A graph of velocity vs time displays the change in an object's velocity in a straight line with regard to time.
What is meant by Velocity-time graph?A plot between speed and time is called a velocity-time graph. It depicts the motion of an object moving straight ahead. At any given time, the magnitude of velocity equals its instantaneous speed.
A graph of velocity vs time displays the change in an object's velocity in a straight line with regard to time. The x-axis is used to measure time, and the y-axis is used to measure velocity. For an item moving at a constant velocity, the height of the velocity-time graph does not change over time.
We can use a velocity vs. time graph to determine position and a time graph to determine velocity. We are aware that v = d/t. Rearranging the equation with a little mathematics reveals that d = v × t.
The complete question is:
Jack throws a rock downward from his deck at time t = 0 and it lands in his yard below. Make velocity-time graph of the rock until it hits the ground? Assume upward is the positive direction.
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A flat coil of wire has an area of 0.020 m2 and contains 50 turns. Initially the coil is oriented so that the normal to its surface is parallel to and in the same direction as a constant magnetic field of 0.18 T. The coil is then rotated through an angle of 30o in a time of 0.10 s. What is the average induced emf? -0.44 V +0.44 V +0.24 V
ANSWER:
3rd option: +0.24 V
STEP-BY-STEP EXPLANATION:
Given:
N = 50
Area = 0.020 m^2
B = 0.18 T
θf = 30°
time = 0.10 s
We can calculate the average induced emf by the following formula
[tex]\epsilon=N\cdot B\cdot A\cdot\left(\frac{\cos\theta_i-\cos\theta_f}{t}\right)[/tex]We replacing:
[tex]\begin{gathered} \epsilon=\left(50\right)\left(0.18\right)\left(0.02\right)\left(\frac{\cos\:0\degree\:-\cos\:30\degree}{\:0.1}\right) \\ \epsilon=0.241\cong0.24\text{ V} \end{gathered}[/tex]The correct answer is 0.24V
Which of the following is not a unit of a force?
Answer: B
Explanation:
One of the common units of force is Newtons, N
Recall,
Force = ma
where
m is the mass of the body
a is the acceleration of the body
If acceleration, a = ms^-2 and mass = kg, then
Force = kgms^-2
Also, joule is the unit of work
recall,
work = force x distance = J
If we divide work by distance(meters), it becomes joule/meter
Thus,
jm^-1 is a unit of force
Thus, the option that is not a unit of force is
B. Js^-1
Emily and Gemma did a Reaction time lab. Emily dropped the ruler while Gemma tried to catch it. She caught the ruler 5 times and her average catching distance is 0.12 m. What is Gemma's reaction time?
The average reaction time of Gemma is 0.1564 seconds.
As we know, Gemma is catching the scale and Emily is dropping the scale.
The whole experiment is taking place under gravity, so the acceleration is constant.
As we know, the scale is dropped, it means that the initial velocity of the scale is zero.
We can use the equation of motion,
The equation is,
S = Ut + 1/2at²
Where,
S is the displacement, which is 0.12 m in our case,
U is initial velocity which is 0m/s because the stone is dropped,
t is the time taken, this is equal tot he reaction time here,
a is the acceleration due to gravity whose value is 9.8m/s.
Now, putting all the values,
0.12 = 1/2(9.8)(t)²
t² = 0.24/9.8
t = 0.1564 seconds.
Gemma reacts in 0.1564 seconds to catch the scale.
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1) Prove that for the adiabatic process of gas pVr =const ; r=Cp/Cv *
2) Write down the expression for the number of molecules of N2 gas having speeds in the interval (100m/s ; 102m/s) for one mole of gas at T=300K
3) Demonstrate how to obtain the Boltzmann Distribution of Energies from the Maxwell-Boltzmann distribution of speeds.*
For an adiabatic process of gas PV(^r)= Constant, where r = Cp/Cv , Cp, and Cv are constants for a particular gas.
What is an ideal gas?It is an imaginary gas for which the volume occupies by it is negligible, this gas does not exist in a practical situation and the concept of an ideal gas is only the theoretical one.
Universal Gas Equation for the ideal gas,
PV=nRT
By differentiating on both sides: P.dV + V.dP = n.R.dT…. (1)
For the adiabatic process, the expansion or compression of gas occurs at a very fast rate which makes no heat transfer. Therefore, dQ= 0
A/c to First Law of TD: dQ=dU+ dW, implies dU= -dW=-P.dV ….. (2)
But dU= n.Cv. dT …. (3) Combining 2 and 3, we get: n.dT= -P.dV/Cv …..(3)
Using R= Cp-Cv, the gas law gives: from 1 and 3.
n.dT= (P.dV + V.dP)/(Cp-Cv) = -PdV/Cv
On rearranging terms we get, (dP/P)+((Cp/Cv)*(dV/V))=0
Cp and Cv are constants for a particular gas, hence integrating,
Ln(P)+ r.Ln(V)=constant (replacing Cp/Cv = r or Gamma)
Hence, P.V(^r)= constant is termed an adiabatic equation.
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In a parallel circuit with a 12 V battery and three 6 Ohm resistors, what is the total current in the entire circuit? Select one:a.36 Ampsb.6 Ampsc.4 Ampsd.2 Amps
First, let's find the equivalent resistance. Since they are in parallel, we can find it as follows:
[tex]\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3} \\ so: \\ \frac{1}{R_{eq}}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2} \\ Hence: \\ R_{eq}=2\Omega \end{gathered}[/tex]Now, we can use ohm's law to calculate the current:
[tex]\begin{gathered} V=IR \\ so: \\ I=\frac{V}{R}=\frac{12}{2}=6 \end{gathered}[/tex]Answer:
b.
6 Amps
What is the electric field strength at a distance of 0.9 m from a charge of 5.71 x 10^-6 C?
Given:
[tex]\begin{gathered} Q=5.71\times10^{-6}\text{ C} \\ r=0.9\text{ m} \end{gathered}[/tex]The electric field strength is given as,
[tex]E=\frac{KQ}{r}[/tex]Here, K is the electrostatic constant.
Putting the values,
[tex]\begin{gathered} E=\frac{9\times10^9\times5.71\times10^{-6}}{(0.9)^2} \\ =63444.44\text{ N/C} \end{gathered}[/tex]Therefore, the electric field strength is 63444.44 N/C.
What is the specific heat c of a 0.500 kg metal sample that rises 5.40 C when 305J of heat is added to it?
ANSWER
[tex]\begin{equation*} 112.96\text{ }J\/kgK \end{equation*}[/tex]EXPLANATION
Parameters given:
Mass of sample, m = 0.5 kg
Temperature change, ΔT = 5.40 °C = 5.40 K
Heat energy, H = 305 J
To find the specific heat capacity of the sample, we have to apply the formula for heat energy:
[tex]H=mc\Delta T[/tex]Where c = specific heat capacity
Therefore, solving for c, the specific heat capacity of the metal sample is:
[tex]\begin{gathered} 305=0.5*c*5.4 \\ c=\frac{305}{0.5*5.4} \\ c=112.96\text{ }J\/kgK \end{gathered}[/tex]That is the answer.
QUESTION 2 (NOVEMBER 2014) Two blocks of masses 20 kg and 5 kg respectively are connected by a light inextensible string, P. A second light inextensible string, Q, attached to the 5 kg block, runs over a light frictionless pulley. A constant horizontal force of 250 N pulls the second string as shown in the diagram below. The magnitudes of the tensions in P and Q are T, and T, respectively. Ignore the effects of air friction.
2.3 Calculate the magnitude of the tension T, in string P. (6)
The magnitude of tension T in string P is 250 N when a constant horizontal force of 250 N pulls the second string.
What is tension and how the tension is calculated out to be 250 N ?Tension is equivalent to pull force , used in most of kinematic questions.Tension can be best explained when you pull a rope or you pull an object during the time period under consideration.Here in this question given, mass of first object is 20 kg , and mass of the second object is 5 kg .Using the equation m1a = T2 - m2a , (m1+m2)a = T2 , (20 +5)250/25 = T2.From this comes the second tension on the string P is 250 N .To know more about tension visit:
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When light goes from a slower medium to a faster medium; what way does the light bend relative to the normal?
Answer: A.
It moves away from the normal.
Explanation: ed mentum or plato
A ball rolls off a table and falls 0.75m to the floor, landing with a speed of 4 m/s. (A) What is the acceleration of the ball just before it strikes the ground? (B) What was the initial speed of the ball? (C) What initial speed must the ball have if it is to land with a speed of 5 m/s?
Answer:
We had this question yesterday, let me check my book quickly
(A) The acceleration of the ball is 9.81 m/s² just before it strikes the ground.
(B) The initial speed of the ball is equal to 1.14 m/s.
(C) The initial speed must be 3.21 m/s if it is to land with a speed of 5 m/s.
What are the equations of motion?The equation of motion is the way to represent the relation between the time, acceleration, initial and final velocity, and distance covered by a moving object.
The three equations of motion are:
[tex]v= u+ at\\v^2 = u^2 +2aS\\S = ut +(1/2) at^2[/tex]
The acceleration of the ball just before it strikes the ground is equal to gravitational acceleration, g = 9.81 m/s².
Given, the final velocity of the ball, v = 4m/s
The height of the table, h = 0.75 m
The initial velocity of the ball, [tex]u = \sqrt{v^2-2gh}[/tex]
[tex]u = \sqrt{(4)^2-2\times 9.8\times 0.75}[/tex]
u = 1.14 m/s
When the final velocity of the ball, v = 5m/s
The initial velocity will be :[tex]u = \sqrt{(5)^2-2\times 9.8\times 0.75}[/tex]
u = 3.21 m/s.
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Which of the following best explains why Venus has a higher temperature than Mercury? And Why?
A. The atmosphere on Venus is thicker than the atmosphere on Mercury.
B. The atmosphere on Venus is thinner than the atmosphere on Mercury.
C. The length of time it takes Venus to revolve around the Sun is shorter than Mercury.
D. The length of time it takes Venus to revolve around the Sun is longer than that of Mercury.
Answer: The reason why Venus is hotter than Mercury is because it has a thick atmosphere primarily made up of carbon dioxide, which is a greenhouse gas that helps retain the heat from the Sun. In comparison to this, Mercury has almost no atmosphere, so any heat that beats down on the planet isn’t retained.
Explanation: Searched
what is velocity ratio
Answer:
Definition of velocity ratio
: the ratio of a distance through which any part of a machine moves to that which the driving part moves during the same time.
pls brainlies me me new pls
Explanation:
Compare(how they are the same) and contrast(how they are different) psychoanalysis and behaviorism as two of the early schools of psychology. Offer the major names associated with each and explain how each explained behavior.
Abstract of American Intercontinental University
This paper will compare and contrast three of the 10 main early psychology views. This assignment's three approaches are behavioral, humanistic, and cognitive. Three Early Psychology Perspectives are compared and contrasted. Psychology, like anything else, offers a plethora of theories and methods. One idea may be beneficial to one patient while being ineffective to another. The trick is to discover the optimal method for each patient. The concept that behaviors arise as a result of conditioning is known as behaviorism. This theory does not acknowledge the presence of interior mental factors such as thoughts, feelings, and moods, nor does it take free will into account.
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draw each of the following vectors, label an angle that specifies the vectors direction, then find its magnitude and direction.a. B= -4.0 I+ 4.0jb. r= (-2.0i-1.0j) cmc. v= (-10-100j) m/s d. a= (20i+10j) m/s^2the I's and j's have the hat and its the vector simble on the letters
A ball rolls off a table and falls 0.75m to the floor, landing with a speed of 4 m/s. (A) What is the acceleration of the ball just before it strikes the ground? (B) What was the initial speed of the ball? (C) What initial speed must the ball have if it is to land with a speed of 5 m/s?
The acceleration of the ball is 21.33m/s²,
We are given that,
The ball fall's from height = d = 0.75m
The final speed of the ball = Vf = 4m/s
So that to know the acceleration of the ball we can calculate by the equation of motion in term of velocity and acceleration i.e. given as,
V = u + at
Where, V is the final velocity , u is the initial velocity , t is the time and a is the acceleration of the object.
t = d/v
t = (0.75m)/(4m/s)
t = 0.1875s
Thus, the value of t, V, and initial velocity is zero putting in equation of motion to get acceleration,
a = (4m/s)/(0.1875s)
a = 21.33m/s²
The acceleration of the ball would be 21.33m/s²
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if an astronaut weighs 981 N on Earth and only 160 N on the moon, then what is the mass on the moon
If an astronaut weighs 981 N on Earth and only 160 N on the Moon, then his mass on the Moon will be 98.1 kg.
Let's calculate the mass of Earth as per the Earth's acceleration due to gravity.
Now, considering the acceleration due to gravity as 10m/s².
Mass = Weight/Acceleration due to gravity
Mass = 981/10
Mass = 98.1 kg
Now, as per the established fact, the mass is independent of acceleration due to the gravity of the planet i.e. The mass of the person on earth and on the moon is same.
Hence, his mass on the Moon will be 98.1 kg.
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Gravitation always does negative work. Is this true or false?
The given statement 'Gravitation always does negative work' is false.
The work done by the gravitation depends upon the reference level of the system. Therefore, according to reference level it is taken as positive or negative.
Two forces F1 = -6.00i + 7.90j and F2 = 6.80i + 5.30j are acting on an object with a mass of m = 4.10 kg. The forces are measured in newtons, i and j are the unit vectors. What is the magnitude of the object's acceleration?
The magnitude of object's acceleration is 3.26m/s².
The mass of the body is 4.10 lg.
The two forces that are acting on the object are F₁ = -6i + 7.9j newton and F₂ = 6.8i + 5.3j Newton.
We know that the force acting on an object is,
F = Ma
Where,
F is the force acting,
M is the mass of the object and,
a is the acceleration of the object.
As we can see, two forces are acting on the body,
We can simplify the forces in x direction and y direction,
The forces are F₁ = -6i + 7.9j N and F₂ = 6.8i + 5.3j N.
So, the total force in x-direction,
Fₓ = (-6+6.8)i
Fₓ = 0.8i
Fᵧ = (7.9+5.3)j
Fᵧ = 13.2j
So, the net force Fₙ on the object is Fₙ = (0.8i + 13.2j) N
Now, putting value of force and mass in the formula,
F = Ma
0.8i + 13.2j = 4.1a
a = 0.19i + 3.21j m/s².
The magnitude of acceleration is,
|a| = √[(0.19)²+(3.21)²]
|a| = 0.361 +10.3
|a| = 3.26m/s².
So, the magnitude of acceleration is 3.26m/s².
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how fast is the angle of depression of the telescope changing when the boat is 190 meters from the shore
ANSWER:
- 0.01943 rad/sec
STEP-BY-STEP EXPLANATION:
The first thing is to make a drawing of what is mentioned in the statement, it would be the following:
Now, we have the following information:
[tex]\begin{gathered} \frac{dy}{dt}=15\text{ m/s} \\ x=50\text{ m} \\ y=190\text{ m} \end{gathered}[/tex]In this right angle triangle formed by telescope of the boat, e can apply the tangent trigonometric ratio, like this:
[tex]\begin{gathered} \tan \theta=\frac{x}{y} \\ \text{ replacing} \\ \theta=\tan ^{-1}\mleft(\frac{x}{y}\mright) \end{gathered}[/tex]Now, we implicitly derive with respect to t:
[tex]\begin{gathered} \frac{d}{dt}(\theta)=\frac{d}{dt}(\tan ^{-1}(\frac{x}{y})) \\ \frac{d}{dt}(\theta)=\frac{1}{1+(\frac{x}{y})^2}\cdot\frac{d}{dt}(\frac{x}{y}) \\ \frac{d}{dt}(\theta)=\frac{y^2}{x^2+y^2}\cdot x\cdot(-\frac{1}{y^2}\cdot\frac{dy}{dt}) \\ \frac{d}{dt}(\theta)=\frac{-x}{x^2+y^2}(\frac{dy}{dt}) \\ \text{ replacing} \\ \frac{d}{dt}(\theta)=\frac{-50}{50^2+190^2}\cdot(15) \\ \frac{d}{dt}(\theta)=-0.01943 \end{gathered}[/tex]The angle of depression is changing at a rate of -0.01943 rad/sec when the boat is 190 m from the shore
A roller coaster car is traveling through a loop at 13 m/s. If the loop has a 23m radius, what centripetal force will the 53kg rider feel?
Given,
The velocity of the car, v=13 m/s
The radius of the loop, r=23 m
The mass of the rider, m=53 kg
The centripetal force is the force that keeps an object in its circular path.
The centripetal force is given by,
[tex]F=\frac{mv^2}{r}[/tex]On substituting the known values,
[tex]\begin{gathered} F=\frac{53\times13^2}{23} \\ =389.43\text{ N} \end{gathered}[/tex]Therefore, the centripetal force on the rider is 389.43 N
When doing a chin-up, a student lifts her body with a force of 400N and a distance of 0.25 meters in 2 seconds. What is the power delivered by the students biceps?
Given:
The student lifts her body with a force F = 400 N
The distance is d = 0.25 N
The time is t = 2 s
To find: The power delivered by student's biceps.
Explanation:
The formula to calculate power is
[tex]P=\frac{F\times d}{t}[/tex]Substituting the values, the power will be
[tex]\begin{gathered} P=\frac{400\times0.25}{2} \\ =\text{ 50 W} \end{gathered}[/tex]Final Answer: The power delivered by student's biceps is 50 W.
Exercise 1 :On a circuit, a pilot covers 600 m in 7.2 s.1. Calculate its speed in m / s.2. Convert this speed to km / h in two different ways.
Given data
*The distance covers by the pilot is d = 600 m
*The given time is t = 7.2 s
(1)
The formula for the speed is given as
[tex]s=\frac{d}{t}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} s=\frac{600}{7.2} \\ =83.3\text{ m/s} \end{gathered}[/tex]Hence, the speed is s = 83.3 m/s
(2)
The speed converted into km/h as
[tex]\begin{gathered} s=83.3\times(\frac{5}{18}) \\ =299.88\text{ km/h} \end{gathered}[/tex]The second way to convert the speed into kilometer per hour as,
[tex]undefined[/tex]How many kilometers does the space shuttle have to travel to complete one orbit? In terms of a circle, what is this distance called? Explain.
40,840.7 kilometers the space shuttle have to travel to complete one orbit. In terms of a circle, this distance is termed as the circumference of the circle.
The circumference of a circle is the length measured around its edge. The diameter of a circle is the distance from the center to the outside.
Here we need to find the distance of the space shuttle that completed one circle, ie, the circumference of the orbit. The Circumference or distance covered by the space shuttle can be denoted by [tex]C_{SS}[/tex] and can be calculated by application of the below formula,
[tex]C_{SS}[/tex] = 2πr
where r is the radius of earth ie, 6500Km
Therefore, the equation becomes:
[tex]C_{SS}[/tex] = 2×π×r
= 2×π×6500
=40,840.7
So, the kilometer required to travel is 40,847.7Km
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Fill in the blanks on machines efficiency
Efficiency = ____________ energy _________ / ____________ energy ______________
Machine Efficiency = output energy/ input energy
The percent of input work that becomes work done by the machine is called efficiency. The output work is always less as compared to the input work because some of the input work gets used in overcoming friction, therefore the efficiency is always less than 100 percent.
Efficiency do not have units. It is written as a decimal or as a percentage. Energy efficiency is the use of less amount of energy to perform the same task. Energy-efficient homes and buildings use lesser energy to heat, cool, and run any appliances and energy-efficient manufacturing facilities use lesser energy to produce goods.
Efficiency measures work or energy that can conserved in a process. We can also say that efficiency is about comparing the output of the energy to the input of the energy.
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Remember the experiment done by Arthur Holly Compton that demonstrated the particle nature of light (X-rays) definitively. The reaction was:γ+e→γ+e (1) where the outgoing gamma was an X-ray of aa. higher?b. lower?frequency than the initial gamma. Circle your choice.
The correct option is (b)
The outgoing gamma rays are of lower frequency than that of the initial gamma-ray. While investigating the scattering of X-rays, Compton observed that the outgoing rays lose some of their energy in the scattering process and emerge with slightly decreased frequency.
The cross product of two vectors (X and Y) is a negative vector when the angle between them is:
A. 0
B. 90
C. 180
D. 270
Answer:
C. 180
Explanation:
Answer:
Explanation:
The cross product:
[ a × b ] = | a | · | b | · sin α
sin α < 0 if α ∈ (180°, 360°)
Answer:
D. 270°
The potential of a 5.0 cm radius conducting sphere is -100 V. What is the charge density on its surface?Group of answer choices-1.8x10-8 C/m23.5x10-8 C/m22.2x10-8 C/m2-2.2x10-8 C/m2-3.5x10-8 C/m2
First lets calculate the surface area
[tex]S=4\cdot\pi\cdot r^2=4\pi\cdot0.05^2m=0.0314m^2[/tex]Now to know the surface charge density we need the next formula:
[tex]CD=\frac{q}{A}[/tex]But we are missing the amount of charge, we only have the potential
So in this case, we going to apply a different formula
[tex]V=\frac{q}{4\cdot\pi\cdot\xi\cdot r}[/tex]q=5.56*10^-10
[tex]CD=-5.56\cdot\frac{10^{-10}}{0.0314}=-1.77\cdot\frac{10^{-8}C}{m^2}[/tex]The anwer might be -1.8x10-8
A spring with spring constant 175 N/m has 20 J of EPE stored in it. How much is it compressed?
Given:
• Spring constant = 175 N/m
,• Energy = 20 J
Let's find by how much it is compressed.
Apply the formula:
[tex]E=\frac{1}{2}kx^2[/tex]Where:
E is the energy = 20 J
k is the sring constant = 175 N/m
x is the compression in meters
Rewrite the formula for x:
[tex]x=\sqrt{\frac{2E}{k}}[/tex]Input values and solve for x:
[tex]\begin{gathered} x=\sqrt[]{\frac{2\times20}{175}} \\ \\ x=\sqrt[]{\frac{40}{175}} \\ \\ x=\sqrt[]{0.2285} \\ \\ x=0.48\text{ m} \end{gathered}[/tex]ANSWER:
0.48 m
Four charges are arranged in a square formation. Take q to be 1 C of charge and a to be 2 cm in length. Four charges are arranged in a square formation. Find the net electric field at the center of the square.
We have to calculate the electric field in the center, so we need the distance to the center R and the interaction of each charge with that point
[tex]E=\sum_{n\mathop{=}0}^{\infty}E_i[/tex]To calculate the distance R we have a triangle
[tex]R=\sqrt{\frac{2a^2}{4}}=\frac{a}{\sqrt{2}}=0.014m=1.41cm[/tex][tex]\begin{gathered} \sum_{n\mathop{=}0}^{\infty}Ex=\frac{9\cdot10^9}{2\cdot10^{-4}}\cdot(cos45)\cdot(6C)=1.91\cdot10^{14}N/C \\ \sum_{n\mathop{=}0}^{\infty}Ey=4.5\cdot10^{13}\cdot(cos45\degree)(2C)=0.636\cdot10^{14}N/C \\ Etot=\sqrt{Ex^2+Ey^2}=2.01\cdot10^{14}N/C \end{gathered}[/tex]Is important to remember that E has direction so you have to calculate each axis, x, and y
The x-component of a force on a 46-g golf ball by a 7-iron versus time is plotted in the following figure: a. Find the x-component of the impulse during the intervals i. [0, 50 ms], and ii. [50 ms, 100 ms] b. Find the change in the x-component of the momentum during the intervals iii. [0, 50 ms], and iv. [50 ms, 100 ms]
The x-component of the impulse during the intervals [0, 50 ms] is 750 Nms.
The x-component of the impulse during the intervals [50, 100 ms] is 1,500 Nms.
The change in the x-component of the momentum during the intervals [0, 50 ms] is 0.75 kgm/s.
The change in the x-component of the momentum during the intervals [50, 100 ms] is 1.5 kgm/s.
What is the impulse experienced by the ball?
The impulse experienced by the ball is calculated from the product of force and time of motion of the ball.
J = Ft
where;
F is the applied forcet is the time of motionThe x-component of the impulse during the intervals i. [0, 50 ms] is calculated as follows;
From the diagram, the impulse between (0, 50 ms) is the area of the triangle.
Jₓ = ¹/₂(b)(h)
where;
b is the base of the triangle = 50 ms h is the height of the triangle = 30 NJₓ = ¹/₂(50 ms)(30 N) = 750 N.ms
The impulse during [50 ms, 100 ms] is the area of the rectangle,
Jₓ = Lb
where;
L is the length = 100 ms - 50 ms = 50 msb is the breadth = 30 NJₓ = 50 ms x 30 N
Jₓ = 1,500 Nms
Impulse is the change in momentum of an object.
The change in the x-component of the momentum during the intervals [0, 50 ms] is calculated as follows;
ΔP = Jₓ = ¹/₂(50 ms)(30 N) = 750 N.ms = 0.75 Ns = 0.75 kgm/s
For interval of [50 ms, 100 ms];
ΔP = 1,500 Nms = 1.5 Ns = 1.5 kgm/s
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