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2 years ago

At which latitudes shown in the image of Earth do people experience the greatest tangential speed? Explain why.

2 answers:

7 0

Tangential speed is dependent on an object's distance from the center of the circle. The farther away from the center an object is, the faster the object has to travel. Therefore, people living at a latitude of 0 degrees experience the greatest tangential speed.

Mademuasel [1]2 years ago

3 0

Sample Response: Tangential speed is dependent on an object's distance from the center of the circle. The farther away from the center an object is, the faster the object has to travel. Therefore, people living at a latitude of 0 degrees experience the greatest tangential speed.

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A system absorbs 52 joules of heat and does work using 25 joules of energy. What is the change in the internal energy of the sys

motikmotik

The change in internal energy is $\Delta U=Q+W$ . Where $\Delta U$ is change in internal energy. $Q$ is heat added to the system (absorbed by the system). $W$ work done on the system. $W$ is taken as positive if work is done on the system and negative if work is done by the system. Here $Q=52$ and $W=-25$ , hence change in internal energy is

$\Delta U=52-25\\ \Delta U=27$

Change in internal energy of the system is 27 J. Internal energy increases.

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2 years ago

Read 2 more answers

Newton's rings are visible when a planoconvex lens is placed on a flat glass surface. For a particular lens with an index of ref

blsea [12.9K]

Answer:

the new diameter of the third ring = 0.607 mm

Explanation:

Consider the radius of $\\m ^{th}\\$ bright ring when air is in between the lens and the plate ;

$r = \sqrt {\frac{(2m+1)\lambda R}{2}}$

Using the expression: $r (n)= \sqrt {\frac{(2m+1)\lambda R}{2n}}$ for the radius of the $\\m ^{th}\\$ bright ring since the liquid (water ) of the refractive index 'n' is now in between the lens and the plate;

where;

$\\m ^{th}\\$ = number of fringe

λ = wavelength

R = radius

n = refractive index of water

Now ;

$r (n)=\frac {r}{n}}$

the radius r of the third bright ring when the air is in between lens and plate = $r= \frac{0.700 \ mm}{2} \\\\$

$r= 0.35 \ mm$

The new radius of the third bright fringe is $r(3) = \frac{0.35}{\sqrt 1.33}$

$r(3) = 0.3035 \ mm$

Calculating the new diameter ; we have:

d(3) = 2(r(3))

d(3) = 2(0.3035)

d = 0.607 mm

Thus, the new diameter of the third ring = 0.607 mm

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2 years ago

A vertical wire carries a current straight up in a region where the magnetic field vector points due north. What is the directio

Elanso [62]

Answer:

The direction of the resulting force on this current is due east.

Explanation:

Given;

direction of the magnetic field to be due north

Applying right hand rule which states that: to determine the direction of the magnetic force on a positive moving charge point the thumb of the right hand in the direction of velocity v, the fingers in the direction of magnetic field B, and a perpendicular to the palm points in the direction of magnetic force.

Since the magnetic force must be perpendicular to the magnetic field, and direction of the magnetic field is due north, then the magnetic force must be due East.

Therefore, the direction of the resulting force on this current is due east.

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2 years ago

What is the mass and density of 237 mL of water

oee [108]

Answer:

<h2><em>V(water)= 237 mL=237×10^-6 m^3</em></h2><h2><em>ρ(water)=1000 kg/m^3</em></h2><h2><em>m=</em><em>ρ×V=(1000)×(237×10^-6)</em></h2><h2><em>m= 237×10^-3 = 0.237 kg</em></h2><h2><em>m= 237 gram.</em></h2>

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2 years ago

A child of mass m is at the edge of a merry-go-round of diameter d. When the merry-go-round is rotating with angular acceleratio

dem82 [27]

Answer:

The torque on the child is now the same, τ.

Explanation:

It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.
In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.
The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of the mass of the child times the square of the distance to the center.
When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:

$I_{to} = I_{d} + m*r^{2} = m*\frac{r^{2}}{2} + m*r^{2} = \frac{3}{2}* m*r^{2} (1)$

So, τ = 3/2*m*r²*α (2)

When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:

$I_{t} = I_{d} + m*\frac{r^{2}}{4} = m*\frac{r^{2}}{2} + m*\frac{r^{2}}{4} = \frac{3}{4}* m*r^{2} (3)$

Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:

τ = 3/4*m*r² * (2α) = 3/2*m*r²

same result than in (2), so the torque remains the same.

7 0

2 years ago