A force of 1000 newtons was necessary to lift a rock. A total of 3000 joules of work was done. How far was the rock lifted?

what is the reaction of zinc nitrate mixed with silver. It's a lab and we are supposed to show what the difference, or any react

Vsevolod [243]

Answer:

b

Explanation:

3 0

3 years ago

Albert is on Earth and measures the speed of light from the sun as 3 E8 m/s. He then gets in a spaceship and speed toward the su

Flura [38]

Choice B, 3 E8 m/s because it will stay the same no matter where he is or how fast he's going

7 0

3 years ago

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A ball is thrown straight up from the ground with speed v0. At the same instant, a second ball is dropped from rest from a heigh

Kazeer [188]

Answer:

a) $t = H/v_0$

b) $H = v_0^2/g$

Explanation:

Let the first ball throw be the point of reference, we can have following the equation of motion:

1st ball: $h_1 = v_0t - gt^2/2$

2nd ball: $h_2 = H - gt^2/2$

a)When the 2 balls collide they are at the same spot at the same time:

$h_1 = h_2$

$v_0t - gt^2/2 = H - gt^2/2$

$v_0t = H$

$t = H/v_0$

b) The first ball is at its highest point when v = 0. That is

$t = v_0/g$

After this time, the 2 balls would have traveled through a distance of

$h_1 = v_0t - gt^2/2 = v_0^2/g - v_0^2/2g$

$h_2 = gt^2/2 = v_0^2/2g$

Since $H = h_1 + h_2$ we can solve for H

$H = v_0^2/g - v_0^2/2g + v_0^2/2g = v_0^2/g$

3 0

3 years ago

A block of mass 500g is pulled from rest on ahorizontal frictionless bench by a steady force F and travels 8m in 2s find the acc

Korvikt [17]

We are Given:

Mass of the block (m) = 500 grams or 0.5 Kg

Initial velocity of the block (u) = 0 m/s

Distance travelled by the block (s) = 8 m

Time taken to cover 8 m (t)= 2 seconds

Acceleration of the block (a) = a m/s²

Solving for the acceleration:

From the seconds equation of motion:

s = ut + 1/2* (at²)

replacing the variables

8 = (0)(2) + 1/2(a)(2)²

8 = 2a

a = 4 m/s²

Therefore, the acceleration of the block is 4 m/s²

3 0

3 years ago

A 1 mW laser beam is incident onto a detector. Determine the fractional fluctuation in number of photons intercepted by the dete

Ugo [173]

Answer:

(a) $625 photons$

(b) $625\times 10^{6}photons$

Explanation:

Given that , the power of the laser beam is,

$P=1 mW\\P=1\times 10^{-3} W$

and time is given is,

$t=1\mu s\\t=10^{-6}s$

Now the energy formula for the laser beam is,

$E=P\times t$

Now,

$E=10^{-3}\times 10^{-6} \\E=10^{-9}J$

(a) The value of energy is given,

$E_{1}=10MeV\\E_{1}=10\times 10^{6}\times 1.6\times 10^{-19}J\\ E_{1}=16\times 10^{-13}J$

Now the no of photons's fraction fluctuation is,

$n=\frac{E}{E_{1} }\\ n=\frac{10^{-9} }{16\times 10^{-13} }\\ E_{1}=625 photons$

Therefore the no of photons is $625 photons$ .

(b)The value of energy is given,

$E_{z}=10eV\\E_{z}=16\times 10^{-19}J$

Now the no of photons's fraction fluctuation is,

$n=\frac{E}{E_{z} }\\ n=\frac{10^{-9} }{16\times 10^{-19} }\\n=625\times 10^{6}photons$

Therefore the no of photons is $625\times 10^{6}photons$ .

5 0

3 years ago