Your boat departs from the bank of a river that has a swift current parallel to its banks. If you want to cross this river in the shortest amount of time, you should direct your boat If you want to cross this river in the shortest amount of time, you should direct your boat:_________
a. perpendicular to the current.
b. upstream.
c. downstream.
d. so that it drifts with the current.
Answer:
D.
Explanation:
Given that your boat departs from the bank of a river that has a swift current parallel to its banks. If you want to cross this river in the shortest amount of time, you should direct your boat: so that it drifts with the current.
If the boat moves perpendicular to the current, the current flow will be the resistance to the movement of the boat. So, it's better for the boat to drifts perpendicularly with the current.
The best answer is therefore option D.
Write one reason why we study motion
Answer:
We study motion becuase it is the best way of determining the best method of doing the work by scrutinizing the motions made by the worker or the machine
Hope this helps :)
A deductive argument is always a stronger argument than an inductive argument
O True
False
A Moving to another question will save this
3. Magnitude of a physical quantity is the number of times a standard quantity is present in it.
Why is math so hard?
Answer:
its just built different
I could give you a big, fancy scientific answer, but it often comes down to the fact that the way that most instructors teach math is not the way the student(s) can most easily understand it.
AV ASSIGNMENT what is the length of wire whose resistivity, 3x10^6 ohm with the r=0.2mm with a given Value of 15.5 ohm's Resistance.
18. The displacement of an object moving 330 km North for 2 hours and an additional 220
km North for 5 hours is
a. 110 km North
b. 220 km North
c. 330 km North
d. 550 km North
19. When something falls to the ground, it accelerates. This acceleration is called the acceleration due to
gravity and is symbolized by the letter g. What is the value of g on Earth's surface?
a. 9.8 m/s2
b. 9.8 m/s
c. - 9.8 m/s2
d. about 20 m/s2
Answer:
Explanation:
1) Displacement of the object will be gotten using simply adding the distances since they are goinf in the same direction (positive y direction)
Displacement= 330km + 220km
Displacement = 550km North
2) For a falling body, the body will possess positive acceleration due to gravity because it is falling under the influence of gravitational force. If a body is not under the influence of gravity and is thrown up, such body wont come back down. Hence the value of acceleration due to gravity of a body falling to the ground is +9.8 m/s² (note that the unit of acceleration is m/s²)
What is the horizontal acceleration of a ball that is launched horizontally with a velocity of 5.6 m/s?
O -9.8 m/s2
O 0 m/s2
O 5.6 m/s2
O 9.8 m/s2
Answer:
A
Explanation:
edge 2020
The horizontal acceleration of a ball that is launched horizontally with a velocity of 5.6 m/s will be zero. Option B is correct.
What is projectile motion?The motion of an item hurled or projected into the air, subject only to gravity's acceleration, is known as projectile motion.
The item is known as a projectile, and the course it takes is known as a trajectory. Falling object motion is a simple one-dimensional kind of projectile motion with no horizontal movement.
The projectile's motion is divided into two parts: horizontal and vertical motion.
The horizontal acceleration of a ball that is launched horizontally with a velocity of 5.6 m/s will be zero.
Hence, option B is correct.
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Bob rides his motorcycle with a constant speed of 72km/h. How far can he travel in 4 hours?
Answer:
288 Km/h
Explanation:
72x4=288
therefore 288 is your answer
1. Small hole is made up of a window
shutter of a room, 12.5m and image
of the car outside the room cast on
the opposite if the image is 6.5m
high and the tree is 32.5cm from
the window. What is the high of
the tree?
Answer:
I don't know sorry I hope you find what your looking
Sophia wanted to estimate the product 73 x 12 so she made an estimate of 70x 10. Would her estimate be greater or near the actual estimate?
A football field lies so one endzone is to the east and one is to the west. A running back gets the ball and starts running 20o north of east. After running 10 meters, he sees open field and starts running straight east for 60 meters at which point a safety closes in on him. He cuts so he is running 35o east of south. After running another 20 meters he gets in the end zone. Ultimately we want to know what is the vector that points from where he started to where he scored.
Answer:
( 80.87 i - 12.96 j ) m
Explanation:
Running back movements :
20⁰ north of east for 10 meters,
Runs straight east for 60 meters
runs 35⁰ east of south for 20 meters
A) show vector pointing from the starting point to the end ( where he scored )
The final vector displacement : ( 80.87 i - 12.96 j ) m
which is : 81.90 m, 9.10⁰ south of east
attached below is the required diagram
A skier accelerates from a stop at a rate of 3 m/s2 for 3 seconds. What is her speed at the end of that time?
Answer:
9 m/s
Explanation:
Given:
v₀ = 0 m/s
a = 3m/s²
t = 3 s
Find: v
v = at + v₀
v = (3 m/s²) (3 s) + 0 m/s
v = 9 m/s
The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from the point where the stone leaves your hand? What is the maximum height of a window you can hit at this distance?
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = [tex]v_{oy}[/tex] t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45 )
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = [tex]v_{oy}^2[/tex] - 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
Two forces,
vector F 1 = (5.90î − 5.60ĵ) N
and
vector F 2 = (4.65î − 5.55ĵ) N,
act on a particle of mass 2.10 kg that is initially at rest at coordinates
(−1.75 m, +4.15 m).
(a) What are the components of the particle's velocity at t=10.4s ?
(b) In what direction is the particle moving at t = 10.4 s? (+counterclockwise from the x axis)
(c) What displacement does the particle undergo during the first 10.4 s?
(d) What are the coordinates of the particle at t = 10.4 s? (in x and y meters)
First compute the resultant force F:
[tex]\mathbf F_1=(5.90\,\mathbf i-5.60\,\mathbf j)\,\mathrm N[/tex]
[tex]\mathbf F_2=(4.65\,\mathbf i-5.55\,\mathbf j)\,\mathrm N[/tex]
[tex]\implies\mathbf F=\mathbf F_1+\mathbf F_2=(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N[/tex]
Then use Newton's second law to determine the acceleration vector [tex]\mathbf a[/tex] for the particle:
[tex]\mathbf F=m\mathbf a[/tex]
[tex](10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N=(2.10\,\mathrm{kg})\mathbf a[/tex]
[tex]\mathbf a\approx(5.02\,\mathbf i-5.31\,\mathbf j)\dfrac{\rm m}{\mathrm s^2}[/tex]
Let [tex]\mathbf x(t)[/tex] and [tex]\mathbf v(t)[/tex] denote the particle's position and velocity vectors, respectively.
(a) Use the fundamental theorem of calculus. The particle starts at rest, so [tex]\mathbf v(0)=0[/tex]. Then the particle's velocity vector at t = 10.4 s is
[tex]\mathbf v(10.4\,\mathrm s)=\mathbf v(0)+\displaystyle\int_0^{10}\mathbf a(u)\,\mathrm du[/tex]
[tex]\mathbf v(10.4\,\mathrm s)=\left((5.02\,\mathbf i-5.31\,\mathbf j)u\,\dfrac{\rm m}{\mathrm s^2}\right)\bigg|_{u=0}^{u=10.4}[/tex]
[tex]\mathbf v(10.4\,\mathrm s)\approx(52.2\,\mathbf i-55.2\,\mathbf j)\dfrac{\rm m}{\rm s}[/tex]
If you don't know calculus, then just use the formula,
[tex]v_f=v_i+at[/tex]
So, for instance, the velocity vector at t = 10.4 s has x-component
[tex]v_{f,x}=0+\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)(10.4\,\mathrm s)=52.2\dfrac{\rm m}{\mathrm s^2}[/tex]
(b) Compute the angle [tex]\theta[/tex] for [tex]\mathbf v(10.4\,\mathrm s)[/tex]:
[tex]\tan\theta=\dfrac{-55.2}{52.2}\implies\theta\approx-46.6^\circ[/tex]
so that the particle is moving at an angle of about 313º counterclockwise from the positive x axis.
(c) We can find the velocity at any time t by generalizing the integral in part (a):
[tex]\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du[/tex]
[tex]\implies\mathbf v(t)=\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j[/tex]
Then using the fundamental theorem of calculus again, we have
[tex]\mathbf x(10.4\,\mathrm s)=\mathbf x(0)+\displaystyle\int_0^{10.4}\mathbf v(u)\,\mathrm du[/tex]
where [tex]\mathbf x(0)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m[/tex] is the particle's initial position. So we get
[tex]\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\displaystyle\int_0^{10.4}\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\right)\,\mathrm du[/tex]
[tex]\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\dfrac12\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=10.4}[/tex]
[tex]\mathbf x(10.4\,\mathrm s)\approx(542\,\mathbf i-570\,\mathbf j)\,\mathrm m[/tex]
So over the first 10.4 s, the particle is displaced by the vector
[tex]\mathbf x(10.4\,\mathrm s)-\mathbf x(0)\approx(270\,\mathbf i-283\,\mathbf j)\,\mathrm m-(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m\approx(272\,\mathbf i-287\,\mathbf j)\,\mathrm m[/tex]
or a net distance of about 395 m away from its starting position, in the same direction as found in part (b).
(d) See part (c).
James yap run fast while dribbling the ball as he crosses the middle part of the basketball court he changes his phase to avoid opponent what specific
component of physical fitness did James Yap?
Answer:
The answer is "Speed".
Explanation:
In the question, the Speed is the correct answer because it can be viewed as its rate from, which a length is covered via an object. It is also divided by time into other components of distance. Its SI unit of speed seems to be the meter/second, but still, the kilometer/hour or, in the Us UK, miles per hour is the most basic fundamental of speed in everyday use.
Physical fitness is the body's ability to efficiently work together to enable good health while performing daily living tasks
The correct option is AgilityThe component of physical fitness James Yap applied is agility
The reason agility is the correct physical fitness component is as follows:
Question: The possible options in the question are;
Agility
Speed
Coordination
Agility is the capability to make nimble movement such as starting, accelerating, decelerating, stopping, change ones direction, quickly and easily while proper posture is being maintainedAgility in basketball enables the athlete to restore their center of gravity to a position vertically above their base constantly as their direction of motion is changed both quickly and at average speed during play, requiring coordination and neuromuscular efficiencyTherefore, the component of physical fitness James Yap applied to run fast while dribbling the ball, changing phase (pace), is agility
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New evidence does not change what scientists already believe to be true. Group of answer choices True False
Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each set of Cartesian coordinates to polar coordinates. The angle should be reported as a positive angle, in degrees, from the positive x-axis. (0° < θ < 360°). What is the θ coordinate for point A? Recall this should be in units of degrees.
Answer: A ([tex]\sqrt{61}[/tex],309.8°)
B (2[tex]\sqrt{2}[/tex], 315°)
C ([tex]3\sqrt{5}[/tex], 26.56°)
Explanation: To transform rectangular coordinates into polar coordinates use:
[tex]r=\sqrt{x^{2}+y^{2}}[/tex] and [tex]\theta=tan^{-1}(\frac{y}{x})[/tex]
For point A:
[tex]r=\sqrt{(-5)^{2}+6^{2}}[/tex]
[tex]r=\sqrt{61}[/tex]
[tex]\theta=tan^{-1}(\frac{6}{-5})[/tex]
[tex]\theta=tan^{-1}(-1.2)[/tex]
[tex]\theta=-50.2[/tex]°
Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:
[tex]\theta=360-50.2[/tex]
[tex]\theta=[/tex] 309.8°
Polar coordinates for point A is ([tex]\sqrt{61}[/tex], 309.8°)
For point B:
[tex]r=\sqrt{2^{2}+(-2)^{2}}[/tex]
[tex]r=\sqrt{8}[/tex]
[tex]r=2\sqrt{2}[/tex]
[tex]\theta=tan^{-1}(\frac{-2}{2} )[/tex]
[tex]\theta=tan^{-1}(1)[/tex]
[tex]\theta=-45[/tex]°
Point B is in IV quadrant, so:
[tex]\theta=360-45[/tex]
[tex]\theta=[/tex] 315°
Polar coordinates for point B is ([tex]2\sqrt{2}[/tex], 315°)
For point C:
[tex]r=\sqrt{(-6)^{2}+(-3)^{2}}[/tex]
[tex]r=\sqrt{45}[/tex]
[tex]r=3\sqrt{5}[/tex]
[tex]\theta=tan^{-1}(\frac{-3}{-6} )[/tex]
[tex]\theta=tan^{-1}(0.5)[/tex]
[tex]\theta=[/tex] 26.56°
Polar coordinates for point C is ([tex]3\sqrt{5}[/tex], 26.56°)
when a driver presses the brake pedal, his car stops with an acceleration of -2.1 m/s. How far will the car travel while coming to a complete stop if its initial speed was 13 m/s?
Answer:
40.2 m
Explanation:
what is the magnitude of the average velocity of the hummingbird between 2 sec and t 4 sec
A scientist observes rock masses that have moved past each other in opposite horizontal directions. Which feature
does the scientist observe?
A. plateau
B. syncline
C. strike-slip fault
D. fault-block mountain
Answer:
C. strike-slip fault
Explanation:
The scientist must have observed a strike- slip fault.
A fault is an evidence of brittle deformation of the crust in the presence of applied stress on earth materials. Here, the earth material is the rock subjected to tension.
Where a fault occurs, there must have been movement between two blocks of rocks. The direction of movement helps us to delineate the fault type.
When two blocks moves past each other horizontally, it is a strike-slip fault like rubbing your palms together. When a block moves in the direction of the dip, it forms a dip-slip fault which results in a fault-block mountain characterized by graben and horst systems.Option A, Plateau is a table landform usually a mountain with flat peak.
Option B is a bowl shaped stratigraphic pattern in which the youngest sequence is at the core of the strata or a fold.
So, the most fitting option is C, a strike-slip fault.
Answer:
Explanation:
c
Which term s the product of force and distance?
O power
O work
Onetforce
Oacceleration
Answer:
The answer should be WORK
(Hope this Helps)
Explanation:
work is equal to the product of force and distance.
W=F*S
W=Work
F=Force applied
S=Distance
A car advertisement states that a certain car can accelerate from rest to
40 km/hr in 5 sec. Find the car's average acceleration.
Answer:
8 km/h²
Explanation:
a = v-u/t
v is final velocity, u is initial velocity
a = 40-0/5
a = 8.0 km/h²
Calculate the mass of air in a room if the floor dimensions are 10 mx
12 m and height 4 m. The density of air is 1.26 kgm.
604.8 kg
Explanation:Hi there !
d = m/V => m = d×V
V = l×w×h = 10m×12m×4m = 480 m³
m = 1.26kg/m³×480m³ = 604.8 kg
Good luck !
Suppose that A’, B’ and C’ are at rest in frame S’, which moves with respect to S at speed v in the +x direction. Let B’ be located exactly midway between A’ and C’. At t’=0 a light flash occurs at B’ and expands outward as spherical wave.
1. According to an observer in S’, do the wave fronts arrive at A’ and C’ simultaneously?
2. According to an observer in S, do the wavefronts arrive at A’ and C’ simultaneously?
3. If you answered no to either 1. or 2., what is the difference in their arrival times and at which point did the front arrive first?
Answer:
1) an observer in B 'sees the two simultaneous events
2)observer B sees that the events are not simultaneous
3) Δt = Δt₀ /√ (1 + v²/c²)
Explanation:
This is an exercise in simultaneity in special relativity. Let us remember that the speed of light is the same in all inertial systems
1) The events are at rest in the reference system S ', so as they advance at the speed of light which is constant, so it takes them the same time to arrive at the observation point B' which is at the point middle of the two events
Consequently an observer in B 'sees the two simultaneous events
2) For an observer B in system S that is fixed on the Earth, see that the event in A and B occur at the same instant, but the event in A must travel a smaller distance and the event in B must travel a greater distance since the system S 'moves with velocity + v. Therefore, since the velocity is constant, the event that travels the shortest distance is seen first.
Consequently observer B sees that the events are not simultaneous
3) let's calculate the times for each event
Δt = Δt₀ /√ (1 + v²/c²)
where t₀ is the time in the system S' which is at rest for the events
3) A 10 ohm resistor and a 5 ohm resistor are connected in a series. A 2 A current
runs through the 10 ohm resistor. What is the potential difference in the 5 ohm resistor?
a) 10 V
c) 30 V
b) 20V
d) 40 V
While helping an astronomy professor, you discover a binary star system in which the two stars are in circular orbits about the system's center of mass. From their color and brightness, you determine that each star has the same mass as our Sun. The orbital period of the pair is 31.3 days , based on the oscillation of brightness observed as one star occludes (hides) the other. From this information you are able to ascertain the distance between the stars.
Required:
Calculate the distance between the stars.
Answer:
The value is [tex]d = 3.66 *10^{10} \ m [/tex]
Explanation:
From the question we are told that
The orbital period is [tex]T = 31.3 \ days = 31.3 * 86400 = 2704320 \ s[/tex]
Generally for the centripetal force acting one of the stars will be equal to the gravitational force between the stars and this can be mathematically represented as
[tex]F_g = F_c[/tex]
=> [tex]\frac{Gm*m}{d^2} = \frac{mv^2}{r}[/tex]
Here m represents the mass of each star given that they have same mass and this equal to the mass of the sun whose value is
[tex]m = 1.989*10^{30} \ kg[/tex]
d(2 * r) is the diameter of the orbit which is distance between each star
r is the radius of the orbit
G is the gravitational constant with value [tex]G= 6.67*0^{-11} \ m^2/kg \cdot s^2[/tex]
v is the velocity of the stars which can be mathematically represented as
[tex]v = \frac{2 \pi r}{T}[/tex]
So
[tex]\frac{Gmm }{(2 * r)^2} = \frac{m (\frac{2\pi r}{T} )^2}{r}[/tex]
=> [tex]r =[ \frac{Gm T^2}{16 \pi ^2} ]^{\frac{1}{3} }[/tex]
=> [tex]r = [\frac{6.67 *10^{-11} * (1.989 *10^{30}) * 2704320^2}{16 * (3.142)^2} ]^{\frac{1}{3} }[/tex]
=> [tex]r = 1.83 *10^{10} \ m[/tex]
Generally the distance between the star is
[tex]d = 2 * r[/tex]
=> [tex]d = 2 * 1.83 *10^{10} [/tex]
=> [tex]d = 3.66 *10^{10} \ m [/tex]
on a time line.
Different categories of events should be
O A) grouped together
OB) indistinguishable from one another
O C) distinguished from one another
D) next to one another
Answer:
D
Explanation:
A 100 kg man stands on bathroom scale inside an elevator, on planet where free fall acceleration is 10 m.s^2. Elevator moves down with constant speed 5 m/s up. Reading of the scale is?
A.1500 N
B.1000 N
C.500 N
D.0 N
Answer: i think its A. Let me knows if its wrong so i can double check
Explanation:
The reading on the scale, if the 100 kg man standing inside the elevator, which is moving with a velocity of 5 m/s is 1000 Newtons, so option B is correct.
What is acceleration?Acceleration describes the speed and direction changes in velocity over time. Acceleration refers to the change in speed or direction of an item or points traveling straight forward. Even though the speed is constant, motion on a circle increases because of the constant change in direction.
Both effects add to the acceleration for all other types of motion. Since acceleration has both a magnitude and a direction, it is a vector quantity. Velocity is a vector.
Given:
The mass of man, m = 100 kg,
The speed of the elevator, s = 5 m/s,
The reading is higher when the person is in a standing elevator (or one moving at a steady speed).
w = mg =100*10 = 1000 N
Now:
If the elevator is going down at 5 m/s and the weight will be
W = m(g -a)
W = 100 (10 -0)
Here the acceleration of lift is zero as it is moving with constant speed,
W = 1000 N
Therefore, the reading on the scale, if the 100 kg man standing inside the elevator which is moving with a velocity of 5 m/s is 1000 Newtons.
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Which of the following statements does not represent assumptions that stage theories share?
A.
Each stage builds on the knowledge of the previous stage.
B.
Each stage corresponds to a specific age.
C.
A child can pass through the stages in any order.
D.
Each stage is qualitatively different than the next.
Each stage corresponds to a specific age. is the statements do not represent assumptions that stage theories share.
What is stage theory?Any speculative construct is used to represent phases or steps in a time-based process, such as a belief that development proceeds in discontinuous phases indicated by functional changes.
Each stage corresponds to a specific age the statements do not represent assumptions that stage theories share.
The assumptions for the stage theory are as follows;
1. Every stage serves as a building block for the next.
2. The phases are followed in the same sequence by everyone.
3. Each step is distinct in terms of quality. It is a change in nature, not merely a change in amount.
Each stage corresponds to a specific age. is the statements do not represent assumptions that stage theories share.
Hence option B is the right option.
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