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3 years ago

Which ammeter will have the biggest reading

Physics

1 answer:

svetlana [45]3 years ago

4 0

Answer:

Explanation:

because it it near to the circuit

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An 80.0 kg man sits on a scale in his car. The car is driving at a speed of 11.0 m/s right as it passes over the top of a semici

Lady_Fox [76]

Answer:

F / g = 138 kg

Explanation:

For this exercise let's use Newton's second law

F- W = m a

the force is equal to the back of the balance

in this case the acceleration is centripetal

a = v² / r

we substitute

F - m g = m v² / r

F = m (g + v²/ r)

calculus

F = 80 (9.8 + 11²/17)

F = 1353 N

the balance reading is this value between gravity

F / g = 1353 / 9.8

F / g = 138 kg

6 0

3 years ago

If we ignore air resistance, a falling body will fall 16t2 feet in t seconds. What is the average velocity between t

erik [133]

Answer:

262.4 m/s

Explanation:

The complete question is

If we ignore air resistance, a falling body will fall 16t^2 feet in t seconds. What is the average velocity between t=8 and t=8.4? Round your answer to two decimal places if necessary.

The distance fallen s = 16t^2

The velocity v = $\frac{ds}{dt}$ = 32t

If we substitute the values of t into the velocity v, we'll have

at t = 8 s, V1 = 32 x 8 = 256 m/s

at t = 8.4 s, V2 = 32 x 8.4 = 268.8 m/s

Average velocity = (V2 - V1)/2 = (268.8 + 256)/2 = 262.4 m/s

7 0

3 years ago

Part A What is the resistance of a 4.4 m length of copper wire 1.3 mm n diameter? The resistivity of copper is 1.68x 10-8 Ω-m Ex

Natalija [7]

Explanation:

It is given that,

Length of the copper wire, l = 4.4 m

Diameter of copper wire, d = 1.3 mm = 0.0013 m

Radius of copper wire, r = 0.00065 m

The resistivity of the copper wire, $\rho=1.68\times 10^{-8}\ \Omega-m$

We need to find the resistance of the copper wire. It is given by :

$R=\rho\dfrac{l}{A}$

$R=1.68\times 10^{-8}\ \times \dfrac{4.4\ m}{\pi (0.00065)^2}$

R =0.055 ohms

So, the resistance of the copper wire is 0.055 ohms. Hence, this is the required solution.

7 0

3 years ago

WILL GIVE BRAINLIEST! QUICK PLEASE HELP!!!

Fantom [35]

Answer:

mass and location

Explanation:

edgenuity test 100%

4 0

3 years ago

A hard-boiled egg of mass 46.0 gg moves on the end of a spring with force constant 25.6 N/mN/m . The egg is released from rest a

soldi70 [24.7K]

Answer:

0.022kg/s

Explanation:

We are given that

Mass of boiled egg=46 g= $\frac{46}{1000} kg=0.046 kg$

$1kg=1000 g$

Constant force=F=25.6 N/m

Initial displacement= $A_1=0.296 m$

Final displacement= $A_2=0.12 m$

Time=t=4.55 s

Damping force= $F_x=-bv_x$

We have to find the magnitude of damping constant b.

We know that the displacement of the oscillator under damping motion is given by

$x=Ae^{-\frac{b}{2m}t}cos(w't+\phi)$

For maximum displacement $cos(w't+\phi)=1$

Therefore , $x=A_2$

Substitute the values

$A_2=A_1e^{-\frac{-b}{2m}t}$

$e^{-\frac{b}{2m}t}=\frac{A_2}{A_1}$

$-\frac{b}{2m}t=ln\frac{A_2}{A_1}$

$lnx=y\implies x=e^y$

Substitute the values

$-\frac{b}{2\times 0.046}\times 4.55=ln\frac{0.12}{0.296}$

$\frac{2\times 0.046}{4.55b}=ln\frac{0.296}{0.12}$

$\frac{2\times 0.046}{4.55}=0.9b$

$b=\frac{2\times 0.46}{4.55\times 0.9}=0.022kg/s$

Hence,the magnitude of damping constant b=0.022kg/s

3 0

3 years ago