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andre [41]
2 years ago
5

Two objects have the same center point of the circle, but are located at different positions away from the center point. Each ob

ject is moving with uniform circular motion.
Which would describe the tangential speed of the objects?

both objects would have the same tangential speed
the object with the smaller radius has a faster tangential speed
the object with the larger radius has a faster tangential speed
both objects would have oscillating tangential speeds
Physics
1 answer:
Sergeeva-Olga [200]2 years ago
5 0

The object with the larger radius has a faster tangential speed. Tangential speed is related to both rotational speed and radial distance from the rotating axis.

<h3>What is uniform circular motion?</h3>

Uniform circular motion is a type of motion of a particle around a circle at a constant speed. The magnitude of the speed of the particle is constant.While the direction is changing continuously.

Tangential speed is related to both rotational speed and radial distance from the rotating axis.

The object with the larger radius has a faster tangential speed.

Hence, option C is correct.

To learn more about the uniform circular motion, refer to the link;

brainly.com/question/2285236

#SPJ1

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A +15 nC point charge is placed on the x axis at x = 1.5 m, and a -20 nC charge is placed on the y axis at y = -2.0m. What is th
IceJOKER [234]

Answer:E=75\ N/m

Explanation:

Given

First charge of q_1=15\ nC is placed at x=1.5\ m

Second charge  q_2=-20\ nC is placed at y=-2\ m

Electric field is given by

E=\frac{kq}{r^2}

Electric field due to q_1 is away from it

E_1=\frac{9\times 10^9\times 15\times 10^{-9}}{(1.5)^2}

E_1=60\ N/m

Electric field due to q_2

E_2=\frac{9\times 10^9\times 20\times 10^{-9}}{2^2}

E_2=45\ N/m

Net electric field will be vector addition of two

\vec{E_{net}}=\vec{E_1}+\vec{E_2}

\vec{E_{net}}=-60\hat{i}-45\hat{j}

Magnitude of Electric field is

E=\sqrt{60^2+45^2}

E=75\ N/m

8 0
3 years ago
Which of the following is prohibited by the 2nd law of thermodynamics?
vitfil [10]

(A) A device that converts heat into work with 100% efficiency

It clearly violates the second law of thermodynamics because it warns that while all work can be turned into heat, not all heat can be turned into work. Therefore, despite the innumerable efforts, the efficiencies of the bodies have only been able to reach 60% at present.

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3 years ago
At 20°C the vapor pressure of dry ice is 56.5 atm. If 10g of dry ice (solid CO2) is placed in an evacuated 0.25 L chamber at a c
jonny [76]

Answer:

Pressure of the gas is greater than the vapor pressure, therefore all the solid will not sublime.

Explanation:

To solve this problem we use ideal gas equation.

PV =nRT

If Presure of the gas (P) is greater than vapor pressure (56.5 atm), all the solid will not sublime. But If P < 56.5 atm, all of it will sublime.

where;

P is pressure of the gas in atm

V is volume of gas in Litre

T is absolute temperature of the gas in Kelvin

n is number of gas moles

R is ideal gas constant or Boltzmann constant = 0.082057 L atm K⁻¹ mol⁻¹

Given T = 20 °C  = 273 + 20 = 293K

Volume = 0.25L

n = Reacting mas (m)/Molar mass(M)

Molar mass of CO₂ = 12 + (16X2) = 12+32 = 44g/mol

Reacting mass = 10g

n = m/M ⇒ 10/44

n = 0.2273

PV =nRT ⇒ P = nRT/V

P = (0.2273X0.082057 X293)/0.25

P = 21.8596 atm

Therefore, since Pressure of the gas is greater than the vapor pressure, all the solid will not sublime.

5 0
4 years ago
If the ball leaves the projectile launcher at a speed of 2.2 m/s at an angle of 30ᴼ, and the projectile launcher is on a table a
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Answer: 1.12 m

Explanation:

This situation is related to parabolic motion, hence we can use the following equations:

y=y_{o}+V_{o}sin \theta t-\frac{g}{2}t^{2} (1)

x=V_{o} cos \theta t (2)

Where:

y=0 m is the ball final height (when it hits the ground)

y_{o}=1.1 m is the ball initial height

V_{o}=2.2 m/s is the initial velocity

\theta=30\° is the angle at which the ball was launched

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the horizontal distance the ball travels

Rewriting (1) with the given values:

0 m=1.1 m+(2.2 m/s)(cos 30\°)t-\frac{9.8 m/s^{2}}{2}t^{2} (3)

Multiplying all the eqquation by -1 and rearranging:

4.9 m/s^{2} t^{2}-1.1 m/s t-1.1 m=0 (4)

So, since we have a quadratic equation here (in the form of0=at^{2}+bt+c,  we will use the quadratic formula to find  t:  

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}   (5)

Where a=4.9, b=-1.1, c=-1.1  

Substituting the known values and choosing the positive result of the equation, we have:  

t=\frac{-(-1.1)\pm\sqrt{(-1.1)^{2}-4(4.9)(-1.1)}}{2(4.9)}  

t=0.59 s (6)

Now, substituting (6) in (2):

x=(2.2 m/s)(cos 30\°)(0.59 s) (7)

x=1.12 m (8) This is the horizontal distance at which the ball hits the ground.

3 0
4 years ago
How does water deposit sediment
uysha [10]

Erosion can move sediment through water, ice, or wind. Water can wash sediment, such as gravel or pebbles, down from a creek, into a river, and eventually to that river's delta. Deltas, river banks, and the bottom of waterfalls are common areas where sediment accumulates. i hope this answers your question

3 0
3 years ago
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