Thank you for posting your question here at brainly. I hope the answer will help. Below are the choices that can be found elsewhere:
<span>A. 1.5 * 10^3 Watts
B. 7.3 * 10^2 Watts
C. 3.5 * 10^2 Watts
D. 2.5 * 10^2 Watts
</span>
<span>Work = force*displacement = 10^2*87 = 8,700 joule
Power = work/time = 8,700/6 = 1.45*10^3 (rounded up to 1.5 kw). The answer is A. </span>
Answer:
Explanation:
When the positively charged half shell is brought in contact with the electroscope, its needle deflects due to charge present on the shell.
When the negatively charged half shell is brought in contact with the positively charged shell , the positive and negative charge present on each shell neutralises each other .So both the shells lose their charges .The positive half shell also loses all its charges
When we separate the half shells , there will be no deflection in the electroscope because both the shell have already lost their charges and they have become neutral bodies . So they will not be able to produce any deflection in the electroscope.
Answer:
9.98 m/s
Explanation:
The force acting on the particle is defined by the equation:
[N]
where x is the position in metres.
The acceleration can be found by using Newton's second law:
where
m = 150 g = 0.150 kg is the mass of the particle. Substituting into the equation,
[m/s^2]
When x = 3.14 m, the acceleration is:
Now we can find the final speed of the particle by using the suvat equation:
where
u = 8.00 m/s is the initial velocity
v is the final velocity
x = 3.14 m is the displacement
Solving for v,
And the speed is just the magnitude of the velocity, so 9.98 m/s.
The source charges' magnitude is signified by the arrows pointing outward. The more arrows there are, the greater is its magnitude. This is because, each arrow represents an electrical force exerted by the source. When you add up all the arrows there is, the electrical force becomes even greater. The answer in descending order would be C > A > B > D.
(a) 3.56 m/s
(b) 11 - 3.72a
(c) t = 5.9 s
(d) -11 m/s
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule.
y = 11t - 1.86t^2
y' = 11 - 3.72t
Now that you have the first derivative, it will give you the velocity as a function of t.
(a) Velocity after 2 seconds.
y' = 11 - 3.72t
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56
So the velocity is 3.56 m/s
(b) Velocity after a seconds.
y' = 11 - 3.72t
y' = 11 - 3.72a
So the answer is 11 - 3.72a
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.
(d) Plug in the value of t calculated for (c) into the velocity function, so:
y' = 11 - 3.72a
y' = 11 - 3.72*5.913978495
y' = 11 - 22
y' = -11
So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
