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NeX [460]
3 years ago
9

A car rounds a 75-m radius curve at a constant speed of 18 m/s. a ball is suspended by a string from the ceiling the car and mov

es with the car. the angle between the string and the vertical is:
Physics
1 answer:
Svetlanka [38]3 years ago
8 0
The angle between the string and the vertical is computed by:
Centripetal acceleration Ac = V²/R = 18²/75 = 4.32 m/s² horizontal 
Gravitational acceleration g = 9.8 m/s² down 

Θ = arctan(Ac/g)
= arctan (4.32m/s^2 / 9.8 m/s^2)
= 23.79° is the answer.
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Joseph studied whether different materials can block certain electromagnetic waves by testing television reception in different
Sedbober [7]
Joseph's experiment could be improved by using the same antenna at each part of the house during each trial instead of using different antenna. By doing so, he can obtain accurate results how is the signal in different part of the house under the same conditions (despite the location). So, he will see the dependence of the signal on the location. If he uses different antenna, than this antenna can also have influence of the signal.
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4 years ago
In one scene in the movie The Godfather II, a solid gold phone is passed around a large table for everyone to see. Suppose the v
dangina [55]
Volume of gold in the phone = 10 cm^3
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1 kg mass = 2.2 pounds
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4 years ago
Who the queen of rap
Inessa [10]

Answer:

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6 0
3 years ago
Read 2 more answers
The mass of 150cm³ of stone is 400g. its density is​
WITCHER [35]

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3 0
3 years ago
One cycle of the power dissipated by a resistor ( R = 800 Ω R=800 Ω) is given by P ( t ) = 60 W , 0 ≤ t &lt; 5.0 s P(t)=60 W, 0≤
OLga [1]

Answer:

42.5W

Explanation:

To solve this problem we must go back to the calculations of a weighted average based on the time elapsed thus,

Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}

We need to calculate the average power dissipated by the 800\Omega resistor.

Our values are given by:

P(t)=60 W, 0\leq t

P(t)=25 W, 5.0\leq t

Aplying the values to the equation we have:

Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}

Power_{avg} = \frac{60(5-0)+25(10-5)}{(5-0)+(10-5)}

Power_{avg} = 42.5W

5 0
3 years ago
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