Answer:
1 kg lead to earth is greater attraction as mass of earth is much more than 1kg lead.
Explanation:
Objects with more mass have more gravity. Gravity also gets weaker with distance. So, the closer objects are to each other, the stronger their gravitational pull is. Earth's gravity comes from all its mass
Answer:
the new resister is 11 ohms.
Explanation:
Set it up like this.
1/x + 1/1.1 = 1 Subtract 1/1.1 from both sides
1/x = 1 - 1/1.1
1 - 1/1.1 = 1/11
1/x = 1/11 Cross multiply
11 = x
If 1/11 bothers you, you could do it it another way.
1 - 1/1.1 = (1.1 - 1 ) / 1.1 = 0.1 / 1.1 Multiply top and bottom by 10
0.1*10/(1.1 * 10 ) = 1 / 11
Answer:
Δ
= 84 Ω, = (40 ± 8) 10¹ Ω
Explanation:
The formula for parallel equivalent resistance is
1 / = ∑ 1 / Ri
In our case we use a resistance of each
R₁ = 500 ± 50 Ω
R₂ = 2000 ± 5%
This percentage equals
0.05 = ΔR₂ / R₂
ΔR₂ = 0.05 R₂
ΔR₂ = 0.05 2000 = 100 Ω
We write the resistance
R₂ = 2000 ± 100 Ω
We apply the initial formula
1 / = 1 / R₁ + 1 / R₂
1 / = 1/500 + 1/2000 = 0.0025
= 400 Ω
Let's look for the error (uncertainly) of Re
= R₁R₂ / (R₁ + R₂)
R’= R₁ + R₂
= R₁R₂ / R’
Let's look for the uncertainty of this equation
Δ / = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’
The uncertainty of a sum is
ΔR’= ΔR₁ + ΔR₂
We substitute the values
Δ / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)
Δ / 400 = 0.1 + 0.05 + 0.06
Δ = 0.21 400
Δ = 84 Ω
Let's write the resistance value with the correct significant figures
= (40 ± 8) 10¹ Ω
Answer:
T=0.372 s, f=2.7 Hz, w=16.9 rad/s, k=179.2 N/m, v= 8.78 m/s, F= 48.4 N
Explanation:
a.)
Period: It is already given in the question "oscillator repeats its motion every 0.372 s".
So T=0.372 s
b)
frequency= f = 1/ T
f = 1/ 0.372
f=2.7 Hz
c).
Angular frequency= w= 2πf
w= 2*π*2.7
w=16.9 rad/s
d)
Spring Constant:
As w=
⇒w²= k/m
⇒k= m*w²
⇒k= 0.628 * 16.9² N/m
⇒k=179.2 N/m
e)
The mass will have maximum speed when it passes through the mean position.
At mean position
Maximum elastic potential energy = Maximum kinetic energy
1/2 k A² = 1/2 m v² ( A is amplitude of oscillation)
⇒ v=
⇒ v= 
\
⇒ v= 8.78 m/s
f)
Maximum force will be exerted on the block when it is at maximum distance.
F= k* A ( A is amplitude of oscillation)
F= 179.2 * 0.27 N
F= 48.4 N
Answer:
a) k = 120 N / m
, b) f = 0.851 Hz
, c) v = 1,069 m / s
, d) x = 0
, e) a = 5.71 m / s²
, f) x = 0.200 m
, g) Em = 2.4 J
, h) v = -1.01 m / s
Explanation:
a) Hooke's law is
F = k x
k = F / x
k = 24.0 / 0.200
k = 120 N / m
b) the angular velocity of the simple harmonic movement is
w = √ k / m
w = √ (120 / 4.2)
w = 5,345 rad / s
Angular velocity and frequency are related.
w = 2π f
f = w / 2π
f = 5.345 / 2π
f = 0.851 Hz
c) the equation that describes the movement is
x = A cos (wt + Ф)
As the body is released without initial velocity, Ф = 0
x = 0.2 cos wt
Speed is
v = dx / dt
v = -A w sin wt
The speed is maximum for sin wt = ±1
v = A w
v = 0.200 5.345
v = 1,069 m / s
d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is
x = A cos wt = 0
x = 0
e) the acceleration is
a = d²x / dt² = dv / dt
a = - Aw² cos wt
The acceleration is maximum when cos wt = ± 1
a = A w²
a = 0.2 5.345
a = 5.71 m / s²
f) the position for this acceleration is
x = A cos wt
x = A
x = 0.200 m
g) Mechanical energy is
Em = ½ k A²
Em = ½ 120 0.2²
Em = 2.4 J
h) the position is
x = 1/3 A
Let's calculate the time to reach this point
x = A cos wt
1/3 A = A cos 5.345t
t = 1 / w cos⁻¹(1/3)
The angles are in radians
t = 1.23 / 5,345
t = 0.2301 s
Speed is
v = -A w sin wt
v = -0.2 5.345 sin (5.345 0.2301)
v = -1.01 m / s
i) acceleration
a = -A w² sin wt
a = - 0.2 5.345² cos (5.345 0.2301)
a = -1.91 m / s²
