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3 years ago

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Alice and Tom dive from an overhang into the lake below. Tom simply drops straight down from the edge, but Alice takes a running

start and jumps with an initial horizontal velocity of 25 m/s. Neither person experiences any significant air resistance. Just as they reach the lake below.A) Alice reaches the surface the lake first. B) Tom of reaches the surface of the first. C) Alice and Tom will reach the surface of the lake at the same time. D) Information is not enough

1 answer:

crimeas [40]3 years ago

3 0

Answer:c

Explanation:

Given

Alice launches with horizontal velocity $u=25\ m/s$

Tom simply drops straight down from the edge

Time taken by both the person is same as they have same initial vertical velocity i.e. zero so the time taken to reach the ground is zero.

Although Alice will travel more horizontal distance compared to Tom.

Thus option c is correct

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A large container contains a large amount of water. A hole is drilled on the wall of the container, at a vertical distance h = 0

hammer [34]

Speed can be found through the application of concepts related to potential energy and kinetic energy, for which you have

$KE = PE$

$\frac{1}{2}mv^2 = mgh$

Where,

m = mass

v = Velocity

g = Gravitational acceleration

h = Height

Re-arrange to find the velocity we have,

$v^2 = 2gh$

$v = \sqrt{2gh}$

$v = \sqrt{2(9.8)(0.54)}$

$v = 3.253m/s$

Therefore the speed at which water squirts out of the hole is .3253m/s

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3 years ago

If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is

viktelen [127]

To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: $V_{f}=V_{i}+gt$
where
$V_{f}$ is the final velocity
$V_{i}$ is the initial velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time
2. Kinematic equation for distance: $d=V_{i}t+ \frac{1}{2} gt^2$
where
$d$ is the distance
$V_{i}$ is the initial velocity
$V_{f}$ is the final velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time

First, we are going to convert 21.82 mi/h to ft/s:
$21.82 \frac{mi}{h} =31.21 \frac{ft}{s}$

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: $V_{f}=0$ . We also know that it initial speed is 31.21 ft/s, so $V_{i}=31.21$ . Lets replace those values in our formula to find $t$ :
$V_{f}=V_{i}+gt$
$0=31.21+(-32)t$
$-32t=-31.21$
$t= \frac{-31.21}{-32}$
$t=0.98$ seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
$d=V_{i}t+ \frac{1}{2} gt^2$
$d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2$
$d=15.22$ ft

Now we can add the height of the building and the maximum height of the object:
$d=160+15.22=175.22$ ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
$d=V_{i}t+ \frac{1}{2} gt^2$
$175.22=31.21t+ \frac{1}{2} (32)t^2$
$16t^2+31.21t-175.22=0$
$t=2.47$ or $t=-4.43$
Since time cannot be negative, $t=2.47$ is the time it takes the object to fall to the ground.

Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
$V_{f}=V_{i}+gt$
$V_{f}=31.21+(32)(2.47)$
$V_{f}=110.25$ ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s

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3 years ago

In a cell, the amount nutrition coming in equals the amount of waste going out. This is an example of _____.

babymother [125]

The answer is B) <span>equilibrium
hope this helps!=-)</span>

5 0

3 years ago

The figure shows a positive charge placed in a uniform electric field. Is the force exerted on the +1 C charge directed up or do

raketka [301]

Answer:

the answer is down on e2020

Explanation:

3 0

3 years ago

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A velocity selector in a mass spectrometer uses a 0.150 T magnetic field. (a) What electric field strength (in volts per meter)

Alekssandra [29.7K]

Answer:

The electric field strength is $6.6\times10^{5}\ V/m$

Explanation:

Given that,

Magnetic field = 0.150 T

Speed $v= 4.40\times10^{6}\ m/s$

We need to calculate the electric field strength

Using formula of velocity

$v=\dfrac{E}{B}$

$E=v\times B$

Where, v = speed

B = magnetic field

Put the value into the formula

$E=4.40\times10^{6}\times0.150$

$E=660000\ V/m$

$E=6.6\times10^{5}\ V/m$

Hence, The electric field strength is $6.6\times10^{5}\ V/m$

4 0

3 years ago