One example of resistive force is the drag force acting on the object falling vertically downward.
The drag force acts in the opposite direction of motion of the object, thus, in the case of a vertically falling object, the drag force acts in the vertically upward direction.
It is known as resistant because it opposes the motion of the object.
Riding a bicycle is an example of resistant force.
Riding a bicycle is an example of resistant force
It is called resistant force because when the bicycle moves forward a force by air occurs in the opposite direction, ultimately slowing down the rider's speed.
Resistant force refers to the force that is felt opposite to the body's movement, affecting the body's speed.
During an equilibrium, there is a balance between both forces which cancels each other out resulting in no motion of an object.
The dishwashing liquid helps in reducing the point of contact between the two blocks and acts as a barrier between the blocks thus the force of friction is reduced.
The Law of Inertia is Newton's 1st law which states that an object that is in motion will tend to remain in motion and an object at rest will tend to remain at rest until no external force is applied. This is why your upper body wants to keep moving even when you have stopped the movement.
For more about Force and Friction refer to the link:https://brainly.com/question/15122221
An object moving with a speed of 8 m/s has a momentum of 56 kg m/s. What is the object's mass?
Given,
The speed of the object v=8 m/s
The momentum of the object, p=56 kg·m/s
The momentum of an object is calculated as the product of the velocity of the object and its mass.
Thus the momentum of the object is,
[tex]p=mv[/tex]Where m is the mass of the object.
Thus,
[tex]m=\frac{p}{v}[/tex]On substituting the known values,
[tex]\begin{gathered} m=\frac{56}{8} \\ =7\text{ kg} \end{gathered}[/tex]Thus the mass of the object is 7 kg.
How much energy is required to changea 48 g ice cube from ice at -14 °C tosteam at 118 °C? The specific heat of iceis 2090 J/kg . °C, the specific heat of wa-ter is 4186 J/kg . ° C, the specific heat ofstream is 2010 J/kg . ° C, the heat of fusionis 3.33 x 10^5 J/kg, and the heat of vaporiza-tion is 2.26 x 10^6 J/kg.Answer in units of J.
Given data
*The given mass is m = 48 g = 48 × 10^-3 kg
*The given initial temperature of the ice is T = -14 °C
*The given temperature of the steam is t = 118 °C
*The specific heat of ice is c = 2090 J/kg °C
*The specific heat of water is s = 4186 J/kg °C
*The specific heat of the stream is 2010 J/kg ° C
*The heat of vaporization is 2.26 x 10^6 J/kg
*The heat of fusion is 3.33 x 10^5 J/kg
The formula for the energy is required to change a 48 g ice cube from ice at -14 °C to
steam at 118 °C is given as
[tex]E=mc\Delta T+mL_f+ms\Delta t_1+mL_v+ms_s\Delta t_2[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} E=(48\times10^{-3}\text{)}\lbrack(2090\times14)+(3.33\times10^5)+(4186\times104)+(2.26\times10^6)+(2010\times14) \\ =148115.7\text{ J} \end{gathered}[/tex]Please help! A jogger runs at a speed of 4 m/s for 600s, slows to 3 m/s for the next 400s and then travels the final 800s at a speed of 2.5 m/s. What is the average speed of the jogger?
Given data:
A jogger runs at a speed of v_1=4 m/s for t_1=600 s.
Again he runs at speed of v_2=3 m/s for t_2=400 s.
Again he runs at speed of v_3=2.5 m/s for t_3=800 s.
The distance is given as,
[tex]s=v\times t[/tex]The distance traveled by the jogger is 600 s is given as,
[tex]\begin{gathered} s_1=v_1\times t_1 \\ =(4\text{ m/s})\times(600\text{ s}) \\ =2400\text{ m} \end{gathered}[/tex]The distance traveled by the jogger in the next 400 s is given as,
[tex]\begin{gathered} s_2=v_2\times t \\ =(3\text{ m/s})\times(400\text{ s}) \\ =1200\text{ m} \end{gathered}[/tex]The distance traveled by the jogger in the last 800 s is given as,
[tex]\begin{gathered} s_3=v_3\times t \\ =(2.5\text{ m/s})\times(800\text{ s}) \\ =2000\text{ m} \end{gathered}[/tex]The average speed is given as,
[tex]\begin{gathered} v_{avg}=\frac{\text{ total distance traveled}}{\text{ total time taken}} \\ =\frac{s_1+s_2+s_3}{t_1+t_2+t_3} \\ =\frac{(2400\text{ m})+(1200\text{ m})+(2000\text{ m})}{(600\text{ s})+(400\text{ s})+(800\text{ s})} \\ \approx3.11\text{ m/s} \end{gathered}[/tex]Therefore, the average speed of the jogger is 3.11 m/s.
(a) What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of 9.57 m/s when going down a slope for 1.97 s? (b) How far does the skier travel in this time?(a)Number_______ Units _________(b)Number_______ Units _________
Given,
The initial velocity of the skier, u=0 m/s
The final velocity of the skier, v=9.57 m/s
The time duration, t=.97 s
(a)
From the equation of motion,
[tex]v=u+at[/tex]Where a is the acceleration of the skier.
On substituting the known values,
[tex]\begin{gathered} 9.57=0+a\times1.97 \\ \Rightarrow a=\frac{9.57}{1.97} \\ =4.86\text{ m/s}^2 \end{gathered}[/tex]Thus the magnitude of average acceleration of the skier is 4.86 m/s²
(b)
The distance traveled by the skier is given by one of the equations of motion as,
[tex]s=ut+\frac{1}{2}at^2[/tex]On substituting the known values,
[tex]\begin{gathered} s=0+\frac{1}{2}\times4.86\times1.97^2 \\ =9.43\text{ m} \end{gathered}[/tex]The skier travels for 9.43 m in the given time.
150 km North, 50 km North, 250 km south,
Answer:
It’s important to be able to visualise these motions.
I can use a scale drawing or just take my finger and trace the motion.
(If I don’t want to go the full 150km, I can go 150mm instead, or trace it on a map. That’s essentially what a map is, a scale drawing.)
The actual distance from where I started to where I end up is the answer.
Two stars have the same luminosity. The first star is one light year away from your telescope. The second star is two light years away from your telescope. The first star’s brightness is:A. two times as great as the second star’sB. four times as great as the second star’s.C. sixteen times as great as the second star’s.D. equal to the second star’s.
f = L / (4pi'd2), where:
f: apparent brightness
L: luminosity
d: distance of star
(p and i' stay constant between the two stars, so let's make them both equal to 1)
Let's say the closer star has a distance of 1 and the further has a distance of 2, and they both have a luminosity of L.
Apparent brightness of further star:
f = L/(4*2^2) = L/16
Apparent brightness of closer star:
f = L/(4*1^2) = L/4
As you can see here, the apparent brightnesses of the two stars differ by a factor of 4.
Answer: B. four times as great as the second star’s.
what is the area of a trapezoid with b1=5=b2=7 h=4
The area of a trapezoid is given by:
[tex]A=\frac{b_1+b_2}{2}h[/tex]Plugging the values given we have:
[tex]\begin{gathered} A=\frac{5+7}{2}\cdot4 \\ A=\frac{12}{2}\cdot4 \\ A=6\cdot4 \\ A=24 \end{gathered}[/tex]Therefore the area is 24 square units.
A calorimeter is used to measure the specific heat of an object. The water bath has an initial temperature of 23.2 ° C. The object with a temperature of 67.8 ° C is placed in the Beaker.After thermal equilibrium is established, the temperature of the water bath is 25.6 ° CWhat is the specific heat of the object (Lead)? We give ; Specific heat of water ce = 4185J.Kg and Density = 1000kg / m
Consider that the mass of water bath is 0.700 kg and the mass of object is 0.200 kg.
The heat required to change the temperature of water is,
[tex]Q_w=m_wC_w(T-T_w)[/tex]The heat required to change the temprature of object is,
[tex]Q_o=m_oC_o(T_o-T)[/tex]Assume that no heat is lost. Therefore,
[tex]Q_w=Q_o[/tex][tex]m_wC_w(T-T_w)=m_oC_o(T_o-T)[/tex][tex]C_o=\frac{m_{w_{}}C_w(T-T_w)}{m_o(T_o-T)}[/tex]Substitute the known values,
[tex]\begin{gathered} C_o=\frac{(0.700\text{ kg)(4185 J/K}.\text{g)(25.6}^{\circ}\text{C-23.2}^{\circ}\text{C)}}{(0.200\text{ kg)(67.8}^{\circ}\text{C-}25.6^{\circ}C)} \\ =\frac{(0.700\text{ kg)(4185 J/K.g)(2.4}^{\circ}\text{C)}}{(0.200\text{ kg)(42.2}^{\circ}\text{C})}(\frac{1\text{ kg}}{1000\text{ g}}) \\ \approx0.833\text{ J/kg.K} \end{gathered}[/tex]Therefore, the specific heat of lead is 0.833 J/kg.K
An airplane starts at rest and accelerates at 5.7 m/s^2 at an angle of 31 south of west after 9 s, how far in the westerly direction has the plane traveled?
Hello! I am currently looking better to understand energy applications from energy and Its Applications. I have a screenshot of the work I am currently doing right now. I know it's pretty straightforward; however, I wanted to understand better since I now don't know how to write the explanation.
Q2
The different types of energy that this system has are mechanical energies. These wans describes the movements and position of objects.
- Kinetic energy, due to speed
- Potential energy, due to high and gravity
- Losses like friction, due to contact between cart and structure. This refearse to the contact between two surface that reduce the speed and generates losses at any system.
Q3
The mechanical energy try to be constant so there is a traid off between kinetic energy and pontential energy. The kinetic E is higher when the potential E is lower, it means that the speed is higher when the high is lower.
Q1
The energy is potential because the high is the maximum possible. On the other hand the speed is 0, is not moving so the kinetic energy is 0 too.
1. Under an apple tree Newton discovered that the reason the apples fall from the tree is the same reason the Moon orbits the Earth. Gravity! (a) Use Newton's law of gravitation to calculate the gravitational attraction betweenthe Earth and the Moon. Explain what measurement you will use. (b) How does the force of gravity change if the Moon got knocked 3 times further away? Would it increase, decrease or remain the same.
To find
(a) Newton's law of gravitation required to calculate the gravitational attraction between the moon and the earth and also the measurements required to calculate.
(b) Change in force of gravity if the moon is knocked 3 times away-(increase/decrease/remain same)
Explanation:
According to Newton's law of gravitation, the gravitational force of attraction between the earth and the moon will be
[tex]F=\frac{GMm}{r^2}[/tex]Here, M is the mass of the earth
m is the mass of the moon
r is the distance between the moon and the earth
G is the universal gravitational constant whose value is
[tex]G=\text{ 6.67}\times10^{-11}\text{ Nm}^2\text{ /kg}^2[/tex]Measurements required to calculate the gravitational force are the mass of the earth, the mass of the moon and the distance between them.
(b)
If the moon is knocked down 3 times further, the distance between the moon and the earth will increase.
As the gravitational force is inversely proportional to the distance between them, so the force will decrease.
How do I find the energy transferred to water ?
We are asked to find the energy transferred to the water for different materials.
Recall that the heat transfer equation is given by
[tex]Q=m\cdot c_p\cdot\Delta T[/tex]Where m is the mass, cp is the specific heat capacity of the material, ΔT is the change in the temperature.
For Iron:
The specific heat capacity of iron is 0.45 J/g°C
Mass of iron = 53.9 g
Change in temperature = 32.3 °C
So, the energy released by the metal (iron) is
[tex]\begin{gathered} Q=m\cdot c_p\cdot\Delta T \\ Q=53.9\cdot0.45\cdot32.3 \\ Q=783.4\; J \end{gathered}[/tex]Therefore, the energy transferred to the water by metal (iron) is 783.4 J
Find the magnitude and direction of an electric field that exerts a 1.38 * 10^-18N northward force on a proton.
Given:
Force = 1.38 x 10⁻¹⁸ N northward.
Let's find the magnitude and direction of the electric field
To find the magnitude, apply the formula:
[tex]E=\frac{F}{q}[/tex]Where:
E is the electric field
F is the force = 1.38 x 10⁻¹⁸ N
q is the charge of proton = -1.60 x 10⁻¹⁹ C
Input the values in the equation and solve for E:
[tex]\begin{gathered} E=\frac{1.38\times10^{-18}}{-1.60\times10^{-19}} \\ \\ E=-8.625\text{ N/C} \end{gathered}[/tex]The magnitude of the electric field is -8.625 N/C.
Since the direction of the force is Northward and the polarity of the charge is negative, the direction of the electric field will be opposite the direction of the force.
Therefore, the direction of the electric field is downward.
ANSWER:
8.625 N/C Southward
Answer:
8.625 N/C northward
Explanation:
The electric force can be related to the electric field and charge with which that field interacts via the equations found in Module 7 Section 1.3 and Chapter 18.4 in College Physics.
https://openstax.org/books/college-physics/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units
(a) What is the resistance (in kΩ) of a 7.50 ✕ 102 Ω, a 2.40 kΩ, and 4.50 kΩ resistor connected in series? kΩ(b) What is the resistance (in Ω) of a 7.50 ✕ 102 Ω, a 2.40 kΩ, and 4.50 kΩ resistor connected in parallel? Ω†
a) The equivalent resistance in series = 7.65 kΩ
b) The equivalent resistance in parallel = 0.51 kΩ
Explanation:The resistances of the given resistors are:
[tex]\begin{gathered} R_1=7.50\times10^2Ω \\ R_1=0.75\times10^3=0.75kΩ \end{gathered}[/tex][tex]\begin{gathered} R_2=2.40kΩ \\ R_3=4.50kΩ \end{gathered}[/tex]The equivalent resistance in series:
[tex]\begin{gathered} R_{eq}=R_1+R_2+R_3 \\ \\ R_{eq}=0.75kΩ+2.40kΩ+4.50kΩ \\ \\ R_{eq}=7.65kΩ \end{gathered}[/tex]b) The equivalent resistance in parallel
[tex]\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} \\ \\ \frac{1}{R_{eq}}=\frac{1}{0.75}+\frac{1}{2.4}+\frac{1}{4.5} \\ \\ \frac{1}{R_{eq}}=1.972 \\ \\ R_{eq}=\frac{1}{1.972} \\ \\ R_{eq}=0.51kΩ \end{gathered}[/tex]Explain Navier-Stokes equations
Navier-Stokes equation are deveped by Claude-Louise Naiver and George Gabriel.
These equatins are partial differential equations They help to describe the motion of viscous fluid substances.
For forces acting on an object, given by A=43.3 N East, B=46.3 N north, C=70.7 N west, D=66.7 N south. B. What is the direction of the force? (enter your answer in degrees counterclockwise from the X axis)
Given:
The forces;
A=43.3 N east
B=46.3 N north
C=70.7 N west
D=66.7 N south
To find:
The direction of the net force.
Explanation:
Let us assume that the east is the positive x-axis and the north is the positive y-axis.
The forces in the vector forms can be written as,
[tex]\begin{gathered} \vec{A}=43.3\hat{i} \\ \vec{B}=46.3\hat{j} \\ \vec{C}=70.7(-\hat{i}) \\ \vec{D}=66.7(-\hat{j}) \end{gathered}[/tex]Where i and j are the unit vectors along the x and y axes respectively.
The vector sum of the forces is,
[tex]\vec{R}=\vec{A}+\vec{B}+\vec{C}+\vec{D}[/tex]On substituting the known values,
[tex]\begin{gathered} \vec{R}=43.3\hat{i}+46.3\hat{j}-70.7\hat{i}-66.7\hat{j} \\ =-27.4\hat{i}-20.4\hat{j} \end{gathered}[/tex]The direction of a vector is,
[tex]\begin{gathered} \theta=\tan^{-1}(\frac{-20.4}{-27.4}) \\ =36.7\degree \end{gathered}[/tex]Final answer:
Thus the direction of the resultant force is 36.7° in counter-clockwise from the x-axis.
Two wooden crates rest on top of one another. The smaller top crate has a mass of M1=22kg and the larger bottom crate has a mass of M2=92kg. There is No friction between the crate and the floor, but the coefficient of static friction between the two crates is Ms=0.87 and the coefficient of kinetic friction between the two crates is MK=0.68. A massless rope is attached to the lower crate to pull it horizontally to the right. (Which should be considered the positive direction for this problem). 1) the rope is pulled with a tension of T = 500N. What is the acceleration of the smaller crate?.2) In the previous situation, what is the frictional force the lower crate exerts on the upper crate?3) What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide?4) The tension is increased in the rope to 1440N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?5) As the upper crate slides, what is the acceleration of the lower crate?
ANSWER
[tex]\begin{gathered} 1)4.39m/s^2 \\ 2)96.58N \\ 3)971.96N \\ 4)6.66m/s^2 \\ 5)14.06m/s^2 \end{gathered}[/tex]EXPLANATION
Parameters given:
Mass of smaller crate, m1 = 22 kg
Mass of larger crate, m2 = 92 kg
Coefficient of static friction between two crates, μs = 0.87
Coefficient of kinetic friction between two crates, μk = 0.68
1) The sum of forces acting in the horizontal direction on the smaller crate is:
[tex]F=m_1a_1[/tex]The sum of forces acting in the horizontal direction on the larger crate is:
[tex]T-F=m_2a_2[/tex]where F = frictional force, T = tension
To find the acceleration of the smaller crate, we have to substitute F from the first equation into the second equation:
[tex]T-m_1a_1=m_2a_2[/tex]The acceleration is equal for both crates since the upper crate is static during the motion, which implies that:
[tex]\begin{gathered} T=m_1a+m_2a \\ T=a(m_1+m_2) \\ \Rightarrow a=\frac{T}{m_1+m_2} \end{gathered}[/tex]Substitute the given values:
[tex]\begin{gathered} a=\frac{500}{22+92} \\ a=4.39m/s^2 \end{gathered}[/tex]That is the acceleration of the smaller crate.
2) To find the frictional force on the smaller crate, substitute the value of acceleration, a, into the first equation:
[tex]\begin{gathered} \Rightarrow F=22\cdot4.39 \\ F=96.58N \end{gathered}[/tex]3) The maximum tension that the rope can be pulled at will occur when there is maximum friction.
That is:
[tex]F_{s(\text{max)}}=\mu_s\cdot m_2\cdot g[/tex]Applying the same principle from the first equation:
[tex]\begin{gathered} F=ma \\ \Rightarrow F_2=m_2a \end{gathered}[/tex]This implies that:
[tex]\begin{gathered} m_2a=\mu_s\cdot m_2\cdot g \\ \Rightarrow a=\mu_s\cdot g \end{gathered}[/tex]To find the tension, apply the already given formula:
[tex]\begin{gathered} T=m_1a+m_2a=(m_1+m_2)a \\ T=(m_1+m_2)\cdot\mu_s\cdot g \end{gathered}[/tex]Therefore, the max tension is:
[tex]\begin{gathered} T=(22+92)\cdot0.87\cdot9.8 \\ T=114\cdot0.87\cdot9.8 \\ T=971.96N \end{gathered}[/tex]4) To find the acceleration of the smaller crate, we apply the formula for kinetic friction:
[tex]F_k=\mu_k\cdot m_1\cdot g[/tex]Recall that:
[tex]F=m_1\cdot a_1[/tex]Therefore:
[tex]\begin{gathered} m_1\cdot a_1=\mu_k\cdot m_1\cdot g \\ \Rightarrow a_1=\mu_k\cdot g \end{gathered}[/tex]Substitute the given values:
[tex]\begin{gathered} a_1=0.68\cdot9.8 \\ a_1=6.66m/s^2 \end{gathered}[/tex]That is the acceleration of the smaller/upper crate.
5) Recall from the second equation that:
[tex]T-F=m_2a_2[/tex]Substitute the formula for kinetic friction:
[tex]\begin{gathered} T-\mu_k\cdot m_1\cdot g=m_2\cdot a_2 \\ \Rightarrow a_2=\frac{T-\mu_k\cdot m_1\cdot g}{m_2} \end{gathered}[/tex]Substitute the given values to solve for a2:
[tex]\begin{gathered} a_2=\frac{1440-(0.68\cdot22\cdot9.8)}{92} \\ a_2=\frac{1440-146.608}{92} \\ a_2=14.06m/s^2 \end{gathered}[/tex]That is the acceleration of the larger/lower crate.
What is the fnet of the 1st fbd on the left
400
Explanation
Step 1
to know the net force, we need to add the forces ( vectors),so vertically we have
positive =1200 N
Negative = 800 N
so, the sum of vertical forces is
[tex]\begin{gathered} \sum ^{\square}_{y-\text{foces}}=1200N-800N \\ \sum ^{\square}_{y-\text{foces}}=400\text{ N} \end{gathered}[/tex]hence, the answer is 400 N
I hope this helps you
x
When will a generator produce a current?A. When the wire inside the generator is coiledB. When a magnet is placed near the generatorC. When it is connected to a batteryD. When a magnet is moving through a coil of wire
A generator uses electromagnetic induction to create current, this means that we need to move the magnet near a conductor to create the current or to move the conductor near the magnet. For a generator we usually move the magnet near a coil; therefore, the answer is D.
The bending ability of a substance is recorded on the index of retraction
To find:
Is the given statement true or false?
Explanation:
The amount of the deviation or the bending of a light when it passes from one object to another depends on the index of refraction of the object. The waves that enter a medium with a lower refractive index will bend towards the normal.
But the index of refraction is not the measure of the ability of the substance to bend the light. It is the ratio of the speed of light in the vacuum to that in the medium.
Final answer:
Thus the given statement is false.
A string length is 8.2 meters and is vibrating as the fifth harmonic. The string vibrates up and down with 12 cycles in 4 seconds. Determine the wavelength for this wave.
ANSWER
3.28 m
EXPLANATION
The length of the string and the wavelength when the string is vibrating as its nth harmonic is related by,
[tex]L=\frac{n}{2}\cdot\lambda[/tex]In this problem, we are given the length of the string L = 8.2 m and we have to find the wavelength for this wave when the string is vibrating as the 5th harmonic,
[tex]8.2m=\frac{5}{2}\cdot\lambda[/tex]Solving for the wavelength λ,
[tex]\lambda=\frac{2}{5}\cdot8.2m=3.28m[/tex]The wavelength of this wave is 3.28 m.
A ball is launched upward with an initial velocity of 20 m/s. What is the speed and direction (hence velocity) ofthe ball 2 seconds into its flight? For this problem, round g=-9.8 m/s2 to g=-10 m/s2 for simplicity, and assumeno air resistance of any kind is present.
Answer:
velocity = 0 m/s
Explanation:
The velocity after t seconds can be calculated as:
[tex]v=v_0+at[/tex]Where v0 is the initial velocity and a is the acceleration. In this case, the acceleration is gravity, so a = g = -10m/s²
Then, replacing v0 by 20m/s and a by -10m/s², we get:
[tex]v=20-10t[/tex]Therefore, the velocity after 2 seconds will be equal to:
[tex]\begin{gathered} v=20-10(2) \\ v=20-20 \\ v=0\text{ m/s} \end{gathered}[/tex]Since the velocity is 0 m/s, there isn't a direction.
So, the answer is:
velocity = 0 m/s
A runaway piano starts from rest and slides down a 22.0° frictionless incline 2.00 m in length.
a) What is the acceleration of the piano (m/s^2)
b) What is the velocity of the piano at the end of the slope (m/s)
Please explain. I cannot figure out where to start with this problem.
Thank you
Answer:
F = M a
M a = M g sin 22
a = g sin 22 = .375 g = .375 * 9.80 m/s^2 = 3.67 m/s^2
(Note M g sin θ is the gravitational force accelerating the piano)
v = (s2 - s1) / t (given S2 - S1 = 2 m)
S = V0 t + 1/2 a t^2 distance traveled in time t
S = 1/2 a t^2 since the piano starts from rest
t = (2 S / a)^1/2
t = (2 * 2 / 3.67)^1/2 = 1.04 sec
v = a * t = 3.67 m/s^2 * 1.04 sec = 3.82 m/s
(Note avg speed = 1.91 m/s * 1.04 sec = 1.99 m)
SUVAT equations are important:
Variable Description SI unit
S displacement m
U initial velocity m/s
V final velocity m/s
A acceleration m/s^2
T total time s
Critique the following statement as True or False, based on Newton's third Law (AKA Action-Reaction):Two students are on ice skates, throwing a ball toward one another. One student, Jessica, is on the left side, and the other student, Jerry, is on the right side. They are separated by 1 meter of distance. After Jessica throws the ball forward (LaTeX: \longrightarrow) towards Jerry, she slides backward (LaTeX: \longleftarrow) , and when Jerry catches the ball (the ball moving toward him), he slides in the direction of the ball hits him (LaTeX: \longrightarrow), so their distance of separation gets larger.Group of answer choicesTRUEFALSE
The given statement is true.
Newton's third law states that for every action there is an equal and opposite reaction. That is for every force there is a reaction force that has the same magnitude but will be acting in the opposite direction.
Thus, when the student throws the ball by applying a certain force, the ball, in turn, applies a force on the student in the opposite direction, which causes the student to slide backwards. And when the other student applies a force to stop the ball, the ball applies a force in the opposite direction. This force causes Jerry to slide in the direction of the ball.
Thus the given statement is true.
Checking my work, I need the acceleration of free fall
After 0.3 seconds,
Horizontal distance, dx = 10
Vertical distance, dy = 8
The horizontal velocity, vx would always be constant
The vertical velocity, vy would change as the object moves upwards and comes own
horizontal velocity, vx = displacement/time = 10/0.3 = 33.33 m/s
vertical velocity, vy = 8/0.3 = 26.67 m/s
Recall the formula for motion,
v = u - at
a = acceleration
t = time = 0.3
v = final velocity = 0
u = initial velocity = 26.67
By substituting these values, we have
0 = 26.67 - a x 0.3
0.3a = 26.67
a = 26.67/0.3
a = 88.9 m/s^2
The acceleration of freefall is 88.9 m/s^2
A body A at 90° and another b at 40°C,if they join together and achieve a temperature of 50°C,this temperature is called :A:melting temperatureB:evaporation temperatureC:equilibrium temperatureD:none of the above
When an object with higher temperature is in contact with another object with lower temperature, the object with higher temperature will release some energy towards the object with smaller temperature, which will receive this energy.
This way, the object with higher temperature will decrease its temperature, while the object with smaller temperature will increase its temperature.
This process continues until both objects have the same temperature.
This final temperature is called equilibrium temperature.
Therefore the correct option is C.
Question 1: Radiation emitted from human skin reaches its peak at A = 940 uma. What is the frequency of this radiation?b. What type of electromagnetic waves are these?C. How much energy (in electron volts) is carried by one quantum ofthis radiation?
Given:
λ = 940 μm
Let's solve for the following:
• (a). What is the frequency of this radiation?
To find the frequency, apply the formula:
[tex]f=\frac{c}{\lambda}[/tex]Where:
f is the frequency
c is the speed of light = 3 x 10⁸ m/s
λ = 940 μm
Thus, we have:
[tex]\begin{gathered} f=\frac{3\times10^8}{940\times10^{-6}} \\ \\ f=3.19\times10^{11\text{ }}Hz. \end{gathered}[/tex]Therefore, the frequency of this radiation is 3.19 x 10¹¹ Hz.
• (b). What type of electromagnetic waves are these?
Since the wave has a frequency of 3.19 x 10¹¹ Hz, the wave will be said to be an infrared.
• (c). How much energy (in electron volts) is carried by one quantum of this radiation?
Apply the formula:
[tex]E=h*f[/tex]Where:
E is the energy in Joules
h is Planck's constant = 6.626 x 10⁻³⁴
f is the frequency gotten in part A.
We have:
[tex]\begin{gathered} E=6.626\times10^{-34}*3.19\times10^{11} \\ \\ E=2.11\times10^{-22}\text{ J} \end{gathered}[/tex]Now, to convert from Joules (J) to electron volts (eV), we have:
[tex]\begin{gathered} 2.11\times10^{-22}*(\frac{1\text{ eV}}{1.602\times10^{-19}}) \\ \\ =1.32\times10^{-3}\text{ eV} \end{gathered}[/tex]Therefore, the energy in electron volts is 1.32 x 10⁻³ eV.
ANSWER:
• (a). 3.19 x 10¹¹ Hz.
• (b). Infrared
• (c). 1.32 x 10⁻³ eV.
There is a current of 0.73 Amps through a light bulb in a 120 Volt circuit. What is the resistance of this light bulb?
ANSWER
[tex]164.38\text{ }\Omega[/tex]EXPLANATION
To find the resistance of the light bulb, we have to apply the formula derived from Ohm's law:
[tex]V=IR[/tex]where V = voltage
I = current
R = resistance
Therefore, solving for R in the formula above, we have that:
[tex]\begin{gathered} R=\frac{V}{I} \\ \\ R=\frac{120}{0.73} \\ \\ R=164.38\text{ }\Omega \end{gathered}[/tex]That is the answer.
The resistance of the light bulb is 164.38 Ω
To find the resistance, We will use the formula:
V = I /R
where,
I = current
V = voltage
R = resistance
Given in the question,
I = 0.73 A
V = 120 v
Now, we will put the values in the formula,
R = V/I
R = 120 / 0.73
R = 164.38 Ω
Therefore, the resistance of the light bulb is 164.38 Ω
To learn more about Resistance:
https://brainly.com/question/4289257
If the rest of the Solar System were todisappear, the Moon wouldA. continue moving in a circle.B. fly off in a straight line.C. start to slow down.D. stop moving.
Given:
The rest of the solar system would disappear
To find:
What happens to the moon
Explanation:
The moon rotates around the Earth in orbit, which means the moon's linear velocity is along the orbit's tangent.
When Earth disappears, the Moon continues its motion due to the inertia and flies off in a straight line.
Hence, when the rest of the solar system disappears, the Moon would fly off in a straight line.
Suppose an astronaut has landed on Mars. Fully equipped, the astronaut has a mass of 135 kg, and when the astronaut gets on ascale, the reading is 500 N. What is the acceleration (in m/s2) due to gravity on Mars?m/s2
ANSWER:
3.7 m/s²
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 135 kg
Force (F) = 500 N
We can determine the value of acceleration as follows:
[tex]\begin{gathered} F=m\cdot a \\ \\ \text{ We solve for a:} \\ \\ a=\frac{F}{m} \\ \\ \text{ We replacing:} \\ \\ a=\frac{500}{135} \\ \\ a=3.7\text{ m/s}^2 \end{gathered}[/tex]The value of the acceleration due to gravity on Mars is 3.7 m/s²