Complete Question
A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?
Answer:
The position of the object at t = 10s is [tex]X = 38.3 \ m[/tex]
Explanation:
From the question we are told that
The acceleration along the x axis is [tex]a_{x}t = -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)[/tex]
The position of the object at t = 0 is x = -14.0 m
The velocity at t = 0 s is [tex]v_{0}x = 7.10 m/s[/tex]
Generally from the equation for acceleration along x axis we have that
[tex]a_x = \frac{dV_{x}}{dt} = -0.032 (15- t)[/tex]
=> [tex]\int\limits {dV_{x}} \, = \int\limits {-0.032(15- t)} \, dt[/tex]
=> [tex]V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]
At t =0 s and [tex]v_{0}x = 7.10 m/s[/tex]
=> [tex]7.10 = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1[/tex]
=> [tex]K_1 = 7.10[/tex]
So
[tex]\frac{dX}{dt} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]
=> [tex]\int\limits dX = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}[/tex]
=> [tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ K_1t +K_2[/tex]
At t =0 s and x = -14.0 m
[tex]-14 = -0.032 [ 15\frac{0^2}{2} - \frac{0^3 }{6} ]+ K_1(0) +K_2[/tex]
=> [tex]K_2 = -14[/tex]
So
[tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ 7.10 t -14[/tex]
At t = 10.0 s
[tex]X = -0.032 [ 15\frac{10^2}{2} - \frac{10^3 }{6} ]+ 7.10 (10) -14[/tex]
=> [tex]X = 38.3 \ m[/tex]
A car is accelerated at a constant rate from 15 m/s to 25 m/s. It takes the car 6 s to reach its final speed. What is the car’s acceleration?
Answer:
1.67 m/s²Explanation:
The car’s acceleration can be found by using the formula
[tex]a = \frac{v - u}{t} \\ [/tex]
where
v is the final velocity
u is the initial velocity
t is the time taken
a is the acceleration
From the question we have
[tex]a = \frac{25 - 15}{6} = \frac{10}{6} = \frac{5}{3} \\ = 1.666666...[/tex]
We have the final answer as
1.67 m/s²Hope this helps you
2. A solenoid. Suppose the south end of a bar magnet was introduced to the right end of this solenoid at a constant velocity. What direction would you expect the induced magnetic field to be in?4. To decrease the magnitude of current induced in the inductor using a bar magnet, you could (select all that apply):a. Decrease the strength of the magnet.b. Decrease the velocity of the magnet going into the solenoid.c. Decrease the number of coils in the solenoid.d. Increase the cross sectional area of the solenoid.6. List the TWO values you need to record to measure a voltage on the oscilloscope
Answer:
2) deflection must be towards the negative side of the voltage.
4) the correct statements are: b and c
Explanation:
2) This question is based on Faraday's law of induction, when we introduce a magnet in a solenoid an induced current is produced that generates a voltage that is given by
E = - N d [tex]\phi_{B}[/tex] / dt
where [tex]\phi_{B}[/tex] = B. A
The bold are vectors
Therefore, when applying this formula to our case, the induction lines of the magnetic field increase as we approach the solenoid, as the South pole approaches the lines are in the direction of the magnet, therefore the normal to the solenoid that has an outgoing direction and the magnetic field has 180º between them and the cos 180 = -1; consequently the deflection must be towards the negative side of the voltage.
4) From the Faraday equation we can see that the inductive electromotive force depends
* The magnitude of B that changes over time
* The area of the loop that changes over time
* The angle between B and the area that changes over time
* A combination of the above
With this analysis we will review the different alternatives given
a) False. It takes a temporary change and an absolute value of B
b) True. As the speed decreases, the change in B decreases, that is, dB / dt decreases
c) True. The current is induced in each turn, if there is a smaller number the total current will be smaller
d) False. A temporary change of area is needed, in addition to increasing the area the current increases
We can see that the correct statements are: b and c
_______mirrors or lenses always produce smaller images.
Answer:
The images produced by a convex mirror are smaller than the object it reflects. The image produced by a concave lens is always virtual. The image in a convex mirror is always upright and is smaller than the object.
Explanation:
:)
what makes the ball stop on rolling after somtime
A rifle of mass M is initially at rest but free to recoil. It fires a bullet of mass m and velocity v (relative to the ground). After firing, the velocity of the rifle (relative to the ground) is:_____.a. –mv.
b. –Mv/m.
c. –mv/M.
d. –v.
e. mv/M.
Answer: Option C) -mv / M is the correct answer
Explanation:
given that;
A rifle of mass M is initially at rest but free to recoil
velocity = v
using conversation of moment
pi = pf
i.e initial moment = final moment
But initial moment was zero (0) since everything was rest but free to recoil
so
pi = pf
0 = mv + MV
V is the recoil velocity of the rifle
MV = -mv
V = -mv / M
Therefore Option C) -mv / M is the correct answer
A stone is dropped from a bridge 45 meters above the surface of a river. What is the time required for the stone to reach the water’s surface?
Answer:
C. 3.0 s
Explanation:
Well, I've taken this question before, so I went back and this was the answer I put, which was correct.
Hope this helps! <3 May I have brainliest if it helps you enough?
It will take 3 s for the stone to get to the surface of the water.
We'll begin by listing out what was given from the question. This includes:
Height (h) = 45 m
Time (t) =?NOTE: Acceleration due to gravity (g) is 10 m/s²
From the data above, we can obtain the time taken for the stone to get to the surface of the water as follow:
H = ½gt²45 = ½ × 10 × t²
45 = 5 × t²
Divide both side by 5[tex]t^{2} = \frac{45}{5} \\\\t^{2} = 9[/tex]
Take the square root of both side[tex]t = \sqrt{9}[/tex]
t = 3 sTherefore, it will take 3 s for the stone to get to the surface of the water.
Learn more: https://brainly.com/question/19339518
A spaceship is accelerating at 1000 m/sec2 . How much force is required from the backthrusters to completely stop the spaceship?
Answer:
1000x Newton
Explanation:
Step one
given data
acceleration= 1000 m/s²
The question did not specify the mass of the mass of the space ship.
So, let's assume the mass is x kg
Step two:
Required is the force F in Newton
From Newtons first law, it states that a body will continue to be at rest or uniform motion unless acted upon by a force.
F=mass x Acceleration
F=ma
Substituting our given data we have
F=1000x Newton
If the particles were moving with a speed much less than c, the magnitude of the momentum of the second particle would be twice that of the first. However, what is the ratio of the magnitudes of momentum for these relativistic particles?
Answer:
p₂ / p₁ = 2 (v₁ / v₂)
Explanation:
The moment is a very useful concept, since it is one of the quantities that is conserved during shocks and explosions, for which it had to be redefined to be consistent with special relativity,
p = m v / √[1+ (v/c)² ]
for the case of speeds much lower than the speed of light this expression is close to
p = m v
In this exercise they indicate that the moment of the second particle is twice the moment of the first, when their velocities are small
p₂ = 2 p₁
p₂/p₁ = 2
in consecuense
m v₂ = 2 m v₁
v₂ = 2 v₁
consider particles of equal mass.
By the time their speeds increase they enter the relativistic regime
p₂ = mv₂ /√(1 + v₂² /c²)
p₁ = m v₁ /√(1 + v₁² / c²)
let's look for the relationship between these two moments
p₂ / p₁ = mv₂ / mv₁ [√ (1+ v₁² / c²) /√ (1 + v₂² / c²)
from the initial statement
p₂ / p₁ = 2 √(c² + v₁²) / (c² + v₂²)
we take c from the root
p₂ / p₁ = 2 √ [(1+ v₁²) / (1 + v₂²)]
this is the exact result, to have an approximate shape suppose that the velocities are much greater than 1
p₂ / p₁ = 2 √ [v₁² / v₂²] = 2 √ [(v₁ / v₂)²]
p₂ / p₁ = 2 (v₁ / v₂)
we see the value of the moment depends on the speed of the particles
A plane is flying due west at 34 m/s. It encounters a wind blowing at 19 m/s south. Find the resultant veloci
Answer:
The resultant velocity has a magnitude of 38.95 m/s
Explanation:
Vector Addition
Given two vectors defined as:
[tex]\vec v_1=(x_1,y_1)[/tex]
[tex]\vec v_2=(x_2,y_2)[/tex]
The sum of the vectors is:
[tex]\vec v=(x_1+x_2,y_1+y_2)[/tex]
The magnitude of a vector can be calculated by
[tex]d=\sqrt{x^2+y^2}[/tex]
Where x and y are the rectangular components of the vector.
We have a plane flying due west at 34 m/s. Its velocity vector is:
[tex]\vec v_1=(-34,0)[/tex]
The wind blows at 19 m/s south, thus:
[tex]\vec v_2=(0,-19)[/tex]
The sum of both velocities gives the resultant velocity:
[tex]\vec v =(-34,-19)[/tex]
The magnitude of this velocity is:
[tex]d=\sqrt{(-34)^2+(-19)^2}[/tex]
[tex]d=\sqrt{1156+361}=\sqrt{1517}[/tex]
d = 38.95 m/s
The resultant velocity has a magnitude of 38.95 m/s
A centripetal force of 210 N acts on a 1,600-kg satellite moving with a speed of 5,500 m/s in a circular orbit around a planet. What is the radius of its orbit?
Answer:
230476.19kmExplanation:
Step one:
given
Force F= 210N
mass m= 1600kg
velocity v=5500m/s
Step two
Required is the radius r
the expression for the force is
[tex]F_c = \frac{mv^2}{r}[/tex]
substitute
210=1600*5500^2/r
cross multiply we have
210r=48400000000
divide both side by 210
r=230476190.476m
r=230476.19km
List Five examples from daily life in which you see periodic motion caused by a pendulum
(Marking Brainliest)
Answer:
by a rocking chair, a bouncing ball, a vibrating tuning fork, a swing in motion, the Earth in its orbit around the Sun, and a water wave.
Explanation:
Akia is balancing the equation Na + H2O NaOH + H2. He tries to find the coefficients that will balance the equation. How could he find the correct coefficients? by counting each individual atom and making sure the number of each kind of atom is the same in the reactants and the products by counting the total atoms and making sure the number of atoms in the reactants is the same as the number of atoms in the products by counting the total mass of each compound and making sure the reactants have more mass than the products by counting the mass of each atom and making sure the reactants are more massive than the products
Answer:
by counting each individual atom and making sure the number of each kind of atom is the same in the reactants and the products. - This is the answer.
Explanation:
Answer:
its A(by counting each individual atom and making sure the number of each kind of atom is the same in the reactants and the products
Explanation:
A copper collar is to fit tightly about a steel shaft whose diameter is 6.0000 cm at 19°C. The inside diameter of the copper collar at that temperature is 5.9800 cm. To what temperature must the copper collar be raised so that it will just slip on the steel shaft, assuming the temperature of both the steel shaft and copper collar are raised simultaneously?
Answer:
T' = 865.15 °C
Explanation:
In order for the copper collar to just slip on the steel shaft the, assuming are heated simultaneously, we must find the final parameters of both and equate them. Because the final diameters of both must be same for the slipping to occur.
FOR COPPER COLLAR:
dc' = dc(1 + ∝c*ΔT)
where,
dc' = final diameter of copper ring
dc = initial diameter of copper ring = 5.98 cm
∝c = coefficient of linear expansion for copper = 16 x 10⁻⁶ °C⁻¹
ΔT = Change in Temperature
Therefore,
dc' = (5.98 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT] ------------- equation (1)
FOR STEEL SHAFT:
ds' = ds(1 + ∝s*ΔT)
where,
ds' = final diameter of steel shaft
ds = initial diameter of steel shaft = 6 cm
∝s = coefficient of linear expansion for steel = 12 x 10⁻⁶ °C⁻¹
ΔT = Change in Temperature
Therefore,
dc' = (6 cm)[1 + (12 x 10⁻⁶ °C⁻¹)ΔT] ------------- equation (2)
Comparing equation (1) and equation (2):
(5.98 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT] = (6 cm)[1 + (12 x 10⁻⁶ °C⁻¹)ΔT]
(5.98 cm/6 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT] = [1 + (12 x 10⁻⁶ °C⁻¹)ΔT]
0.9967 + (1.59 x 10⁻⁵ °C⁻¹)ΔT = 1 + (12 x 10⁻⁶ °C⁻¹)ΔT
1 - 0.9967 = [(15.9 -12) x 10⁻⁶ °C⁻¹]ΔT
0.0033/3.9 x 10⁻⁶ °C⁻¹ = ΔT
ΔT = 846.15 °C
but,
ΔT = T' - T = T' - 19°C = 846.15°C
T' = 846.15 °C + 19 °C
T' = 865.15 °C
A body of moment of inertia I=0.80kgm2 about a fixed axis, rotating with constant angular velocity of 100 rad s-1, then torque acting on it will be:
a)80Nm
b)zero
c)160Nm
d)120Nm
In Figure 1 , an engine that delivers power P is attached to a pulley that lifts a box of mass m0 a vertical distance of y0 at a constant speed in time t0 . If, as shown in Figure 2 the engine again delivers power P , how long would it take the engine to raise a box of mass 3m0 the same vertical distance y0 at a constant speed?
Options:
1/3 t0
t0
root(3 t0
3t0
9t0
Answer:
t = 3 t₀
Explanation:
The formula for the power delivered by the engine can be given in terms of the change in the potential energy of the box as follows:
P = P.E/t = mgy/t
where,
P = Power delivered by the engine
m = mass of box lifted = m₀
g = acceleration due to gravity
y = vertical distance or height box covered = y₀
t = time taken = t₀
Therefore,
P = m₀gy₀/t₀
t₀ = m₀gy₀/P -------------- equation (1)
Now, if we change the mass of box as:
m = 3m₀
y = y₀ (constant)
P = Constant
Then,
P = 3m₀gy₀/t
t = 3m₀gy₀/P
using equation (1):
t = 3 t₀
Answer:
t=3to
Explanation:
I took the quiz it is right
Which phrase desenbes an irregular galaxy ?
has a round shape
contains many young stars
has arms that extend from the center
Is larger than other types of galaxies
Answer:
contains many young stars
Explanation:
Irregular galaxies have no definite shape, which means that the first option is incorrect. They are definitely not round.
However, they contain many young stars because the degree of star formation is fast. They also contain old stars. Thus, the second choice is correct.
The "spiral galaxy" is the type of galaxy that has arms that extend from the center. These arms look "spiral," which influenced its name. This makes the last choice incorrect.
They are actually smaller than the other types of galaxies. This makes them prone to collisions. This makes the last choice incorrect.
Answer:
Contains many young stars
Explanation:
the resistance of a heater element is 1200 ohms and draws a circuit of 0.4 amperes calculate the voltage of the circuit
Answer:
0.00033333333
Explanation:
A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.90 m above the pool.
Required:
a. What is her highest point above the board?
b. How long a time are her feet in the air?
c. What is her velocity when her feet hit the water?
Answer:
a = 1.152s
b = 0.817 m
c = 7.29m/s
Explanation: let the following
From the first equation of linear motion
V = u+at..........1
parameters be represented as :
t = Time taken
v = Final velocity
a = Acceleration due to gravity = 9.8m/s²
u = Initial velocity = 4 m/s
s = Displacement
V = 0
Substitute the values into equation 1
0 = 4-9.8(t)
-4 = -9.8t
t = 4/9.8
t = 0.408s
From : s = ut+1/2at^2.........2
S = 4×0.408+0.5(-9.8)×0.408^2
S= 1.632-4.9(0.166)
S = 1.632-0.815
S = 0.817m
Her highest height above the board is 0.817 m
Total height she would fall is 0.817+1.90 = 2.717 m
From equation 2
s = ut+1/2at^2
2.717 m = 0t+0.5(9.8)t^2
2.717 m = 0+4.9t^2
2.717 m = 4.9t^2
2.717/4.9 = t^2
0.554 =t^2
t =√0.554
t = 0.744s
Hence, her feet were in the air for 0.744+0.408seconds
= 1.152s
Also recall from equation 1
V= u+at
V = 0+9.8(0.744)
V = 7.29m/s
Hence, the velocity when she hits the water is 7.29m/s
Finally,
a = 1.152s
b = 0.817 m
c = 7.29m/s
A distant galaxy has a redshift z = 5.82 and a recessional velocity vr = 287,000 km/s (about 96% of the speed of light.) Notice that the equation z=vrcz=vrc does not hold true for recessional velocities approaching the speed of light. What is the distance to the galaxy in light years?
Answer: 4100 Mpc
Explanation:
Since H o = 70 km/s/Mpc
Redshift z = 5.82
Recessional velocity vr = 287,000 km/s
Then, the distance to the galaxy in light years will be:
= Recessional velocity / H o
= 287000 / 70
= 4100 Mpc
A scientist decides to replicate an experiment completed by another scientist. Which statement describes something that would not affect the results of the replicated experiment
Hi, you've asked an incomplete question. However, I provided some explanation about the replication process that scientists do.
Explanation:
Replication in research involves carefully repeating an original experiment to see whether the same result would be arrived at as in previous research experiments.
For most scientists today, in other to avoid anything that would erroneously affect the results of the replicated experiment they usually follow the same procedures as carried by the previous researchers.
What average force is required to stop an 1100-kg car in 8.0s if the car is travelling at 95km/h?
Answer: 13062.5 N
What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 95 km/h? i got the answer to be 13062.
Answer:
The answer is 13062.5 Newtons (N).
Explanation:
The mass of the car is 1100 kg.
The acceleration is 95 km/h.
Using this information, we can use Newton's 2nd Law, F=MA.
1100 kg * 95km/h = 104500 N
Because the answer wants average force, we need to divide the answer by 8 seconds, giving us 13062.5 N.
104500 N / 8 seconds = 13062.5 N (avg force).
A projectile is fired with an initial speed of 36.6 m/s at an angle of 42.2° above the horizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total horizontal distance covered (that is, the range), and (d) the speed of the projectile 1.50 s after firing.
Answer:
Explanation:
a) Maximum height id expressed as;
H = u²sin² theta/2g
H = 36.6²(sin42.2)²/2(9.8)
H = 1,339.56(0.6717)/19.6
H = 899.81/19.6
H = 45.91m
Hence the maximum height is 45.91m
b) The Time of flight is the total time in air expressed as;
T = 2usin theta/g
T = 2(36.6)sin42.2/9.8
T = 73.2sin42.2/9.8
T = 49.17/9.8
T = 5.017secs
Hence the total time in air is 5.017scs
c) Range = U²sin2(theta)/g
Range = 36.6²sin2(42.2)/9.8
Range = 1,339.5(0.9952)/9.8
Range = 1,333.11/9.8
Range = 136.03m
Hence the range is 136.03m
d) USing the rime of flight formula;
T =2Usintheta/g
1.5 = 2Usin42.2/9.8
2Usin42.2 =1.5*9.8
2Usin42.2 = 14.7
U = 14.7/2sin42.2
U = 14.7/1.3434
U = 10.94m/s
Hence the speed of the projectile is 10.94m/s
A cheetah can maintain a maximum constant velocity of 34.2 m/s for 8.70 s. What is
the displacement the cheetah covered at that velocity?
Answer:
297.54mExplanation:
step one:
given data
velocity v=34.2m/s
time t= 8.7s
Step two
Required is the distance the cheetah has covered on the condition
we know that speed= distance/time
make distance subject of formula we have
distance= velocity *time
distance= 34.2*8.7
distance = 297.54m
Therefore the displacement the cheetah covered at that velocity
is 297.54m
A velocity vector has magnitude 100.0 m/s and make an angle of 160o with the positive x-axis. Determine the x- and y-components of the vector.
Answer:
Vₓ = -93.96 m/s
Vy = 34.2 m/s
Explanation:
The x-component of the velocity vector can be given as follows:
Vₓ = V Cos θ
where,
Vₓ = x-component of velocity vector = ?
V = Magnitude of Velocity Vector = 100 m/s
θ = Angle with positive x-axis = 160°
Therefore,
Vₓ = (100 m/s)Cos 160°
Vₓ = -93.96 m/s
The y-component of the velocity vector can be given as follows:
Vy = V Sin θ
where,
Vy = y-component of velocity vector = ?
V = Magnitude of Velocity Vector = 100 m/s
θ = Angle with positive x-axis = 160°
Therefore,
Vy = (100 m/s)Sin 160°
Vy = 34.2 m/s
Question 5 of 5
What is a definite sign of overtraining?
O A. Depression
B. Declining athletic performance
O C. A ravenous appetite
O D. Big bursts of energy
( the answer is B. i just put it on here because i didn’t see it and i got it correct :)
Suppose a popular FM radio station broadcasts radio waves with a frequency of 96. MHz. Calculate the wavelength of these radio waves. Be sure your answer has the correct number of significant digits.
Answer:
3.125 meters.
Explanation:
(3.0*10^8)/(96*10^6)
= 3.125 meters.
Hope this helped!
Assuming the acceleration due to gravity on the moon is exactly one-sixth of the acceleration due to gravity on Earth, what is the weight of the object on the moon?
I have assumed a weight of 120 N on Earth.
Answer:
The object weighs 20 N on the moon
Explanation:
Weight
The weight of an object depends on the mass m of the object and the acceleration of gravity g of the place they are in.
The formula to calculate the weight is:
W = m.g
If g_e is the acceleration of gravity on Earth, and g_m is the acceleration of gravity on the moon, we know:
[tex]g_m=1/6 g_e[/tex]
Dividing by ge:
[tex]g_m/g_e = 1/6[/tex]
An object of weight We=120 N on planet Earth has a mass of:
[tex]m = 120 / g_e[/tex]
Multiplying by gm:
[tex]m.g_m=120 g_m/g_e[/tex]
Substituting the ratio of accelerations of gravity:
[tex]m.g_m=120 * 1/6[/tex]
Since m.gm is the weight on the Moon Wm:
[tex]W_m=20~N[/tex]
The object weighs 20 N on the moon
Assuming the acceleration due to gravity on the moon is exactly one-sixth of the acceleration due to gravity on Earth, the weight of the object on the moon would be one-sixth of its weight on Earth.
Knowing an object's weight on Earth and applying the ratio of the acceleration caused by gravity on the moon to that on Earth will allow us to determine an object's weight on the moon.
Assume that the object's weight on Earth is W (in newtons).
On Earth, the acceleration due to gravity = 9.8 [tex]m/s^2[/tex].
On the moon, = (1/6) * 9.8 [tex]m/s^2[/tex] = 1.6333 [tex]m/s^2[/tex]
Weight = Mass * Acceleration due to gravity
Weight on the moon = W * (1/6)
Thus, the weight of the object on the moon would be one-sixth of its weight on Earth.
For more details regarding gravity, visit:
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Two solid spheres are made from the same material, but one has twice the diameter of the other. Which sphere will have the greater bulk modulus?
Answer:
It will be the same for both
Explanation:
from this question we have one similarity between these two spheres.
- they are both made from the same material,
The difference between both spheres is that:
- one of the spheres has its diameter to be twice as large as that of the other one.
We are to say the sphere with the greater bulk modulus.
If the material is the same thenthe Bulk modulus is also the same. It is not dependent on the material since it is a constant for that materia
Therefore the correct answer is:
It will be the same for both spheres.
What happens when calcium reacts with chlorine?
A. Electrons become delocalized among the atoms
B. Electrons are shared between the calcium atoms and the chlorine
atoms
C. Electrons move from the chlorine atoms to the calcium atoms,
D. Electrons move from the calcium atoms to the chlorine atoms,
When calcium reacts with chlorine. Electrons move from the calcium atoms to the chlorine atoms. Option D is correct.
What is a chemical reaction?When one or more chemicals are transformed into one or more other compounds, a chemical reaction occurs.
In an ionic compound, calcium chloride is present. To build a complete outer shell of electrons, the calcium atom loses two electrons and each chlorine atom obtains one.
When calcium combines with chlorine, it forms calcium chloride. From the calcium atoms to the chlorine atoms, electrons travel.
Hence, option D is correct.
To learn more about the chemical reaction refer to the link;
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An ant can run at an average speed of 0.083 meters per second, calculate how long it would take an ant to run 100 meter dash in seconds and in minutes
Answer:
The time taken by the ant is 1204.82 seconds or 20.08 minutes
Explanation:
Given:
Speed of the ant (s) = 0.083 m/s
Distance traveled by the ant (D) = 100 m
We know that distance traveled by a body is equal to the product of the speed of the body and the time taken for its travel.
Let the time taken by the ant be 't' seconds.
Now, as per the formula:
Distance = Speed × Time
⇒ 100 m = 0.083 m/s × t
⇒ [tex]t = (100\ m)/(0.083\ m/s)[/tex]
[tex]\therefore t=1204.82\ s[/tex]
Now, we know that,
60 seconds = 1 minute
So, 1 second = [tex]\frac{1}{60}\ min[/tex]
Therefore, [tex]1204.82\ s = \frac{1204.82}{60}=20.08\ min[/tex]
Hence, the time taken by the ant is 1204.82 seconds or 20.08 minutes