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3 years ago

Someone please help me

Physics

2 answers:

tresset_1 [31]3 years ago

8 0

Answer:

The answer should be 3tsp

Explanation:

3tsp

Schach [20]3 years ago

3 0

Answer:

300x480 teaspoons

Explanation:

when converting cups to teaspoons just multiply by 48

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A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a

UNO [17]

Answer:

$P_{avg} = 6.283*10^{-9} \ W$

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

$\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})$

where;

$E= \frac{d \phi}{dt}$

$E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t$

$E_{rms} = \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}$

Replacing our values into above equation; we have:

$E_{rms} = \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}$

$E_{rms} = \frac {8.98909588*10^{-4} }{\sqrt{2}}$

$E_{rms} = 6.356*10^{-4} \ V$

Then the $P_{avg$ is calculated as:

$P_{avg} = \frac{E^2}{R}$

$P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}$

$P_{avg} = 6.283*10^{-9} \ W$

6 0

3 years ago

Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach

Svetlanka [38]

Answer:

$\Delta V = 1.8 \times 10^7 V$

Explanation:

GIVEN

diameter = 15 fm = $15 \times 10^{-15}$ m

we use here energy conservation

$K_{i}+U_{i} =K_{f}+U_{f}$

there will be some initial kinetic energy but after collision kinetic energy will zero

$K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}$

on solving these equations we get kinetic energy initial

$KE_{i} = 5.65\times 10 ^{-12} \times \frac {1 eV}{1.6 \times 10^{-19}}$

$KE_{i} = 35.33$ J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e

and gains kinetic energy K =e∆V ..........(ii)

by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the ${238}_U$ nucleus after being accelerated through a potential difference ∆V

equating (i) and second equation we get

e∆V = 35.33 Me V

$\Delta V = \frac{35.33}{2} MV\\\Delta V = 1.8 \times 10^7 V$

7 0

3 years ago

An ice skater spins at 2.5 rev/s when his arms are extended. He draws his arms in and spins at 10.0 rev/s. By what factor does h

Rainbow [258]

Answer:

The moment of inertia decreased by a factor of 4

Explanation:

Given;

initial angular velocity of the ice skater, ω₁ = 2.5 rev/s

final angular velocity of the ice skater, ω₂ = 10.0 rev/s

During this process we assume that angular momentum is conserved;

I₁ω₁ = I₂ω₂

Where;

I₁ is the initial moment of inertia

I₂ is the final moment of inertia

$I_2 = \frac{I_1 \omega_1}{\omega_2} = \frac{I_1*2.5}{10} \\\\I_2 = 0.25I_1 = \frac{1}{4}I_1$

Therefore, the moment of inertia decreased by a factor of 4

4 0

2 years ago

The force you have to overcome to start an object moving is ____. A. Rolling friction c. Sliding friction b. Static friction d.

finlep [7]

Air resistance Please select the best

5 0

3 years ago

35. (II) A 28-g rifle bullet traveling 190 m/s embeds itself in a 3.1-kg pendulum hanging on a 2.8-m long string, which makes th

pentagon [3]

Answer:

Explanation:

Let mass of bullet = m1 = 28g= 0.028 kg

mass of pendulum = m2 = 3.1 kg

8 0

3 years ago