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Anvisha [2.4K]
3 years ago
11

A ball is rolled uphill a distance of 4 meters before it slows, stops, and begins to roll back. The ball rolls downhill 28 meter

s before coming to rest against a tree. What is the magnitude of the ball’s displacement?
Physics
1 answer:
MrRissso [65]3 years ago
4 0
The main assumption is that the ball followed the same way uphill and back, so it was at the same initial point when it rolled back 4 m.

At this moment the displacement was zero (remember that the displacement is the difference between the final position and the initial position).

From that point the ball displaced other 24 m down (28m - 4m = 24 m). And, that is the magnitude of the ball's displacement.

Answer: 24 m.
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Answer:

The self and world construct

Explanation:

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3 0
4 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

6 0
3 years ago
A motor exerts a force of 12,00N to lift an elevator 8.0m in 7.0secs what is the power produced by the motor?
SashulF [63]

Answer:

1371.4watt

Explanation:

from power=energy/time

BUT energy=force times distance

6 0
3 years ago
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