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3 years ago

Grilled cheese or Cheese toasty??

Physics

2 answers:

LUCKY_DIMON [66]3 years ago

6 0

Answer:

<em>Toasties are British in origin and were usually made in ovens. They were the same as a grilled cheese but the bread was not buttered. Some Brits would make them in a pan in the oven, too, which is even closer to the stove-top grilled cheese.</em>

jarptica [38.1K]3 years ago

4 0

Answer: grilled cheese lol

Explanation:

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A bodybuilder loads a bar with 550 Newtons (~125 pounds) of weight and pushes the bar over her head 10 times. Each time she lift

mafiozo [28]

Work = (force) x (distance)

Each time she lifts the weight, she does

(550 N) x (0.5 m) = 275 joules of work against gravity.

Each time she lets the bar down gently, gravity does

(550 N) x (0.5 m) = 275 joules of work against her muscles.

If the human physical muscular system were 100% efficient, and capable
of absorbing work as well as spending it, then the bodybuilder would do
exactly zero work in the process of 1-up followed by 1-down.

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3 years ago

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Explain what the ionsphere is and how it interacts with some radio waves

mrs_skeptik [129]

Answer:

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Explanation:

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3 years ago

A car traveling with constant speed travels 150 km in 7200s what is the speed of the car

Oksana_A [137]

Velocity is distance/time

so 150/7200=.0208km/s

unless you have to convert it to miles or something else. but use the formula!

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3 years ago

) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha

vaieri [72.5K]

(a) 0.0021 s, 2926.5 rad/s

The frequency of the B note is

$f= 466 Hz$

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s$

The angular frequency instead is given by

$\omega = 2\pi f$

And substituting

f = 466 Hz

We find

$\omega = 2\pi (466 Hz)=2926.5 rad/s$

(b) 20 Hz, 125.6 rad/s

In this case, the period of the sound wave is

T = 50.0 ms = 0.050 s

So the frequency is equal to the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz$

While the angular frequency is given by:

$\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s$

(c) $4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s$

The minimum angular frequency of the light wave is

$\omega_1 = 2.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s$

The maximum angular frequency of the light wave is

$\omega_2 = 4.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s$

(d) $2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s$

In this case, the frequency is

$f=5.0 MHz = 5.0 \cdot 10^6 Hz$

So the period in this case is

$T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6 Hz}=2.0 \cdot 10^{-7} s$

While the angular frequency is given by

$\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s$

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3 years ago

How is voltmeter connected in the circuit to measure the potential difference between two points?

34kurt

Voltmeter is used to find the potential difference between two points.

We always connect it in parallel to the points where we need the potential difference.

Here in order to make the reading accurate we can increase the resistance of voltmeter so that it can not withdraw any current from the circuit.

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4 years ago