Answer:
Gordon Allport’s
Explanation:
edge2o2o
The cardinal, central, and secondary traits are all part of Gordon Allport’s categorized traits. The Correct option is A
Who was Gordon Allport ?
Gordon Willard Allport was born on 11 November 1897 and died 9 October 1967. He was an American psychologist. Allport was first psychologists who studied on personality. he has developed theory of personality. which was one of the greatest finding in the study of personality psychology. He was Appointed as a social science instructor at Harvard University in 1924,
Gordon Allport was a great trait theorist who categorized personality traits into three categories cardinal, central, and secondary.
Hence option A is Correct.
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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:
what makes the ball stop on rolling after somtime
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
_______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:
:)
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 small object moves along the x-axis with acceleration ax(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.
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 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
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:
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
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
A circular conducting loop with a radius of 1.00 m and a small gap filled with a 10.0 Ω resistor is oriented in the xy-plane. If a magnetic field of 2.0 T, making an angle of 30º with the z-axis, increases to 11.0 T, in 2.5 s, what is the magnitude of the current that will be caused to flow in the conductor?
Answer:
ill get back to this question once i find the answer to it
plzzzzzzzzzzzzzzzzzzzzzzzzzz help 20 points
Answer:
1.23
Explanation:
[tex]{\underline{\pink{\textsf{\textbf{ Answer : }}}}}[/tex]
➩ 1.23 feet
[tex]{\underline{\purple{\textsf{\textbf{Explanation : }}}}}[/tex]
Given :
Simon cuts a pipe that was 4.92 feet long Then he cuts it into four equal pieces.To find :
What is the length of the each piece.Solution :
As it is told that it's divided into four equal pieces
Therefore,
We must divide it by 4 to get the length of each piece.
So,
[tex] \sf \to \: \frac{4.92}{4} \\ \sf \to \: 1.23 \: feet \: ans.[/tex]
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 :)
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 box of books weighing 319 N is shoved across the floor by a force of 485 N exerted downward at an angle of 35 degres below the horizontal.a) If the coeficent of friction between the box and the floor is 0.57, how long does it take to move the box 4 meters, starting from rest?b) If If the coeficent of friction between the box and the floor is 0.75, how long does it take to move the box 4 meters, starting from rest?
Let w, n, p, and f denote the magnitudes of the 4 forces acting on the box.
• w = weight = 319 N
• n = normal force
• p = pushing force = 485 N
• f = friction = µ n, where µ is the coefficient of static friction
The net force on the box points in the direction that the box moves, which is to the right. In particular, this means the box is vertically in equilibrium. Split up the vectors into their vertical and horizontal components, and apply Newton's second law. (I take up and right to be the positive vertical and horizontal directions, respectively.)
• vertical:
p sin(-35°) + n - w = 0
and solving for n,
- (485 N) sin(35°) + n - 319 N = 0
n ≈ 597 N
• horizontal:
p cos(-35°) - f = m a
where a is the magnitude of the net acceleration on the box. Solve for a. Since f = µ n and m = w / g (where g = 9.80 m/s² is the mag. of the acc. due to gravity) we get
p cos(35°) - µ n = (w / g) a
(485 N) cos(35°) - µ (597 N) = (319 N) / (9.80 m/s²) a
a ≈ (12.2 - 18.3 µ) m/s²
(a) If µ = 0.57, then the net acceleration on the box is
a ≈ (12.2 - 18.3 • 0.57) m/s² ≈ 1.75 m/s²
so that the time t required to move the box 4 m is
4 m = 1/2 a t ²
t ≈ √((8 m) / (1.75 m/s²))
t ≈ 2.14 s
(b) The box does not move.
If µ = 0.75, then
a = (12.2 - 18.3 • 0.75) m/s² ≈ -1.55 m/s²
but a negative acc. here means the applied acc. points *opposite* the direction of movement, thus making the box move backward which doesn't make sense. The coefficient of friction is too large for the given applied force to get the box moving. With µ = 0.75, the frictional force to overcome has mag. f ≈ 448 N. But the given push contributes a horizontal force of (485 N) cos(-35°) ≈ 397 N. This mag. needs to be increased in order to get the box moving.
(a) The time taken to move the box 4 meters is 2.14 s.
(b) The box will decelerate when the coefficient of friction is 0.75 and cannot be moved to 4 meters forward.
The given parameters;
weight of the book, W = 319 Napplied force, F = 485 Nangle of inclination, θ = 35 ⁰The mass of the books is calculated as;
[tex]m = \frac{W}{g} \\\\m = \frac{319}{9.8} \\\\m = 32.55 \ kg[/tex]
The normal force on the box is calculated as follows;
[tex]F_n = -W - Fsin\theta\\\\F_n = -319 - (485\times sin35)\\\\F_n = -597.18 \ N[/tex]
The frictional force when the coefficient of friction is 0.57;
[tex]F_f = \mu F_n\\\\F_f = 0.57 \times -597.18\\\\F_f = -340.39 \ N[/tex]
The acceleration of the box is calculated as follows;
[tex]F cos \theta - F_f = ma\\\\(485)cos(35) \ - 340.39 = 32.55 a\\\\56.899 = 32.55a\\\\a = \frac{56.899}{32.55} \\\\a = 1.75 \ m/s^2[/tex]
The time taken to move the box 4 meters is calculated as;
[tex]s = v_0t + \frac{1}{2} at^2\\\\s = 0 + \frac{1}{2} at^2\\\\t = \sqrt{\frac{2s}{a} } \\\\t = \sqrt{\frac{2\times 4}{1.75} } \\\\t = 2.14 \ s[/tex]
(b) The frictional force when the coefficient of friction is 0.75;
[tex]F_f = \mu F_n\\\\F_f = 0.75 \times -597.18\\\\F_f = -447.885 \ N[/tex]
The acceleration of the box is calculated as follows;
[tex]F cos \theta - F_f = ma\\\\(485)cos(35) \ -447.885 = 32.55 a\\\\-50.596 = 32.55a\\\\a = \frac{-50.596}{32.55} \\\\a = -1.55\ m/s^2[/tex]
Thus, the box will decelerate when the coefficient of friction is 0.75 and cannot be moved to 4 meters forward.
Learn more here:https://brainly.com/question/21684583
A 12-V battery is connected across a 100-Ω resistor. How many electrons flow through the wire in 1.0 min?
Answer:
The quantity of charge or electron flowing the wire in the given time is 4.5 x 10¹⁹ electrons.
Explanation:
Given;
emf of the battery, V = 12 V
resistance of the resistor, R = 100-Ω
time of current flow, t = 1 min
charge of 1 electron = 1.602 x 10¹⁹ C
The current through this circuit is given by;
I = V / R
I = (12) / (100)
I = 0.12 A
The quantity of charge or electron flowing the wire in the given time is calculated as;
Q =It
where;
I is the current flowing through the wire
t is the time of current flow = 1 x 60s = 60 s
Q = 0.12 x 60
Q = 7.2 C
1.602 x 10⁻¹⁹ C --------------- 1 electron
7.2 C -----------------------------? electron
[tex]= \frac{7.2 }{1.602*10^{-19}} \\\\= 4.5*10^{19} \ electrons[/tex]
Therefore, the quantity of charge or electron flowing the wire in the given time is 4.5 x 10¹⁹ electrons.
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.
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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 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 50kg boy stands on rough horizontal ground. The coefficient
of static friction, us, is 0.68. The maximum static friction
between the boy and the ground is __N.
Given :
A 50 kg boy stands on rough horizontal ground. The coefficient of static friction, us, is 0.68.
To Find :
The maximum static friction between the boy and the ground is _ N.
Solution :
We know maximum static friction is given by :
[tex]F = \mu mg \\\\F= 0.68\times 50\times 9.8\\\\F = 333.2\ N[/tex]
Therefore, maximum static friction is 333.2 N.
Hence, this is the required solution.
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
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!
Earth is about 93 million miles from the Sun but only 239,000 miles from the Moon. Likewise, the Sun has a mass that is over 20 million times that of the Moon. Earth experiences a strong gravitational attraction to both of these bodies, but for two different reasons. Using what you know about the factors that influence gravity, what specific characteristic or characteristics contribute to the attraction between Earth and the Moon? What specific characteristic or characteristics contribute to the attraction between Earth and the Sun?
Answer:
Gravity is dependent on the mass of two bodies and the distance between them. There is a strong gravitational attraction between Earth and the Moon because they’re relatively close to one another. There is a strong gravitational attraction between Earth and the Sun because the Sun is so massive
Answer:
Gravity is determined by the mass of two bodies and their separation. Because the Earth and the Moon are so close to one another, they have a tremendous gravitational attraction. Because the Sun is so large, there is a significant gravitational force between Earth and it.
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 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
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
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.
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
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
Suppose the electric field between two points separated by 4 meters is 37 Volt/m. What is the electric potential (in Volt) between the two points? Use exact numbers; do not estimate.
Answer:
148 V
Explanation:
The electric field can be obtained using the formula;
E= V/d
where;
E is the electric field intensity = 37 volt/m
V is the potential difference (the unknown)
m is the distance of separation = 4m
V = Ed
V = 37 volt/m * 4 m
V= 148 V