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Drupady [299]
3 years ago
5

An attacker at the base of a castle wall 3.60 m high throws a rock straight up with speed 8.00 m/s from a height of 1.70 m above

the ground. Does the change in speed of the downward-moving rock agree with the magnitude of the speed change of the rock moving upward between the same elevations?
Physics
1 answer:
ryzh [129]3 years ago
6 0

Answer:

we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances

Explanation:

given data

base = 3.60 m

speed u = 8 m/s

height = 1.70 m

to find out

check change in speed

solution

we know here formula for v  that is

v² = u² - 2gh      ............1        for upward speed

v² = u² + 2gh     ............2        for projected speed

so here put all value and find v with h = 3.60 - 1.70 = 1.9 m

v² = 8² - 2(9.8) 1.9  = 26.76

v² = 8² + 2(9.8) 1.9   = 101.24

v = 5.173  m/s    ..............3

v = 10.061 m/s   ...................4

so change in speed form 3 and 4 equation

change in speed = v - u = 8 - 5.173  = 2.827 m/s     .................5

change in speed = v - u = 10.061 - 8 = 2.061 m/s     ..................6

so now we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances

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Inga [223]

Answer: Find the answer in the explanation

Explanation: Given the Roman numeral and the representation

I. part of a coal-fired power plant

II. part of a nuclear power plant

III. part of a coal-fired power plant and part of a nuclear power plant

a.) Boiler : I

b.) Combustion chamber: I

c.) Condenser: I

d.) Control rod: II

e.) Generator: III

f.) Turbine: III

Toward the end processes part of both coal fire and nuclear power, they both make use of turbine and generator to generate electricity.

7 0
3 years ago
A car traveling at 50 m/s comes to a stop in 5 seconds. What is the<br> acceleration of this car?
Alexandra [31]

Answer:10

Explanation:

You have to do speed divided by time so your answers 10

4 0
2 years ago
Shay reacts solid zinc and aqueous copper sulfate to form aqueous zinc sulfate and solid copper. If he reacts 10.1 grams of zinc
zhenek [66]

Answer:

Now since mass of reactant is equal to mass of the product after the reaction so we can say that mass conservation is applicable here

Explanation:

As we know that zinc reacts with copper sulfate

so the reaction is given as

Zn + CuSO_4 --> ZnSO_4 + Cu

so here we have

Zn = 10.1 g

CuSO_4 = 18.6 g

ZnSO_4 = 20 g

Cu = 8.7 g

Now total mass of reactant is given as

M_1 = 10.1 + 18.6 = 28.7 g

Mass of the product is given as

M_2 = 20 + 8.7 = 28.7 g

Now since mass of reactant is equal to mass of the product after the reaction so we can say that mass conservation is applicable here

7 0
3 years ago
The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.180 × 10-3 m3. Into this cavity is placed 1.100 × 10-
nikdorinn [45]

Answer:

The answer is "60.74^{\circ}".

Explanation:

Cavity and benzene should be extended in equal quantities.

\to 1.18 \times 10^{-3}\times (1+ \Delta T \times 0.000051) = 1.1\times 10^{-3} \times (1+ \Delta T \times 0.00124)\\\\\to  (\frac{1.18}{1.1})\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 - 1- \Delta T \times 0.00124=0\\\\

\to 0.072+ \Delta T \times 0.000054672 - \Delta T \times 0.00124=0\\\\ \to 0.072+ \Delta T ( 0.000054672 -0.00124)=0\\\\ \to \Delta T ( 0.000054672 -0.00124)= -0.072\\\\ \to \Delta T = -\frac{0.072}{( 0.000054672 -0.00124)}\\\\ \to \Delta T = -\frac{0.072}{-0.001185328 }\\

\to \Delta T = \frac{0.072}{0.001185328 }\\\\ \to \Delta T = 60.74^{\circ}\\

5 0
3 years ago
The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the
tigry1 [53]

Answer:

A) The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) The maximum speed of the plastic sphere will be 2.2 m/s

Explanation:

Hi there!

I´ve found the complete problem on the web:

<em>A toy launcher that is used to launch small plastic spheres horizontally contains a spring with a spring constant of 50. newtons per meter. The spring is compressed a distance of 0.10 meter when the launcher is ready to launch a plastic sphere.</em>

<em>A) Determine the elastic potential energy stored in the spring when the launcher is ready to launch a plastic sphere.</em>

<em>B) The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the plastic sphere will be launched. [Neglect friction.] [Show all work, including the equation and substitution with units.]</em>

<em />

A) The elastic potential energy (EPE) is calculated as follows:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compressing distance

EPE = 1/2 · 50 N/m · (0.10 m)²

EPE = 0.25 J

The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) Since there is no friction, all the stored potential energy will be converted into kinetic energy when the spring is released. The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the sphere.

v = velocity

The kinetic energy of the sphere will be equal to the initial elastic potential energy:

KE = EPE = 1/2 · m · v²

0.25 J = 1/2 · 0.10 kg · v²

2 · 0.25 J / 0.10 kg = v²

v = 2.2 m/s

The maximum speed of the plastic sphere will be 2.2 m/s

6 0
3 years ago
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