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3 years ago

14

The photoelectric effect proved:

1 answer:

Jobisdone [24]3 years ago

6 0

Answer:

i think c)light has a charge

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A 5.0kg toolbox is raised from the ground by a rope. If the upward acceleration of the bucket is 2.5 m/s^2, find the force exert

swat32

Answer:

62 N

Explanation:

Sum of the forces on the toolbox:

∑F = ma

T − mg = ma

T = mg + ma

T = m (g + a)

T = (5.0 kg) (9.8 m/s² + 2.5 m/s²)

T = 61.5 N

Rounded to two significant figures, the force exerted by the rope is 62 N.

8 0

3 years ago

What is Newton first law?<br>

Anit [1.1K]

Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force

7 0

3 years ago

Two parallel plates are a distance apart with a potential difference between them. A point charge moves from the negatively char

IgorC [24]

Answer:

K' = 1200 J

Explanation:

To find the kinetic energy you first take into account the formula for the kinetic energy of the charge:

$K=\frac{1}{2}mv^2$ = 800J (1)

m: mass of the charge

v: final speed of the charge when it reaches the positively charged plate.

Furthermore, you have that the acceleration of the charge is obtained by using the second Newton law:

$F=ma=qE\\\\a=\frac{qE}{m}$ (2)

a: acceleration

E: electric field

q: charge

The electric field between two parallel plates is V/d, being V the potential difference and d the separation between plates. You replace E in (2) and obtain:

$a=\frac{qV}{md}$

Next, you take into account the following formula for the calculation of the final speed of the charge:

$v^2=v_o^2+2ad\\\\v_o=0m/s\\\\v=\sqrt{\frac{2qVd}{md}}=\sqrt{\frac{2qV}{m}}$

Next, you replace this value of v in (1):

$K=\frac{1}{2}mv^2=\frac{1}{2}m(\frac{2qV}{m})=qV$ = 880J (3)

If the distance between plates is tripled, and the potential difference is halved, you have for the new final speed:

$v'^2=v'_o^2+2a(3d)\\\\v_o=0m/s\\\\v'=\sqrt{6ad}=\sqrt{6(\frac{q}{md})\frac{V}{2}d}=\sqrt{\frac{3qV}{m}}$

And the kinetic energy becomes:

$K'=\frac{1}{2}mv^2=\frac{1}{2}m(\frac{3qV}{m})=\frac{3}{2}qV$ (4)

You calculate the ratio between both kinetic energies K and K', that is, you divide equations (3) and (4), in order to find the new kinetic energy:

$K=qV=800J\\\\K'=\frac{3}{2}qV\\\\\frac{K}{K'}=\frac{qV}{3/2\ qV}=\frac{2}{3}\\\\K'=\frac{3}{2}K=\frac{3}{2}(800J)=1200J$

hence, the kinetic energy of the charge incresases to 1200J

4 0

3 years ago

If a flea can jump straight up to a height of 0.410 m , what is its initial speed as it leaves the ground?

aivan3 [116]

Initial velocity = $v_0$

acceleration in the downward direction = -9.8 $\frac {m}{s^2}$

Final velocity at the highest point = 0

Maximum height reached = 0.410 m

Now, Using third equation of motion:

$\(v^2 = {v_0}^{2} + 2aH$

$\(0^2 = {v_0}^{2} - 2 \times 9.8 \times 0.410$

${v_0}^{2} = 2 \times 9.8 \times 0.410$

$v_0 = 2.834 \frac {m}{s}$

Speed with which the flea jumps = $2.834 \frac {m}{s}$

4 0

3 years ago

If a plane has an airspeed of 40 m/s and is experiencing a crosswind of 30 m/s, what is its ground speed in m/s?

Anon25 [30]

<h2>Answer:50 $ms^{-1}$ </h2>

Explanation:

Let $v_{a}$ be the airspeed.

Let $v_{w}$ be the cross wind speed.

We know that,ground speed is the vector sum of airspeed and cross wind speed and airspeed is perpendicular to cross wind speed.

If $v_{1}$ and $v_{2}$ are two perpendicular vectors,the resultant vector has the magnitude $\sqrt{|v|_{1}^{2}+|v|_{2}^{2}}$

Given,

$v_{a}=40ms^{-1}\\v_{c}=30ms^{-1}$

So,the ground speed is $\sqrt{40^{2}+30^{2}}=\sqrt{2500}=50ms^{-1}$

6 0

3 years ago