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2 years ago

10

I need help ASAP

1 answer:

dolphi86 [110]2 years ago

7 0

Answer:

The horizontal applied force is 150.67 N.

Explanation:

We have,

Distance covered by the applied force on the couch is 0.75 m

Work done by the couch is 113 J

It is required to find the magnitude of the horizontal force applied. Due to the application of force work is done by an object. It can be given by :

$W=Fd$

F is horizontal force applied

$F=\dfrac{W}{d}\\\\F=\dfrac{113}{0.75}\\\\F=150.67\ N$

So, the horizontal applied force is 150.67 N.

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An astronaut holds a rock 100m above the surface of Planet X . The rock is then thrown upward with a speed of 15m/s , as shown i

Butoxors [25]

Answer: $5 m/s^{2}$

Explanation:

The described situation is is related to vertical motion (and free fall). So, we can use the following equation that models what happens with this rock:

$y=y_{o}+V_{o}sin\theta t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0m$ is the rock's final height

$y_{o}=100 m$ is the rock's initial height

$V_{o}=15 m/s$ is the rock's initial velocity

$\theta=90\°$ is the angle at which the rock was thrown (directly upwards)

$t=10 s$ is the time

$g$ is the acceleration due gravity in Planet X

Then, isolating $g$ and taking into account $sin(90\°)=1$ :

$g=(-\frac{2}{t^{2}})(y-y_{o}-V_{o}t)$ (2)

$g=(-\frac{2}{(10 s)^{2}})(0 m-100 m-(15 m/s)(10 s))$ (3)

Finally:

$g=5 m/s^{2}$ (4) This is the acceleration due gravity in Planet X

7 0

2 years ago

A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, take

adell [148]

Answer:

$F_{x}=2.31N$ to the right.

$F_{y}=2.1N$ to in the upwards direction.

Explanation:

In order to solve this problem, we must first start by drawing a diagram of the situation. (See attached diagram).

So, remember that a force is determined by multiplying the mass of the parcticle by its acceleration:

F=ma

so in order to find the components of the force, we need to start by finding its acceleration.

Acceleration is found by using the following formula:

$a=\frac{V_{f}-V{0}}{t}$

so we can subtract the two vectors, like this:

$a=\frac{(6.00i+4.0j)m/s-1.60i m/s}{8s}$

which yields:

$a=\frac{(4.4i+4.0j)m/s}{8s}$

or:

$a=(0.55i + 0.5j) m/s^{2}$

so now I can find the components of the force:

$F=(4.2kg)(0.55i + 0.5j) m/s^{2}$

which yields:

F=(2.31i+2.1j)N

so the components of the force are:

$F_{x}=2.31N$ to the right.

$F_{y}=2.1N$ to in the upwards direction.

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2 years ago

The A-string (440 HzHz) on a piano is 38.9 cmcm long and is clamped tightly at both ends. If the string tension is 667N, what's

Mice21 [21]

Answer:

Mass, m = 2.2 kg

Explanation:

It is given that,

Frequency of the piano, f = 440 Hz

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The frequency in the spring is given by :

$f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$

$\mu=\dfrac{m}{L}$ is the linear mass density

On rearranging, we get the value of m as follows :

$m=\dfrac{T}{4Lf^2}$

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or

m = 2.2 grams

So, the mass of the object is 2.2 grams. Hence, this is the required solution.

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erastova [34]

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