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34kurt
3 years ago
13

A 3kg object moving at 4 m/s encounters a 20 N resistive force over a duration of 0.20 seconds. The impulse experienced by this

object is _____
Physics
1 answer:
WITCHER [35]3 years ago
3 0

Answer:

the impulse experienced by this object is 4 Ns

Explanation:

Given;

mass of the object, m = 3 kg

velocity of the object, v = 4 m/s

resistive force, F = 20 N

duration of impact, t = 0.2 s

The impulse experienced by this object is calculated as follows;

J = F x t

J = 20 x 0.2

J = 4 Ns

Therefore, the impulse experienced by this object is 4 Ns

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An FM radio station broadcasts at 9.23 × 107 Hz. Given that the radio waves travel at 3.00 × 108 m/s, what is the wavelength of
STALIN [3.7K]

Answer:

3.2m

Explanation:

Given parameters:

Frequency of the FM radio = 9.23 x 10⁷Hz

Velocity of the waves = 3  x 10⁸m/s

Unknown:

Wavelength of the wave = ?

Solution:

To solve for the wavelength of the wave, we need the velocity equation;

       Velocity = frequency x wavelength.

Radio waves are all electromagnetic radiations produced by both electrical and magnetic fields perpendicularly oriented to one another.

 Since the unknown is wavelength, we solve for it:

       

    3  x 10⁸  = 9.23 x 10⁷ x wavelength

          wavelength = \frac{ 3 x 10^{8} }{9.23 x  10^{7} }

        wavelength = 3.2m

7 0
3 years ago
A 4.4 kg marble (really big heavy marble) is accelerating down an incline. When it reaches level ground it slows down to a stop
KengaRu [80]

#A

Mass=4.4kg

acceleration=-1.74m/s^2

Use newtons second law

\\ \rm\longmapsto Force=ma

\\ \rm\longmapsto Force=4.4(-1.74)

\\ \rm\longmapsto Force=-7.656N

#B

initial velocity=u

Final velocity=v=0

Acceleration=a=-1.74m/s^2

Time=t=1.27s

\\ \rm\longmapsto a=\dfrac{v-u}{t}

\\ \rm\longmapsto u=v-at

\\ \rm\longmapsto u=0-(-1.74)(1.27)

\\ \rm\longmapsto u=1.74(1.27)

\\ \rm\longmapsto u=2.2m/s

4 0
2 years ago
A syringe containing 1.54 ml of oxygen gas is cooled from 92.8 ∘c to 0.4 ∘c. what is the final volume vf of oxygen gas? (assume
Andrews [41]
Givens
=====
Ti = 92.8 oC = 92.8 + 273 = 365.8
T2 = 0.4 oC = 0.4 + 273 = 273.4

Vi = 1.54 mL
Vf = ?????

Formula
======
Vi/Ti = Vf / Tf
1.54 / 365.8 = Vf / 273.4

Calculations
=========
1.54 * 273.4 / 355.8 = Vf
vf = 1.15 mL

Comment
=======
if your teacher is concerned about sig digs, the answer is 1 mL. I'm not sure how to show that it is 1 sig dig. You can ask about this.

5 0
3 years ago
During a chemical reaction, Bromine (Br) would be expected to
Andrew [12]

Answer:

During a chemical reaction, Bromine (Br) would be expected to  <u><em>gain 1 valence electron to have a full octet.</em></u>

Explanation:

In the periodic table the elements are ordered so that those with similar chemical properties are located close to each other.

The elements are arranged in horizontal rows, called periods, which coincide with the last electronic layer of the element. That is, an element with five electronic shells will be in the fifth period.

The columns of the table are called groups. The elements that make up each group coincide in their electronic configuration of valence electrons, that is, they have the same number of electrons in their last.

The elements tend to resemble the closest noble gases in terms of their electronic configuration of the last layer, that is, having eight electrons in the last layer to be stable.

Bromine belongs to group 17 (VII A), which indicates that it has 7 electrons in its last shell. So bromine requires more energy to lose all 7 electrons and generate stability, than it does to gain 1 electron and fill in 8 electrons to be stable. So:

<u><em>During a chemical reaction, Bromine (Br) would be expected to  gain 1 valence electron to have a full octet.</em></u>

6 0
3 years ago
The orbital motion of Earth around the Sun leads to an observable parallax effect on the nearest stars. For each star listed, ca
Murrr4er [49]

Answer:

Following are the answer to this question:

Explanation:

Formula:

D(PC) =\frac{1}{parallax}\\\\D(av)=D(PC) \times 20.626\ J

Calculating point A:

when the value is 0.38

\to 0.38 \toD(PC)= \frac{1}{0.38}\\\\

                   =2.632

\to D(a.v) = \frac{1}{0.38} \times 206265\\

               =542,802.6

Calculating point B:

when the value is 0.75

\to D(PC)=\frac{1}{0.75}

                =1.33

\to D(a.v) = \frac{1}{0.75} \times 206265\\

             =275,020

Calculating point C:

when the value is 0.28

\to D(PC)=\frac{1}{0.28}

                =3.571

\to D(a.v) = \frac{1}{0.28} \times 206265\\

               =736660.7

Calculating point D:

when the value is 0.42

\to D(PC)=\frac{1}{0.42}

                =2.38

\to D(a.v) = \frac{1}{0.42} \times 206265\\

               =490910.7

Calculating point E:

when the value is 0.31

\to D(PC)=\frac{1}{0.31}

                =3.226

\to D(a.v) = \frac{1}{0.31} \times 206265\\

               =665370.97

6 0
3 years ago
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