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3 years ago

Someone please help me with this

Physics

1 answer:

AnnZ [28]3 years ago

5 0

Answer: I think 1 is 300 metres per second.

Explanation: wave speed= wave length x wave frequency

wave speed = 3 x 100

wave speed is 300 m/s

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Power is strength and _______ combined. <br> weight<br> Speed<br> height<br> length

bogdanovich [222]

When speed is combined with strength it is power

5 0

4 years ago

Read 2 more answers

The charge entering the positive terminal of an element is q = 5 sin 4πt mC while the voltage across the element (plus to minus)

Artemon [7]

Answer:

(a). The power delivered to the element is 187.68 mW

(b). The energy delivered to the element is 57.52 mJ.

Explanation:

Given that,

Charge $q=5\sin4\pi t\ mC$

Voltage $v=3\cos4\pi t\ V$

Time t = 0.3 sec

We need to calculate the current

Using formula of current

$i(t)=\dfrac{dq}{dt}$

Put the value of charge

$i(t)=\dfrac{d}{dt}(5\sin4\pi t)$

$i(t)=5\times4\pi\cos4\pi t$

$i(t)=20\pi\cos4\pi t$

(a).We need to calculate the power delivered to the element

Using formula of power

$p(t)=v(t)\times i(t)$

Put the value into the formula

$p(t)=3\cos4\pi t\times20\pi\cos4\pi t$

$p(t)=60\pi\times10^{-3}\cos^2(4\pi t)$

$p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi t}{2})$

Put the value of t

$p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi\times0.3}{2})$

$p(t)=30\pi\times10^{-3}(1+\cos8\pi \times0.3)$

$p(t)=187.68\ mW$

(b). We need to calculate the energy delivered to the element between 0 and 0.6 s

Using formula of energy

$E(t)=\int_{0}^{t}{p(t)dt}$

Put the value into the formula

$E(t)=\int_{0}^{0.6}{30\pi\times10^{-3}(1+\cos8\pi \times t)}$

$E(t)=30\pi\times10^{-3}\int_{0}^{0.6}{1+\cos8\pi \times t}$

$E(t)=30\pi\times10^{-3}(t+\dfrac{\sin8\pi t}{8\pi})_{0}^{0.6}$

$E(t)=30\pi\times10^{-3}(0.6+\dfrac{\sin8\pi\times0.6}{8\pi}-0-0)$

$E(t)=57.52\ mJ$

Hence, (a). The power delivered to the element is 187.68 mW

(b). The energy delivered to the element is 57.52 mJ.

6 0

3 years ago

An accelerometer is a device that uses the extension of a spring to measure acceleration in terms of Earth's gravitational accel

ELEN [110]

In your question explains that an accelerometer is a device that uses the extension spring to measure acceleration in terms of Earth's Gravitational accelerations. So if this device is extended just 0.43 centimeters the possible accelerations is letter B. 0.80 g

3 0

4 years ago

Read 2 more answers

Need help on question b

ZanzabumX [31]

The answer is A.

A positive charge’s electric field pushes out.

Hope this helps! -Avenging

8 0

3 years ago

A mass with a charge of 4.60 x 10-7 C rests on a frictionless surface. A compressed spring exerts a force on the mass on the lef

stira [4]

Answer:

The compression of the spring is 24.6 cm

Explanation:

magnitude of the charge on the left, q₁ = 4.6 x 10⁻⁷ C

magnitude of the charge on the right, q₂ = 7.5 x 10⁻⁷ C

distance between the two charges, r = 3 cm = 0.03 m

spring constant, k = 14 N/m

The attractive force between the two charges is calculated using Coulomb's law;

$F = \frac{kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(4.6\times 10^{-7})(7.5\times 10^{-7})}{(0.03)^2} \\\\F= 3.45 \ N$

The extension of the spring is calculated as follows;

F = kx

x = F/k

x = 3.45 / 14

x = 0.246 m

x = 24.6 cm

The compression of the spring is 24.6 cm

4 0

3 years ago