When a car with bright headlights comes toward you at night, you should:?

A block of density pb = 9.50 times 10^2 kg/m^3 floats face down in a fluid of density pt = 1.30 times 10^3 kg/m^3. The block has

Nookie1986 [14]

Answer:

The depth and acceleration are 0.1919291 ft and 3.61 m/s².

Explanation:

Given that,

Density of block $\rho_{b} =9.50\times10^2\ kg/m^3$

Density of fluid $\rho_{t} =1.30\times10^3\ kg/m^3$

We need to calculate the depth

Using balance equation

$mg=\rho g V$ ....(I)

We know that,

The density is

$\rho=\dfrac{m}{V}$

$m=\rh0\times V$

Put the value of m in equation (I)

$\rho_{b}\times V\times g=\rho_{t}\times g\times V$

$\rho_{b}\times A\times H\times g=\rho_{t}\times g\times A\times h$

$h=\dfrac{\rho_{b}H}{\rho_{t}}$

Put the value into the formula

$h=\dfrac{9.50\times10^2\times8.00\times10^{-2}}{1.30\times10^3}$

$h= 5.85\ cm$

$h=0.1919291\ ft$

We need to calculate the acceleration

Using formula of net force

$F_{net}=\rho_{t}\times g\times V- \rho_{b}\times g\times V$

$ma=\rho_{t}\times g\times V- \rho_{b}\times g\times V$

$\rho_{b}\times V\times a=\rho_{t}\times g\times V- \rho_{b}\times g\times V$

$a=(\dfrac{\rho_{t}}{\rho_{b}}-1)g$ ....(II)

Put the value in the equation (II)

$a=(\dfrac{1.30\times10^3}{9.50\times10^2}-1)\times9.8$

$a=3.61\ m/s^2$

Hence, The depth and acceleration are 0.1919291 ft and 3.61 m/s².

4 0

3 years ago

Why is a control group important?

Likurg_2 [28]

Answer:

A. It changes all the time, allowing us to see if the experimental group stays the same.

control group mean the thing the stay the same during the Experiment.

I hope this help

I hope this is correct

4 0

4 years ago

Three 3-ohm resistors are connected in parallel. The total resistance for the resistors is

Anna11 [10]

Explanation:

I think the total resistance of the resistors is 1 ohm . it is also known as equivalent resistance.

hope it helps.

3 0

3 years ago

A flywheel with radius of 0.300 m starts from rest and accelerates with a constant angular acceleration of 0.400 rad/s2.For a po

oksian1 [2.3K]

Answer:

$0.12\ m/s^2$

Explanation:

Given that,

The radius of a flywheel, r = 0.3 m

Angular acceleration of a flywheel, $\alpha =0.4\ rad/s^2$

We need to find the magnitude of the tangential acceleration after 2.00 s of acceleration.

The relation between the tangential and angular acceleration is given by :

$a_t=r\alpha \\\\a_t=0.3\times 0.4\\\\a_t=0.12\ m/s^2$

So, the required magnitude of tangential acceleration is $0.12\ m/s^2$ .

6 0

3 years ago

A golf club rotates 215 degrees and has a length (radius) equal to 29 inches. The time it took to swing the club was 0.8 seconds

vichka [17]

Answer:

The average linear velocity (inches/second) of the golf club is 136.01 inches/second

Explanation:

Given;

length of the club, L = 29 inches

rotation angle, θ = 215⁰

time of motion, t = 0.8 s

The angular speed of the club is calculated as follows;

$\omega = (\frac{\theta}{360} \times 2\pi, \ rad) \times \frac{1}{t} \\\\\omega = (\frac{215}{360} \times 2\pi, \ rad) \times \frac{1}{0.8 \ s} \\\\\omega = 4.69 \ rad/s$

The average linear velocity (inches/second) of the golf club is calculated as;

v = ωr

v = 4.69 rad/s x 29 inches

v = 136.01 inches/second

Therefore, the average linear velocity (inches/second) of the golf club is 136.01 inches/second

8 0

3 years ago