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4 years ago

List four ways a person can withdraw money from a checking account.

1 answer:

8 0

1. By using an ATM to withdraw money with a debit card
2. By using an ATM to withdraw money with a credit card
3. Going in the bank and requesting a withdrawal
4. Using a bank app to withdraw money

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A mass executes SHM at the end of a light spring. (a) What fraction of the total energy of the system is potential and what frac

PolarNik [594]

Answer:

Explained

Explanation:

A) The total energy of the system is defined by the energy at maximum amplitude, which we'll call A. At that point, the energy of the system is

E = 1/2×m×A^2;

since energy is conserved, this is also the total amount of energy that the system ever has.

So at x=1/2A,

the potential energy of the system is 1/8×m×A^2

which is one-fourth of the system's total energy. Therefore, the remaining three-fourths is kinetic.

B) (i) Doubling the maximum amplitude will quadruple the total energy:

$E= \frac{1}{2}m(2A)^2$

(ii) Doubling the maximum amplitude will double the maximum velocity

$\frac{1}{2}m(2A)^2= \frac{1}{2}mV^2$

(iii) Doubling the maximum amplitude will double the maximum acceleration: m×a = -k(2A)

(iv) Doubling the maximum amplitude leaves the period unchanged:

$T= 2\pi\sqrt{\frac{m}{k} }$

(neither m nor k has changed).

6 0

3 years ago

What does a branching tree diagram show? The evolution in the size of organisms The order in which specific traits may have evol

Anarel [89]

Answer:

a) Species change over time; some traits become more common, others less. This process of change is driven by natural selection. The traits that become more common are the ones that are “adaptive” or “increase fitness” (that is, a creature's chances of living longer and producing more offspring)

Explanation:

because it is

8 0

4 years ago

It is 5.0 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10 km/h

Ostrovityanka [42]

Answer:

Walking burns up more energy,1740000J

Explanation:

Given that the displacement is 5.0km, and running at 10km/h and uses, walking at 3km/hr and uses 290watts:

Energy consumption for running is calculated as:

$700watts=700j/s,d=5000m,v=\frac{25}{9}m/s\\\\\therefore E_c=\frac{d}{v}\times Energy \ usage\\\\=\frac{5000}{\frac{25}{9}}\times 700j/s\\\\=1260000J$

Energy consumption for walking is calculated as:

$290watts=290j/s,d=5000m,v=\frac{5}{6}m/s\\\\\therefore E_c=\frac{d}{v}\times Energy \ usage\\\\=\frac{5000}{\frac{5}{6}}\times 290j/s\\\\=1740000J$

Walking is a slower process hence the need for more energy over longer periods raltive to running the same distance.

Hence walking burns more energy; 1,740,000J. It burns more because you walk for a greater period of time.

5 0

3 years ago

Please help will give branliest

USPshnik [31]

Nabr is the answer bcoz in a chemical reaction reactants react to form new substances and according to your options NaBr should be the answer

6 0

3 years ago

Traveling 221 miles from Boston Back Bay Station to NYC Penn Station takes 3 hours

Nostrana [21]

Answer:

Approximately $116\; \text{miles}$ for the train from Boston to NYC Penn Station.

Approximately $105\; \text{miles}$ for the train from NYC Penn Station to Boston.

Explanation:

Convert minutes to hours:

$\begin{aligned}t(\text{BOS $\to$ NYC}) &= 3\; {\text{hour}} + 40\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &=\left(3 + \frac{2}{3}\right)\; \text{hour}\\ &= \frac{11}{3}\; \text{hour} \end{aligned}$ .

$\begin{aligned}t(\text{NYC $\to$ BOS}) &= 4\; {\text{hour}} + 5\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &= \frac{49}{15}\; \text{hour} \end{aligned}$ .

Calculate average speed of each train:

$\begin{aligned}v(\text{BOS $\to$ NYC}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{11}{3}\; \text{hour}} \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}$ .

$\begin{aligned}v(\text{NYC $\to$ BOS}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{49}{15}\; \text{hour}} \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}$

Assume that it takes a time period of $t$ for the trains to pass by each other after departure. Distance each train travelled would be:

$s(\text{NYC $\to$ BOS}) = v(\text{NYC $\to$ BOS})\, t$ .

$s(\text{BOS $\to$ NYC}) = v(\text{BOS $\to$ NYC})\, t$ .

Since the trains have just passed by each other, the sum of the two distances should be equal to the distance between the stations:

$v(\text{NYC $\to$ BOS})\, t + v(\text{BOS $\to$ NYC})\, t = 221\; \text{mile}$ .

Rearrange and solve for $t$ :

$(v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC}))\, t = 221\; \text{mile}$ .

$\begin{aligned}t &= \frac{221\; \text{mile}}{v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC})} \\ &= \frac{221\; \text{mile}}{\displaystyle \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} + \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}} \\ &= \frac{539}{279}\; \text{hour}\end{aligned}$ .

Distance each train travelled in $t = (539 / 279)\; \text{hour}$ :

$\begin{aligned}s(\text{BOS $\to$ NYC}) &= v\, t \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 116\; \text{mile}\end{aligned}$ .

$\begin{aligned}s(\text{NYC $\to$ BOS}) &= v\, t \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 105\; \text{mile} \end{aligned}$ .

8 0

2 years ago