120CD 165

A 2.20-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 29.0 N is require

g100num [7]

Answer:

a. 145 N/m b. 1.29 Hz c. 1.62 m/s d. 0 m e. 13.2 m/s² f. ± 0.2 m g. 2.9 J h. 0.54 m/s i. 4.39 m/s²

Explanation:

a. The force constant of the spring

The spring force F = kx and k = F/x where k is the spring constant. F = 29.0 N and x = 0.200 m

k = 29.0 N/0.200 m = 145 N/m

b. The frequency of oscillations, f

f = 1/2π√(k/m) m = mass = 2.20 kg

f = 1/2π√(145 N/m/2.20 kg) = 1.29 Hz

c. maximum speed of the object

The maximum elastic potential energy of the spring = maximum kinetic energy of the object

1/2kx² = 1/2mv²

v = (√k/m)x where v is the maximum speed of the object

v = (√145/2.2)0.2 = 1.62 m/s

d Where does the maximum speed occur?

The maximum speed occurs at 0 m

e. The maximum acceleration

a = kx/m = 145 × 0.2/2.2 = 13.2 m/s²

f. The maximum acceleration occurs at x = ± 0.2 m

g. The total energy of the system is the maximum elestic potential energy of the system

E = 1/2kx² = 1/2 × 145 × 0.2² = 2.9 J

h. When x = x₀/3

1/2k(x₀/3)² = 1/2mv²

kx₀²/9 = mv²

v = 1/3(√k/m)x₀ = 1/3(√145/2.2)0.2 = 0.54 m/s

i When x = x₀/3

a = kx₀/3m = 145 × 0.2/(2.2 × 3)= 4.39 m/s²

8 0

3 years ago

Prisms separate <br> light, such as that from the Sun, by wavelength

SVETLANKA909090 [29]

Sorry I'm so late, but I just took this test and the answer is white (for people who didn't study well ;) )

6 0

3 years ago

PLEASE HELP ASAP!!!! What displacement did the object undergo in the time interval between t= 2.0s and t=8.0s?

Katyanochek1 [597]

Displacement s = (u+v)*t/2 (t refers to delta time)
= (0.45 + 2.7)*6/2
= 3.15*3
= 9.45 m

3 0

3 years ago

Consider two wheels with fixed hubs. The hub cannot move, but the wheel can rotate about it. The hubs are fixed to a stationary

kykrilka [37]

Answer:

Magnitude of force on wheel B is 4 N

Explanation:

Given that

$I=mr^2$

For wheel A

m= 1 kg

d= 1 m,r= 0.5 m

F=1 N

We know that

T= F x r

T=1 x 0.5 N.m

T= 0.5 N.m

T= I α

Where I is the moment of inertia and α is the angular acceleration

$I_A=1 \times 0.5^2\ kg.m^2$

$I_A=0.25\ kg.m^2$

T= I α

0.5= 0.25 α

$\alpha = 2\ rad/s^2$

For Wheel B

m= 1 kg

d= 2 m,r=1 m

$I_B=1 \times 1^2\ kg.m^2$

$I_B=1 \ kg.m^2$

Given that angular acceleration is same for both the wheel

$\alpha = 2\ rad/s^2$

T= I α

T= 1 x 2

T= 2 N.m

Lets force on wheel is F then

T = F x r

2 = F x 1

So F= 2 N

Magnitude of force on wheel B is 2 N

3 0

3 years ago

Calculate the heat required, in calories, to raise

Goshia [24]

Answer: $Q=4450.74\ J$

Explanation:

Given

mass of tin $m=100 gm$

Also we know that

specific heat of tin $c_{tin}=0.21 J/gm-^{\circ}C$

Melting Point of Tin $T=231.94^{\circ}C$

Heat required to raise the temperature of 100 gm tin is given by

$Q=mc\Delta T$

where $Q=heat\ added$

$c=specific\ heat\ of\ tin$

$\Delta T=change\ in\ temperature$

$Q=100\times 0.21\times (231.94-20)$

$Q=4450.74\ J$

5 0

3 years ago