When two objects are near each other, how would increasing one object’s mass affect it?

A 2600-m-high mountain is located on the equator. how much faster does a climber on top of the mountain move than a surfer at a

puteri [66]

The climber move 0.19 m/s faster than surfer on the nearby beach.

Since both the person are on the earth, and moves with the constant angular velocity of earth, however there linear velocity is different.

Number of seconds in a day, t=24*60*60=86400 sec

The linear speed on the beach is calculated as

V1= $\frac{2πr}{t}$

Here, t is the time

Plugging the values in the above equation

V1= $\frac{2π*6.4*10^6}{86400}$ =465.421 m/s

Velocity on the mountain is calculated as

V2= $\frac{2π(r+h)}{t}$

Plugging the values in the above equation

V2= $\frac{2π(6.4*10^6+2600}{86400}$ =465.61 m/s

Therefore person on the mountain moves faster than the person on the beach by 465.61-465.421=0.19 m/s

5 0

3 years ago

Please i need your help w/h/w you make the following measurements if an object:42kg, and 22m

Rufina [12.5K]

Answer:

$1.9 kg/m^3$

Explanation:

The density of an object is given by

$d=\frac{m}{V}$

where

m is the mass of the object

V is its volume

In this problem,

m = 42 kg

V = 22 m^3

Substituting into the equation, we find the object's density:

$d=\frac{42 kg}{22 m^3}=1.9 kg/m^3$

8 0

3 years ago

A Cat is on a balcony floor (90cm below the railing), keenly eyeing a butterfly hovering 60 cm above the railing. With what spee

GenaCL600 [577]

We will have the following:

First, the equation to use is the following:

$d=v_ot+\frac{1}{2}at^2$

Now, we transform the total distance the cat would need to travel:

$90\operatorname{cm}+60\operatorname{cm}=150\operatorname{cm}\cdot\frac{1m}{100\operatorname{cm}}=1.5m$

So, the cat would need to travel 1.5 meters. ("d" in the equation).

Now, using the speed given we determine the time it would take the cat to traverse the 1.5 meters:

$t=\frac{1.5m\cdot1s}{0.45m}\Rightarrow t=\frac{10}{3}\Rightarrow t=3.333\ldots$

So, the time it would take the cat to traverse the distance will be approximately 3.33 seconds.

Now, we know that the acceleration will be given by Earth's gravity, so:

$1.5m=v_0(\frac{10}{3}s)+\frac{1}{2}(-\frac{9.8m}{s^2})(\frac{10}{3}s)^2\Rightarrow1.5m=v_0(\frac{10}{3}s)+(-\frac{490}{9}m)$ $\Rightarrow\frac{1007}{18}m=v_0(\frac{10}{3}s)\Rightarrow v_0=\frac{1007}{60}\frac{m}{s}\Rightarrow v_0=16.78333\ldots$

So, the initial vvelocity the cat must leave the floor in order to arrive at the butterfly with the optimum pouncing speed of 0.45 m/s is approximately 16.78 m/s or exactly 1007/60 m/s.

5 0

2 years ago

Which statement describes between a cup if cold water and a cup of boiling water

marishachu [46]

The thermal energy keeps these two changing their temperature over time.

The surrounding temperature (environment) clearly effects the cold water AND the boiling water.

Eventually, both would reach an equilibrium in temperature. <span />

4 0

3 years ago

The Sears Tower is nearly 400 m high. How long would it take a steel ball to reach the ground if dropped on the top? What will b

kipiarov [429]

Answers:

a) 9.035 s

b) -88.543 m/s

Explanation:

The described situation is related to vertical motion (especifically free fall) and the equations that will be useful are:

$y=y_{o}+V_{o}t+\frac{1}{2}gt^{2}$ (1)