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AlladinOne [14]
3 years ago
7

Si unit of small g and G​

Physics
1 answer:
Mila [183]3 years ago
5 0

Answer:

G

Explanation:

because the Si unite of g is G mean gram

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A body is thrown up with a velocity of 78.4 m per second.How high will it rise and how much time it will take to return to its p
a_sh-v [17]

Answer:

The maximum height reached by the body is 313.6 m

The time to return to its point of projection is 8 s.

Explanation:

Given;

initial velocity of the body, u = 78.4 m/s

at maximum height (h) the final velocity of the body (v) = 0

The following equation is applied to determine the maximum height reached by the body;

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u²/2g

h = (78.4²) / (2 x 9.8)

h = 313.6 m

The time to return to its point of projection is calculated as follows;

at maximum height, the final velocity becomes the initial velocity = 0

h = v + ¹/₂gt²

h = 0 + ¹/₂gt²

h =  ¹/₂gt²

2h = gt²

t² = 2h/g

t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 313.6}{9.8} }\\\\t = 8 \ s

4 0
3 years ago
An object weighs 63.8 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 16.8 N.
I am Lyosha [343]

Answer:

The density of this object is approximately 1.36\; {\rm kg \cdot L^{-1}}.

The density of the oil in this question is approximately 0.600\; {\rm kg \cdot L^{-1}}.

(Assumption: the gravitational field strength is g =9.806\; {\rm N \cdot kg^{-1}})

Explanation:

When the gravitational field strength is g, the weight (\text{weight}) of an object of mass m would be m\, g.

Conversely, if the weight of an object is (\text{weight}) in a gravitational field of strength g, the mass m of that object would be m = (\text{weight}) / g.

Assuming that g =9.806\; {\rm N \cdot kg^{-1}}. The mass of this 63.8\; {\rm N}-object would be:

\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}.

When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

When this object was suspended in water, the buoyancy force on this object was (63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}. Hence, the weight of water that this object displaced would be 47.0 \; {\rm N}.

The mass of water displaced would be:

\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}.

The volume of that much water (which this object had displaced) would be:

\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}.

Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately 4.793\; {\rm L}.

The mass of this object is 6.50\; {\rm kg}. Hence, the density of this object would be:

\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}.

(Rounded to \text{$3$ sig. fig.})

Similarly, since this object was fully immersed in oil, the volume of oil displaced would be equal to the volume of this object: approximately 4.793\; {\rm L}.

The weight of oil displaced would be equal to the magnitude of the buoyancy force: 63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}.

The mass of that much oil would be:

\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{28.2\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 2.876\; {\rm kg}\end{aligned}.

Hence, the density of the oil in this question would be:

\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{2.876\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 0.600\; {\rm kg \cdot L^{-1}} \end{aligned}.

(Rounded to \text{$3$ sig. fig.})

7 0
2 years ago
A slender rod of length 80.0 cm and 0.600 kg has its center of gravity at its geometrical center. But its density is not uniform
Vlad [161]

Answer:

Icm = 0.0701 kgm^2

Explanation:

7 0
2 years ago
How can all matter on Earth be classified?
Dvinal [7]

Answer:

by helping

Explanation:

u see the answer is by helping cause cleaning the earth help and helpingb animals so it would be helping

3 0
3 years ago
Read 2 more answers
Solve ~<br><img src="https://tex.z-dn.net/?f=13x%20-%2026%20%2B%2039%20%3D%200" id="TexFormula1" title="13x - 26 + 39 = 0" alt="
Natalka [10]

Explanation:

13x - 26 + 39 = 0

13x=26-39

13x=-13

x= -1

4 0
2 years ago
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