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3 years ago

Which is an action that can cause an object to move or change its state of motion? energy, force, equilibrium, or inertia?

Physics

2 answers:

Licemer1 [7]3 years ago

8 0

<span>an action that can cause an object to move or change its state of motion is force</span>

Black_prince [1.1K]3 years ago

8 0

Answer:

Force

Explanation:

Force is a push or pull which can change the state of body whether it is rest or in motion or tries to change the state of body.

Force is equal to the rate o change of momentum of body.

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A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and

iogann1982 [59]

Answer:

$p_2 = 1.76 atm$

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

p_2 = ?

t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

$\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}$

Putting the values

$\frac{1.15*2.37}{280} =\frac{p_2 *1.68}{304}$

$9.733*10^{-3} = \frac{p_2 *1.68}{304}$

$9.733*10^{-3}*304 = p_2*1.68$

$\frac{9.733*10^{-3}*304}{1.68} =p_2$

$p_2= 1.76 atm$

7 0

3 years ago

A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch

Lina20 [59]

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

$x=V_{o}cos \theta t$ ----------------------(1)

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

$V=V_{o}-gt$ ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

$V_{o}$ is the golf ball's initial velocity

$\theta=0\°$ is the angle (it was a horizontal shot)

t is the time

y is the final height of the ball

$y_{o}$ is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

$t=\sqrt{\frac{2 y_{o}}{g}}$

$t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}$

$t=1.597$ s

Substituting (6) in (1):

$67.1 =V_{o} cos(0\°) 1.597$ -------------------(4)

Step 2: Finding $V_{o}$ :

From equation(4)

$67.1 =V_{o}(1) 1.597$

$V_0 = \frac{6.71}{1.597}$

$V_{o}=42.01$ m/s (8)

Substituting $V_{o}$ in (3):

$V=42.01 -(9.8)(1.597)$

v =42 .01 - 15.3566

V=26.359 m/s

5 0

3 years ago

A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of

Nastasia [14]

Answer:

Amplitude = 0.058m

Frequency = 6.25Hz

Explanation:

Given

Amplitude (A) = 8.26 x 10-2 m

Frequency (f) = 4.42Hz

Conversation of energy before split

½mv² = ½KA²

Make A the subject of formula

A = $v\sqrt{\frac{m}{k} }$

Conversation of energy after split

½(m/2)V'² = ½(m/2)V² = ½KA'²

½(m/2)V² = ½KA'²

Make A the subject of formula

First divide both sides by ½

(m/2)V² = KA'²

Divide both sides by K

$\frac{m}{2K}$ V² = A'²

$v\sqrt{\frac{m}{2k} }$ = A'

Substitute $v\sqrt{\frac{m}{k} }$ for A in the above equation

A' = A/√2

A' = 8.26 x 10^-2/√2

A' = 0.05840702012600882

Amplitude after split = 0.058 (Approximated)

Frequency (f') = f√2

f' = 4.42√2

f' = 6.25082394568908011

Frequency after split = 6.25Hz (approximated)

8 0

3 years ago

What experiment should I make using Gravitational Force? <br><br>PLEASE HELP ME :)

Bogdan [553]

You could try the "Spinning Bucket" or the "Center Of Gravity" experiment. There are plenty more that you could research! Hope this helped :)

8 0

3 years ago

What is the maximum angular momentum Lmax that an electron with principal quantum number n = 2 can have? Express your answer in

butalik [34]

Answer:

$L_{max} = 1.414$ ℏ

Given:

Principle quantum number, n = 2

Solution:

To calculate the maximum angular momentum, $L_{max}$ , we have:

$L_{max} = \sqrt {l(1 + l)}$ (1)

where,

l = azimuthal quantum number or angular momentum quantum number

Also,

n = 1 + l

2 = 1 + l

l = 1

Now,

Using the value of l = 1 in eqn (1), we get:

$L_{max} = \sqrt {1(1 + 1)} = \sqrt 2$

$L_{max} = 1.414$ ℏ

4 0

3 years ago