Let's ask this question step by step:
Part A)
a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
ab = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
Part (c)
(a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
(a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
(a + b) b = 10 + 36
(a + b) b = 46
Part (d)
comp (ba) = (a.b) / lbl
a.b = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
lbl = root ((2.0) ^ 2 + (4.0) ^ 2) = root (20)
comp (ba) = 26 / root (20)
answer
2k
26
46
26 / root (20)
Here in this question as we can see there is no air friction so we can use the principle of energy conservation
now here we know that
now plug in all values in above equation
divide whole equation by mass "m"
so height of the ball from ground will be 1.35 m
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Answer:
Change in potential energy of the block-spring-Earth
system between Figure 1 and Figure 2 = 1 Nm.
Explanation:
Here, spring constant, k = 50 N/m.
given block comes down eventually 0.2 m below.
here, g = 10 m/s.
let block be at a height h above the ground in figure 1.
⇒In figure 2,
potential energy of the block-spring-Earth
system = m×g×(h - 0.2) + 1/2× k × x². where, x = change in spring length.
⇒ Change in potential energy of the block-spring-Earth
system between Figure 1 and Figure 2 = (m×g×(h - 0.2)) - (1/2× k × x²)
= (1×10×0.2) - (1/2×50×0.2×0.2) = 1 Nm.
Answer:
0.05 W
Explanation:
The power dissipated by a device can be written as
where
P is the power dissipated
V is the voltage drop on the device
I is the current flowing through the device
In this problem, we have
V = 5.0 V is the voltage drop across the device
I = 10.0 mA = 0.01 A is the current through it
By applying the formula, we find the power dissipated:
