Given a = 10 cm/s²
u = 0 cm/s
v = 50 cm/s
we know that
v²=u²+2aS
2500=2×10×S
2500÷20 = S
S= 125 cm
The ramp is 125 cm
Force required is 100 N
<u>Given that;</u>
Rate of acceleration = 5 m/s²
Mass of object = 20kg
<u>Find:</u>
Force required
<u>Computation:</u>
Force = Mass × Acceleration
Force required = Rate of acceleration × Mass of object
Force required = 20 × 5
Force required = 100 N
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The voltage<span> difference between the two plates can be expressed in terms of the </span>work<span> done on a positive test charge q when it moves from the positive to the negative plate.</span><span>
E=V/d
where V is the voltage and d is the distance between the plates.
So,
E=6.0V/1mm= 6000 V/m. The electric field between the plates is 6000 V/m.</span>
Answer:
Explanation:
Thinking about the logics it can but it may be dim because 1.12 is lower than 2,5v so this will mean u lamp may not work or may work very dimely due to the low voltage it is receiving.
