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Natali [406]
3 years ago
12

When you whirl a small rubber stopper on a cord in a vertical circle, you find a critical speed at the top for which the tension

in the cord is zero. At this speed. the force of gravity on the object 1S itselfisufficient to supply the necessary centripetal force. How slowly can you swing a 15-g stopper like this so that It will just follow a circle with a radius of m?​
Physics
1 answer:
Sauron [17]3 years ago
6 0

Answer:

  √(9.8/m) radians per second

Explanation:

The centripetal acceleration required for a radius of m is ...

  a = mω²

so, the angular speed required is ...

  ω = √(a/m) =  √(g/m) . . . . . where g is the acceleration due to gravity

  ω = √(9.8/m) . . . radians per second

__

The speed is independent of the mass. For a 1 m radius, the speed is about 29.9 RPM.

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A motorboat maintained a constant speed of 15 miles per hour relative to the water in going 10 miles upstream and then returning
Lesechka [4]

Answer:

speed of current is 5 mile/hr

Explanation:

GIVEN DATA:

speed of motorboat = 15 miles/hr relative with water

let c is speed of current

15-c is speed of boat at  upstream

15+c is speed of boat at downstream

we know that

travel time=distance/speed

\frac{10}{15-c} +\frac{10}{15+c} = 1.5

150+10c+150-10c=1.5(15-c)(15+c)

300=1.5(225-c^2)

300=337.5-1.5c^2

200=225-c^2

c^2=25

c = 5

so speed of current is 5 mile/hr

6 0
3 years ago
The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t
dem82 [27]

Answer:

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

T = 2\pi \sqrt{\frac{r^3}{GM}}

now for the time period of moon around the earth we can say

T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}

here we know that

T_1 = 0.08 year

r_1 = 0.0027 AU

M_e = mass of earth

Now if the same formula is used for revolution of Earth around the sun

T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}

here we know that

r_2 = 1 AU

T_2 = 1 year

M_s = mass of Sun

now we have

\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}

\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}

12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

4 0
3 years ago
The half-life of the reaction gets shorter as the initial concentration is increased. True or False
Art [367]

True

The half-life isn’t applicable to a first order reaction because it does not rely on the concentration of reactant present. However the 2nd order reaction is dependent on the concentration of the reactant present.

The relationship between the half life and the reactant is an inverse one.

The half life is usually reduced or shortened with an increase in the concentration and vice versa.

3 0
3 years ago
Which electromagnetic wave has the lowest frequencies (less than 3×109 hertz)?
ICE Princess25 [194]

Answer:

Radio waves

Explanation:

The electromagnetic spectrum includes all different types of waves, which are usually classified depending on their frequency. Ordering them from the highest frequency to the lowest frequency, they are:

- Gamma rays

- X-rays

- Ultraviolet

- Visible light

- Infrared radiation

- Microwaves

- Radio waves

Radio waves are the electromagnetic waves with lowest frequency, their frequency is lower than 300 GHz (3\cdot 10^{11} Hz) and therefore they are the electromagnetic waves with lowest energy (in fact, the energy of an electromagnetic wave is proportional to its frequency). They are generally used for radio and telecommunications since this type of waves can travel up to long distances.

5 0
3 years ago
Read 2 more answers
Will increasing length and cross-sectional area increase the resistance of an object?
Klio2033 [76]

Answer:

Increasing length increases resistance

increasing cross sectional area reduces the resistance

.

Explanation:

The formula for resistance of an object is

r = d \frac{l}{a}

where r is resistance, d is resistivity of the material, l is length of material and a is cross sectional area of the object. This equation shows us that resistance is directly proportional to length and inversely proportional to cross sectional area. Hence, increasing length increases resistance while increasing cross sectional area reduces the resistance.

If these 2 variables are varied to the same extent, the net effect can be zero on the resistance.

5 0
2 years ago
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