Answer:
i think it should be to make life easier
''The Evolution Of Sakura And Hinata''
Who looks better?! Who do you think has improved a lot in the Naruto Series?
Answer:
hinata
Explanation:
unlike sakura she has not change at all
In my opinion, Sakura looks prettier and she developed way more than Hinata. Personally, I dislike the both of them because of their Tsundere side, but there's no doubt Sakura developed way more than Hinata. In the original Naruto series, Sakura was honestly a burden. There were some missions where Team 7 had to rescue Sakura. However, after the training with Tsunade, she gained monster strength and healing abilities. Additionally, throughout Naruto Shippuden she learned new jutsu like Yin Seal: Release.
However with Hinata, it's a bit more different. In the original Naruto series, I feel believe she was stronger than Sakura. She relied on her infamous visual prowess--the Byakugan. However in Naruto Shippuden, she improved slightly compared to Sakura. In my opinion, she would have been the most useless shinobi out of the eleven characters (YES, more useless than Tenten) if it weren't for her visual prowess. She's fortunate that she was born into the Hyuga clan. Every jutsu that's in her arsenal relies heavily on the Byakugan. And if Sakura and Hinata were to battle each other, it's most likely that Sakura would win. It's true that Hinata can block Sakura's chakra points, but I think she could dodge her attacks. And even if her chakra points are blocked, Sakura can rely on Taijutsu and raw strength to win.
Though both kunoichi are honorable, I believe Sakura is more prettier and stronger than Hinata.
Why do stretching exercises increase flexibility more than cardio exercises?
An electron is accelerated from rest through a potential difference of 300V. it then passes through a uniform 0.001-T magnetic field, oriented perpendicular to the electrons velocity. what is the magnitude of the magnetic force on the electron?
Given :
Potential difference, V = 300 V.
Magnetic Field, B = 0.001 T.
To Find :
The magnitude of the magnetic force on the electron.
Solution :
We know, for perpendicular orientation, force is given by :
[tex]F = qVB\\\\F = 1.6 \times 10^{-19} \times 300\times 0.001\ N\\\\F = 4.8\times 10^{-20}\ N[/tex]
Hence, this is the required solution.
A 8.45μC particle with a mass of 6.15 x 10^-5 kg moves perpendicular to a 0.493-T magnetic field in a circular path of radius 34.1 m. How much time will it take for the particle to complete one orbit?
a. 92.7 s
b. 0.0927 s
c. 9.27 s
d. 927 s
This question is incomplete, the complete question is;
A 8.45μC particle with a mass of 6.15 x 10⁻⁵ kg moves perpendicular to a 0.493-T magnetic field in a circular path of radius 34.1 m.
How much time will it take for the particle to complete one orbit?
a. 92.7 s
b. 0.0927 s
c. 9.27 s
d. 927 s
Answer:
it will take 92.7 seconds for the particle to complete one orbit.
Option a) 92.7 s is the correct option
Explanation:
Given that;
mass m = 6.15 x 10⁻⁵ kg
q = 8.45μC = 8.45 × 10⁻⁶ C
B = 0.493
we know that
Time period T = 2πr / V
where r = mv/qB
so T = 2πm/qB
we substitute
T = (2 × 3.14 × 6.15 x 10⁻⁵) / ( 8.45 × 10⁻⁶ × 0.493)
T = 0.0003862 / 0.000004165
T = 92.7 sec
Therefore it will take 92.7 seconds for the particle to complete one orbit.
Option a) 92.7 s is the correct option
A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?
Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.
A crossbow is fired horizontally off a cliff with an initial velocity of 15 m/s. If the arrow takes 4s to hit the ground, what is the range of the projectile?
Answer:
The range of the projectile is 60 m
Explanation:
Horizontal Motion
When an object is thrown horizontally with a speed vo from a height h, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.
The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:
[tex]v_x=v_o[/tex]
The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:
[tex]v_y=g.t[/tex]
The horizontal distance is calculated as a constant speed motion:
[tex]x = v_x.t[/tex]
Knowing the crossbow is fired horizontally at vo=vx=15 m/s and it takes t=4 s to hit the ground, thus the range of the projectile is:
x = 15*4 = 60
The range of the projectile is 60 m
Which of the following is NOT a type of variable?
A. Burette
B. Ordered
C. Continuous
D. Categoric
Answer:
A
Explanation:
because its not trust me :)
Elevator is accelerating upward 3.5 M/S2 and has a mass of 300 KG. The force of gravity is 2940 N. What is the tension force pulling elevator up?
Answer:
T = 3990 N
Explanation:
The free body diagram for the elevator consists of a tension force pointing up, and its weight pointing down. So the elevator's net force is:
F = T - 2940N
ad at the same time, using Newton's second law, we have that this net force should equal the elevator's mass (300 kg) times its acceleration (a):
T - 2940N = 300kg (3.5m/s^2)
then
T = 2940 N + 1050 N
T = 3990 N
A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) (4.00 m/s2 ) .At that instant and in unit-vector notation, what is the acceleration of the wallet
Complete Question:
A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.
They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet
Answer:
aw = 3 i + 6 j m/s2
Explanation:
Since both objects travel in uniform circular motion, the only acceleration that they suffer is the centripetal one, that keeps them rotating.It can be showed that the centripetal acceleration is directly proportional to the square of the angular velocity, as follows:[tex]a_{c} = \omega^{2} * r (1)[/tex]
Since both objects are located on the same radial line, and they travel in uniform circular motion, by definition of angular velocity, both have the same angular velocity ω.∴ ωp = ωw (2)
⇒ [tex]a_{p} = \omega_{p} ^{2} * r_{p} (3)[/tex]
[tex]a_{w} = \omega_{w}^{2} * r_{w} (4)[/tex]
Dividing (4) by (3), from (2), we have:[tex]\frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}[/tex]
Solving for aw, we get:[tex]a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2[/tex]
F = 5 Newtons
W = 75 Joules
d = ?
ANSWER
What is the average power consumption in watts of an appliance that uses 4.69 kW · h of energy per day?
Answer:
The average power is [tex]P = 195 .42 \ W[/tex]
Explanation:
From the question we are told that
The energy of the appliance is [tex]E = 4.69 \ kWh = 4.69 *10^{3} \ \ Wh[/tex]
The time considered is [tex]t = 1 \ day = 24 \ hours[/tex]
Generally the average power consumption is mathematically represented as
[tex]P = \frac{E}{t}[/tex]
=> [tex]P = \frac{4.69 *10^{3}}{24 }[/tex]
=> [tex]P = 195 .42 \ W[/tex]
Mr. Jones starts from rest and begins to accelerate straight to the bathroom at a rate of 0.5 m/s for 10 seconds. What kind of motion is this
A. linear
b centripetal
c.free fall
d projectile
Answer:
A. linear
Explanation:
because they are going in a straight line
Discuss two ways to determine your muscular strength explain the advantages
Answer:
Two ways to determine ones muscular strength is to either lift the heaviest weight possible or by doing one repetition. The advantages these two methods is that it is easy to determine and does not require calculations or estimating.
Explanation:
Answer:
Two ways to determine ones muscular strength is to either lift the heaviest weight possible or by doing one repetition. The advantages these two methods is that it is easy to determine and does not require calculations or estimating.
Explanation:
A football player runs down the field at a speed of 8 m/s how long will it take him to run 20 m?
When a mass of a cart is 10 kg, and an applied force is 5 N, The acceleration of the cart is
5.0 m/s2
2.0 m/s2
0.5 m/s2
0.2 m/s2
Answer:
0.5 m/s2
Explanation:
F = ma
5 = 10a
a = 5/10
a =0.5
Objects falling through air are slowed by the force of air resistance. Which objects were slowed the most by air resistance?
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, it should be noted that when objects of different sizes fall in absence of air resistance, the objects will get to the ground at the same time. But with the presence of air resistance, the heaviest object gets to the ground first; meaning it has the least air resistance while the lightest object will arrive at the ground last because it has the greatest air resistance and is slowed down the most by the air resistance.
Thus, the lightest object in the completed question is the answer.
Two stars in a faraway part of the Milky Way are orbiting each other as a binary star system. By
careful measurement we find out that they are separated by 1.7 AU. We also determine that
their orbital period is 594 Earth-days. The total mass of the two-object system is
Answer: [tex]M_{total}=[/tex] 1.85
Explanation: Estimate the total mass of a binary system is done by a reformulation of Kepler's Third Law, which states that the square of the period of a planet's orbit is proportional to the cube of its semimajor axis, i.e.:
[tex]a^{3}=(M_{1}+M_{2})P^{2}[/tex]
where
a is semimajor axis in astronomical units (AU);
P is period measured in years;
[tex]M_{1}+M_{2}[/tex] is total mass of the two-stars system;
For the two stars faraway in the Milky Way:
1 year is equivalent of 365 days, so period in years:
[tex]P=\frac{594}{365}[/tex]
P = 1.63 years
Calculating total mass:
[tex]a^{3}=(M_{total})P^{2}[/tex]
[tex]M_{total}=\frac{a^{3}}{P^{2}}[/tex]
[tex]M_{total}=\frac{1.7^{3}}{1.63^{2}}[/tex]
[tex]M_{total}=[/tex] 1.85
The total mass of the two-object system is 1.85 mass units.
This is a short question can anyone help me please
Thank you
Picture Above
Answer:
I thinks it's
deficit spending
Explanation:
cause When a government spends more than it collects in taxes, it is said to have a budget deficit.
Net force causes motion
Answer:
yes
Explanation:
9. In the graph below, what is the force being exerted on
the 16-kg cart?
A. 4N
C. 16N
B. 8N
D. 32 N
Answer:
D
Explanation:
A bullet of mass 4.2 g strikes a ballistic pendulum of mass 1.0 kg. The center of mass of the pendulum rises a vertical distance of 18 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.
Answer:
449.020m/s
Explanation:
We have been provided with the following information:
Mass of bullet = Mb = 4.2g
Mass of pendulum = 1.0 = Md
Vertical distance = h = 18cm
Gravitational force = g = 9.8
We have kinetic energy converted to potential energy for the entire system
1/2(Mb+Md)V² = (Mb + Md)gh
We have V as the speed of system during collision
Mbv = (Mb+Md)V
We divide through by Mb
v = (Mb+Md/Mb)√2gh
4.2x10^-3 = 0.0042
18cm = 0.18m
(0.0042+1.0/0.0042)√2x9.8x0.18
= 1.0042/0.0042√3.528
= 239.095x1.878
= 449.020m/s
This is the answer to the question
Thank you!
A car is moving at an average speed of 20 meters per second. This is equivalent to
Answer:
44.73 MP/H or 71.98 KM/H
Explanation:
A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force) acts on her. The student has a mass of 71.0 kg, and the height of the water slide is 12.7 m. If the kinetic frictional force does -5.10 103 J of work, how fast is the student going at the bottom of the slide
Answer:
[tex]10.27m/s[/tex]
Explanation:
Given data
work W= -5.10 10^3 J
mass m= 71kg
final height of slide h2= 12.7m
initial height of slide h1=0m
initial velocity v1= 0m/s
final velocity v2=?
Step two:
required
Final velocity
The work-energy theorem is expressed as'
[tex]W=1/mv_2^2 +mgh_2-(1/mv_1^2+mgh_1)[/tex]
make V2 subject of formula we have final speed
[tex]v_2=\sqrt{\frac{2W}{m}+v_1^2-2g(h_1-h_2) } \\\\[/tex]
substitute our given data we have
[tex]v_2=\sqrt{\frac{2*(-5.1*10^3)}{71}+0^2-2*9.81(12.7) } \\\\v_2=\sqrt{143.66-249.174 } \\\\v_2=\sqrt{105.514 } \\\\v_2=10.27m/s[/tex]
The student going at 10.27m/s
How do unbalanced forces acting on an object affect its motion when the object is at rest? What if it is moving?
Answer:
It pushes it because an unbalanced force is pushing more newtons than something that isn't even moving. Even if it is moving, it depends which side is pushing/pulling the most force.
Explanation:
Answer:
It pushes it because an unbalanced force is pushing more newtons than something that isn't even moving. Even if it is moving, it depends which side is pushing/pulling the most force.
Are the refractive index and the speed of light in a vacuum direct propotional or inversley
The refractive index of the medium is inversely proportional to the velocity of light in it. As the refractive index of a medium increases, the speed of light going through that medium decreases.
A rack of seven spherical bolwing balls (each 7.00 kg, radius of 9.50 cm) is positioned along a line located a distance =0.850 m from a point ,
Calculate the gravitational force
the bowling balls exert on a ping-pong ball of mass 2.70 g, centered at point .
Answer:
8.72*[tex]10^{-12}[/tex] N
Explanation:
force of attraction f = G m1m2/ r^2 = [tex]\frac{6.67*10^{-11}*7*5*2.70 }{.85*.85*1000}[/tex] = 8.72*[tex]10^{-12}[/tex] N
The gravitational force the bowling balls exert on a ping-pong is given as 8.72*[tex]10^{-12}[/tex] N.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
force of attraction
f = G [tex]m_1m_2[/tex]/r²
= [tex]6.67*10^{-11}[/tex]*7*5*2.70/.85²*1000
= 8.72*[tex]10^{-12}[/tex] N
The gravitational force the bowling balls exert on a ping-pong is given as 8.72*[tex]10^{-12}[/tex] N.
To learn more about force refer to the link:
brainly.com/question/13191643
#SPJ3
In the concluding paragraph, how does the author make the conflicting point of view that vegetarianism is strange and unnatural seem less reasonable?
Answer:
By pointing out that vegetarianism isn't contagious
Explanation:
I got this right on iready!
Answer:
by point out how food isn't contagious
Explanation:
got it right
A 500-N box is at rest on the floor. Dennis Elbo makes several
attempts to move the box, pushing against the box with varying
amounts of horizontal force. Yet the box never does move. In this
situation, the amount of static friction force experienced by the box
Select all that apply.
-
0 is 500 N
O is equal to the force with which Dennis exerts on the
box
has an upper limit and Dennis O has not yet exceeded the upper limit
Ois always the coefficient of friction multiplied by the normal force value
Answer:
Select the second and the third options you listed.
Explanation:
Select the answer options:
"is equal to the force with which Dennis exerts on the box."
and
"has an upper limit and Dennis has not yet exceeded the upper limit."
In fact, this upper limit of the static friction force is the product of the coefficient of static friction ([tex]\mu[/tex]) times the weight of the box.
when you stir a cup of tea the floating clip collect at the centre of the Cup rather than in the outer Rim why
Answer:
hsvshxansjusjsnwjwisks
Explanation:
When we stir a cup of tea we create a force in the center which pulls out all the particles towards it this is the basic reason for collection of tea leaves at the center of the cup rather than at the rim of the cup, it is similar to the the case of tornado where it takes all the particles present on it way to its ..
find the fundamental units involved in derived units
newton
watt
joule
pascal
cubic meter
9ycy8c8t 7f fixfuozofuxt8lsrupsurpaurae6pUeoUe6eoUeFipzuroz6d0, 7d0z6e0z7e0zurpz6e0z
Explanation:
force newton N - m·kg·s-2
pressure, stress pascal Pa N/m2 m-1·kg·s-2
energy, work, quantity of heat joule J N·m m2·kg·s-2
power, radiant flux watt W J/s m2·kg·s-3
volume cubic meter m3