All categories

13 days ago

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

2 answers:

7 0

The tangential speed of an object is influenced by its distance from the center of the circle. The further away an object is from the center, the quicker it must move. Consequently, individuals residing at a latitude of 0 degrees experience the highest tangential speed.

ValentinkaMS [1.1K]13 days ago

3 0

Sample Response: The tangential speed is determined by how far the object is from the center of the circle. The farther from the center, the faster the object needs to go. Thus, people who live at 0 degrees latitude experience the highest tangential speed.

You might be interested in

A compressed spring has 16.2J of elastic potential energy when it is compressed 0.30m . What is the spring constant of the sprin

ValentinkaMS [1144]

I hope this provides the assistance you need.

3 0

10 days ago

On a caterpillars map all distances are marked in kilometers . The caterpillars map shows the distance between two milkweed plan

ValentinkaMS [1144]

Answer:

The equivalent distance in kilometers is 4012 × $10^{-6}$ km.

Explanation:

It's known that 1 millimeter converts to $10^{-3}$ meters. Then, 1 meter converts to $10^{-3}$ kilometers. Therefore, the conversion for 1 millimeter to kilometers can be stated as

1 mm = $10^{-3}$ m

1 m = $10^{-3}$ km

Thus, 1 mm = $10^{-3}$ × $10^{-3}$ km = $10^{-6}$ km.

Given the distance of 4012 mm, the corresponding distance in kilometers will be

4012 mm = 4012 × $10^{-6}$ km.

The distance therefore is 4012 × $10^{-6}$ km.

5 0

13 days ago

Compressed air is used to fire a 60 g ball vertically upward from a 0.70-m-tall tube. The air exerts an upward force of 3.0 N on

Yuliya22 [1153]

Answer:

2.87 m

Explanation:

Given parameters:

Mass of the ball (m) = 60 g = 0.06 kg

Height of the tube (h) = 0.70 m

Force applied on the ball by compressed air (F) = 3.0 N

Initial velocity of the ball (u) = 0 m/s (Assumed)

Final velocity of the ball at the tube's exit (v) =?

Acceleration of the ball (a) =?

The ball's weight is derived from multiplying mass and gravity. Therefore,

Weight (W) = $mg=0.06\times 9.8=0.588\ N$

Thus, the total force acting on the ball equals the net of upward force minus the weight.

Net force = Air force - Weight

$F_{net}=F-mg\\F_{net}=3.0-0.588 = 2.412\ N$

According to Newton's second law, net force equals the mass multiplied by acceleration.

$F_{net}=ma\\\\a=\frac{F_{net}}{m}=\frac{2.412\ N}{0.06\ kg}=40.2\ m/s^2$

Acceleration (a) is calculated as 40.2 m/s².

Using the motion equation, we find:

$v^2=u^2+2ah\\\\v^2=0+2\times 40.2\times 0.7\\\\v=\sqrt{56.28}=7.5\ m/s$

Let’s denote the maximum height achieved as 'H'.

Next, we apply the principle of energy conservation from the pipe's peak to the maximum height.

A decrease in kinetic energy equals an increase in potential energy.

$\frac{1}{2}mv^2=mgH\\\\H=\frac{v^2}{2g}$

Substituting the values, we solve for 'H', yielding:

$H=\frac{56.28}{2\times 9.8}\\\\H=\frac{56.28}{19.6}=2.87\ m$

Hence, the ball ascends to a height of 2.87 m above the top of the tube.

6 0

4 days ago

Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele

ValentinkaMS [1144]

Answer:

0.018 J

Explanation:

The work required to bring the charge from infinity to the point P is equal to the change in its electric potential energy. This can be expressed as

$W = q \Delta V$

where

$q=3.0 \mu C = 3.0 \cdot 10^{-6} C$ represents the charge's magnitude

and $\Delta V = 6.0 kV = 6000 V$ signifies the potential difference between point P and infinity.

After substituting into the formula, we arrive at

$W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J$

4 0

7 days ago

While dangling a hairdryer by its cord, you observe that the cord is vertical when the hairdryer isoff and, once it is turned on

Yuliya22 [1153]

Answer:

The air exiting from the hairdryer is moving at a speed of 10 m/s.

Explanation:

The thrust generated by the hairdryer enables it to maintain an elevation angled at 5° from vertical; thus, we derive from the force diagram

$(1).\: tan (5^o) = \dfrac{F_t}{Mg}$

by substituting $M =0.420kg$ , $g = 9.8m/s^2$ into the equation and resolving for $F_t$ we find:

$F_t = Mg\:tan(5^o)$

$F_t = (0.420kg)(9.8m/s^2)\:tan(5^o)$

$\boxed{F_t = 0.3601N.}$

This thrust is linked to the speed of air ejection $v$ through the equation

$(2).\: F_t = v\dfrac{dM}{dt}$

where $dM/dt$ signifies the rate of air ejection, which is known to be

$0.06m^3/2s = 0.03m^3/s$

and since $1m^3 = 1.2kg$ ,

$0.03m^3/s \rightarrow 0.036kg/s$

$\dfrac{dM}{dt} = 0.036kg/s,$

by inserting these values into equation (2), we obtain the value of $F_t$ as:

$0.3601N = 0.036v$

resulting in

$v= \dfrac{0.3601}{0.036}$

$\boxed{v =10m/s.}$

which indicates the air velocity discharged from the hairdryer.

6 0

14 days ago