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2 years ago

4) A force of 500 N acts on an area of 0.05m2. Find the pressure in Pascal.

Physics

1 answer:

snow_tiger [21]2 years ago

6 0

Answer:

pressure = force ie 500 N divided by area ie 0.05m².

p=f by a

p= 500n divided by 0.05 m²

p= 10,000 pascal

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A person throws a pumpkin at a horizontal speed of 4.0 — off a cliff. The pumpkin travels 9.5 m horizontally

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Complete Question

A person throws a pumpkin at a horizontal speed of 4.0 m/s off a cliff. The pumpkin travels 9.5m horizontally before it hits the ground. We can ignore air resistance.What is the pumpkin's vertical displacement during the throw? What is the pumpkin's vertical velocity when it hits the ground?

Answer:

The pumpkin's vertical displacement is $H = 27 .67 \ m$

The pumpkin's vertical velocity when it hits the ground is $v_v__{f}} = 23.298 \ m/s$

Explanation:

From the question we are told that

The horizontal speed is $v_h = 4 m/s$

The horizontal distance traveled is $d = 9.5 \ m$

The horizontal distance traveled is mathematically represented as

$S = v_h * t$

Where t is the time taken

substituting values

$9.5 = 4 * t$

=> $t = \frac{9.5}{4}$

$t = 2.38 \ sec$

Now the vertical displacement is mathematically represented as

$H = v_v t + \frac{1}{2} a_v t^2$

now the vertical velocity before the throw is zero

$H = 0 + \frac{1}{2} (9.8) * (2.375)^2$

$H = 27 .67 \ m$

Now the final vertical velocity is mathematically represented as

$v_v__{f}} = v_v + at$

substituting values

$v_v__{f}} = 0 + (9.8)* (2.375)$

$v_v__{f}} = 23.298 \ m/s$

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3 years ago

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Answer:

(A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

Explanation:

Given that,

Magnetic field = 0.45 T

The loop's diameter changes from 17.0 cm to 6.0 cm .

Time = 0.53 sec

(A). We need to find the direction of the induced current.

Using Lenz law

If the direction of magnetic field shows into the paper then the direction of the induced current will be clockwise.

(B). We need to calculate the magnetic flux

Using formula of flux

$\phi_{1}=BA\cos\theta$

Put the value into the formula

$\phi_{1}=0.45\times(\pi\times(8.5\times10^{-2})^2)\cos0$

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We need to calculate the magnetic flux

Using formula of flux

$\phi_{2}=BA\cos\theta$

Put the value into the formula

$\phi_{2}=0.45\times(\pi\times(3\times10^{-2})^2)\cos0$

$\phi_{2}=0.00127\ Wb$

We need to calculate the magnitude of the average induced emf

Using formula of emf

$\epsilon=-N(\dfrac{\Delta \phi}{\Delta t})$

Put the value into t5he formula

$\epsilon=-1\times(\dfrac{0.00127-0.01021}{0.53})$

$\epsilon=0.016867\ V$

$\epsilon=16.87\ mV$

(C). If the coil resistance is 2.5 Ω.

We need to calculate the induced current

Using formula of current

$I=\dfrac{\epsilon}{R}$

Put the value into the formula

$I=\dfrac{0.016867}{2.5}$

$I=0.00675\ A$

$I=6.75\ mA$

Hence, (A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

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