You push on a 35 N object for 3 seconds. find work

A 0.500-kg object connected to a light spring with a spring constant of 20.0 N/m oscillates on a frictionless horizontal surface

givi [52]

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

4 0

3 years ago

A brick is dropped with zero initial speed from the roof of a building and strikes the ground in 1.90 s. How tall is the buildin

irinina [24]

Answer:

17.69 m

Explanation:

The time it takes the brick to strike the ground is 1.90 seconds.

We can apply one of Newton's equation of linear motion to find the height of the building:

$s = ut + 0.5gt^2$

where s = distance (in this case height)

u = initial velocity = 0 m/s

t = time = 1.90 s

g = acceleration due to gravity = 9.8 m/s^2

Therefore:

s = (0 * 1.9) + (0.5 * 9.8 * 1.9 * 1.9)

s = 0 + 17.68

s = 17.69 m

The height of the building is 17.69 m.

6 0

4 years ago

Recall that if a line is parallel to the vector v and passes through the point P0, which is the tip of the position vector r0, t

Elenna [48]

Answer:

The vector equation of the line is

$\overrightarrow{r}=+t$

Parametric equations for given line are

$x=7+t\\y=-8+6t\\z=3-13t$

Explanation:

The vector equation of the line is given by

$r(t) = r_{o} + tv$

r₀ = (7, -8, 3)

v = (1, 6, -13)

At these points the vector equation for this line is:

$\overrightarrow{r}=\overrightarrow{r_{o}}+t\overrightarrow{v}\\\overrightarrow{r}=+t$

Parametric equations for given line are

$x=7+t\\y=-8+6t\\z=3-13t$

8 0

3 years ago

If the temperature of a volume of ideal gas increases for 100°C to 200°C, what happens to the average kinetic energy of the mole

mina [271]

Answer: $\Delta E=2.0715\times 10^{-21} J$

Explanation:

Given

Temperature of the gas is increased from 100 to 200

Also we know that average kinetic energy of the molecules is

$E=\frac{3}{2}\cdot \frac{R}{N_A}T$

Where

R=Gas constant

$N_A$ =Avogadro's number

T=Temperature in kelvin

$\frac{R}{N_A}=1.381\times 10^{-23}$

So kinetic energy increases by

$\Delta E=\frac{3}{2}\times 1.381\times 10^{-23}\left ( 200-100\right )$

$\Delta E=2.0715\times 10^{-21} J$

8 0

3 years ago

The acceleration of a train moving from rest and its speed reaches 36m/sec in 9 sec

jenyasd209 [6]

Answer:

4 m/s^2

Explanation:

The acceleration is defined as: Δv/Δt (the difference of the velocity over a time period in which happens that difference).

Remember that a difference is calculated by subtracting the initial value of a physical quantity from its final value.

In our case:

Δv = Vfinal - Vinitial = 36m/s - 0 m/s = 36m/s

Δt = 9s

a = Δv/Δt = 36m/s / 9s = 4m/s^2

4 0

2 years ago