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2 years ago

suppose larry stands with more weight on one scale than the other if one scale reads 400 N. what does the other read?

2 answers:

8 0

The other scale will either have a higher or lower number because 400 could be the bigger number or the smaller number but we don't know that so you cant exactly answer it.<span />

siniylev [52]2 years ago

8 0

You will take 400N divide by 2. The answer will be 200N.

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An engineer is designing a process for a new transistor. She uses a vacuum chamber to bombard a thin layer of silicon with ions

adelina 88 [10]

Answer:

$E=5.7\times 10^{-3}\ V/m$

Explanation:

Given that

$m_p=5.18\times 10^{-26}\ kg$

re= 46 cm

Vp= 180 m/s

We know that

$E=\dfrac{\Delta V}{r}$

$\Delta V=\dfrac{1}{4}\dfrac{m_pv_p^2}{e}$

$E=\dfrac{1}{4}\dfrac{m_pv_p^2}{e.r_e}$

Now by putting the all given values in the questions

$E=\dfrac{1}{4}\times \dfrac{5.18\times 10^{-26}\times 180^2}{1.6\times 10^{-19}\times 0.46}$

$E=5.7\times 10^{-3}\ V/m$

So the average electric field is $E=5.7\times 10^{-3}\ V/m$ .

6 0

2 years ago

A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'

lbvjy [14]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3

= 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

$Y =\frac{mgL}{A\Delta L}$

$\Delta L =\frac{mgL}{A\times Y}$

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

5 0

1 year ago

A) The current theory of the structure of the Earth, called plate tectonics, tells us that the continents are in constant motion

suter [353]

A) The mass of the continent is $2.5\cdot 10^{21} kg$

B) The kinetic energy is 2016 J

C) The speed of the jogger should be 7.1 m/s

Explanation:

The mass of the continent can be calculated as

$m = \rho V$

where

$\rho = 2800 kg/m^3$ is its density

V is its volume

We have to calculate its volume. We know that the continent is represented as a slab of side 5900 km (so its surface is 5900 x 5900, assuming it is a square) and depth of 26 km, so its volume is:

$V=(5900 km)^2 (26 km)=9.05\cdot 10^8 km^3 =9.05 \cdot 10^8 \cdot (10^9 m^3/k^3)=9.05\cdot 10^7 m^3$