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3 years ago

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If a pendulum clock keeps perfect time at the base of a mountain, will it also keep perfect time when it is moved to the tip of

the mountain? Explain.

1 answer:

vazorg [7]3 years ago

8 0

Answer:

No

Explanation:

The period of a pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration due to gravity

We see that the period of the pendulum depends on the value of g. However, the value of the gravitational acceleration is different at different locations on Earth. In particular, at the top of the mountain the value of g is slightly lower than the value of g at the base of the mountain; in fact, g is given by

$g=\frac{GM}{r^2}$

where

G is the gravitational constant

M is the Earth's mass

r is the distance from the Earth's center

so since r is greater at the top of the mountain, g is lower, and therefore the period of the pendulum will be slightly longer.

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A cart of mass 6.0 kg moves with a speed of 3.0 m/s towards a second stationary cart with a mass of 3.0 kg. The carts move on a

IgorLugansk [536]

Answer:1.5

Explanation:

Given

mass of first cart $m_1=6 kg$

initial Velocity $u_1=3 m/s$

mass of second cart $m_2=3 kg$

$u_2=0 m/s$

In the absence of External Force we can conserve momentum

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\frac{m_1u_1+m_2u_2}{m_1+m_2}$

$v=\frac{6\times 3+3\times 0}{6+3}$

$v=2 m/s$

Final kinetic Energy of two masses

$K.E._2=\frac{1}{2}(m_1+m_2)v^2$

$K.E._2=\frac{1}{2}\cdot (3+6)\cdot (2)^2$

$K.E._2=18 J$

Initial Kinetic Energy

$K.E._1=\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2$

$K.E._1=\frac{1}{2}6\times 3^2+0$

$K.E._1=27 J$

$ratio =\frac{K.E._1}{K.E._2}=\frac{27}{18}=1.5$

5 0

4 years ago

A woman can row a boat at 5.60 km/h in still water. (a) If she is crossing a river where the current is 2.80 km/h, in what direc

katrin2010 [14]

Answer:

a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.

Explanation:

So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

$cos \theta = \frac {V_{river}}{V_{boat}}$

When solving for theta, we get that:

$\theta =cos^{-1} ( \frac {V_{river}}{V_{boat}})$

so now we can substitute the corresponding values:

$\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})$

Which yields:

$\theta = 60^{o}$

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

$V_{y}=5.60km/hr*cos(210^{o})$

$V_{y}=-4.85km/hr$

The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:

$t=\frac{y}{V_{y}}$

$t=\frac{5.60km}{4.85km/hr}=1.155hr$

part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

$V_{ds}=V_{river}+V{boat}$

$V_{ds}=2.80km/hr+5.60km/hr=8.40km/hr$

and now up stream:

$V_{us}=V_{boat}-V{river}$

$V_{us}=5.60km/hr-2.80km/hr=2.80km/hr$

Once we got these two velocities we will now need to find the time to take each trip:

time down stream:

$t_{ds}=\frac{x}{v_{ds}}$

$t_{ds}=\frac{2.80km}{8.40km/hr}=0.333hr$

and the time up stream:

$t_{us}=\frac{x}{v_{us}}$

$t_{us}=\frac{2.80km}{2,80km/hr}=1hr$

so the total time will be:

$t_{ds}+t_{us}=0.333hr+1hr=1.333hr$

d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:

t=1.333hr

e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)

f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

$t=\frac{d}{v_{boat}}$

$t=\frac{5.60km}{5.60km/hr}$

so

t= 1hr

4 0

4 years ago

Read 2 more answers

emmainna [20.7K]

Answer:

Mistakes, chaos theory.

Explanation:

Weather forecast refers to the application of technology and science in predicting what the condition's of an atmosphere would be at a particular time in a given location. Keyword here is "Predicting."

To accurately "predict" the weather, observation data needs to be collected from weather stations across the world. The data collected from the stations are too complex such that they can only be processed by supercomputers.

It is important to note that supercomputers used to predict weather are liable to make mistakes because the enormous challenge of weather forecasting is more than their capabilities.

In addition, there is a concept known as chaos theory. It explains how very small changes in initial weather conditions can change the final results. No thanks to chaos, meteorologists would not be able to predict weather accurately.

This is the reason why the world still suffers from the incidence of surprise rain, storms and other weather events that could be devastating.

4 0

4 years ago

Suppose you have two small pith balls that are 5.5 cm apart and have equal charges of -29 nc?

zysi [14]

The question is missing, however, I guess the problem is asking for the value of the force acting between the two balls.

The Coulomb force between the two balls is:
$F= k_e \frac{ q_1 q_2}{r^2}$
where $k_e=8.99\cdot10^9~N m^2 C^{-2}$ is the Coulomb's constant, $q_1=q_2=29~nC=29\cdot 10^{-9}~C$ is the intensity of the two charges, and $r=5.5~cm=0.055~m$ is the distance between them.

Substituting these numbers into the equation, we get
$F=2.5~10^{-3}~N$

The force is repulsive, because the charges have same sign and so they repel each other.

6 0

3 years ago

A paintball is fired horizontally from a tower 45 m above the ground. If the paintball gun fires at 90 m/s… How long does it tak

ankoles [38]

Answer:

The time taken for the paint ball to hit the ground is $t = 2.143 \ s$

The distance of the landing point from the tower is $d = 192.86 \ m$

Explanation:

From the question we are told that

The height of the tower is $h = 45 \ m$

The speed of the paintball in the horizontal direction is $v_x = 90 \ m/s$

Generally from kinematic equation we have that

$h = ut + \frac{1}{2} at^2$

Here u is the initial velocity of the paintball in the vertical direction and the value is 0 m/s , this because the ball was fired horizontally

a is equivalent to $g = 9.8\ m/s^2$

t is the time taken for the paintball to hit the ground

So

$45 = 0* t + \frac{1}{2} 9.8 * t^2$

=> $t = 2.143 \ s$

Generally the distance of its landing position from the tower is

$d = v * t$

=> $d = 90 * 2.143$

=> $d = 192.86 \ m$

3 0

3 years ago