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2 years ago

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A 4-N object object swings on the end of a string as a simple pendulum. At the bottom of the swing, the tension in the string is

9 N. What is the magnitude of the centripetal acceleration of the object at the bottom of the swing? Give the answer as a function of the acceleration due to gravity, g.

1 answer:

Amanda [17]2 years ago

6 0

Answer:

Explanation:

Given mg = 4N .

m = 4 / g

At the bottom of the swing let centripetal acceleration be a

T - mg = ma

9 - 4 = ma

5 = 4 a / g

a = 5g / 4

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Shows an object suspended from two ropes. The weight of the object is 360 N. The magnitude of the tension

konstantin123 [22]

Answer:

Tension T in each rope will be 254.56 N.

Explanation:

From the picture attached,

Weight suspended by the two ropes will be supported by the vertical components of the tension in the ropes.

Vertical components of both the ropes = Tsin(45)° + Tsin(45)°

= 2Tsin(45)°

= $2T(\frac{1}{\sqrt{2}})$

= $T\sqrt{2}$

Since, $T\sqrt{2}=360$

T = $\frac{360}{\sqrt{2} }$

T = 254.56 N

Therefore, tension T in each rope will be 254.56 N.

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2 years ago

what will be the resistivity of a metal wire of 2m length and 0.6mm in a diameter ,if the resistance of the wire is 50ohm . find

Maru [420]

Answer:Resistivity of the wire is 7.065 × 10^-8 Ω m.

Explanation:

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2 years ago

In this lab you will use a cart and track to explore various aspects of motion. You will measure and record the time it takes th

Keith_Richards [23]

It is given that by using track and cart we can record the time and the distance travelled and also the speed of the cart can be recorded. With all this data we can solve questions on the laws of motion.

Like using the first law of motion we can determine the force of gravity acting on the cart that has moved a certain distance and the velocity or the speed of card has already been registered and since time is known putting the values in formula would help us calculate the gravitational pull acting on cart.

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2 years ago

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What is the value of g on the surface of Saturn? Assume M-Saturn = 5.68×10^26 kg and R-Saturn = 5.82×10^7 m.Choose the appropria

Likurg_2 [28]

Answer:

Approximately $\rm 11.2 \; N \cdot kg^{-1}$ at that distance from the center of the planet.

Option A) The low value of $g$ near the cloud top of Saturn is possible because of the low density of the planet.

Explanation:

The value of $g$ on a planet measures the size of gravity on an object for each unit of its mass. The equation for gravity is:

$\displaystyle \frac{G \cdot M \cdot m}{R^2}$ ,

where

$G \approx 6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2$ .
$M$ is the mass of the planet, and
$m$ is the mass of the object.

To find an equation for $g$ , divide the equation for gravity by the mass of the object:

$\displaystyle g = \left.\frac{G \cdot M \cdot m}{R^2} \right/\frac{1}{m} = \frac{G \cdot M}{R^2}$ .

In this case,

$M = 5.68\times 10^{26}\; \rm kg$ , and
$R = 5.82 \times 10^7\; \rm m$ .

Calculate $g$ based on these values:

$\begin{aligned} g &= \frac{G \cdot M}{R^2}\cr &= \frac{6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2\times 5.68\times 10^{26}\; \rm kg}{\left(5.82\times 10^7\; \rm m\right)^2} \cr &\approx 11.2\; \rm N\cdot kg^{-1} \end{aligned}$ .

Saturn is a gas giant. Most of its volume was filled with gas. In comparison, the earth is a rocky planet. Most of its volume was filled with solid and molten rocks. As a result, the average density of the earth would be greater than the average density of Saturn.

Refer to the equation for $g$ :

$\displaystyle g = \frac{G \cdot M}{R^2}$ .

The mass of the planet is in the numerator. If two planets are of the same size, $g$ would be greater at the surface of the more massive planet.

On the other hand, if the mass of the planet is large while its density is small, its radius also needs to be very large. Since $R$ is in the denominator of $g$ , increasing the value of $R$ while keeping $M$ constant would reduce the value of $g$ . That explains why the value of $g$ near the "surface" (cloud tops) of Saturn is about the same as that on the surface of the earth (approximately $9.81\; \rm N \cdot kg^{-1}$ .

As a side note, $5.82\times 10^7\rm \; m$ likely refers to the distance from the center of Saturn to its cloud tops. Hence, it would be more appropriate to say that the value of $g$ near the cloud tops of Saturn is approximately $\rm 11.2 \; N \cdot kg^{-1}$ .

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2 years ago

Two roads intersect at right angles, one going north-south, the other east-west. an observer stands on the road 60 meters south

Sloan [31]

observer is standing at distance d = 60 m south from the intersection

cyclist is travelling at speed v = 10 m/s

now after t = 8 s its displacement from intersection is given by

$x = 10*8 = 80 m$

so the position of cyclist makes an angle with the observer

$\theta = tan^{-1}\frac{80}{60} = 53 degree$

now the component of velocity of cyclist along the line joining its position with the observer is given as

$v = v_o cos\phi$

here

$\phi = 90 -\theta$

$\phi = 90 - 53 = 37 degree$

$v = 10 cos37 = 8 m/s$

so at this instant cyclist is moving away with speed 8 m/s

7 0

2 years ago

Read 2 more answers