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Y_Kistochka [10]
3 years ago
6

A 0.25-kg ball attached to a string is rotating in a horizontal circle of radius 0.5m. Assume the tension force is directed hori

zontally.
If the ball revolves twice every second, what is the tension in the string?
Physics
1 answer:
borishaifa [10]3 years ago
7 0

Answer:

T= 19.74 N

Explanation:

The tension in the string will be equal to centripetal force on the ball. (Newton's 3rd law, equal acting and reacting forces acts on a body)

T = Fc

Fc = m w ²r

here:

m = 0.25 kg, r = 0.5 m

w = 2 rps = 2 * 2π rad/s      ( ∴ 1 revolution = 2π radians )

w = 4π rad/s

Fc = m w ²r = 0.25 * ( 4π )² * 0.5 N

Fc = T = 19.74 N

T= 19.74 N

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__________ research is designed to measure the association between variables that are not manipulated by the researcher.
Sliva [168]

Answer:

Option c. Correlational

Explanation:

Correlational research is basically a method of carrying out a research where an experiment can not be conducted and where a researcher has to figure out that whether the two variable are in association with each other and if these are related then in what way the association between these two exists without any manipulation from the researcher's end.

It seeks to understand, measure and make assessment of the relationship or  

association between these variables but can not find out if a variable is caused by the other variable.

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4 years ago
How does the energy moves from the suns core to the photosphere?
Minchanka [31]

Answer: Energy from the core travels by radiation through the radiative

zone, then by convection through the convection zone.

Explanation:

4 0
3 years ago
Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth
mylen [45]

Answer: 3.7 \times 10^{-4} N

Explanation:

The gravitational pull between two object is given by:

F = G\frac{Mm}{r^2}

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N

4 0
3 years ago
Read 2 more answers
g A box having mass 0,5kg is placed in front of a 20 cm compressed spring. When the spring released, box having mass m1, collide
vagabundo [1.1K]

Answer:Expression given below

Explanation:

Given mass of spring\left ( m_1\right )=0.5 kg

Compression in the spring\left ( x\right )=20 cm

Let the spring constant be K

Using Energy conservation

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\frac{1}{2}Kx^2=\frac{1}{2}m_1v^2

v=x\sqrt{\frac{k}{m_1}}

now conserving momentum

m_1v=\left ( m_1+m_2\right )v_0

v_0=\frac{m_1}{m_1+m_2}v

where v_0 is the final velocity

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nydimaria [60]

Answer:

no the moon does not rotate it only goes in circle just like the sun so I disagree with your friend

3 0
3 years ago
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