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3 years ago

How to convert 1.5 days into minutes?

Physics

2 answers:

beks73 [17]3 years ago

8 0

Answer:

2160 minutes

Explanation:

1 day = 24 hours

1.5 days = 24 * 1.5 = 36hours

1 hour = 60mins

therefore,

36 hours = 60 * 36 = 2160mins

yan [13]3 years ago

3 0

Answer:

This conversion of 1.5 days to minutes has been calculated by multiplying 1.5 days by 1,440 and the result is 2,160 minutes.

Explanation:

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Calculate the (absolute) pressure at the bottom of a neighborhood swimming pool 30.0 m by 8.0 m whose uniform depth is 2.0 m. Th

Alika [10]

Answer:

Total pressure= 120945[Pa]

Force exerted = 29026800 [N] or 29.02*10^6 [N]

Explanation:

We know that the total pressure is the result of the sum of the atmospheric pressure plus the manometric pressure. The equation is:

$Ptotal=Patm + Pman$

In this problem we know the atmospheric pressure 101.325x10^3 [Pa], therefore we need to find the manometric pressure.

The manometric pressure in the bottom of the swimming pool depends only on the water column of water generated (depth of the swimming pool)

$Pman = density*g*h$

where:

density = density of the water 1000 [kg/m^3]

g= gravity [m/s^2]

h= column of water (meters)

replacing the values:

$Pman= 1000 *9.81* 2 = 19620 [Pa]\\\\$

The total pressure will be:

$Ptotal= 101325+19620 = 120945 [Pa]\\\\$

The force exerte on the bottom is defined by the following expression:

$Pressure=Force/area\\\\Force= Pressure*Area\\\\Area = 30m*8m= 240 m^2Force= 120945*240\\Force= 29026800N or 2958 Ton$

4 0

4 years ago

Four identical resistors have 12 ohm each. Find the equivalent resistance Req when : (a) all four of them are connected in serie

Elanso [62]

Answer:

(a): When the four resistors are connected in series the equivalent resistor value is Req= 48Ω

(b): when the four resistors are connected in parallel the equivalent resistor value is Req=3Ω

Explanation:

R=R1=R2=R3=R4= 12Ω

(a)

Req= R1+R2+R3+R4

Req= 48 Ω

(b)

Req= (1/12 * 4)⁻¹

Req= 3 Ω

6 0

3 years ago

A truck traveling at a constant speed of 24 m/s passes a more slowly moving car. The instant the truck passes the car, the car b

Ymorist [56]

Answer:

7.15 m/s

Explanation:

We use a frame of reference in which the origin is at the point where the trucck passed the car and that moment is t=0. The X axis of the frame of reference is in the direction the vehicles move.

The truck moves at constant speed, we can use the equation for position under constant speed:

Xt = X0 + v*t

The car is accelerating with constant acceleration, we can use this equation

Xc = X0 + V0*t + 1/2*a*t^2

We know that both vehicles will meet again at x = 578

Replacing this in the equation of the truck:

578 = 24 * t

We get the time when the car passes the truck

t = 578 / 24 = 24.08 s

Before replacing the values on the car equation, we rearrange it:

Xc = X0 + V0*t + 1/2*a*t^2

V0*t = Xc - 1/2*a*t^2

V0 = (Xc - 1/2*a*t^2)/t

Now we replace

V0 = (578 - 1/2*1.4*24.08^2) / 24.08 = 7.15 m/s

6 0

3 years ago

Explain how to measure the weight of an object

erastova [34]

Answer:

The formula for calculating weight is F = m × 9.8 m/s2, where F is the object's weight in Newtons (N) and m is the object's mass in kilograms.

3 0

3 years ago

A big olive (* - 0.50 kg) lies at the origin of an xy coordinate system, and a big BrazlI nut (M - 1.5^kg) lie^s at the point (1

Afina-wow [57]

The <em>estimated</em> displacement of the center of mass of the olive is $\overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]$ .

<h3>Procedure - Estimation of the displacement of the center of mass of the olive</h3>

In this question we should apply the definition of center of mass and difference between the coordinates for <em>dynamic</em> ( $\vec r$ ) and <em>static</em> conditions ( $\vec r_{o}$ ) to estimate the displacement of the center of mass of the olive ( $\overrightarrow{\Delta r}$ ):

$\vec r - \vec r_{o} = \left[\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot(m_{i}\cdot g + F_{i, x})}{\Sigma \limits_{i =1}^{2}(F_{i,x}+m_{i}\cdot g)} ,\frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot(m_{i}\cdot g + F_{i, y})}{\Sigma \limits_{i =1}^{2}(F_{i,y}+m_{i}\cdot g)} \right]-\left(\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}, \frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}\right)$ (1)

Where:

$r_{i, x}$ - x-Coordinate of the i-th element of the system, in meters.
$r_{i,y}$ - y-Coordinate of the i-th element of the system, in meters.
$F_{i,x}$ - x-Component of the net force applied on the i-th element, in newtons.
$F_{i,y}$ - y-Component of the net force applied on the i-th element, in newtons.
$m_{i}$ - Mass of the i-th element, in kilograms.
$g$ - Gravitational acceleration, in meters per square second.

If we know that $\vec r_{1} = (0, 0)\,[m]$ , $\vec r_{2} = (1, 2)\,[m]$ , $\vec F_{1} = (0, 3)\,[N]$ , $\vec F_{2} = (-3, -2)\,[N]$ , $m_{1} = 0.50\,kg$ , $m_{2} = 1.50\,kg$ and $g = 9.807\,\frac{kg}{s^{2}}$ , then the displacement of the center of mass of the olive is:

<h3>Dynamic condition $\vec{r} = \left[\frac{(0)\cdot (0.50)\cdot (9.807)+(0)\cdot (0) + (1)\cdot (1.50)\cdot (9.807) + (1)\cdot (-3)}{(0.50)\cdot (9.807) + 0 + (1.50)\cdot (9.807)+(-3)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (0)\cdot (3) + (2)\cdot (1.50)\cdot (9.807) +(2) \cdot (-2)}{(0.50)\cdot (9.807) + (3)+(1.50)\cdot (9.807)+(-2)} \right]$ $\vec r = (0,704, 1.233)\,[m]$ </h3>

<h3>Static condition</h3><h3> $\vec{r}_{o} = \left[\frac{(0)\cdot (0.50)\cdot (9.807) + (1)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807) + (1.50)\cdot (9.807)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (2)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807)+(1.50)\cdot (9.807)} \right]$ </h3><h3> $\vec r_{o} = \left(0.75, 1.50)\,[m]$ </h3><h3 /><h3>Displacement of the center of mass of the olive</h3>

$\overrightarrow{\Delta r} = \vec r - \vec r_{o}$

$\overrightarrow{\Delta r} = (0.704-0.75, 1.233-1.50)\,[m]$

$\overrightarrow{\Delta r} = (-0.046, -0.267)\,[m]$

The <em>estimated</em> displacement of the center of mass of the olive is $\overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]$ . $\blacksquare$

To learn more on center of mass, we kindly invite to check this verified question: brainly.com/question/8662931

3 0

2 years ago