Which lists three organic biological molecules?

A light spring of constant 179 N/m rests vertically on the bottom of a large beaker of water. A 5.32 kg block of wood of density

Digiron [165]

Answer:

Compression of the spring: 0.18 m (downward)

Explanation:

The forces acting on the block of wood are:

- The force of gravity, acting downward, of magnitude $mg$ , where m = 5.32 kg is the mass of the block and $g=9.8 m/s^2$ is the acceleration due to gravity

- The force exerted by the spring, downward, of magnitude $kx$ , where $k=179N/m$ is the spring constant and $x$ is the elongation of the spring

- The buoyant force, upward, of magnitude $\rho V g$ , where $\rho=1000 kg/m^3$ is the water density and V the volume of the block

Since the block is in equilibrium, the net force is zero, so we can write

$mg+kx-\rho V g=0$ (1)

We have to find the volume of the block first. We have:

m = 5.32 kg (mass)

$\rho_w = 622 kg/m^3$ (wood density)

So, the volume is

$V=\frac{m}{\rho_w}=\frac{5.32}{622}=0.0086 m^3$

So now we can re-arrange eq.(1) to find the elongation of the spring, x:

$x=\frac{-mg+\rho Vg}{k}=\frac{-(5.32)(9.8)+(1000)(0.0086)(9.8)}{179}=0.18 m$

So, the spring is compressed by 0.18 m.

7 0

3 years ago

At an outdoor market, a bunch of bananas is set on a spring scale to measure the weight. The spring sets the full bunch of banan

Elina [12.6K]

Answer:

2.67kg

Explanation:

The maximum velocity, $v _ {max}$ of a body experiencing simple harmonic motion is given by equation (1);

$v_{max}=\omega A............(1)$

where $\omega$ is the angular velocity and A is the amplitude.

The problem describes the oscillation of a loaded spring, and for a loaded spring the angular velocity is given by equation (2);

$\omega=\sqrt{\frac{k}{m}}.................(2)$

where k is the force constant of the spring and m is the loaded mass.

We can make $\omega$ the subject of formula in equation (1) as follows;

$\omega=\frac{v_{max}}{A}.................(3)$

We then combine equations (2) and (3) as follows;

$\frac{v_{max}}{A}=\sqrt{\frac{k}{m}}.................(4)$

According to the problem, the following are given;

$v_ {max }=1.92m/s\\A=0.21m\\k=223N/m$

We then substitute these values into equation (4) and solve for the unknown mass m as follows;

$\frac{1.92}{0.21}=\sqrt{\frac{223}{m}}$

$9.143=\sqrt{\frac{223}{m}}$

Squaring both sides, we obtain the following;

$9.143^2=\frac{223}{m}\\9.143^2*m=223\\83.592m=223\\therefore\\m=\frac{223}{83.592}\\m=2.67kg$

8 0

3 years ago

Which of the following displays has the highest hz frequency

Ganezh [65]

Answer:

Plasma

Explanation:

3 0

3 years ago

If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will

Anastasy [175]

Assuming that the voltage is constant, and the resistance was doubled, since I=V/R, the current through the circuit will be halved. As P=IV, with the same voltage and halved current, the power dissipated by the circuit will be halved

3 0

3 years ago

A machine has a mechanical advantage of 5. if 300 newtons of input force is used to produce 3,000 newton meters of work what is

Arte-miy333 [17]

F(in) x d(in)= F(out)x d(out)
A machine has a mechanical advantage of 5.
F(out )= 5xF(in)300 newtons

F(in) = 300 N
F(out)= 5xF(in) = 5x300 N
= 1500 N

Work in = Work out = 3000 N x m3000
N x m = F_in x d_ind_in = 3000 Nxm / F_in = 3000 N x m / 300 N = 10 m3000 Nxm = F_out x d_out d_out = 3000 Nx m / F_out = 3000 N x m / 1500 N = 2 m

8 0

4 years ago