Answer:
a = 0.009 J
b = 0.19 m/s
c = 0.005 J and 0.004 J
Explanation:
Given that
Mass of the object, m = 0.5 kg
Spring constant of the spring, k = 20 N/m
Amplitude of the motion, A = 3 cm = 0.03 m
Displacement of the system, x = 2 cm = 0.02 m
a
Total energy of the system, E =
E = 1/2 * k * A²
E = 1/2 * 20 * 0.03²
E = 10 * 0.0009
E = 0.009 J
b
E = 1/2 * k * A² = 1/2 * m * v(max)²
1/2 * m * v(max)² = 0.009
1/2 * 0.5 * v(max)² = 0.009
v(max)² = 0.009 * 2/0.5
v(max)² = 0.018 / 0.5
v(max)² = 0.036
v(max) = √0.036
v(max) = 0.19 m/s
c
V = ±√[(k/m) * (A² - x²)]
V = ±√[(20/0.5) * (0.03² - 0.02²)]
V = ±√(40 * 0.0005)
V = ±√0.02
V = ±0.141 m/s
Kinetic Energy, K = 1/2 * m * v²
K = 1/2 * 0.5 * 0.141²
K = 1/4 * 0.02
K = 0.005 J
Potential Energy, P = 1/2 * k * x²
P = 1/2 * 20 * 0.02²
P = 10 * 0.0004
P = 0.004 J
Answer:
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Explanation:
The wheel and axle is a simple machine consisting of a wheel attached to a smaller axle so that these two parts rotate together in which a force is transferred from one to the other.
Answer:
Ft = 17.48°C
Explanation:
Ft is the final temperature. However, ice absorbs heat during two process of melting and cooling and as such, there is no loss of heat to or from the surrounding hence by conservation of energy.
Therefore,
Heat absorbed by water of 20g = heat rejected by water of 265g.
So; M(ice)[C(ice) [(ΔT) + LH(ice) + C(water)(ΔT)] = C(water) M(water) (ΔT)
So, 20[(2.108) [0 - (-20)] + 333.5 + 4.187(Ft - 0)]] = (285)(4.187) (25 - Ft)
To get;
7513 + 83.74 Ft = 29832.4 - 1193.3 Ft
So factorizing, we get;
83.74 Ft + 1193.3 Ft = 29832.4 - 7513
So; 1277.04 Ft = 22319.4
So; Ft = 22319.4/1277.04 = 17.48°C
Answer:
Tungsten wire is used as the filament in light bulbs. It glows white-hot as current passes through it.
Tungsten is used in light bulbs because its high <u>RESISTANCE</u> converts electric energy into light and heat.
Explanation:
It is given that for the convex lens,
Case 1.
u=−40cm
f=+15cm
Using lens formula
v
1
−
u
1
=
f
1
v
1
−
40
1
=
15
1
v
1
=
15
1
−
40
1
v=+24.3cm
The image in formed in this case at a distance of 24.3cm in left of lens.
Case 2.
A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens i.e. the source is placed at the focus of mirror, then the rays after reflection becomes parallel for the lens such that
u=∞
f=15cm
Now, using mirror’s formula
v
1
+
u
1
=
f
1
v
1
+
∞
1
=
15
1
v=+15cm
The image is formed at a distance of 15cm in left of mirror
