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Alchen [17]
3 years ago
6

4: Which of the following is true of an object in a circular motion at a constant speed?

Physics
1 answer:
kaheart [24]3 years ago
8 0
C. The object is accelerating as the direction of the object is changing
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What is the magnitude and direction (right or left) of the
Sunny_sXe [5.5K]

Answer: 12 N to the right

Explanation:

If we calculate the net force acting on the box, we will have:

<u>In y-component:</u>

Fy_{net}=F_{n}+F_{g} (1)

Where F_{n}=12 N is the Normal force, directed upwards and F_{g}=-12 N is the weight of the box (gravity force), directed downwards.

Fy_{net}=12 N-12 N (2)

Fy_{net}=0 N (3) Hence the net force in the vertical component is zero

<u>In x-component:</u>

Fx_{net}=F_{left}+F_{right} (4)

Where F_{left}=-3 N and F_{right}= 15 N

Fx_{net}=-3 N + 15 N (5)

Fx_{net}=12 N (6) This is the net force in the horizontal component

Therefore, the total net force acting on the box is 12 N directed to the right

5 0
3 years ago
2. We know that the Earth is moving around the sun. When we look at the sky we see
marin [14]

Answer:

Answer: The Sun, the Moon, the planets, and the stars all rise in the east and set in the west. And that's because Earth spins -- toward the east. ... Earth rotates or spins toward the east, and that's why the Sun, Moon, planets, and stars all rise in the east and make their way westward across the sky.

6 0
3 years ago
In the Bohr model of the hydrogen atom, an electron({rm mass};m=9.1; times 10^{ - 31;}{rm kg}) orbits a proton at a distance of
max2010maxim [7]

Answer:

n=6.56×10¹⁵Hz

Explanation:

Given Data

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

To find

Revolutions per second

Solution

Let F be the force of attraction

let n  be the number of revolutions per sec made by the electron around the nucleus then the centripetal force is given by

F=mω²r......................where ω=2π  n

F=m4π²n²r...............eq(i)

as the values given where

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

we have to find n from eq(i)

n²=F/(m4π²r)

n^{2} =\frac{8.2*10^{-8} }{9.11*10^{-31}* 4\pi^{2} *5.3*10^{-11}  }\\ n^{2}=4.31*10^{31}\\ n=\sqrt{4.31*10^{31}}\\ n=6.56*10^{15}Hz

8 0
3 years ago
Peak to peak value of voltage is
hram777 [196]

Answer: Peak-to-peak voltage is the distance from the lowest negative amplitude, or trough, to the highest positive amplitude, or crest, of the AC voltage waveform. In other words, peak-to-peak voltage is equal to the full height of the waveform. Peak-to-peak voltage can be found using peak voltage or RMS voltage.

Explanation:hope tht gave u a clue have a wonderful Christmas Eve time with your family!❄️

3 0
2 years ago
Read 2 more answers
The sound wave passes from the sea-water into the air. State what happens to; the speed of the waves
dusya [7]
Answer. Explanation: Frequency of the sound decreases and the speed of sound becomes 346m/s from near about 1500 m/s.
6 0
3 years ago
Read 2 more answers
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