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marysya [2.9K]
2 years ago
12

A block rests on a flat plate that executes vertical simple harmonic motion with a period of 0.74 s. What is the maximum amplitu

de of the motion for which the block does not separate from the plate? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.
Physics
1 answer:
Mumz [18]2 years ago
8 0

Answer:

maximum amplitude  = 0.13 m

Explanation:

Given that

Time period T= 0.74 s

acceleration of gravity g= 10 m/s²

We know that time period of simple harmonic motion given as

T=\dfrac{2\pi}{\omega}

0.74=\dfrac{2\pi}{\omega}

ω = 8.48 rad/s

ω=angular frequency

Lets take amplitude = A

The maximum acceleration given as

a= ω² A

The maximum acceleration should be equal to g ,then block does not separate

a= ω² A

10= 8.48² A

A=0.13 m

maximum amplitude  = 0.13 m

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A 2.0 g metal cube and a 4.0 g metal cube are 6.0 cm apart, measured between their centers, on a horizontal surface. For both, t
emmasim [6.3K]

Answer:

a) t=10.2s

b) The 2g-cube moves first

Explanation:

Since the electric force is the same on both cubes and so is the coefficient of static friction, the first one to move will be the one with less mass.

So, on the 2g-cube the sum of forces are:

\left \{ {{Ff-Fe=0} \atop {N-m*g=0}} \right.

Replacing the friction on the first equation:

\mu*m*g-Fe=0  Thus   Fe=\mu*m*g=12.74*10^{-3}N

The electric force is:

Fe = \frac{K*q^2}{d^2}  Solving for q:

q=71.44nC

This amount divided by the rate at which they are being charged:

t = 71.44nC / 7nC/s = 10.2s

7 0
1 year ago
The velocity component with which a projectile covers certain horizontal distance is maximum at the moment of? a) Hitting the gr
Crazy boy [7]
A) hitting the ground
3 0
2 years ago
When were Earth’s landmasses first recognizable as the continents we know today? 10 million years ago 135 million years ago 180
Bess [88]

Answer:

b

Explanation:

i took the test

6 0
1 year ago
Read 2 more answers
Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for
Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0
2 years ago
Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d
Step2247 [10]
Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that F = k \frac{Q_1Q_2}{d^2} so if we double the charge on Q_1 and double the distance to (2d) we plug these into the equation to find

<span>F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}</span>

So we see the new force is exactly 1/2 of the old force so your answer should be \frac{1}{2}F if I can remember my physics correctly.

9 0
2 years ago
Read 2 more answers
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