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Goryan [66]
3 years ago
13

Challenge: Determine the speed of the waves at each tension setting (high, medium and low). Explain what measurements you made t

o calculate the speed. Settings: amplitude: 0.75 cm damping: zero high tension: medium tension: low tension:

Physics
1 answer:
Svetradugi [14.3K]3 years ago
4 0

Answer:

high tension: 4.2 × 1.5 = 6.3 cm/s

medium tension: 2.8 ×1.5 = 4.2 cm/s

low tension: 0.8 × 1.5 = 1.2 cm/s

Explanation: Given Settings:

amplitude: 0.75 cm

damping: zero

Using

Speed = frequency ×wavelength

Where

Wavelength = 0.75 × 2 = 1.5 cm

Therefore:

high tension: 4.2 × 1.5 = 6.3 cm/s

medium tension: 2.8 ×1.5 = 4.2 cm/s

low tension: 0.8 × 1.5 = 1.2 cm/s

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How much energy is transferred electrically by a 2000 W cooker in half an hour?
Kryger [21]

Responder:

E = 1440 kJ

Explicación:

Se da que,

La potencia de un horno de cocción es de 800 W

El voltaje al que se opera es de 230 V

Tiempo, t = 30 minutos = 1800 segundos

Necesitamos encontrar la energía eléctrica utilizada por el horno de cocción. El producto de la potencia y el tiempo es igual a la energía consumida. Entonces,

5 0
3 years ago
Read 2 more answers
square root A 1400 kg car is coasting on a horizontal road with a speed of 18 m/s . After passing over an unpaved, sandy stretch
Lina20 [59]

Answer:

The net force on the car is 2560 N.

Explanation:

According to work energy theorem, the amount of work done is equal to the change of kinetic energy by an object. If 'W' be the work done on an object to change its kinetic energy from an initial value 'K_{i}' to the final value 'K_{f}', then mathematically,

W = K_{f} - K_{i} = \dfrac{1}{2}~m~(v_{f}^{2} - v_{i}^{2})........................................(I)

where 'm' is the mass of the object and 'v_{i}' and 'v_{f}' be the initial and final velocity of the object respectively. If 'F_{net}' be the net force applied on the car, as per given problem, and 's' is the displacement occurs then we can write,

W = F_{net}~.~s.......................................................(II)

Given, m = 1400~Kg,~v_{i} = 18~m~s^{-1}~v_{f} = 14~m~s^{-1}~and~s = 35~m.

Equating equations (I) and (II),

&& - F_{net} \times 35~m = \dfrac{1}{2} \times 1400~Kg~\times(14^{2} - 18^{2})~m^{2}~s^{-2}\\&or,& F_{net} = \dfrac{\dfrac{1}{2} \times 1400~Kg~\times(14^{2} - 18^{2})}{35}~N\\&or,& F_{net} = 2560~N

6 0
3 years ago
7. A car travels a certain distance with a speed of
goblinko [34]
Let distance be x.

Time in going, t 1 = x/50

Time in coming, t 2 = x/40

Total time T = x/40 +x/50 = 9x/200

Average speed = total distance/ total time

=( x+x)/(9x/200)= 2x ×200/9x = 400/9 = 44.44 km/h
5 0
3 years ago
Which statement about electromagnetic waves is true?
masya89 [10]
D. Is the right answer
7 0
3 years ago
A playground carousel is rotating counterclockwise about its center on frictionless bearings. A person standing still on the gro
Shtirlitz [24]

Answer:

1.94601 rad/s

Explanation:

I_1 = Moment of inertia of carousel = 124 kgm²

\omega_1 = Angular speed of carousel = 3.5 rad/s

\omega_2 = Angular speed of person

r = Radius of carousel = 1.53 m

m = Mass of person = 42.3 kg

Moment of inertia of person

I_2=mr^2\\\Rightarrow I_2=42.3\times 1.53^2\\\Rightarrow I_2=99.02007\ kgm^2

As the angular momentum is conserved in the system

I_1\omega_1=(I_1+I_2)\omega_2\\\Rightarrow \omega=\frac{I_1\omega_1}{(I_1+I_2)}\\\Rightarrow \omega_2=\frac{124\times 3.5}{(124+99.02007)}\\\Rightarrow \omega_2=1.94601\ rad/s

The angular speed of the carousel after the person climbs aboard is 1.94601 rad/s

5 0
3 years ago
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