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4 years ago

9

You throw a football straight up. Air resistance can be neglected. (a) When the football is 4.00 m above where it left your hand

, it is moving upward at 0.500 m/s. What was the speed of the football when it left your hand

1 answer:

Oksana_A [137]4 years ago

4 0

Answer:

The speed of the football when it lift your hand is 8.86 m/s.

Explanation:

Given that,

Final speed of the football, v = 0.5 m/s

Height above the ground, s = 4 m

We need to find the speed of the football when it left your hand. The ball will move under the action of gravity. Using third equation of motion as:

$v^2-u^2=2gs\\\\u^2=v^2-2gs\\\\u^2=(0.5)^2-2\times (-9.8)\times 4\\\\u=8.86\ m/s$

So, the speed of the football when it lift your hand is 8.86 m/s. Hence, this is the required solution.

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M (b) Calculate the speed of an electron that is accelerated through the same electric potential difference.

MaRussiya [10]

The speed of a electron that is accelerated from rest through an electric potential difference of 120 V is $6.49\times10^6m/s$

<h3>How to calculate the speed of the electron?</h3>

We know, that the energy of the system is always conserved.

Using the Law of Conservation of energy,

$\triangle K+\triangle$ U=0

Here, K is the kinetic energy and U is the potential energy.

Now, substituting the formula of U and K, we get:

$(\frac{1}{2} mv^2)+(-q\triangle V)$ =0------(1)

Here,

m is the mass of the electron

v is the speed of the electron

q is the charge on the electron

V is the potential difference

Let $v_f$ and $v_i$ represent the final and initial speed.

Here, $v_i$ =0

Solving for $v_f$ , we get:

$v_f=\sqrt{\frac{-2q\triangle V}{m} }$

$=\sqrt{\frac{2\times1.602\times10^{-19}\times 120}{9.109\times10^{-31}} }$

=6.49 $\times10^6$ m/s

To learn more about the conservation of energy, refer to:

brainly.com/question/2137260

#SPJ4

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1 year ago

1) Determine the magnitude of energy for each of the blanks on the diagram. Give the correct values for 1A, 1B, and 1C.

morpeh [17]

Answer:

Explanation: y’all taking the same test as me hahahahah I got the answers but I can’t attach the picture here so hit me up on snap daniela_0789

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3 years ago

Recording the location of a star requires a measurement of:

Scorpion4ik [409]

Answer:

It requires a measurement of altitude azimuth time.

Hope this helps, if it did, please give it a brainliest.

7 0

3 years ago

A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist

Paha777 [63]

Answer:

Part a)

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

Explanation:

As we know by energy conservation the total energy at the bottom of the bowl is given as

$\frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy

now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

$\frac{1}{2}mv^2 = mgh'$

now we know that

$I = \frac{2}{5}mr^2$

$\omega = \frac{v}{r}$

so we have

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh$

$\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh$

$\frac{7}{10}mv^2 = mgh$

$\frac{1}{2}mv^2 = \frac{10}{14}mgh$

so the height on the smooth side is given as

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

6 0

4 years ago

Read 2 more answers

What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL

lozanna [386]

Answer:

$\displaystyle \Delta x=1.74\ m$

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

$\displaystyle PE = \frac{1}{2}k(\Delta x)^2$

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

$\displaystyle \Delta x=\swrt{\frac{2PE}{k}}$

Substituting:

$\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}$

Calculating:

$\displaystyle \Delta x=\sqrt{3.0175}$

$\boxed{\displaystyle \Delta x=1.74\ m}$

6 0

3 years ago