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jenyasd209 [6]
3 years ago
7

Two ways that weak nuclear forces and electromagnetic forces are simaliar

Physics
1 answer:
STALIN [3.7K]3 years ago
6 0
Have very short ranges
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A dock worker loading crates on a ship finds that a 27 kg crate, initially at rest on a horizontal surface, requires a 80 N hori
Aleksandr-060686 [28]

Answer:0.302

Explanation:

Given

mass of crate m=27 kg

Force required to set crate in motion is 80 N

Once the crate is set  in motion 56 N is require to move it with constant velocity

i.e. 80 N is the amount of force needed to just overcome static friction and 56 is the kinetic friction force

thus  

f_s(static\ friction)=\mu \cdot N

where \muis the coefficient of static friction and N is Normal reaction

N=mg

f_s=\mu mg

\mu mg=80

\mu =\frac{80}{27\times 9.8}

\mu =0.302

7 0
4 years ago
When sugar or another substance is dissolved in water, it disappears from view and forms a homogeneous mixture with the water, a
adoni [48]
You can tell if the sugar is still there by boiling off the water and leaving the sugar behind in the container. Sugar is a solid, and therefore cannot evaporate, so when the water reaches boiling point, it will evaporate at a quicker rate than before (water evaporates at any temperature in liquid form; just not enough to be noticeable) and leave the container to become water vapour
6 0
3 years ago
Where would a bowling ball and a napkin fall with the same acceleration?
schepotkina [342]
That would happen at any place where they don't have to
fall through air or anything else.

Examples:

-- on the moon
-- on an asteroid
-- on a comet
-- on Mercury
-- on Earth, in a vacuum chamber with all the air pumped out of it
5 0
3 years ago
Read 2 more answers
A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi
vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

V=\frac {C_2V_2-C_1V_1}{C_1+C_2}

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}

Therefore

V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437

Charge, Q is given by CV hence for the first capacitor charge will be Q_1=C_1V

Here, Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C

8 0
3 years ago
How many 90-W, 120-V light bulbs can be connected to a 20-A, 120-V circuit without tripping the circuit breaker? (Note: This des
spayn [35]

Answer:

26

Explanation:

Total power of the circuit

P = V × I = 120 × 20 = 2400 W

Power of each bulb = 90 W

Number of bulbs which work without tripping

N = power of circuit / power of each bulb

N = 2400 / 90

N = 26.66

Number of bulbs should be integer so the number of bulbs be 26.

8 0
3 years ago
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