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3 years ago

What type of charge will an object have if the object contains less protons than electrons?

Physics

1 answer:

Burka [1]3 years ago

6 0

Answer:

Hello, I believe it would have a negative charge considering protons have a positive charge while elctrons have a negative charge

Explanation:

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Consider a glider flying at 400 meters altitude, when suddenly all its static ports become blocked by volcanic ash. The pressure

Furkat [3]

Answer:

the airspeed indicated by the pitot-tube driven airspeed indicator is 91.23m/s

Explanation:

Pitot tube

$U = \sqrt{\frac{(p_t - p_s)2}{d} }$

U = velocity(m/s)

$p_t$ = stagnation pressure (pa)

$p_s$ = static pressure (pa)

d = fluid density(kg/m³)

$p_t = p_a_t_m + \frac{1}{2} dv^2$

v = true velocity

= 101325 + 1/2(1.225)(25)²

$p_t = 101,707.8125pa$

$p_s = 96,610pa$

d = 1.225kg/m³

$U = \sqrt{\frac{2(101,707.8125 - 96,610)}{1.225} } \\\\U = 91.23m/s$

the airspeed indicated by the pitot-tube driven airspeed indicator is 91.23m/s

5 0

3 years ago

On his honeymoon, James Joule attempted to explore the relationships between various forms of energy by measuring the rise of te

Marizza181 [45]

Answer:

The maximum temperature rise = 0.047 °C

Explanation:

Potential Energy, P = mgh

Energy transfered, Q=mcΔT

Potential energy = Energy transfered

mgh = mcΔT

gh = cΔT

ΔT = gh/c

ΔT = (9.81 * 20) / 4186

ΔT = 0.047 °C

8 0

3 years ago

For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is

ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

A₀ is the amplitude at t=0 and

$w'$ is the angular frequency of damped SHM, which is given by,

$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀ $e^{\frac{-bt}{2m}}$ =A₀/8

⇒ $e^{\frac{bt}{2m}}=8$

applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

b) $w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

$w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}$

$T'=\frac{2\pi}{w'}$ , where $T'$ is time period of damped SHM

⇒ $T'=\frac{2\pi}{2.86}=2.2s$

let $n$ be number of oscillations made

then, $nT'=t$

⇒ $n=\frac{24}{2.2}=11(approx)$

8 0

3 years ago

Why it is difficult to run fast in sand

balu736 [363]

Running on sand requires 1.6 times more energy spent than running on hard surface, so the force applied by our foot on sand is less.

8 0

3 years ago

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A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis (Fig. 11-48).A toy train

masya89 [10]

Answer:

0.166 rad/s

Explanation:

See attachment for calculations

5 0

3 years ago

Read 2 more answers