Ask question

Ask question

All categories

3 years ago

13

Match the block to its correct density. 1. Block A 0.64 Kg/L 2. Block B 0.917 kg/L 3. Block C 0.70 Kg/L 19.3 kg/L 4. Block D 3.5

3 Kg/L 5. Block E

1 answer:

Sedbober [7]3 years ago

4 0

Answer:

quit school aghahahha

Explanation:

xD

You might be interested in

The intensity I of light varies inversely as the square of the distance D from the source. If the intensity of illumination on a

emmasim [6.3K]

The intensity on a screen 20 ft from the light will be 0.125-foot candles.

<h3>What is the distance?</h3>

Distance is a numerical representation of the length between two objects or locations.

The intensity I of light varies inversely as the square of the distance D from the source;

I∝(1/D²)

The ratio of the intensity of the two cases;

$\rm \frac{I_1}{I_2} =(\frac{D_2}{D_1} )^2\\\\ \rm \frac{2}{I_2} =(\frac{20}{5} )^2\\\\ \frac{2}{I_2} =4^2 \\\\ I_2= \frac{2}{16} \\\\ I_2= 0.125 \ foot-candles$

Hence, the intensity on a screen 20 ft from the light will be 0.125 foot-candles

To learn more about the distance refer to the link;

brainly.com/question/26711747

#SPJ1

6 0

2 years ago

When a 3.0 kg mass is hung from a vertical massless spring, the spring is stretched 40 cm. What is the spring constant of the sp

Dmitry_Shevchenko [17]

Answer:

0.74 N/cm

Explanation:

The following data were obtained from the question:

Mass (m) = 3 Kg

Extention (e) = 40 cm

Spring constant (K) =?

Next, we shall determine the force exerted on the spring.

This can be obtained as follow:

Mass (m) = 3 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = mg

F = 3 × 9.8

F = 29.4 N

Finally, we shall determine the spring constant of the spring. This can be obtained as follow:

Extention (e) = 40 cm

Force (F) = 29.4 N

Spring constant (K) =?

F = Ke

29.4 = K × 40

Divide both side by 40

K = 29.4 / 40

K = 0.74 N/cm

Therefore, the spring constant of the spring is 0.74 N/cm

5 0

2 years ago

What is the function of the electron transport chain in cellular respiration. why is oxygen needed for the electron transport ch

SpyIntel [72]

1) <span>The function of the electron transport chain is to pump protons in the mitochondrion inter-membrane, thus building up a proton gradient. This gradient will allow the ATP syntheses</span><span>.</span>

2) Why we need oxygen for the electron transport chain:
At the end of the electron transport chain is the Oxygen that will accept electrons and picks up protons to form water. If the oxygen molecule is not there the electron transport chain will stop running, and ATP will no longer be produced. Basically, we need the oxygen to produce more ATP.

8 0

3 years ago

Four point masses of 3.0 kg each are arranged in a square on masslessrods. The length of a side of the square is 0.50m. What is

Zigmanuir [339]

Answer:

Part a)

$I = 1.5 kg m^2$

Part b)

$I = 0.75 kg m^2$

Part c)

$I = 1.5 kg m^2$

Explanation:

Part a)

Moment of inertia of the system about an axis passing through B and C is given as

$I = mL^2 + mL^2 + m(0) + m(0)$

$I = 2mL^2$

$I = 2(3 kg)(0.50^2)$

$I = 1.5 kg m^2$

Part b)

Moment of inertia of the system about an axis passing through A and C is given as

$I = m(0^2) + m(\frac{L}{\sqrt2})^2 + m(0) + m(\frac{L}{\sqrt2})^2$

$I = 2m\frac{L^2}{2}$

$I = (3 kg)(0.50^2)$

$I = 0.75 kg m^2$

Part c)

Moment of inertia of the system about an axis passing through the center of the square and perpendicular to the plane of the square

$I = m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2$

$I = 4m\frac{L^2}{2}$

$I = 2(3 kg)(0.50^2)$

$I = 1.5 kg m^2$

8 0

2 years ago

Hooke’s law states that the distance that a spring is stretched by hanging object varies directly as the mass of the object. If

trasher [3.6K]

Answer:

$d_{2}$ = 33.33 cm

Explanation:

Given:

When mass, $m_{1}$ =21 kg

distance travelled is $d_{1}$ = 140 cm

When mass, $m_{2}$ =5 kg

distance travelled is $d_{2}$ = ?

Hooke's law state that within elastic limit, when an external force is applied to a body, the body gets deformed and when the force is released the gets back to its original form.

Therefore according to the question,

$\frac{d_{1}}{m_{1}}=\frac{d_{2}}{m_{2}}$

$\frac{140}{21}=\frac{d_{2}}{5}$

$d_{2}$ = 33.33 cm

Distance travelled is 33.33 cm when mass is 5 kg.

8 0

2 years ago