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adelina 88 [10]
2 years ago
9

Weigt A box sitting still on the ground by itself has a weight of 700N, what is the mass? (gravity's acceleration is 9.80 m/s2)​

Physics
1 answer:
Alex2 years ago
7 0
  • Force=700N
  • Acceleration=9.8m/s^2

Using newtons law

\\ \rm\longmapsto F=ma

\\ \rm\longmapsto m=\dfrac{F}{a}

\\ \rm\longmapsto m=\dfrac{700}{9.8}

\\ \rm\longmapsto m=71.4kg

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Answer:

I hope this answer is correct

Explanation:

The mechanical advantage of a machine is 4. Mechanical advantage MA is the ratio of output (generated by the machine) force to input (applied to the machine) force. So MA = 4 means that for example if you apply 100 N then your machine will multiply that force and generate 400 N.

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The fact that current is uniform anywhere in a series circuit is an application of the principle of:
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Due to the conservation of electric charge, the current is uniform throughout an electrical circuit.
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Water is entering the prism at a rate of A m^3/hr. The prism is empty at time 0. Express the depth d of the water in meters in t
zavuch27 [327]

This question is incomplete, the complete question is;

The picture shows a triangular prism. The end of prism are equilateral triangles with x meters. the other dimension of the prism is L meters

a) Find the volume V in terms of x and L

b) Water is entering the prism at a rate of A m³/hr. The prism is empty at time 0. Express the depth d of the water in meters in terms of A, the length of time t the water has been entering the trough, and the length L of the prism.  

Answer:

a) the volume V in terms of x and L is  ((√3/4)x²L) m³

b) required expression is (2/(3)^(1/u))√(At/L)

Explanation:

Given that;

form the question and image below;

triangular prism ends are equilateral triangle

side length = x meter

Dimension of the prism = L meter

Area of the equilateral triangle = √3/4 (side)² = √3/4 (x)² meter

Volume of the triangular prism = Area × height

= √3/4 (x)² × L

V = ((√3/4)x²L) m³

Therefore, the volume V in terms of x and L is  ((√3/4)x²L) m³

b)

Rate of water entering = A m³/hr

Depth of water tank = d meter

Time = t

Length of prism = L

now Rate of water entering is A m³/hr

dv/d = A                             [  V = ((√3/4)x²L) m³ ]

and

dv/dt = √3/4 [2x dx/dt ] L                   { L is constant }

so

A = √3/4 [2x dx/dt ] L  

∫A dt = √3/2 [ Lx dx ]                   { Integrate both sides}

At = √3/2 × Lx × x²/2

x² = uAt / √3L                              { we find square root of both sides}

x = √( uAt / √3L )

x = (2/(3)^(1/u))√(At/L)

Therefore; required expression is (2/(3)^(1/u))√(At/L)

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Answer:

A and C

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The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the Sun, namely 1.496x10^(11) m. The p
Rudiy27

Answer:

a) How many parsecs are there in one astronomical unit?

4.85x10^{-6}pc

(b) How many meters are in a parsec?

3.081x10^{16}m

(c) How many meters in a light-year?

9.46x10^{15}m

(d) How many astronomical units in a light-year?

63325AU

(e) How many light-years in a parsec?

3.26ly

Explanation:

The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun (1.496x10^{11} m) is defined as 1 astronomical unit:

\tan{p} = \frac{1AU}{d}

Where d is the distance to the star.

Since p is small it can be represent as:

p(rad) = \frac{1AU}{d}  (1)

Where p(rad) is the value of in radians

However, it is better to express small angles in arcseconds

p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}

p('') = 2.06x10^5 p(rad)

p(rad) = \frac{p('')}{2.06x10^5} (2)

Then, equation 2 can be replace in equation 1:

\frac{p('')}{2.06x10^5} = \frac{1AU}{d}  

\frac{d}{1AU} = \frac{2.06x10^5}{p('')}  (3)

From equation 3 it can be see that 1pc = 2.06x10^5 AU

<em>a) How many parsecs are there in one astronomical unit? </em>

1AU . \frac{1pc}{2.06x10^5AU} ⇒ 4.85x10^{-6}pc

<em>(b) How many meters are in a parsec? </em>

2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU} ⇒ 3.081x10^{16}m

<em>(c) How many meters in a light-year? </em>

To determine the number of meters in a light-year it is necessary to use the next equation:

x = c.t

Where c is the speed of light (c = 3x10^{8}m/s) and x is the distance that light travels in 1 year.

In 1 year they are 31536000 seconds

x = (3x10^{8}m/s)(31536000s)

x = 9.46x10^{15}m

<em>(d) How many astronomical units in a light-year?</em>

9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m} ⇒ 63325AU

<em>(e) How many light-years in a parsec?</em>

2.06x10^{5}AU . \frac{1ly}{63235AU} ⇒ 3.26ly

5 0
3 years ago
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