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miss Akunina [59]

2 years ago

10

Si tengo 4litros de agua a 20C, calcula cuantos litros de agua a 32C tengo que agregar para que su temperatura de equilibrio sea

21C.

1 answer:

elena55 [62]2 years ago

6 0

Answer:

V = 0.364 litros

Explanation:

4 ltr (1000 gm/ltr) = 4000 gm

4000(20) + m(32) = (4000 + m)(21)

4000(20) + m(32) = 4000(21) + m(21)

m(32 - 21) = 4000(21 - 20)

11m = 4000

m = 363.6363...gm

363.6363 gm / 1000gm/ltr = 0.3636363... ltr

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Why will a struck tuning fork sound louder when it is held against a table?.

QveST [7]

Answer:
1.)the surface of the table is set into vibration
2.)the initial vibration was larger
3.)The entire table vibrates

Explanation:<u><em>hope this helps :3 have a good day xlXCherryColaXlx..(pls mark me brainliest)</em></u>

7 0

2 years ago

What are the sign and magnitude in coulomb's of a point charge that produces a potential of -1.50 V at a distance of 2.00 mm

Charra [1.4K]

Answer:

The sign of the charge is negative

The magnitude of the charge is 3.33 x 10⁻¹³ C

Explanation:

Given;

potential difference, V = -1.5 V

distance of the point charge, r = 2 mm = 2 x 10⁻³ m

The magnitude of the charge is calculated as follows;

$V = \frac{kq}{r} \\\\q = \frac{Vr}{k} \\\\where;\\\\k \ is \ coulomb's \ constant = 9\times 10^9 \ Nm^2/C^2\\\\q = \frac{-1.5 \times 2\times 10^{-3}}{9\times 10^9 } \\\\q = -3.33 \times 10^{-13} \ C\\\\Magnitude \ of \ the\ charge, q = 3.33 \times 10^{-13} \ C$

8 0

3 years ago

A solid circular shaft and a tubular shaft, both with the same outer radius of c=co = 0.550 in , are being considered for a part

Norma-Jean [14]

Answer:

The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp

Explanation:

<u>Polar moment of Inertia</u>

$(I_p)s = \frac{\pi(0.55)4}2$

= 0.14374 in 4

<u>Maximum sustainable torque on the solid circular shaft</u>

$T_{max} = T_{allow} \frac{I_p}{r}$

= $(14 \times 10^3) \times (\frac{0.14374}{0.55})$

= 3658.836 lb.in

= $\frac{3658.836}{12}$ lb.ft

= 304.9 lb.ft

<u>Maximum sustainable torque on the tubular shaft</u>

$T_{max} = T_{allow}( \frac{Ip}{r})$

= $(14 \times10^3) \times ( \frac{0.13101}{0.55})$

= 3334.8 lb.in

= $(\frac{3334.8}{12} )$ lb.ft

= 277.9 lb.ft

<u>Maximum sustainable power in the solid circular shaft</u>

$P_{max} = 2 \pi f_T$

= $2\pi(2.1) \times 304.9$

= 4023.061 lb. ft/s

= $(\frac{4023.061}{550})$ hp

= 7.315 hp

<u>Maximum sustainable power in the tubular shaft</u>

$P _{max,t} = 2\pi f_T$

= $2\pi(2.1) \times 277.9$

= 3666.804 lb.ft /s

= $(\frac{3666.804}{550})$ hp

= 6.667 hp

7 0

3 years ago

Space-shuttle astronauts experience accelerations of about 35 m/s2 during takeoff. What force does a 75 kg astronaut experience

amm1812

Answer:

<h3>The answer is 2625 N</h3>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 75 × 35

We have the final answer as

<h3>2625 N</h3>

Hope this helps you

5 0

3 years ago

An object is at rest in front of a compressed spring. It travels over a surface that exerts a kinetic frictional force on it and

PolarNik [594]

Answer:

the object will travel 0.66 meters before to stop.

Explanation:

Using the energy conservation theorem:

$E_i+K_i+W_f=K_f+U_f$

The work done by the friction force is given by:

$W_f=F_f*d\\W_f=\µ*m*g*d\\W_f=0.35*4*9.81*d\\W_f=13.7d[J]$

so:

$\frac{1}{2}1800*(10*10^{-2})+0-13.7d=0+0\\d=0.66m$

3 0

3 years ago