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4 years ago

A sample has a mass of 15 g and a volume of 3mL

Physics

1 answer:

bezimeni [28]4 years ago

7 0

Thanks for sharing. The density of the sample
is 5 g/cm³. Do you have a question to ask ?

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Find the net force and acceleration. 15 points. Will give brainliest!

gladu [14]

Answer:

$\boxed{F_{net} = 28.7 \ N}$

$\boxed{a = 2.1 \ m/s^2}$

Explanation:

Finding the net force:

Firstly , we'll find force of Friction:

$F_{k} = (micro)_{k}mg$

Where $(micro)_{k}$ is the coefficient of friction and m = 13.6 kg

$F_{k} = (0.16)(13.6)(9.8)\\$

$F_{k} = 21.32 \ N$

Now, Finding the net force:

$F_{net} = F - F_{k}\\F_{net} = 50 - 21.32\\$

$F_{net} = 28.7 \ N$

Finding Acceleration:

$a = \frac{F_{net}}{m}$

$a = \frac{28.7}{13.6}$

$a = 2.1 \ m/s^2$

8 0

4 years ago

Scientists take scientific measurements carefully in order to ensure that reliability and validity. What is the difference betwe

Gala2k [10]

Accuracy describes how close a measurement is to the truest value while precision shows how close the measurements are to one another when repeated analysis of the same measurements are performed. Precision most likely is free from any any variation.

7 0

3 years ago

How are frequency and wave period related?

dedylja [7]

-- The unit of frequency is "per second" (Hz), which is [reciprocal time].

-- The unit of period is "second", which is [time].

Do you see where this is going ?

'Frequency' and 'period' are reciprocals of each other.

For any wave ...

Period = (1) / (frequency) .

Frequency = (1) / (period) .

6 0

3 years ago

A truck of mass 200kg rests on an inclined plane hindered from rolling down the surface by a storing sprint whose force constant

mixas84 [53]

Answer:

1.92 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 200 Kg

Spring constant (K) = 10⁶ N/m

Workdone =?

Next, we shall determine the force exerted on the spring. This can be obtained as follow:

Mass (m) = 200 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = m × g

F = 200 × 9.8

F = 1960 N

Next we shall determine the extent to which the spring stretches. This can be obtained as follow:

Spring constant (K) = 10⁶ N/m

Force (F) = 1960 N

Extention (e) =?

F = Ke

1960 = 10⁶ × e

Divide both side by 10⁶

e = 1960 / 10⁶

e = 0.00196 m

Finally, we shall determine energy (Workdone) on the spring as follow:

Spring constant (K) = 10⁶ N/m

Extention (e) = 0.00196 m

Energy (E) =?

E = ½Ke²

E = ½ × 10⁶ × (0.00196)²

E = 1.92 J

Therefore, the Workdone on the spring is 1.92 J

3 0

3 years ago

A 25 kg child is riding on a swing. If the child travels 8.9 m/s at the bottom of their swing, how high into the air is the chil

Setler [38]

Answer:

h = 4.04 m

Explanation:

Given that,

Mass of a child, m = 25 kg

The speed of the child at the bottom of the swing is 8.9 m/s

We need to find the height in the air is the child is able to swing. Let the height is h. Using the conservation of energy such that,

$mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}$

Put all the values,

$h=\dfrac{(8.9)^2}{2\times 9.8}\\\\h=4.04\ m$

So, the child is able to go at a height of 4.04 m.

7 0

3 years ago