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snow_tiger [21]

4 years ago

10

An 80 kg man is standing in an elevator which is moving down and slowing down at a rate of 2.0 m/s2. how hard must the floor of

the elevator push up on the man?

1 answer:

V125BC [204]4 years ago

8 0

F = mass * acceleration
F = 80kg * 2m/s^2
F = 160N

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Can someone help? Thanks.

Lilit [14]

The average speed of Frank's car is 2 m/s and the average speed of Noah's car is 1.8 m/s.

<u>Explanation:</u>

Average speed is the measure of total distance covered at different time period. Since, the formula used for calculating the average speed is the ratio of total distance covered by each car to the time taken to cover that distance.

As there is only one set of data i.e., distance and time, the average speed will be equal to the speed of the car.

So in this case, the total distance covered by franks car is 300 cm = 3 m in 1.5 s. Then, the average speed will be

$Avg. S_{frank} = \frac{Total\ distance\ rolled\ by\ his\ car}{Time\ taken}$

$Avg.S_{frank} =\frac{300 \times 10^{-2} }{1.5}=2\ m/s$

Similarly, the average speed of Noahs car which rolled a distance of 360 cm = 3.6 m in time 2 s, will be

$Avg.S_{Noah} =\frac{360 \times 10^{-2} }{2}=1.8\ m/s$

Thus, the average speed of Frank's car is 2 m/s and the average speed of Noah's car is 1.8 m/s.

4 0

3 years ago

A computer is purchased for $2816 and depreciates at a constant rate to $0 in 8 years. Find a formula for the value, V , of the

marusya05 [52]

Answer:

The formula its $f(t) \ = \ - \ 352 \ \frac{\$ }{years} \ t \ + \ \$ \ 2816$
After 5 years, the computer value its $ 1056

Explanation:

<h3>Obtaining the formula</h3>

We wish to find a formula that

Starts at 2816. $f(0 \ years) \ = \ \$ \ 2816$
Reach 0 at 8 years. $f( 8 \ years) \ = \ \$ \ 0$
Depreciates at a constant rate. m

We can cover all this requisites with a straight-line equation. (an straigh-line its the only curve that has a constant rate of change) :

$f(t) \ = \ m\ t \ + \ b$ ,

where m its the slope of the line and b give the place where the line intercepts the <em>y</em> axis.

So, we can use this formula with the data from our problem. For the first condition:

$f ( 0 \ years ) = m \ (0 \ years) + b = \$ \ 2816$

$b = \$ \ 2816$

So, b = $ 2816.

Now, for the second condition:

$f ( 8 \ years ) = m \ (8 \ years) + \$ \ 2816 = \$ \ 0$

$m \ (8 \ years) = \ - \$ \ 2816$

$m = \frac{\ - \$ \ 2816}{8 \ years}$

$m = \frac{\ - \$ \ 2816}{8 \ years}$

$m = \ - \ 352 \frac{\$ }{years}$

So, our formula, finally, its:

$f(t) \ = \ - \ 352 \ \frac{\$ }{years} \ t \ + \ \$ \ 2816$

<h3>After 5 years</h3>

Now, we just use <em>t = 5 years</em> in our formula

$f(5 \ years) \ = \ - \ 352 \ \frac{\$ }{years} \ 5 \ years \ + \ \$ \ 2816$

$f(5 \ years) \ = \ - \$ \ 1760 + \ \$ \ 2816$

$f(5 \ years) \ = $ \ 1056$

4 0

4 years ago

Which of the following is a correct step in the process of adding 8.0 x 10^ -2 and 6.0 x 10^ -3

dybincka [34]

Explanation:

In order to compute correctly the sum of the two terms, we have to rewrite one of them such that they have the same exponent.

The two terms are:

$8.0 \cdot 10^{-2}$

$6.0 \cdot 10^{-3}$

For instance, we can re-write the second term such as it also has a power $10^{-2}$ . In order to do that, we have to move the decimal point one place to the left, therefore:

$6.0 \cdot 10^{-3} = 0.6\cdot 10^{-2}$

At this point, the two numbers have the same exponent, so we can just add them together by adding the bases and keeping the same exponent, -2:

$8.0\cdot 10^{-2} + 0.6 \cdot 10^{-2} = (8.0 + 0.6) \cdot 10^{-2} = 8.6\cdot 10^{-2}$

8 0

3 years ago

To water the yard you use a hose with a diameter of 3.2 cm. Water flows from the hose with a speed of 1.1 m/s. If you partially

cupoosta [38]

Answer:

The speed of water flow inside the pipe at point - 2 = 34.67 m / sec

Explanation:

Given data

Diameter at point - 1 = 3.2 cm

Velocity at point - 1 = 1.1 m / sec = 110 cm / sec

Diameter at point - 2 = 0.57 cm

Velocity at point - 2 = ??

We know that from the continuity equation the rate of flow is constant inside a pipe between two points.

Thus

⇒ $A_{1}$ × $V_{1}$ = $A_{2}$ × $V_{2}$

⇒ $\frac{\pi }{4}$ × $d_{1} ^{2}$ × $V_{1}$ =

⇒ $d_{1} ^{2}$ × $V_{1}$ = $d_{2} ^{2}$ × $V_{2}$

⇒ $(3.2)^{2}$ × 110 = $(0.57)^{2}$ × $V_{2}$

⇒ $V_{2}$ = 3467 cm / sec

⇒ $V_{2}$ = 34.67 m / sec

Thus the speed of water flow inside the pipe at point - 2 = 34.67 m / sec

3 0

3 years ago

Examine the scenario. Two neutral objects, a balloon and a sweater, are rubbed against each other. Which choice most accurately

Ne4ueva [31]

Electrons move from the sweater to the balloon. The sweater becomes positively charged, while the balloon becomes negatively charged.

Explanation:

Sweater is a conductive material, which means it readily gives away its electrons.

Consequently, when you rub a balloon on Sweater, this causes the electrons to move from the Sweater to the balloon's surface.
The rubbed part of the balloon now acquired a negative charge. Objects made of rubber, such as the balloon, are basically electrical insulators, meaning that they resist electric charges flowing through them.
This is why only part of the balloon may have a negative charge (where the wool rubbed it) and the rest may remain neutral after electrostatic process.

4 0

3 years ago