Ask question

Ask question

All categories

3 years ago

10

A new business should be based on an entrepreneur's individual interests because the entrepreneur must have sufficient confidenc

e to succeed. be willing to take personal responsibility. have enough determination to work alone. dedicate as many hours as needed to the work.

2 answers:

hammer [34]3 years ago

8 0

Answer: D) dedicate as many hours as needed to the work.

Explanation:

edge 2020

Elza [17]3 years ago

7 0

Answer:

D

Explanation:

You might be interested in

Help me ultra sound vs infrasound

hodyreva [135]

Answer:

ultrasound: high pitch, above the human range of hearing, reflects back

infrasound: longitudinal waves, in the human range, low pitch

Explanation:

4 0

3 years ago

A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3

Advocard [28]

Answer:

r = 5.07 m

Explanation:

given,

velocity of the man , v = 3.43 m/s

centripetal acceleration, a = 2.32 m/s²

magnitude of position of = ?

using centripetal acceleration formula

$a_c =\dfrac{v^2}{r}$

$2.32 =\dfrac{3.43^2}{r}$

$r =\dfrac{3.43^2}{2.32}$

r = 5.07 m

The magnitude of the position vector relative to rotational axis is equal to 5.07 m.

5 0

3 years ago

A Christmas light is made to flash via the discharge of a capacitor. The effective duration of the flash is 0.25 s (which you ca

Sonbull [250]

Answer:

The correct solution is:

(a) $1.375\times 10^{-2} \ J$

(b) $4.43\times 10^{-3} \ C$

(c) $1.42\times 10^{-3} \ F$

(d) $178.57 \ \Omega$

Explanation:

The given values are:

Effective duration of the flash,

ζ = 0.25 s

Average power,

$P_{avg}=55 \ mW$

$=55\times 10^{-3} \ W$

Average voltage,

$V_{avg}=3.1 \ V$

Now,

(a)

⇒ $E=P_{avg}\times \zeta$

On substituting the values, we get

⇒ $=55\times 10^{-3}\times 0.25$

⇒ $=1.375\times 10^{-2} \ J$

(b)

⇒ $E=Q\times V_{avg}$

then,

⇒ $Q=\frac{E}{V_{avg}}$

On substituting the values, we get

⇒ $=\frac{1.375\times 10^{-2}}{3.1}$

⇒ $=4.43\times 10^{-3} \ C$

(c)

⇒ $C=\frac{Q}{V}$

⇒ $=\frac{4.43\times 10^{-3}}{3.1}$

⇒ $=1.42\times 10^{-3} \ F$

(d)

As we know,

⇒ $R=\frac{1}{4C}$

⇒ $=\frac{1}{4\times 1.42\times 10^{-3}}$

⇒ $=\frac{1000}{5.6}$

⇒ $=178.57 \ \Omega$

5 0

3 years ago

A temperature of 20°C is equivalent to approximately?

umka21 [38]

The answer is A.68 degrees

7 0

3 years ago

Read 2 more answers

The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equa

Sonja [21]

Answer:

$m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)$

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

Explanation:

For this case we know the following info:

$c_{pA} > c_{pB}$

Where c means specific heat for the substance A and B.

We also know that the initial temperatures for both sustances are equal:

$T_{iA}= T_{iB}$

We assume that we don't have melting or vaporization in the 2 substances. So we just have presence of sensible heat given by this formula:

$Q = m c_p \Delta T$

And for this case we know that Both A and B are at the same initial temperature when equal amounts of energy are added to them, so then we have this:

$Q_A = Q_B$

And if we replace the formula for sensible heat we got:

$m_A c_{pA} \Delta T_A = m_B c_{pB} \Delta T_B$

And if we replace for the change of the temperature we got:

$m_A c_{pA} (T_{fA} -T_{iA}) = m_B c_{pB} (T_{fB}- T_{iB})$

And since $T_{iA}= T_{iB}= T$ we have this:

$m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)$

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

4 0

3 years ago