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2 years ago

When a pair of 10-n forces act on a box of candy, the net force on the box is?

Physics

2 answers:

Grace [21]2 years ago

8 0

Any of the above depending on the direction of forces

Gnesinka [82]2 years ago

7 0

It can be anything from zero to 20N. It depends on the directions of the forces.

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What is the most likely effect on the life of a plant if a student cuts it’s flowers?

goblinko [34]

Answer:

Explanation:

The reproduction parts of the plant are in the flower and around it so this would eliminate the plants ablity to reproduce.

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3 years ago

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The distance between two slits is 1.50 *10-5 m. A beam of coherent light of wavelength 600 nm illuminates these slits, and the d

Fed [463]

Answer: y = 2.4×10^-6m or y= 2.4μm

Explanation: The formulae for the distance between the central bright fringe to any other fringe in pattern is given as

y = R×mλ/d

Where y = distance between nth fringe and Central bright spot fringe.

m = position of fringe = 4

λ = wavelength of light= 600nm = 600×10^-9 m

d = distance between slits = 1.50×10^-5m

R = distance between slit and screen = 2m

y = 2 × 4 × 600×10^-9/2

y = 4800×10^-9/2

y = 2400 × 10^-9

y = 2.4×10^-6m or y= 2.4μm

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2 years ago

The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg astronaut inside the station pushes off one wall of the st

Aleonysh [2.5K]

Answer:

$a = 5.83 \times 10^{-4} m/s^2$

Explanation:

Since the system is in international space station

so here we can say that net force on the system is zero here

so Force by the astronaut on the space station = Force due to space station on boy

so here we know that

mass of boy = 70 kg

acceleration of boy = $1.50 m/s^2$

now we know that

$F = ma$

$F = 70(1.50) = 105 N$

now for the space station will be same as above force

$F = ma$

$105 = 1.8 \times 10^5 (a)$

$a = \frac{105}{1.8 \times 10^5}$

$a = 5.83 \times 10^{-4} m/s^2$

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3 years ago

Two froghoppers sitting on the ground aim at the same leaf, located 35 cm above the ground. Froghopper A jumps straight up while

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2 years ago

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Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and

strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance $R_e$ defined by the formula:

$\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}$

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

$\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\$

and when this is combined with the third resistor in series, the equivalent resistance ( $R''_e$ ) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

$R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}$

The problem states that the difference between the equivalent resistances in both circuits is given by:

$R''_e=R_e+630 \,\Omega$

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

$\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega$

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3 years ago