Answer:
Using sold paper is better than to burn it because burning paper may cause pollution.
Explanation:
Hope it will help ^_^
whats fater, 1m/sor 1 mph
Answer:
1m/s should be the answer
Answer:
[tex]\Huge \boxed{\mathrm{1 \ m/s }}[/tex]
[tex]\rule[225]{225}{2}[/tex]
Explanation:
We will convert 1 m/s into mph.
1 m/s = 2.237 mph
2.237 mph > 1 mph
1 m/s is faster than 1 mph.
[tex]\rule[225]{225}{2}[/tex]
The x-component of vector R is Rx = −28.2 units and its y-component is Ry = 19.6 units. What are its magnitude and direction? Give the direction as an angle measured counterclockwise from the +x-direction.
Answer:
Explanation:
Rx = -28.2 units
Ry = 19.6 units
magnitude of R = √ [( - 28.2 )² + ( 19.6 ) ]
= √ ( 795.24 + 384.16 )
= 34.34 units
If θ be the angle measured counterclockwise from the +x-direction
Tanθ = 19.6 / - 28.2 = -0.695
θ = 180 - 34.8
= 145.2° .
Is there any induced current in the pickup coil when the polarity of the bar magnet is flipped?
Answer:
Faraday's law , he direction of the magnetic field changes by 180º, in the polarity inversion processes, induces a voltage.
Explanation:
For this exercise let's use Faraday's law
E = - dФ / dt
Ф = B.A = B A cos θ
where B is the magnetic field, A is the area and θ is the angle between the field line and the normal to the area.
We can see that an electromotive force (voltage) is indexed when there is a variation of the field B, a variation of the area and change of the angle or when there is a combinational of them.
In this case, the magnitude of the field is constant, as the wire is rigid metal, the area is constant, but the direction of the magnetic field changes by 180º, in the polarity inversion processes, for which reason each change induces a voltage.
If a voltage is created in the ring, which has a resistance, a current is also generated in it.
Therefore the answer is If a current is created in the hoop
Find the magnetic field strength at 1.50 m from the center of the circular region (e.g., outside the electric-field region).
Answer:
[tex]$ 1.81 \times 10^{-7} \ T$[/tex]
Explanation:
Given :
r = 1.50 m
R = 1 m
[tex]$\frac{dE}{dt}$[/tex] = [tex]$ 4.88 \times 10^{10} $[/tex] V / m s
Therefore the displacement current is
[tex]$ I_d = \epsilon_0. \frac{dE}{dt} . A $[/tex]
= [tex]$ \epsilon_0. \frac{dE}{dt} . \pi R^2 $[/tex]
Now according to law
[tex]$ B= \frac{\mu_0I_d.\frac{dE}{dt}.\pi R^2}{2 \pi r}$[/tex]
= [tex]$ \frac{\mu_0I_d.\frac{dE}{dt}. R^2}{2 r}$[/tex]
= [tex]$ \frac{4 \pi \times 10^{-7} \times 8.85 \times 10^{-12} \times 4.88 \times 10^{10} \times 1^2}{2 \times 1.5} $[/tex]
= [tex]$ 1.81 \times 10^{-7} \ T$[/tex]
Therefore, the magnetic field strength at 1.50 m from the center of the ring is [tex]$ 1.81 \times 10^{-7} \ T$[/tex].
A-0.06 C charge that moves downward is in a uniform electric field with a strength of 200 N/C. What is the
magnitude and direction of the force on the charge?
O 12 Nup
O 12 N down
O 3333 Nup
O 3333 N down
Answer:
its B
Explanation:
i just took the test
The correct option is B. The magnitude of the force on the charge is 12N and the direction is downward.
The electric field is a region of space where electric force can be felt. The formula for calculating the magnitude of a force in an electric field is expressed as:
[tex]\overline F=qE[/tex] where:
q is the charge measured in CoulombsE is the electric field intensity measured in N/CGiven the following parameters:
q = -0.06C
E = 200N/C
Substitute the given parameters into the formula as shown:
[tex]\overline F = -0.06 * 200\\\overline F = -12N\\[/tex]
Since the modulus of the force is in the negative sense, hence the direction will be downward.
Learn more here: https://brainly.com/question/22735564
You are piloting a small airplane in which you want to reach a destination that is 750 km due north of your starting location. Once you are airborne, you find that (due to a strong but steady wind) to maintain a northerly course you must point the nose of the plane at an angle of 22 west of true north. From previous flights on this route in the absence of wind, you know that it takes you 3.14 h to make the journey. With the wind blowing, you find that it takes 4.32 h. A fellow pilot calls to ask you about the wind velocity (magnitude and direction). What is your report? FYI so you can check yourself the answers are 101 km/h at 62 degrees east of south.
Answer:
v_wind = 101.46 km / h , θ = 61.8
Explanation:
This is a velocity composition exercise.
Let's do the problem in parts. Let's start by knowing the speed of the plane without air.
v = d / t
v = 750 / 3.14
v = 238.85 km / h
This is the speed of the plane relative to the Earth and it does not change.
In the second part, when there is wind, the travel time is greater than when there is no wind, therefore the wind delays the plane. To be more general, suppose that the wind has two components vₓ and [tex]v_{y}[/tex]
Let's use trigonometry to find the components of the plane's speed
cos θ = v_N / v
sin θ = v_W / v
v_N = v cos θ
v_W = v sin θ
let's calculate
V _N = 238.85 cos 22 = 221.46 km / h
v_W = -238.85 sin 22 = -89.47
the negative sign is because the plane is going west and the positive sign is the east direction.
As it indicates that the destination of the avine is towards the north, the x component of the wind must be
vₓ - v_W = 0
vₓ = v-w
vₓ = 89.47 km / h
in the direction to the East.
Now let's analyze the component of the wind in the Nort-South direction,
Indicate the travel time, let's calculate the speed that the component must have the speed of the plane
v_total = d / t
v_total = 750 / 4.32
v_total = 173.61 km / h
This is the final speed of the plane, which can be written
v_total = v_n - vy
vy = v_n - v_total
vy = 221.46 - 173.61
vy = 47.85 km
this component is directed towards the south
Let's use the Pythagorean Theorem, to find the magnitude
v_wind² = vₓ² + vy²
v_wind = √ (89.47² + 47.85²)
v_wind = 101.46 km / h
the address will then be found using trigonometry
θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹1 (47.85 / 89.47)
θ = 28.14
Therefore, the magnitude of the wind speed is 101.5 km / h and its direction is 28º south of the East, to give this value
90- θtea = 90- 28.2
θ = 61.8
East of South
You wish to watch TV at exactly 85 dB and no louder to avoid long term damage to your hearing. You record the sound intensity level to be 125 dB from a 1,200 W amplifier. The source is way too loud so you hit the remote and reduce the power to make the new sound intensity level be 107 dB.
1. What is the new power coming from the amplifier?
2. How far away are you from the amplifier now?
3. The sound is still too loud so you decide to cut amplifier power in half. How much farther/closer do you need to move to get down to 85 dB?
Answer:
1) the new power coming from the amplifier is 19.02 W
2) The distance away from the amplifier now is 5.50 m
3) u₁ = 69.24 m
Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther
Explanation:
Lets say that I am at a distance "u" from the TV,
Let I₁ be the corresponding intensity of the sound at my location when sound level is 125dB
SO
S(indB) = 10log (I₁/1₀)
we substitute
125 = 10(I₁/10⁻¹²)
12.5 = log (I₁/10⁻¹²)
10^12.5 = I₁/10^-12
I₁ = 10^12.5 × 10^-12
I₁ = 10^0.5 W/m²
Now I₂ will be intensity of sound when corresponding sound level is 107 dB
107 = 10log(I₂/10⁻²)
10.7 = log(I₂/10⁻¹²)
10^10.7 = I₂ / 10^-12
I₂ = 10^10.7 × 10^-12
I₂ = 10^-1.3 W/m²
Now since we know that
I = P/4πu² ⇒ p = 4πu²I
THEN P₁ = 4πu²I₁ and P₂ =4πu²I₂
Therefore
P₁/P₂ = I₁/I₂
WE substitute
P₂ = P₁(I₂/I₁) = 1200 × ( 10^-1.3 / 10^0.5)
P₂ = 19.02 W
the new power coming from the amplifier is 19.02 W
2)
P₁ = 4πu²I₁
u =√(p₁/4πI₁)
u = √(1200/4π × 10^0.5)
u = 5.50 m
The distance away from the amplifier now is 5.50 m
3)
Let I₃ be the intensity corresponding to required sound level 85 dB
85 = 10log(I₃/10⁻¹²)
8.5 = log (I₃/10⁻¹²)
10^8.5 = I₃ / 10^-12
I₃ = 10^8.5 × 10^-12
I₃ = 10^-3.5 w/m²
Now, I ∝ 1/u²
so I₂/I₃ = u₁²/u²
u₁ = √(I₂/I₃) × u
u₁ = √(10^-1.3 / 10^-3.5) × 5.50
u₁ = 69.24 m
Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther
What is the value of the magnetic flux through the coil due to the wire with a current I passing through it
Answer:
Magnetic flux is zero
Explanation:
We know that
magnetic flux (B) =
magnetic field x area x cosစ
And စ being the angle between magnetic field and area vector
So we are given စ= 90°
So
B = magnetic field x area x cos(90)
B= 0
What is the approximate radius of a nitrogen nucleus that has seven protons and seven neutrons?
Answer:
3.5E-15m
Explanation:
Using
R= Ro A^ 1/3
Where Ro= 1.2fm
and A= 7
= 1.2*10^-15 x (7)^1/3
= 3.5*10^-15m
Alice and Bob are bicycle-racing, on a two-mile track. Bob rides the first mile at 50 mph,
and the second mile at 71 mph. Calculate the time in minutes) that he travels the two
miles.
Answer:
2.04 minutes
Explanation:
He travels the first mile at 50 mph so:
[tex]60/50=1.2[/tex]
so it takes 1.2 minutes to travel each mile when going 50 miles per hour
and for the second mile he travels at 71 mph so:
[tex]60/71=0.84[/tex] (this answer is rounded)
so it takes him 0.84 minutes to travels each mile while going 71 mph.
If you ad them together his total times while biking two miles was 2.04 minutes.
What challenges are faced by scientist investigating effects of radiation in Chernobyl and Eniwetok Atoll in the Chernobyl accident? 3 sentences long.
Answer:
The health hazard of directly going to these places to take direct readings and observations.
Explanation:
The Chernobyl accident was a catastrophic radioactive accidents, with an immediate fatal effect on some victims (mostly firemen and law enforcement officers). Scientists investigating the effects of the radiation left on the venues of the accident face a health risk if they intend to carry out a full scale on-field testing and experiment. Some of the relocated occupants of these places, that have long been relocated, have shown some health effect of the radiation. And scientist studying and working on these places for too long stand a risk of developing health complications if they are exposed to the radiations for too long.
Acceleration of 1.5 ms expressed in km /hr2?
A worker pushes horizontally on a large crate with a force of 265 n and the crate is moved 4.5 M how much work was done
Answer:
the work done is 1192.5Nm
Explanation:
work done = force × distance
w.d = 265N × 4.5m
w.d = 1192.5Nm
mark me as brainliest pllyyyzzz
g summarize what you know about how changes in frequency and amplitude determine the characteristics
Answer:
Loudness and pitch are the characteristics of a sound related to amplitude and frequency.
Explanation:
Loudness and pitch are the characteristics of a sound that are related with the frequency and amplitude of a sound. If any change occurs in the frequency and amplitude of a sound, change also occurs in the loudness and pitch of a sound. If a source has larger amplitude so it produces large loudness or volume of a sound. On the other hand when the frequency is higher than the pitch will also be higher and the sound produce will be high.
Effect of thunder storm
Answer:
Some of the most severe weather occurs when a single thunderstorm affects one location for an extended time. Thunderstorms can bring heavy rains (which can cause flash flooding), strong winds, hail, lightning, and tornadoes. Severe thunderstorms can cause extensive damage to homes and property.
Explanation:
A grating has 460 rulings/mm. What is the longest wavelength for which there is a 6.0th-order diffraction line
Answer:
λ = 3.62 x 10⁻⁷ m = 362 nm
Explanation:
The grating equation gives the relationship between the wavelength, the diffraction line order and the diffraction angle. The grating equation is written as follows:
mλ = d Sinθ
where,
m = order of diffraction = 6
λ = longest wavelength = ?
d = 1/(460 rulings/mm)(1000 mm /1 m) = 2.17 x 10⁻⁶ m/ruling
θ = Diffraction angle = 90° (for longest wavelength)
(6)λ = (2.17 x 10⁻⁶ m/ruling) Sin 90°
λ = (2.17 x 10⁻⁶ m/rulings)/6
λ = 3.62 x 10⁻⁷ m = 362 nm
think of a teacher you have enjoyed/liked, and tell me what made you think they were a good teacher.
Answer:
a teacher that ive enjoyed is my 5th grade reading teacher. I enjoyed her because she always made class fun
Explanation:
Find the tension between the 4 kg and the 6 kg blocks once the system is released from rest and the 9 kg block accelerates downward.
Answer:
Tension on the string is 48 N
Explanation:
Mass of A = 6 kg
mass of B = 4 kg
If the 9 kg accelerates downwards then the equation for the 6 kg mass side of the pulley will be
6g - T = 6a ....1
where
g is acceleration due to gravity = 10 m/s^2
a is the acceleration of the body
T is the tension in the string.
For the 4 kg mass side of the string, we have
T - 4g = 4a ....2
transposing, we have
T = 4a + 4g ....3
substitute equation 3 in equation 1, we have
6g - (4a + 4g) = 6a
6g - 4a - 4g = 6a
6g - 4g = 6a + 4a
2g = 10a
but g = 10 m/s^
2 x 10 = 10a
20 = 10a
acceleration a = 20/10 = 2 m/s^2
substitute value of a in equation 1
6g - T = 6a
6(10) - T = 6(2)
60 -T = 12
60 - 12 = T
T = 48 N
How much power is used by a hair dryer if it does 40,000 J of work in 40 seconds?
Answer:
40000÷40=1000 joules is required to work in 40 seconds
A student bikes to school by traveling first dN = 0.800 miles north, then dW = 0.300 miles west, and finally dS = 0.100 miles south.
Similarly, let d--> W be the displacement vector corresponding to the second leg of the student's trip. Express d⃗ W in component form.
Express your answer as two numbers separated by a comma. Be careful with your signs.
Answer:
66.8°
Explanation:
Calculating for tthe first leg
dNx = 0;
dNy = + 0.8 mi ......................(0 , 0.8)
then for the second
dWx = - 0.3 mi;
dWy = 0 ...................(-0.3 , 0)
then the third leg
dSx = 0;
dSy = - 0.1 mi ......................(0 , -0.1)
So the displacement vector is
db = ( -0.3 , (0.8 - 0.1) )
db = ( -0.3 , 0.7)
Finding the magnitude i= √(0.3^2 + 0.7^2 )
= 0.76
So the direction is
စ= tan^-1( 0.7 / 0.3 ) = 66.8°
what is the acceleration of the object?
______ m/s2
Answer:
-2.5
Explanation:
Edge 2020
By observing the given graph The acceleration of the object would be -2.5 m/s², in other words, one can say the deacceleration of the object is 2.5 m/s²
What is acceleration?The rate of change of the velocity with respect to time is known as the acceleration of the object. Generally, the unit of acceleration is considered as meter/seconds².
acceleration =change in velocity/change in time
The slope of the velocity time diagram represents the acceleration of the moving object
As we can see from the graph the Y axis is representing the velocity and the X axis is representing the time
The Y coordinate changes from 40 m/s to 10 m/s and the X coordinates change from 0 s to 12 seconds
slope = (10 -40 )/(12-0)
acceleration = 30/12
acceleration = -2.5 m/s²
Thus, By looking at the provided graph In other terms, the object's acceleration would be -2.5 m/s2, or its deacceleration would be 2.5 m/s².
Learn more about acceleration from here
brainly.com/question/2303856
#SPJ5
How long would it take Exena to close the door if she doubled the magnitude of the force applied to the door
Answer:
y /y₀ = 1 /√2 = 0.70
Explanation:
To close the door, a torque must be applied in such a way that there is an angular acceleration and after a period of time the angle of the door moves.
If the force is doubled keeping the other parameters without altering angular acceleration is
F r = I α
α = F r / I
if we double the force
α = 2F r / I
α=2 α₀
let's use the angular kinematics relations
θ = w₀ t + ½ α t²
we assume the door was still before pushing it
θ = ½ a t²
t = √ (2θ / a)
as we saw the angular acceleration is double
t = √ (2θ / 2a₀)
t = 1 / √2 a₀
the relationship between the two times is
y /y₀ = 1 /√2 = 0.70
This time that has been reduced closes the door.
It takes Serina 4.44 hours to drive to school. Her route is 15 km long. What is Serina’s average speed on her drive to school?
Answer: 3.38 km per hour
Explanation:
Do your electric field and equipotential lines appear as expected?
Answer:
The question is incomplete. Let me explain the relationship between electric field and equipotential lines.
The electric field lines point symmetrically away from the positive point charge, this clearly indicates that they are perpendicular to the equipotential lines.
Explanation:
The potential for a point charge is V = [tex]\frac{kQ}{r}[/tex]
where k is a constant with value of 8.99 x 109 N m2/ , and r is the distance from point charge.
This shows that the potential for a point charge is the same anywhere on an imaginary sphere of radius r surrounding the charge.
The attached image is an isolated positive point charge Q and its electric field lines.
In the image, the electric field lines radiate out from a positive charge and terminate on negative charges.
The blue lines indicates the magnitude and the direction of the electric field.
This clearly helps us to see that the electric field lines point away from the positive point charge, which means that they are perpendicular to the equipotential lines.
describe the motion of an object for which the direction of the acceleration is the same as the direction of motion of the object
Answer:
motion of an object for which the direction of the acceleration is the same as the direction of motion of the object occurs when the body or object travels with an increase in velocity
Explanation:
Because for direction of object to be the same with direction of acceleration velocity must be increasing
A 30cm long tendon was found to stretch 9.0mm by a force of 11.2N.The tendon was approximately round with an average diameter of 8.9 mm.Calculate Young's modulus of this tendon.
Answer:
Young's Modulus = 6 x 10⁶ Pa = 6 MPa
Explanation:
First we need to find the stress on tendon:
Stress = Force/Area
Stress = 11.2 N/πr²
Stress = (11.2 N)/π(4.45 x 10⁻³ m)²
Stress = 180 x 10³ Pa
Now, we calculate the strain:
Strain = Change in length/Original Length
Strain = (9 x 10⁻³ m)/(0.3 m)
Strain = 0.03
Now, we calculate the Young's Modulus:
Young's Modulus = Stress/Strain
Young's Modulus = (180 x 10³ Pa)/0.03
Young's Modulus = 6 x 10⁶ Pa = 6 MPa
A particle moves along a line so that its velocity at time t is ( | |. Find the displacement of the particle during the time period .
Complete question is;
A particle moves along a line so that its velocity at time t is
v(t) = t² − t − 6
(measured in meters per second).
(a) Find the displacement of the particle during the time period 1 ⩽ t ⩽ 4.
(b) Find the distance traveled during this time period.
Answer:
A) -4.5 m
B) 10.17 m
Explanation:
We are given;
v(t) = t² − t − 6
We know that v = ds/dt
Thus,
S = v dt
s(4) - s(1) = (4,1)∫t² − t − 6 dt
= (4,1)[(t³/3) - (t²/2) - 6t]
= ((4³/3) - (4²/2) - 6(4)) - ((1³/3) - (1²/2) - 6(1))
= 64/3 - 8 - 24 - 1/3 + 1/2 + 6
= -4.5 m
B) Since v(t) = t² − t − 6, then factorizing we can write it as;
v(t) = (t + 2)(t - 3)
Thus, v(t) ⩽ 0 at the interval (1, 3) and v(t) ≥ 0 at the interval (3, 4)
Thus;
-x1 = (1, 3)∫t² − t − 6 dt
-x1 = (1,3)(t³/3 - t²/2 - 6t)
-x1 = (3³/3 - 3²/2 - 6(3)) - (1³/3 - 1²/2 - 6(1))
-x1 = 9 - 9/2 - 18 - 1/3 + 1/2 + 6
-x1 = -3 - 4 - 1/3
-x1 = -22/3 m
x1 = 22/3 m
x2 = (3, 4)∫t² − t − 6 dt
x2 = (3, 4)(t³/3 - t²/2 - 6t)
x2 = ((4³/3) - (4²/2) - 6(4)) - (3³/3 - 3²/2 - 6(3))
x2 = 64/3 - 8 - 24 - 9 + 9/2 + 18
x2 = -23 + 64/3 + 9/2
x2 = (-138 + 128 + 27)/6
x2 = 17/6 m
Thus, total distance = x1 + x2 = 22/3 + 17/6 = 10.17m
Review the Four Social Errors and Biases in the Highlights area on page 127 in Ch. 4 of THiNK: Critical Thinking and Logic Skills for Everyday Life. Which of the social errors/biases in the book are you most affected by? How can you overcome this social error/bias? Ch. 5 describes fallacies (when an argument seems to be correct but isn't). What is one fallacy you have personally used or seen in an argument? Discuss how critical thinking skills will make you less likely to be influenced by arguments that are based on fallacies and faulty reasoning. Reflect on the learning activities, concepts, ideas, and topics covered this week. What is the most interesting activity or concept you learned this week? Mention any concepts that are still a bit confusing to you or that you have questions about.
Answer: Provided in the explanation
Explanation:
I have understood that I have been influenced by the 'affinity bias' for quite a while. It caused me to feel fascination and feel better and right about individuals who had comparable intrigue and thought designs. I sort of began to feel this is the one for me based on those likenesses yet later used to be miserable when I comprehended that those similitudes are not many and insufficient consistently. I have begun to beat this inclination by rehearsing self reflection. I have begun to think about what causes me to feel pulled in to somebody and on the off chance that I introspect that it's exclusively founded on likenesses, at that point I cause myself to get that and monitoring that helps in escaping the inclination.
I make an effort not to get into the snare of paradoxes in my own announcements yet I have been forced to bear deceptions during contentions. One deception which I have encountered most is the 'Foul play' error as there have been a great deal of occurrences when individuals began to tear down me and my family when they couldn't win on a contention regarding balanced and rationale.
Basic reasoning causes us in understanding our defects and those issues of our own which keeps us from taking better choices and furthermore forestalls us tackling issues in more successful manners. Through basic reasoning, one can reflect and recognize utilization of deceptions in the contentions and that will help him in deciphering the circumstance from an alternate perspective. He will assess the circumstance better and through appropriate induction, he will have the option to get over those issues in thinking and subsequently, have the option to take better and more powerful choices for taking care of an issue.
The most fascinating idea which I have learned is that of the utilization of reflection or explicitly self reflection to comprehend ourselves better and furthermore to display basic reasoning appropriately. Being said that, I am a still somewhat confused about the manners by which we can reflect appropriately at each circumstance in a successful manner. This disarray remains in light of the fact that most deceptions and predispositions are oblivious in nature while the basic considering aptitudes reflection is cognizant and I meander whether a cognizant ability will have the option to break down and distinguish each oblivious inclinations or whether a few guards will keep a few inclinations covered up.
The north pole of a bar magnet points towards a thin circular coil of wire containing 40 turns. The magnet is moved away from the coil, so that the ux through one turn inside the coil decreases by ∆Φ = 0.3 T.m2 in a time ∆t = 0.2 s. What is the average EMF induced in the (whole) coil during this time interval? Viewed from the side opposite of the bar magnet (from the right), does the induced current run clockwise or counterclockwise? Explain briey. [2 p.]
Answer:
60 V
The direction will be anticlockwise from the right side of the magnet.
Explanation:
The change in magnetic flux = ∆Φ = 0.3 T.m^2
The change in time = ∆t = 0.2 s
number of turns = 40 turns
The induced emf E = N∆Φ/∆t =
E = (40 x 0.3)/0.2 = 60 V
If the magnet is moved away from the coil, the induced current on the coil will try to oppose the motion of the magnet by attracting the magnet towards the coil. For the magnet to be attracted towards the coil, it must possess the equivalent of a magnetic south pole. For the equivalent of a magnetic south pole, the current on the coil will flow in the clockwise direction when viewed from the left side of the magnet. When viewed from the right side of the magnet, the direction will appear as anticlockwise.
A cannonball falls at a rate of about 10 m/s2 due to gravity. It has a mass of 200 kg. What is the total force of the
cannonball?
[tex]\star[/tex][tex]\blue\star[/tex]
[tex]\blue\star[/tex]Givenmass = 200kgacceleration = 10m\s^2Star coding
[tex]\blue\star[/tex]As we know that force = mass × accelerationF = m×aF = 200 ×10F = 2000N[tex]\blue\star[/tex]so mate here is ur ans force is equal to 2000N.Hope it helps