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3 years ago

Salmon often jump waterfalls to reach their

Physics

1 answer:

Y_Kistochka [10]3 years ago

4 0

Answer:

5.38 m/s

Explanation:

Given (in the x direction):

Δx = 2.45 m

v₀ = v cos 42.5°

a = 0 m/s²

Δx = v₀ t + ½ at²

(2.45 m) = (v cos 42.5°) t + ½ (0 m/s²) t²

2.45 = (v cos 42.5°) t

t = 3.32 / v

Given (in the y direction):

Δy = 0.373 m

v₀ = v sin 42.5°

a = -9.8 m/s²

Δx = v₀ t + ½ at²

(0.373 m) = (v sin 42.5°) t + ½ (-9.81 m/s²) t²

0.373 = (v sin 42.5°) t − 4.905 t²

0.373 = (v sin 42.5°) (3.32 / v) − 4.905 (3.32 / v)²

0.373 = 2.25 − 54.2 / v²

v = 5.38

Graph:

desmos.com/calculator/5n30oxqmuu

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3 years ago

A toaster using a Nichrome heating element operates on 120 V. When it is switched on at 28 ∘С, the heating element carries an in

sukhopar [10]

Answer:

The final temperature of the element = 262.67°C

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The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W

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Our given parameters include;

A Nichrome heating element operates on 120 V.

Voltage (V) = 120V

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Initial Temperature (T₁) = 28°C

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$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$ ---------------------- (1)

in which R₁ is the resistance at temperature T₁

$R_{T(2)$ is the resistance at temperature T₂

Given that V= IR

R = $\frac{V}{I}$

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$R_{28}= \frac{120V}{1.36A}$

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$R_{T(2)$ = $\frac{120V}{1.23A}$

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From (1) above;

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$

97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]

$\frac{97.56}{88.24}= 1+(4.5*10^{-4})(T-28^0C)$

1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)

0.1056 = 4.5×10⁻⁴(T₂-28°C)

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(b)

What is the power dissipated in the heating element initially and when the current reaches 1.23 A

The power dissipated in the heating element initially can be calculated as:

P = I²₁R₂₈

P = (1.36A)²(88.24Ω)

P = 163.209 W

P ≅ 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:

$P= I^2_2R_{T^0C$

P = (1.23)²(97.56Ω)

P = 147.598524

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