find the gravitational force the rock exert on the Sun. It is a case of Newton's third law of motion which states that the force acting from one object to another is equal to the force acing from second object to the first object and the two forces must be in opposite direction. Hence, the gravitational force the rock exert on the Sun is same i.e. 29.9 N.

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4 years ago

Prove law of conservation of energy for a stone moving vertically down ( explain energy at B & C )

STALIN [3.7K]

As per given condition of point B we can see that height at point B is "h/2" from the ground

So we know that potential energy is given as

U = mgh

so here we have to put height h = h/2

so potential energy is U = mgh/2

now for kinetic energy we need to find the speed of it after falling the distance h/2

now by kinematics we will have

$v_f^2 - 0^2 = 2(g)(h/2)$

now for kinetic energy

$KE = \frac{1}{2}mv^2 = \frac{1}{2}m(gh)$

$KE = 1/2mgh$

now total energy will be given as

$E = mgh/2 + mgh/2 = mgh$

now for point C we can say that it is the point near to ground

So here height is ZERO

now potential energy will also be zero

U = 0

now for kinetic energy we need to find speed

$v^2 - 0^2 = 2(g)(h)$

now kinetic energy

$KE = \frac{1}{2}mv^2 = \frac{1}{2}m(2gh)$

$KE = mgh$

now again we have total energy

$E = 0 + mgh = mgh$

6 0

4 years ago

One end of a horizontal spring with force constant 76.0 N/m is attached to a vertical post. A 5.00-kg can of beans is attached t

Jobisdone [24]

Answer:

a. The speed is 2.39 m/s

b. The acceleration of the block is 10.2 $\frac{m}{s^{2} }$

Explanation:

First, we have to do the energy balance where we consider two states, the first where the spring remains still and the second when it is stretched 0.400m:

$K_{1} +U_{1}+W_{ext}=K_{2}+U_{2}\\K_{1}=0\\U_{1}=\frac{1}{2} kx^{2} _{1} =0\\W_{ext}=F$ Δx= $(0.400m)\\$

W_{ext}=20.4 Nm

$U_{2} =\frac{1}{2} kx^{2} =\frac{1}{2} (76.0N/m)0.400^{2}=6.08Nm\\k_{2} =\frac{1}{2}mv^{2} _{2} \\\frac{1}{2} mv^{2} _{2}=W_{ext}-U_{2}\\v_{2}=\sqrt{\frac{W_{ext}-U_{2}}{m} } \\v_{2}=\sqrt{\frac{20.4Nm-6.08Nm}{2.5kg} } \\v_{2}=2.39 \frac{m}{s}$

To determine, the acceleration we solve the following equation for a:

$F=ma\\a=\frac{F}{m} =\frac{51.0N}{5.00kg}\\a=10.2\frac{m}{s^{2} }$

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4 years ago

If you tried to use a diverging lens as a magnifier by placing the object at the focal point, what type of image would be produc

gladu [14]

Answer:

Upright and smaller than the object

Explanation:

Diverging lens as the name suggests that the rays diverge after the refraction and do not meet in reality. A concave lens is called diverging lens. When there is refraction of light through a concave lens then the light bends away from the principal axis and hence never meet in reality but on tracing the rays backwards the rays appear to meet leading to the formation of a virtual image, which is erect and smaller than the object for an object placed at the focus of the lens.

Similar image is formed for any case when the object is between optical center and infinity.

7 0

4 years ago