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3 years ago

6

A shopping cart is initially at rest when it undergoes a constant acceleration. After

1 answer:

Luba_88 [7]3 years ago

4 0

The answer is A.3.3 m/s2

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(20 points) You are at the center of a boat and have been rowing the boat for a long time. You weigh only 80 kg and your 120 kg

valina [46]

Answer:

Explanation:

From the given information:

Let the first weight be $m_ 1$ = 80 kg

The weight of the buddy be $m_2$ = 120 kg

The weight of Bubba be $m_3$ = 60 kg

Also, since you and Budda are a distance of 4m to each other, then the length to which both meet buddy will be:

$x_1 = x_3 = \dfrac{4}{2} \\ \\ = 2$

The length of the boat be $x_2$ = 4 m

∴

We can find the center of mass of the system by using the formula:

$X_{CM} = \dfrac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\ \\ X_{CM} = \dfrac{(80 \times 2)+(120\times4)+(60\times2)}{80+120+60} \\ \\ X_{CM} = \dfrac{160+480+120}{260} \\ \\ \mathbf{X_{CM} = 2.923}$

4 0

2 years ago

How much time does it take for a car to accelerate from 3.44 m/s to 20.9 m/s if the average acceleration is 6.00 m/s^3

miv72 [106K]

Do you go to GA Connexus Academy??

4 0

3 years ago

Read 2 more answers

What type of electromagnetic radiation is being used in the picture?

pashok25 [27]

Answer: b i think

Explanation:

6 0

3 years ago

Darcy is going to make raspberry jam for the county fair.

valentinak56 [21]

Answer:

She can make have 30 jars with raspberries in them with 50 left over.

Explanation:

1,700 divided by 55

30 equally

but 50 left over

This means that she can make have 30 jars with raspberries in them with 50 left over.

5 0

3 years ago

A cart with mass m vibrating at the end of a spring has an extra block added to it when its displacement is x=+A. What should th

Pavel [41]

Answer:

The block's mass should be $3m$

Explanation:

Given:

Cart with mass $m$

From the conservation of energy before mass is added,

$\frac{1}{2} mv^{2} = \frac{1}{2} kA^{2}$

Where $A =$ amplitude of spring mass system, $k =$ spring constant

$A = v\sqrt{\frac{m}{k} }$

Now new mass $M$ is added to the system,

$\frac{1}{2} (m +M ) v^{2} = \frac{1}{2} k A^{2}$

$A = v \sqrt{\frac{m +M }{k} }$

Here, given in question frequency is reduced to half so we can write,

$f' = \frac{f}{2}$

Where $f =$ frequency of system before mass is added, $f' =$ frequency of system after mass is added.

$\omega ' = \frac{\omega}{2}$

$\sqrt{\frac{k}{m +M} } = \frac{\sqrt{\frac{k}{m} } }{2}$

$\frac{k}{m +M } = \frac{k}{4m}$

$M = 3m$

Therefore, the block's mass should be $3m$

8 0

3 years ago