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maxonik [38]
3 years ago
12

A wooden block has a mass of 562 g and a volume of 72 cm3. what is the density

Physics
1 answer:
Gnom [1K]3 years ago
7 0

Answer:

7\frac{29}{36}g/cm³

Explanation:

The formula for density is \frac{mass}{volume}. Let's apply the formula to the question:

\frac{562}{72} = 7\frac{29}{36}g/cm³

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It should be 4. An alternator
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While skateboarding at 19 km/h, Alana throws a tennis ball at 11 km/h to her friend Oliver. If Alana is the reference frame, the
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Haven't taken physics but I would assume if her friend is standing in front of her that you would add up the speeds and get 30 km/hr.
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A research submarine can withstand an external pressure of 62 megapascals (million pascals) all the while maintaining a comforta
Alex

Answer:

<em>The depth will be equal to</em> <em>6141.96 m</em>

<em></em>

Explanation:

pressure on the submarine P_{sea} = 62 MPa = 62 x 10^6 Pa

we also know that P_{sea} = ρgh

where

ρ is the density of sea water = 1029 kg/m^3

g is acceleration due to gravity = 9.81 m/s^2

h is the depth below the water that this pressure acts

substituting values, we have

P_{sea} = 1029 x 9.81 x h = 10094.49h

The gauge pressure within the submarine P_{g} = 101 kPa =  101000 Pa

this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.

Equating the pressure P_{sea}, we have

62 x 10^6 = 10094.49h

depth h = <em>6141.96 m</em>

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3 years ago
Which Table of ordered pairs represent proportional relationship
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Answer:

Ratio table of ordered pairs represent proportional relationship .

<em>Hope </em><em>it</em><em> is</em><em> helpful</em><em> to</em><em> you</em>

5 0
3 years ago
The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh
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In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

v^2=v^2_o-2ax

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}

Next, consider that the formula for the force is:

F=ma

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

F=(820kg)(0.66\frac{m}{s^2})=542.20N

Hence, the required force to stop the car is 542.20N

4 0
1 year ago
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