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3241004551 [841]

3 years ago

15

A cannon elevated at 40 degrees is fired at a wall 300m away on level ground. The initial speed of the cannonball is 89m/s How l

ong does it take for the ball to hit the wall? At what height does the ball hit the Wall?

1 answer:

lianna [129]3 years ago

3 0

For the answer to the question above, we'll have to use these formulas.

A) to find time to travel the 300m,
just find horizontal component of the velocity and divide.
ie x=89 x t x cos 40, t=x/89 x cos 40

<span>B) y=vtsin 40 - gt^2/2, just sub in
</span>
I believe you can do the rest.

I hope I helped you with my answers.

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the plane of a 5.0 cm by 8.0 cm rectangular loop wire is parallel to a 0.19 t magnetic field. if the loop carries a current of 6

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Here is the state ment :0.2 loop

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3 years ago

Allen Aby sets up an Atwood Machine and wants to find the acceleration and the tension in the string. Please help him. Two block

Vitek1552 [10]

<u>Answers:</u>

In order to solve this problem we will use Newton’s second Law, which is mathematically expressed after some simplifications as:

<h2> $F=ma$ (1) </h2>

This can be read as: The Net Force $F$ of an object is equal to its mass $m$ multiplied by its acceleration $a$ .

We will also need to <u>draw the Free Body Diagram of each block</u> in order to know the direction of the acceleration in this system and find the Tension $T$ of the string (<u>See figure attached). </u>

We already know<u> $m_{2}$ is greater than $m_{1}$ </u>, this means the weight of the block 2 $P_{2}$ is greater than the weight of the block 1 $P_{1}$ ; therefore <u>the acceleration of the system will be in the direction of $P_{2}$ </u>, as shown in the figure attached.

We also know by the information given in the problem that <u>the pulley does not have friction and has negligible mass</u>, and <u>the string is massless</u>.

This means that the tension will be the same along the string regardless of the difference of mass of the blocks.

Now that we have the conditions clear, let’s begin with the calculations:

1) Firstly, we have to find the weight of each block, in order to verify that block 2 is heavier than block 1.

This is done using equation (1), where the force of the weight $P$ is calculated using the <u>acceleration of gravity</u> $g=9.8\frac{m}{s^{2}}$ acting on the blocks:

<h2> $P=mg$ (2) </h2>

<u>For block 1: </u>

$P_{1}=m_{1}g$ (3)

$P_{1}=1.5kg(9.8\frac{m}{s^{2}})$

<h2> $P_{1}=14.7N$ (4) </h2>

<u>For block 2: </u>

$P_{2}=m_{2}g$ (5)

$P_{2}=2.4kg(9.8\frac{m}{s^{2}})$

<h2> $P_{2}=23.52N$ (6) </h2>

Then, we are going to <u>find the acceleration $a$ of the whole system: </u>

$F_{r}=P_{1}+P_{2}$ (7)

<h2> $P_{1}+P_{2}=(m_{1}+m_{2})a$ (8) </h2>

Where the Resulting Force $F_{r}$ is equal to the sum of the weights $P_{1}$ and $P_{2}$ .

In the figure attached, note that $P_{1}$ is in opposite direction to the acceleration $a$ , this means it must <u>have a negative sing</u>; while $P_{2}$ is in the same direction of $a$ .

Here we only have to isolate $a$ from equation (8) and substitute the values according to the conditions of the system:

$-14.7N+23.52N=(1.5kg+2.4kg)a$

$8.82N=(3.9kg)a$

Then:

$a=\frac{8.82N }{3.9kg}$

<h2> $a=2.26\frac{m}{ s^{2}}$ </h2><h2>This is the acceleration of the system. </h2>

2) For the second part of the problem, we have to find the tension $T$ of the string.

We can choose either the Free Body Diagram of block A or block B to make the calculations, <u>the result will be the same</u>.

Let’s prove it:

For $m_{1}$

we see in the free body diagram that the <u>acceleration is in the same direction of the tension of the string</u>, so:

$F_{r}=T-P_{1}$ (9)

$T-P_{1}=m_{1}a$ (10)

$T-14.7N=(1.5kg)( 2.26\frac{m}{ s^{2}})$

Then;

<h2> $T=18.09N$ This is the tension of the string </h2><h2> </h2>

For $m_{2}$

we see in the free body diagram that the acceleration is in opposite direction of the tension of the string and must <u>have a negative sign,</u> so:

$F_{r}=T-P_{2}$ (9)

$T-P_{2}=m_{2}a$ (10)

$T-23.52N=(2.4kg)(-2.26\frac{m}{ s^{2}})$

Then;

<h2> $T=18.09N$ This is the same tension of the string </h2>

6 0

3 years ago

Olegator [25]

Answer:

v2^2 - v1^2 = 2 g s fundamental formula

v2 = v1 + 2 g = v1 + 19.8 increase in velocity in 2 sec

v1^2 + 39.6 v1 + 392 - v1^2 = 2 * 9.8 * 123.1 = 2412.76

v1 = (2412.76 - 392) / 39.6 = 51.03

v2 = 51.03 + 19.6 = 70.63

T = 70.63 / .8 = 7.207 sec time to fall height of tower

S = 1/2 g T^2 = 4.9 * 7.207^2 = 254.5 m

(Note v2^2 - v1^2 = 70.63^2 - 51.03^2 = 2385 m

2385 / (2 * 9.8) = 122 m (close to 123.1 as was given

6 0

3 years ago

When forming a bond, an atom that has 3 electrons in its second shell and a filled first shell will?

Butoxors [25]

Macromolecule polymers are assembled by the connecting of monomers. An -OH group is detached from one monomer and a hydrogen atom is detached from an additional in a procedure named dehydration synthesis in the monomers bond. For every subunit supplementary to a macromolecule in which one water molecule is detached. Macromolecule polymers are broken down by breaking bonds among subunits. This procedure is named hydrolysis and is the opposite of dehydration. During hydrolysis the hydrogen atom is supplementary to one monomer and a hydroxyl cluster to the other and by breaking the covalent bond in the middle of the monomers.

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3 years ago

Find the wavelength of a photon which has 500.0 eV of energy.

jekas [21]

Answer:

2.5 x 10⁻⁹ m

Explanation:

E = Energy of photon = 500 eV = 500 x 1.6 x 10⁻¹⁹ J

c = speed of photon = 3 x 10⁸ m/s

λ = wavelength of photon = ?

Energy of photon is given as

$E = \frac{hc}{\lambda }$

inserting the values

$500\times 1.6\times 10^{-19} = \frac{(6.63\times 10^{-34})(3\times 10^{8})}{\lambda }$

λ = 2.5 x 10⁻⁹ m

5 0

4 years ago