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EleoNora [17]
2 years ago
7

A rock is pulled back in a slingshot as shown in the diagram below. The elastic on the slingshot is displaced 0.2 meters from it

s initial position. The rock is pulled back with a force of 10 newtons.
When the rock is released, what is its kinetic energy?
Physics
1 answer:
irakobra [83]2 years ago
4 0

Answer:

id

Explanation:

i don't know

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Seema knows the mass of a basketball. What other information is needed to find the ball’s potential energy? the volume and heigh
Wewaii [24]

 It is because the potential energy is similar to MgH.

 When it comes to MgH, it means mass, gravity and height respectively.

 By using the value of acceleration, seema will find the potential energy of a ball.



7 0
3 years ago
Read 3 more answers
An airplane accelerates from a speed of 88m/s to a speed of 132 m/s during a 15 second time interval. How far did the airplane t
Gelneren [198K]

Answer:

1650\:\mathrm{m}

Explanation:

We can use the following kinematics equations to solve this problem:

v_f=v_i+at,\\{v_f}^2={v_i}^2+2a\Delta x.

Using the first one to solve for acceleration:

132=88+a(15),\\15a=44,\\a=\frac{44}{15}=2.9\bar{3}\:\mathrm{m/s^2}.

Now we can use the second equation to solve for the distance travelled by the airplane:

132^2=88^2+2\cdot2.9\bar{3}\cdot \Delta x,\\\Delta x= \frac{9680}{2\cdot2.9\bar{3}},\\\Delta x =\fbox{$ 1650\:\mathrm{m}$}(three significant figures).

6 0
3 years ago
onsider 1000 mL of a 1.00 × 10-4 M solution of a certain acid HA that has a Ka value equal to 1.00 × 10-4. Water was added or re
solmaris [256]

Answer:

The volume of the final solution, V = 0.0305L

Explanation:

Number of moles = Concentration * volume

Concentration of HA = 1.00 * 10⁻⁴M

Volume of HA = 1000mL = 1 L

Number of moles of HA =  1.00 * 10⁻⁴ * 1

Number of moles of HA =  1.00 * 10⁻⁴ mols

Equation of reaction:

HA → H⁺  +  A⁻

If 1 mol of HA produces 1 mol of H⁺  and  A⁻, 1.00 * 10⁻⁴ mol of HA will produce 1.00 * 10⁻⁴ mol of  H⁺  and  A⁻.

Since only 16% dissociation occurs = 0.16

Number of moles of  H⁺ produced = 0.16 *  1.00 * 10⁻⁴

Number of moles of  H⁺ produced = 1.6 * 10⁻⁵mols

Number of moles of  A⁻ produced = 0.16 *  1.00 * 10⁻⁴

Number of moles of  A⁻ produced = 1.6 * 10⁻⁵mols

Since 16% of HA dissociated into  H⁺  and  A⁻, 84% of HA is left

Number of mols of HA left = 0.84 *  1.00 * 10⁻⁴

Number of mols of HA left =  8.4 * 10⁻⁵mols

Concentration = num of moles/volume

Let the volume of the final solution be V

Conc of HA = 8.4 * 10⁻⁵/V

Conc of H⁺ = 1.6 * 10⁻⁵/V

Conc of A⁻ =  1.6 * 10⁻⁵/V

To calculate the dissociation constant

k_{a} = [H^{+} ][A^{-} ]/[HA]

k_{a}= [1.6 * 10^{-5} /V][1.6 * 10^{-5} /V]/[8.4 * 10^{-5} /V]\\k_{a}= 3.05 * 10^{-6} /V\\k_{a} = 1.00 * 10^{-4}\\ 1.00 * 10^{-4} = 3.05 * 10^{-6} /V\\V= 3.05 * 10^{-6}/ 1.00 * 10^{-4}\\V=3.05 * 10^{-2}\\V=0.0305 L

3 0
3 years ago
The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine th
Luba_88 [7]

Here is the complete question

The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine the horizontal shear stress at point H, which is located L  = 20 mm below the centriod

The missing image which is the remaining part of this question is attached in the image below.

Answer:

The horizontal shear stress at point H is  \mathbf{\tau_H \approx  42.604 \ N/mm^2}

Explanation:

Given that :

The internal shear force V  =  80 kN = 80 × 10³ N

The moment of inertia = 64,900,000

The length = 20 mm below the centriod

The horizontal shear stress  \tau can be calculated by using the equation:

\tau = \dfrac{VQ}{Ib}

where;

Q = moment of area above or below the point H

b = thickness of the beam = 10  mm

From the centroid ;

Q = Q_1 + Q_{2}

Q = A_1y_1 + A_{2}y_{2}  

Q = ( ( 70 × 10) × (55) + ( 210 × 15) (90 + 15/2) ) mm³

Q = ( ( 700) × (55) + ( 3150 ) ( 97.5)  ) mm³

Q = ( 38500 +  307125 ) mm³

Q = 345625 mm³

\tau_H = \dfrac{VQ}{Ib}

\tau_H = \dfrac{80*10^3  * 345625}{64900000*10 }

\tau_H = \dfrac{2.765*10^{10}}{649000000 }

\tau_H = 42.60400616 \ N/mm^2

\mathbf{\tau_H \approx  42.604 \ N/mm^2}

The horizontal shear stress at point H is  \mathbf{\tau_H \approx  42.604 \ N/mm^2}

7 0
3 years ago
Do heavier objects fall more slowly than lighter objects?
vredina [299]

Depending on the surface area and weight, either can fall slower.

<u>Explanation: </u>

If you were to drop a feather and a brick off a building at the same height, the brick would reach the ground first. This is due to air resistance and how it slows the feather down compared to the brick. The surface area of the feather creates more resistance compared to its weight, therefore it hits the ground slower. It really all depends on the surface area, weight and the air resistance it creates.

If all air resistance were to be removed, then what would happen is that the two objects would fall and reach the ground at the same time. This is because the objects are in free fall, and because they are in free fall, the two objects would be falling at an acceleration of 9.81 m/s². This is true for all objects in free fall.

4 0
3 years ago
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