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2 years ago

A tight knot can be easily opened by using a long spanner. Give reason.

Physics

2 answers:

olga nikolaevna [1]2 years ago

5 0

Answer:

Torque

Explanation:

loosening loosening a boat requires a certain amount of torque which is the product of the force acting on the wrench and the moment arm, or distance from the center of rotation at which the force is applied. the longer the handle, the last force it takes to generate the same amount of torque

erma4kov [3.2K]2 years ago

4 0

Answer:

It is because the effort distance is greater than the load distance

Explanation:

As we know, Effort×effort distance = load × load distance

So when effort distance is increases,

The effort decreases

So when the spanner's handle is long

A tight knot can easily be opened by less effrot

I hope it helped

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A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 270 ω at room temperature, what is

vova2212 [387]

solution:

$consider the following data\\
length of slicon bar with circular cross section is 4cm or 0.04m\\
at room temperature resistance of the slicon bar is 270\Omega \\
represent the resistance in mathematical from\\
r=p\frac{1}{A}---1\\
where r is resistance and l is the length \\
A is cross sectional area\\
it is clear that resistivity of the silicon meterial is 6.4\times^2 \Omega.m\\
substitute 6.4\times10^2 for p,270\Omega for R and 0.04m for l i equation (1).\\$ $270=(604\times10^2)\frac{0.04}{A}\\
rewrite the equation\\
a=(6.4\times10^2)\frac{(0.04)}{270}\\
=0.9481m^2\\
write the formula for the circular cross sectional area of silicon bar.\\
A=\pi r^2\\
substitute 0.9481 for A in the above equation\\
\pi r^2=0.9481
r^2=\frac{0.9481}{3.14},since \pi =3.14\\
0.30194\\
further simplified\\
r^2=0.30194\\
\sqrt{0.30194}\\
\cong 0.1509m\\
\cong 150.1mm$

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2 years ago

How do some businesses believe VR is affecting their training for employees?

katen-ka-za [31]

They think everything is involved with tech , some business like going old school , it’s artificiaciality is off meaning consequences aren’t actually accurate like in real life , lack of flexibility can’t change anything because it’s programmed so you can’t actually act question or change scenarios ,it costs too much as well , can be health risk can cause stress and anxiety,

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2 years ago

A solar heated house loses about 5.4 × 107 cal through its outer surfaces on a typical 24-h winter day.

mojhsa [17]

Answer:

Explanation:

Q=mcΔθ

Q=quantity of heat , m= mass of the storage rock

Δθ= temperature change.

m= Q/(cΔθ)

Q=5.4 $10^{7}$

Δθ=62°C-20°C

=42°C

c=0.21cal/g.°C

$m=\frac{5.4*10^{7} }{0.21*42} \\\\m=6122448.98g\\$

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2 years ago

Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller

kobusy [5.1K]

Answer:

σ₁ = $3.167 * 10^{-6}$ C/m²

σ₂ = $7.6 * 10 ^{-6}$ C/m²

Explanation:

The given data :-

i) The radius of smaller sphere ( r ) = 5 cm.

ii) The radius of larger sphere ( R ) = 12 cm.

iii) The electric field at of larger sphere ( E₁ ) = 358 kV/m. = 358 * 1000 v/m

$E_{1} = (\frac{1}{4\pi\epsilon }) (\frac{Q_{1} }{R^{2} } )$

$358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }$

Q₁ = 572.8 $* 10^{-9}$ C

Since the field inside a conductor is zero, therefore electric potential ( V ) is constant.

V = constant

∴ $\frac{Q_{1} }{R} = \frac{Q_{2} }{r}$

$Q_{2} = \frac{r}{R} *Q_{1}$

$Q_{2} = \frac{5}{12} *572.8*10^{-9}$ = $238.666 *10^{-9}$ C

Surface charge density ( σ₁ ) for large sphere.

Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁ = $\frac{Q_{1} }{A_{1} }$ = $\frac{572.8 *10^{-9} }{0.180864}$ = $3.167 * 10^{-6}$ C/m².

Surface charge density ( σ₂ ) for smaller sphere.

Area ( A₂ ) = 4 * π * r² = 4 * 3.14 * 0.05² =0.0314 m².

σ₂ = $\frac{Q_{2} }{A_{2} }$ = $\frac{238.66 *10^{-9} }{0.0314}$ = $7.6 * 10 ^{-6}$ C/m²

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2 years ago

Platinum (pt) has the fcc crystal structure, an atomic radius of 0.1387 nm, and an atomic weight of 195.08 g/mol. what is its th

prohojiy [21]

The equation to be used is written as:

ρ = nA/VcNₐ
where
ρ is the density
n is the number of atoms in unit cell (for FCC, n=4)
A is the atomic weight
Vc is the volume of the cubic cell which is equal to a³, such that a is the side length (for FCC, a = 4r/√2, where r is the radis)\
Nₐ is Avogradro's constant equal to 6.022×10²³ atoms/mol

r = 0.1387 nm*(10⁻⁹ m/nm)*(100 cm/1m) = 1.387×10⁻⁸ cm
a = 4(1.387×10⁻⁸ cm)/√2 = 3.923×10⁻⁸ cm
V = a³ = (3.923×10⁻⁸ cm)³ = 6.0376×10⁻²³ cm³

ρ = [(4 atoms)(195.08 g/mol)]/[(6.0376×10⁻²³ cm³)(6.022×10²³ atoms/mol)]
ρ = 21.46 g/cm³

7 0

2 years ago

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