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liberstina [14]
2 years ago
10

3.7. A dog searching for a bone walks 3.50 m south, then 8.20 m at an angle 23.1 degrees north of east, and finally 15.0 m west.

(a) What is the magnitude of the dog’s total displacement? (b) What is the direction of the dog’s total displacement where directly east is taken as zero degrees and counter-clockwise is positive?
Physics
1 answer:
andriy [413]2 years ago
5 0

Answer:

Explanation:

We shall represent displacement of dog in vector form , in terms of i , j , i representing east  and  j representing north .

Dog travels 3.5 m south .

Displacement D₁ = - 3.5 j

then dog travels 8.2 m , 23.1 degree north of east

Displacement D₂ = 8.2 cos23.1 i + 8.2 sin23 j

D₂ = 8.2 cos23.1 i + 8.2 sin23.1  j

= 7.54 i + 3.22 j  

Third displacement

D₃ = - 15i

Total displacement = D₁ + D₂ + D₃

= - 3.5 j + 7.54 i + 3.22 j  -15i

= - 7.46 i - 0.28 j

Magnitude of displacement = √ ( 7.46² + .28²)

= √(55.65 + .08 )

= 7.46 m

b ) Direction of displacement

If Ф be angle , displacement makes with west direction

TanФ =  .08 / 55.65 = .00143

Ф = .082 degree south of west or almost west .

From east , this angle = 180 + .082 = 180.082 , counterclockwise .

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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
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Answer:

The final temperature of the element = 262.67°C

The power dissipated in the heating element initially = 163.21 W

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

Explanation:

Our given parameters include;

A Nichrome heating element operates on 120 V.

Voltage (V) = 120V

Initial Current (I₁) = 1.36 A

Initial Temperature (T₁) = 28°C

Final Current (I₂) = 1.23 A

Final Temperature (T₂) = unknown ????

Temperature dependencies of resistance is given by:

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}

Therefore, the resistance at temperature 28°C is;

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

= 88.24Ω

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

= 97.56Ω

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)

\frac{0.1056}{4.5*10^{-4}}= T-28^0C

T - 28° C = 234.67

T = 234.67 + 28° C

T = 262.67 ° C

(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

P ≅ 147.60 W

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