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krek1111 [17]
3 years ago
10

A 26.5 g object is thrown straight up into the air. If the object's initial speed is 1.60 m/s, determine how high the object wil

l rise.
Physics
1 answer:
fenix001 [56]3 years ago
8 0

Answer:

0.131 m

Explanation:

Applying,

v² = u²+2gh................... Equation 1

Where, v = final velocity, u = initial velocity, g = acceleration due to gravity, h =  maximum height of the object

make h the subject of the equation

h = (v²-u²)/2g........... Equation 2

From the question,

Given: v = 0 m/s (at the maximum height), u = 1.60 m/s

Constant: g = -9.8 m/s² (thrown up)

Substitute these values into equation 2

h = (0²-1.6²)/[2×(-9.8)]

h = -2.56/-19.6

h = 0.131 m

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<span> C is the specific heat of a compound 
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Which statement best describes the acceleration of a baseball?
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C the baseball began at rest and rolls at a rate of 14.7 m/s after 1.5 seconds

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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

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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When crossing this potential barrier we can observe an increase in current but the voltage does not go higher than 0.65\ V. Even with the high input voltage.

Hence, diode maintains proper voltage in circuits.

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