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1 year ago

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Add the following numbers, using scientific notation and the correct amount of significant digits. 3.4⋅10 8m + 4.56⋅10 9m = ____

_.
A 4.9 ⋅10 9m
B 4.9 ⋅10 8m
C 7.96 ⋅10 9m
D 7.96 ⋅10 8m

2 answers:

Gnoma [55]1 year ago

7 0

The answer is B
<span>B 4.9 ⋅10 8m
</span>

Artist 52 [7]1 year ago

5 0

Answer:

The answer is A) 4.9 × 10^9

Explanation:

I literally just did the calculation, the other dude is wrong.

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Look at these two sentences about Undeposited Funds.1. By posting to Undeposited Funds, you can create a single bank deposit for

antiseptic1488 [7]

The question with the complete options:

Look at these two sentences about Undeposited Funds. 1. By posting to Undeposited Funds, you can create a single bank deposit for multiple payments, making it easy ___________. 2. When receiving a payment, make sure _________________. Which of the options below correctly fills in the blanks? A.)1. To match your bank register with your bank statement; 2. the Deposit to account is Undeposited Funds

B) 1. To match your bank register with your bank statement; 2. the Deposit to account is Checking

C)1. To match your expenses with your bank statement; 2. the Deposit to account is Uncategorized asset

D)1. To match your bank register with your bank statement; 2. the Deposit to account is Uncategorized funds.

Answer: The correct option is A (1. To match your bank register with your bank statement; 2. the Deposit to account is Undeposited Funds)

Explanation: Undeposited funds is a type of account created to keep funds that are not yet deposited in the individuals account. It's a default account which is used by online marketers to keep funds until they are ready to be paid.

By posting to Undeposited Funds, you can create a single bank deposit for multiple payments, making it easy to match your bank register with your bank statement. When receiving a payment, make sure the Deposit to account is Undeposited Funds.

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4 0

1 year ago

A 0.50 kilogram ball is held at a height of 20 meters. What is the kinetic energy of the ball when it reaches halfway after bein

Margarita [4]

Potential energy at any point is (M G H). On the way down, only H changes. So halfway down, half of the potential energy remains, and the other half has turned to kinetic energy. Half of the (M G H) it had at the tpp is (0.5 x 9.8 x 10) = 49 joules.

6 0

1 year ago

Read 2 more answers

A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.

Sedaia [141]

Answer:

335°C

Explanation:

Heat gained or lost is:

q = m C ΔT

where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Heat gained by the water = heat lost by the copper

mw Cw ΔTw = mc Cc ΔTc

The water and copper reach the same final temperature, so:

mw Cw (T - Tw) = mc Cc (Tc - T)

Given:

mw = 390 g

Cw = 4.186 J/g/°C

Tw = 22.6°C

mc = 248 g

Cc = 0.386 J/g/°C

T = 39.9°C

Find: Tc

(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)

Tc = 335

7 0

1 year ago

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit

inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses $x_1$ distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1$

When speed doubles

$\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2$

divide 1 and 2

$\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}$

$x_2=2x_1$

Therefore spring compresses twice the initial value

7 0

1 year ago

A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn

EastWind [94]

(a) 12.0 V

In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, $C=5.00 \mu F$ , the charged stored on the capacitor at the end of the process is

$Q=CV=(5.00 \mu F)(12.0 V)=60 \mu C$

When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship

$V=\frac{Q}{C}$

and since neither Q nor C change, the voltage V remains the same, 12.0 V.

(b) (i) 24.0 V

In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:

$C=\frac{\epsilon_0 \epsilon_r A}{d}$

where:

$\epsilon_0$ is the permittivity of free space

$\epsilon_r$ is the relative permittivity of the material inside the capacitor

A is the area of the plates

d is the separation between the plates

As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: $C'=\frac{C}{2}$ . The new voltage across the plate is given by

$V'=\frac{Q}{C'}$

and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is

$V'=\frac{Q}{C'}=\frac{Q}{C/2}=2 \frac{Q}{C}=2V$

So, the new voltage is

$V'=2 (12.0 V)=24.0 V$

(c) (ii) 3.0 V

The area of each plate of the capacitor is given by:

$A=\pi r^2$

where r is the radius of the plate. In this case, the radius is doubled: r'=2r. Therefore, the new area will be

$A'=\pi (2r)^2 = 4 \pi r^2 = 4A$

While the separation between the plate was unchanged (d); so, the new capacitance will be

$C'=\frac{\epsilon_0 \epsilon_r A'}{d}=4\frac{\epsilon_0 \epsilon_r A}{d}=4C$

So, the capacitance has increased by a factor 4; therefore, the new voltage is

$V'=\frac{Q}{C'}=\frac{Q}{4C}=\frac{1}{4} \frac{Q}{C}=\frac{V}{4}$

which means

$V'=\frac{12.0 V}{4}=3.0 V$

3 0

1 year ago