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2 years ago

13

I agree to hire you for 30 days. You can decide between two methods of payment: either (1) $1000 a day, or (2) one penny on the

first day, two pennies on the second day and continue to double your daily pay each day up to day 30. Use quick estimation to make your decision, and justify it

1 answer:

sertanlavr [38]2 years ago

3 0

Answer:

The second deal.

Explanation:

I had that before, if u pick the second one you will get twice as much in a year as the first one.

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SOMEONE HELP ME PLEASE!!!!!

enyata [817]

Answer:

I think it would be the 3rd or 4th one.

8 0

3 years ago

A basketball player standing under the hoop shoots the ball straight up with an initial velocity of v0Part (a) What is the maxim

Mama L [17]

Answer:

Maximum height will be $h=\sqrt{\frac{v_0^2}{2g}}$

Explanation:

We have given initial velocity through which basketball is thrown $=v_0$

Acceleration due to gravity $=g\ m/sec^2$

At maximum height velocity will be zero

So final velocity v = 0 m /sec

According to third equation of motion $v^2=u^2+2gh$

$0^2=v_0^2-2gh$ ( Negative sign is due to upward acceleration )

$h=\sqrt{\frac{v_0^2}{2g}}$

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3 years ago

Kepler’s third law states that for all objects orbiting a given body, the cube of the semimajor axis (A) is proportional to the

GarryVolchara [31]

Explanation:

Kepler’s third law states that for all objects orbiting a given body, the cube of the semimajor axis (A) is proportional to the square of the orbital period (P).

For each of our planets orbiting the Sun, the relationship between the orbital period and semimajor axis can be represented by the equation as:

$P^2=kA^3$

k is constant of proportionality

It is required to solve the above equation for k

$k=\dfrac{P^2}{A^3}$

8 0

3 years ago

Chapter 16, Problem 63. A person is standing in a room at 18 ◦C. The exposed surface area and skin temperature of the person are

Juliette [100K]

Answer: $Q=248.011 W$

Explanation:

Given

Temperature of Room $T_{\infty }=18^{\circ}\approx 291 K$

Area of Person $A=1.7 m^2$

Temperature of skin $T=32^{\circ}\approx 305 K$

Heat transfer coefficient $h=5 W/m^2.k$

Emissivity of the skin and clothes $\epsilon =0.9$

$\Delta T=32-18=14^{\circ}$

Total rate of heat transfer=heat Transfer due to Radiation +heat transfer through convection

Heat transfer due radiation $Q_1=\epsilon \sigma A(T^4-T_{\infty }^4 )$

where $\sigma =stefan-boltzman\ constant$

$Q_1=0.9\times 5.687\times 10^{-8}\times 1.7\times (305^4-291^4)$

$Q_1=129.01 W$

Heat Transfer due to convection is given by

$Q_2=hA(\Delta T)$

$Q_2=5\times 1.7\times 14=119 W$

$Q=Q_1+Q_2$

$Q=129.01+119=248.011 W$

7 0

3 years ago

How does the density of water change when: (a) it is heated from 0o

Radda [10]

Answer:

[b] it id heated from 4o

Explanation:

3 0

3 years ago