<span>B.Shape of continents look like puzzle pieces that can fit into one another</span>
<span>What I have here is exactly the same problem, however, with the time changed to 19 mins:
metabolic energy = metabolic power*time = 1.150*19*60 = 1.311 kJ..corresponding to 1.311/4.186 = 313,2 Cal or kcal
If we reasonably assume a metabolic eff.cy of 20%, it means we need to assume food for 1500 Cal approx.
Just plug the value t=15min to the equation and you will surely get the correct answer.
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
</span>
The light coming out of a concave lens will never meet.
So, the answer is A. will never meet.
Happy Studying! ^^
Answer:
33.6 Ns backward.
Explanation:
Impulse: This can be defined as the product of force and time. The S.I unit of impulse is Ns.
From Newton's second law of motion,
Impulse = change in momentum
I = mΔv................................. Equation 1
Where I = impulse, m = mass of the skater, Δv = change in velocity = final velocity - initial velocity.
Given: m = 28 kg, t = 0.8 s, Δv = -1.2-0 = -1.2 m/s (Note: the initial velocity of the skater = 0 m/s)
Substituting into equation 1
I = 28(-1.2)
I = -33.6 Ns
Thus the impulse = 33.6 Ns backward.
<span>1) The differential equation that models the RC circuit is :
(d/dt)V_capacitor </span>+ (V_capacitor/RC) = (V_source/<span>RC)</span>
<span>Where the time constant of the circuit is defined by the product of R*C
Time constant = T = R*C = (</span>30.5 ohms) * (89.9-mf) = 2.742 s
2) C<span>harge of the capacitor 1.57 time constants
1.57*(2.742) = 4.3048 s
The solution of the differential equation is
</span>V_capac (t) = (V_capac(0) - V_capac(∞<span>))e ^(-t /T) + </span>V_capac(∞)
Since the capacitor is initially uncharged V_capac(0) = 0
And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery V_capac(∞) = 9V
This means,
V_capac (t) = (-9V)e ^(-t /T) + 9V
The charge in a capacitor is defined as Q = C*V
Where C is the capacitance and V is the Voltage across
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T) + 9V
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V = -1.87V +9V
V_capac (4.3048 s) = 7.1275 V
Q (4.3048 s) = 89.9mF*(7.1275V) = 0.6407 C
3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V
Q (∞) = 89.9mF*(9V) = 0.8091 C
