To determine the focal length of a convex lens, if we are using sun as distant object, then we get a value which is closer to the accurate value of convex lens.
To find the answer, we need to know more about the focal length of a convex lens.
Why we get a focal length which is closer to the accurate value if we focusing sun?The focal length of an optical system can be defined as the measure of how sharply it converges or diverges the light.Focal length of convex lens is the distance between the pole and the focus of the convex lens.Convex lens converges all the rays coming from the distant object at its focal point to form a real image.If the object is at more distance from the lens, then we get sharper image.If we focus on a nearby window, then image will form away from the focal point.In the case of a distant tree, we need to move the screen towards the lens to get the focal point.Thus, we can conclude that, the if we are using sun as distant object, then we get a value which is closer to the accurate value of convex lens.
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If we use the sun as a distant object to calculate the convex lens's focal length, we obtain a number that is more closely related to the actual convex lens value.
We need to learn more about a convex lens's focal length in order to locate the solution.
Why, if we concentrate the sun, do we obtain a focal length that is closer to the correct value?An optical system's focal length is a measurement of how sharply the light converges or diverges.The distance between the convex lens's pole and its focus is known as its focal length.Convex lenses focus all of the rays from the distant object at their focal point to create a true image.We receive a crisper image if the object is farther away from the lens.Away from the focal point, an image will form if we concentrate on a neighboring window.To get the focal point on a distant tree, we must move the screen in the direction of the lens.We can therefore draw the conclusion that if we use the sun as a distant object, we will obtain a value that is closer to the convex lens's actual value.
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Calculate the time needed for a 0.600 kg hammer to reach the surface of the Earth
if dropped from 10.0 m on Earth and on the Moon. The gravitational strength on
the Moon is 1.6 N/kg.
The time needed for the hammer to reach the surface of the Earth is 3.54 s.
Time of motion of the hammer
The time of motion is calculated as follows;
t = √(2h/g)
where;
h is height of fallg is acceleration due to gravityt = √(2 x 10 / 1.6)
t = 3.54 s
Thus, the time needed for the hammer to reach the surface of the Earth is 3.54 s.
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Inez uses hairspray on her hair each morning before going to school. The spray spreads out before reaching her hair partly because of the electrostatic charge on the hairspray droplets. If two drops of hairspray repel each other with a force of 9 x 10^-9 N at a distance of 0.07 cm ( 7 x 10^ -4 m), what is the charge on each of the equally charged drops of hairspray?
Please show the steps as well. Written out, if possible.
The charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C
What is Columb's law?The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Similar charges repel each other, whereas charges that are opposed attract each other.
Given data;
Electric force,F = 9 × 10 ⁻⁹ N
Distance between charges,d = 7 × 10⁻⁴ m
Chrge,q₁ = q₂ =q C
From Columb's law;
[tex]\rm F = K \frac{q_1q_2}{d^2} \\\\ 9 \times 10^{-9} = 9 \times 10^9 \frac{q^2}{(7 \times 10^{-4})^2} \\\\ q^2 = 4.9 \times 10^{-25} \\\\ q = 7 \times 10^{-13} \ C[/tex]
Hence the charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C
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What was the initial speed of a car if its speed is 40 m/s after 5 seconds of accelerating at -4 m/s²?
A. 50 m/s
B. 60 m/s
C. 25 m/s
D. 20 m/s
Answer:
[tex]\huge\boxed{\sf V_i=60 \ m/s}[/tex]
Explanation:
Given Data:Final speed = [tex]V_f[/tex] = 40 m/s
Time = t = 5 s
Acceleration = a = -4 m/s²
Required:Initial velocity = [tex]V_i[/tex] = ?
Formula:[tex]\displaystyle a = \frac{V_f-V_i}{t}[/tex]
Solution:Put the givens in the above formula
[tex]\displaystyle -4=\frac{40 - V_i}{5} \\\\Multiply \ -5 \ to \ both \ sides\\\\-4 \times 5 = 40 - V_i\\\\-20 =40-V_i\\\\Subtract \ 40 \ to \ both \ sides\\\\-20-40=-V_i\\\\-60\ m/s=-V_i\\\\60 \ m/s = V_i\\\\V_i=60 \ m/s\\\\\rule[225]{225}{2}[/tex]
A stone of mass 150g is rotated in a horizontal circle at 10m/s which is attached to the end of a 1m long. what will be the acceleration of the stone and it's centripetal force?
force is mass multiply by acceleration so it will be 150 multiply by 10 is 1500N
Answer:
Acceleration: [tex]100\; {\rm m\cdot s^{-2}}[/tex] assuming that the radius of the rotation is [tex]1\; {\rm m}[/tex].
Centripetal force: [tex]15\; {\rm N}[/tex].
Explanation:
In a circular motion, if the tangential velocity is [tex]v[/tex] and the radius of the motion is [tex]r[/tex], the centripetal acceleration of the motion would be [tex]a = (v^{2} / r)[/tex].
In this question, it is implied that for this circular motion, [tex]v = 10\; {\rm m\cdot s^{-1}}[/tex] while [tex]r = 1\; {\rm m}[/tex]. Thus, the (centripetal) acceleration would be:
[tex]\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(10\; {\rm m\cdot s^{-2}})^{2}}{1\; {\rm m}} \\ &= 100\; {\rm m \cdot s^{-2}}\end{aligned}[/tex].
Note that the unit of mass in this question is gram, whereas the standard unit for mass should be [tex]{\rm kg}[/tex] (so as to leverage the fact that [tex]1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}[/tex].) Apply unit conversion: [tex]m = 150\; {\rm g} = 0.150\; {\rm kg}[/tex].
Using that fact that [tex](\text{net force}) = (\text{mass}) \, (\text{acceleration})[/tex]:
[tex]\begin{aligned} (\text{net force}) &= (\text{mass}) \, (\text{acceleration}) \\ &= 0.150\; {\rm kg} \times 100\; {\rm m\cdot s^{-2}} \\ &= 15\; {\rm kg \cdot m \cdot s^{-2}} \\ &= 15\; {\rm N}\end{aligned}[/tex].
Suppose the carts collided on a surface that had a slight incline to it. Would you expect the momentum to be conserved
Momentum is conserved when carts are collided on a slanting plane.
To find the answer, we need to know about the conversation of momentum.
What's the conversation of momentum?Conservation of linear momentum says the total momentum before the collision and after the collision remains the same. Mathematically, m1u1+m2u2 = m1v1+m2v2How is the momentum conserved when collision occurs on a slanting plane?On a slanting plane, the velocity has two components, horizontal component horizontal component Vertical componentSo, its momentum has also similar two components. The momentum is conserved along horizontal direction and vertical direction separately.Thus, we can conclude that the momentum is conserved when carts are collided on a slanting plane.
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Question 17
The driver of a car traveling along a straight road with a
speed of 72KM ph observes a signboard which give the
speed limit to be 54KM ph. The signboard is 70m ahead
when the driver applies the brakes, calculate the
acceleration of the car which will cause the car to pass the
signboard at the stated speed limit?
So, the acceleration of the car is -1.25 m/s². In other word, the car is also decelerating by 1.25 m/s².
IntroductionHi ! I will help you to discuss about "deceleration in a straight line movement". Please note in advance that deceleration is acceleration which has a negative value. When an object decelerates, the object will continue to move until it reaches a certain speed (which is less than before) or until it stops. The higher the deceleration value, an object that is moving will stop faster and cover a shorter distance.
Formula UsedIn this opportunity, I will give you the following equation to express the relationship between final velocity and initial velocity, acceleration, and distance.
[tex] \boxed{\sf{\bold{(v_t)^2= (v_0)^2 + 2 \times a \times s}}}[/tex]
With the following condition:
[tex] \sf{v_t} [/tex] = final velocity of an object (m/s)[tex] \sf{v_0} [/tex] = initial velocity of an object (m/s)a = acceleration that happen (m/s²)s = the shift or distance of the object (m)Problem SolvingWe know that:
[tex] \sf{v_0} [/tex] = initial velocity of an object = 72 km/h = 20 m/s[tex] \sf{v_t} [/tex] = final velocity of an object = 54 km/h = 15 m/ss = the shift or distance of the object = 70 mNote :
1 m/s = 3.6 km/h. So 10 m/s = 36 km/hWhat was asked ?
a = acceleration that happen = ... m/s²Step by step :
[tex] \sf{(v_t)^2 = (v_0)^2 + 2 \times a \times s} [/tex]
[tex] \sf{15^2 = 20^2 + 2 \times a \times 70} [/tex]
[tex] \sf{225 = 400 + 140 \times a} [/tex]
[tex] \sf{140 a = -175} [/tex]
[tex] \sf{a = \frac{-175}{140}} [/tex]
[tex] \boxed{\sf{\bold{a = -1.25 \: m/s^2}}} [/tex]
ConclusionHere, we see that the acceleration is -1.25 m/s². In other words, the car is also decelerating by 1.25 m/s².
Zero, a hypothetical planet, has a mass of 4.5 x 1023 kg, a radius of 3.2 x 106 m, and no atmosphere. A 10 kg space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial kinetic energy of 5.0 x 107 J, what will be its kinetic energy when it is 4.0 x 106 m from the center of Zero
a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J
b.) K 2 +GmM( r 11− r 21)=6.9×10 7 J
Applying Law of Energy conservation :
K 1+U 1
=K 2+U 2
⇒K 1− r 1GmM
=K 2− r 2 GmM
where M=5.0×10 23kg,r1
=> R=3.0×10 6m and m=10kg
(a) If K 1
=5.0×10 7J and r 2
=4.0×10 6 m, then the above equation leads to
K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J
(b) In this case, we require K 2
=0 and r2
=8.0×10 6m, and solve for K 1:K 1
=K 2 +GmM (r 11− r 21)=6.9×10 7 J
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Write an argument of any one
Do you think parallel universes exist?
or
Do you think portals exist?
A force of 15 n is applied to a spring, causing it to stretch 0.3 m. what is the spring constant for this particular spring? n/m
Answer:
-50N/m
Explanation:
Force , F = 15N
Displacement , x = 0.3m
Spring constant , K = ?
K = -F/x
K = -15N/0.3m
K = -50N/m
Answer: 50 N/m
Explanation:
Edge 2022
What is the net force (magnitude and direction) of the system below?
The net force (magnitude and direction) of the system given in the question is 40 N horizontal to the right
How to determine the net forceCase 1 (Net force between up and downward force)
Force up (Fu) = 50 NForce down (Fd) = 30 NNet force 1 (F1) = ?F1 = Fu - Fd
F1 = 50 - 30
F1 = 20 N up
Case 2 (Net force between right and left)
Force right (Fr) = 60 NForce left (Fl) = 20 NNet force 2 (F2) = ?F2 = Fr - Fl
F2 = 60 - 20
F2 = 40 N right
SUMMARY
Net force between up and down = 20 N upNet force between right and left = 40 N rightFrom the above, the net force between right and left (i.e 40 N) is greater than the net force between up and down (i.e 20 N)
Thus, the net force of the system will be 40 N horizontal to the right
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The constant G in Newton's equation _______. View Available Hint(s)for Part A The constant G in Newton's equation _______. was measured by Newton shows gravity to be a relatively huge force produces equilibrium makes the units of measurement consistent
The constant G in Newton's equation makes the units of measurement consistent was measured by Newton shows gravity to be a relatively huge force produces equilibrium makes the units of measurement consistent.
Newton's law:The three fundamental laws of classical mechanics known as Newton's laws of motion describe how an object's motion and the forces acting on it interact.
Newton's First lawNewton's Second lawNewton's Third lawNewton's first law, an object will not change its motion unless a force acts on it.
Newton's second law, the force on an object is equal to its mass times its acceleration.
Newton's third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.
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A uniform stick has a mass of 4.42 kg and a length of 1.23 m. It is initially lying flat at rest on a frictionless horizontal surface and is struck perpendicularly by a puck imparting a horizontal impulsive force of impulse 12.8 N s at a distance of 46.4 cm from the center. Determine the subsequent motion of the stick.
10.7 rad/s is the final angular velocity of the stick.
Given:
Mass of the stick = 4.42 kg
Length of the stick = 1.23m
Force of impulse (I) = 12.8 N s
The linear velocity of the stick, [tex]v=\frac{I}{m}[/tex]
[tex]v=\frac{12.8 N.s (\frac{1 kg m/s^2}{1 N}) }{4.42 kg}[/tex]
[tex]v[/tex] [tex]= 2.89 m/s[/tex]
Therefore, the final linear velocity of the stick is 2.89 m/s
∴[tex]w=\frac{12 Ir}{ml^{2} }[/tex]
[tex]w=\frac{12 ( 12.8 N.s ) ( 46.4 cm)}{(4.42 kg) (1.23 m)^2}[/tex]
[tex]w= \frac{12 (12.8 N.s) (46.4 cm) (\frac{10^-^2 m}{1 cm} )}{(4.42 kg) (1.23m)2}[/tex]
[tex]w=10.7 rad/s[/tex]
Therefore, 10.7 rad/s is the final angular velocity of the stick.
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If a baseball pitcher throws a fastball at a horizontal speed of 160 km/h, how long does the ball take to reach home plate 18.4 m away
Answer: 0.000115 seconds
Explanation: how I got this is by first converting the meters to kilometers then I had 160 km/h and 0.0184 km away so then you do 0.0184 divided by 160 (0.0184 / 160) and get 0.000115 seconds.
An object weighing 2.7n while in air and 1.2n when completely immersed in water.find the relative density of the object
Answer:
1.8 = relative density (there are no units for relative density)
Explanation:
It displaces water equal to it's volume and gets buoyancy equal to that amount of water
2.7 - 1.2 = 1.5 N of buoyancy
density of water = 1 gm /cc
1.5 N = m (9.81)
m of water displaced = .1529 kg
152.9 cc of water will produce this buoyancy....this is the volume of the object
find mass of object 2.7 = m (9.81) shows m = .2752 kg = 272.5 gm
density = mass/ volume = 272.5 / 152.9 = 1.8 gm / cc
Relative to water (which is 1 gm / cc) the relative density is 1.8
====> it is 1.8 times denser than water and will sink when in water....
A cord of mass 0.65 kg is stretched between two supports 8.0 m apart. If the tension in the cord is 120 N, how long will it take a pulse to travel from one support to the other
The time taken by the pulse to travel from one support to the other is 0.208 s.
Given:The mass of the cord is m = 0.65 kg.
The distance between the supports is, d = 8.0 m.
The tension in the cord is T = 120 N.
The time taken by the pulse to travel from one support to the other is given as,
[tex]v=\frac{d}{t}[/tex]
[tex]t=\frac{d}{v}[/tex]
Here, v is the linear velocity of a pulse. Its value is,
[tex]v=\sqrt{\frac{T d }{m} }[/tex]
[tex]v=\sqrt{\frac{120 * 8}{0.65} }[/tex]
[tex]v= 38.43 m/s[/tex]
Then,
[tex]t=\frac{8}{38.43}[/tex]
[tex]t=0.208 s[/tex]
Thus, the time taken by the pulse to travel from one support to the other is 0.208 s.
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Determine the launch speed of a horizontally launched projectile that lands 26.3m from the base of a 19.3m high cliff.
The launch velocity of the projectile is 13.28 m/s.
What is projectile motion?
The motion of an object thrown in the air under the force of gravity is known as projectile motion.
Since the object is launched horizontally, its initial velocity along the vertical direction is zero. From the second kinematic equation,
s=u*t+(1/2)at^2.
where s is the displacement, t is the time, u is the initial velocity and a is the acceleration. Since the height is decreasing, so it will be taken negative.
For the vertical motion, s=-19.3 m, a=-9.8 m/s^2 and u=0. Put the values in the above equation and solve it.
-19.3 = (0)*t+(1/2)*(-9.8)*t^2
19.3 = (1/2)*(9.8)*t^2
t=1.98 s
Since the velocity along the horizontal direction is constant, the displacement along the horizontal direction is given by the formula,
X=vt
where X is the horizontal displacement, v is the initial horizontal velocity and t is the time.
For the horizontal motion, X=26.3 m and t=1.98 s. Put the values in this equation and solve it.
26.3=v*(1.98)
v=13.28 m/s
The launch velocity is equal to the initial horizontal velocity, so it is equal to 13.28 m/s.
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You travel in a circle, whose circumstances is 8 kilometers, at an average speed 8 kilometers/hour. If you stop at the same point you started from what is your average velocity
Answer:
The Average Velocity is 2.22m/s.
Explanation:
Average Velocity = displacement(m) ÷ time(s)
8km = 8000m
8km/hr = 2.22m/s
Finding Time
Time = distance÷speed
Time = 8000m ÷ 2.22m/s = 3603.6s = 3604s
Finding Average Velocity
Average Velocity = displacement ÷ time
Average Velocity = 8000m ÷ 3604s
Average Velocity = 2.22m/s
Therefore the Average Velocity is 2.22m/s.
In a test run, a rocket-powered car is driving down a test track at a constant speed when the rockets are fired. it then accelerates at 4 m/s2 for 8.0 s covering a distance of 224 m. at what speed was the car travelling when the rockets were fired?
With the use of the second equation of motion formula, the speed the car travelled when the rockets were fired is 12 m/s.
Speed in a Linear MotionLinear motion depicts movement in a straight line. While speed is the distance travelled per time taken.
Given that In a test run, a rocket-powered car is driving down a test track at a constant speed when the rockets are fired. it then accelerates at 4 m/s2 for 8.0 s covering a distance of 224 m.
The given parameters are;
Accelerates a = 4 m/s2Time t = 8sDistance s = 224m
The speed the car travelled when the rockets were fired can be calculated by using the below formula
s = ut + 1/2a[tex]t^{2}[/tex]
Substitute all the parameters into the formula
224 = 8u + 1/2 x 4 x [tex]8^{2}[/tex]
224 =8u + 2 x 64
224 = 8u + 128
224 - 128 = 8u
8u = 96
u = 96/8
u = 12 m/s
Therefore, the speed the car travelled when the rockets were fired is 12 m/s.
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A rocket takes off from Earth's surface, accelerating straight up at 31.2 m/s2. Calculate the normal force (in N) acting on an astronaut of mass 90.4 kg, including her space suit.
The normal force is 2820.48N in the negative y direction.
According to Newton's second law of motion,
Force = mass × acceleration
F = m×A
Note that rocket takes off from Earth's surface, accelerating straight up at 31.2 m/s² .
The rocket accelerates upwards, hence the acceleration will be negative because it defies gravity's law (it keeps going into space without coming down)
Acceleration of the rocket = -31.2m/s²
Mass of the astronaut = 90.4kg
Normal force acting on the astronaut = -31.2 × 90.4kg
= -2820.48N
Therefore, the normal force is 2820.48N in the negative y direction.
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A box weighing 660 N is sliding across a cement floor and the coefficient of sliding
friction between the box and the floor is 0.15. If the force pushing the box is 500 N, what
is the acceleration of the box?
O 6 m/s²
O 2 m/s²
O 10 m/s²
pts
O 9 m/s²
The acceleration of the box is 6 m/s²; option A.
What is the acceleration of the box?The acceleration of the box is determined as follows:
Frictional force = 660 * 0.15 = 99 N
Net force = 500 - 99 = 401 N
Force = mass * accelerationMass of the box = 660/9.8 = 67.3 kg
Acceleration = net force/mass
Acceleration = 401/67.3
Acceleration = 6 m/s²
In conclusion, the acceleration of the box is determined from the net force and the mass of the box.
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Verification of laws of reflection of sound (Record Work)
The Angle of Incidence is equal to the Angle of Reflection.
What is the Verification of laws of reflection of sound?Take a wooden drawing board and fix a white sheet of paper on it. In the middle of paper draw a straight line and Mark a point B on it. Draw a perpendicular . Place a mirror on line such that the flat side of the mirror is along the line. Hold the mirror in the mirror holder.
Fix two steel pins P and Q on the straight line AB at least 10 cm apart. Look for the images of the pins P and Q and fix two pins P a such that P', Q', and images of P and Q are all in the same straight line. Remove the pins and draw small circles around the pinpricks.
Remove the mirror and Join P'Q' which produce the straight line to meet at B. Measure ∠ABN = i and ∠CBN = r. It is found that ∠i=∠r. This proves that the Angle of Incidence is equal to the Angle of Reflection.
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A dentist holds a small mirror 1.9
cm from a patient’s tooth. The image
formed is upright and 5
times as largeas the object. (a) Is
the image real or virtual? (b) What
is the focal length of the mirror? Is it
concave or convex? (c) If the mirror
is moved closer to the tooth, does the
image get larger or smaller? (d) For
what range of object distances does
the mirror produce an upright image?
Answer:
follow me for legitimate answer… oh? U don’t like the feeling of people stealing ur points huh?! Well theirs a taste of your own medicine
Explanation:
25 points please help PLEASE
Journal prompt to be answered in 2 fully developed paragraphs
Prompt: What are some products (or programs) that you could purchase to help your performance in your current physical activity? How would the product (or program help)? Do you really think it is effective? Use specific examples from your experience.
Answer:
você tem um camping de em Rio do Rio de Janeiro ou em lojas em SP e SP e RJ e região metropolitana de Porto Velho de Porto Velho Rio
Long distance running is the physical activity I engage in now, and there are a number of items and programs that can greatly improve my performance.
A GPS powered watch is one such item which I think is quite useful. I can monitor my speed, distance, heart rate and other important information while running with this device. This data lets me change my pace, track my effort, and make sure I stay within my desired heart rate range. The watch also provides post-run analysis, which I can use to identify my weaknesses and modify my training appropriately. In my opinion, GPS running watches have changed the game by allowing me to run harder and push my physical limits.
A strength training program designed for runners has also been very helpful in enhancing my performance. The program's primary emphasis is on functional workouts that target the major running muscle groups, such as the core, glutes and legs.
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What is weight??????????????????
Answer:
Weight is a body's relative mass, or the quantity of matter contained by it.
Explanation:
Hope this helps
A toy car starts from the rest and accelerates at 1.50m/s2 [E] for 2.25s. What is the final velocity, of the car
3.375m/s is the final velocity of the car.
How do you find final velocity?The final velocity depends on how large the acceleration is and the distance over which it acts.
Initial velocity of an object, you can multiply the acceleration due to a force by the time the force is applied and add it to the initial velocity to get the final velocity.
According to the question,
A toy car starts from the rest and accelerates
So the acceleration = 1.50m/s²
Time = 2.25s
[tex]x=x_{0} + vt[/tex]
[tex]x = 0 + ( 1.50m/s^2*2.25s)[/tex]
[tex]x = 3.375m/s[/tex]
The final velocity, of the car is 3.375 m/s.
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Select odd one out and also give the reason:
a) Gravitational force , magnetic force , electrostatic force
Gravitational force is not related to other force because it is related to gravity.
How magnetic force and electrostatic force are related?Electrostatic force and magnetic force are link to each other because both have charges. Electrostatic force is type of force that is present between two electrically charged particles. They can either be a repulsive or attractive force.
So we can conclude that Gravitational force is not related to other force because it is related to gravity.
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The engine of a 1560-kg automobile has a power rating of 75 kW. Determine the time required to accelerate this car from rest to a speed of 100 km/h at full power on a level road. Is your answer realistic
The time required to accelerate this car from rest to a speed of 100 km/h at full power on a level road is 7.73 seconds.
Mass of the engine = 1500 kg
Power rating = 75 kW = 75,000 W
Final speed = 100 km/hr = = 27.78 m/s
v₁ = 0
Power = Work done ÷ Time
Work done = Final energy - Initial energy
=1/2 x 1560 x 27.78^2 = 1/2 x 1560 x 0^2\
= 578703.72 J
Thus,
75,000 = 578703.70 ÷ time=7.72 seconds.
Work is done every time a force moves something over a distance. By multiplying the force by the distance traveled in the direction of the force, you can calculate the energy transferred or the work done. Energy transmitted = work completed = force x distance traveled in the direction of the force.
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A 450-kg sports car accelerates from rest to 100 m/s in 20.0 s. What magnitude force does a 90.0 kg passenger experience during the acceleration
Magnitude of force experienced by the 90Kg passenger is 450N.
To find the answer, we need to know about the pseudo force.
What's the pseudo force experienced by a passenger traveling in a vehicle?When a passenger inside a vehicle, it experiences pseudo force in the opposite direction of motion of the vehicle.The pseudo force= mass of the passenger × acceleration of the vehicleWhat's the acceleration of the vehicle when it achieves a speed of 100 m/s from rest in 20s?Acceleration= Change in velocity/timeHere, change in velocity= 100m/s and time = 20sSo, acceleration= 100/20= 5m/s²What's the pseudo force experienced by the 90Kg passenger?Pseudo force= 90×5=450N.
Thus, we can conclude that the force experienced by the passenger is 450N.
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The Viking spacecraft observed several features on Mars that, at first, appeared to be artificially crafted. After further study the features were primarily naturally occurring phenomena resulting from _____ on the Martian surface.
Answer:
Erosion of the surface of Mars.
Explanation:
The surface of Mars have features which were formed as a result of the erosion of the surface of the Mars.
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which two forms of energy do burning substances produce
Answer:thermal (heat) energy and radiant (light) energy.
Explanation:
Answer:
Answer: We use the chemical energy in fuels by burning them and transforming them into other types of energy: thermal energy, as when we burn fuel for heat; and kinetic energy, as when we burn gasoline to power our car's motion
Explanation:
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