We know that Period of a wave is the inverse of its Frequency
So, Period = 1 / Frequency
From the above, we can say that Period is inversely proportional to Frequency and hence, any change in Frequency will be the inverse change in the period
Therefore, we can say that if the frequency is increased by 5 times, the period will increase by 1/5 times
show all work and round to nearest 100th thank you and you will get braineist
Answer: 16.78 miles
Explanation:
distance = rate * time
we're given the rate ( 5 m/s) and we're given the time it takes to get home (1.5 hrs). But notice how the units of hours don't match the seconds of the rate so we need to convert the hours into seconds.
1 hr = 3600 seconds so 1.5 hours = 5400 seconds
now we can plug it in and solve for the distance
distance = (5 m/s) * 5400 seconds
distance = 27000 m
now we have to convert meters to miles, so we divide our answer by 1609 and get 16.78
A potential difference of 1.20 V will be applied to a 33.0 m length of 18-gauge copper wire (diameter = 0.0400 in.). Calculate (a) the current, (b) the magnitude of the current density, (c) the magnitude of the electric field within the wire, and (d) the rate at which thermal energy will appear in the wire.
Answer:A) Current = 1.739A, B)current density, J = 2.147x10^6 A/m2
magnitude of electric field , E = 0.036 N/C
)rate of thermal energy, P =2.086W
Explanation:
Resistance = R = ρL/A
But the cross-section area of the wire. is given as
Diameter / 2 = 0.04/2 =0.02in to m = 0.02 / 39.37= 0.000508
A = πr^2 = π x 0.000508^2 = 8.10 x 10^-7
since resistivity of copper,ρ= 17x10-9 ohm.m
so resistance is R = ρL/A
17x10-9 x 33 / 8.1x10-7
= 0.69 ohm.
A) Current = I = Voltage /Resistance =1.20/0.69 =1.739A
B)current density, J = Current /Area
= 1.739/8.1x10-7
= 2.147x10^6 A/m2
c)magnitude of electric field , E = Current density x resistivity =J ρ
E = 2.147 x 10^6 x 17 x 10^-9
E = 0.036 N/C
D)rate of thermal energy, P = I² R =1.739² X 0.69
=2.086W
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:
If a penny is dropped from rest from a tower takes 2 seconds to hit the ground, how far did it travel?
a 29.4 m
b 19.6 m
с 6.8 m
d 9.8 m
Answer:
B
Explanation:
t = 2s
u = 0m/s (released from rest)
a = +g = 9.8m/s²
s = H = ?
using,
s = ut + 1/2at²
H = 0(2) + 1/2(9.8)(2²)
H = 0 + 9.8(2)
s = H = 19.6m
''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.
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]
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 40-cm-diameter, 300 g beach ball is dropped with a 4.0 mg ant riding on the top. The ball experiences air resistance, but the ant does not. What is the magnitude of the normal force exerted on the ant when the ball's speed is 4.0 m/s?
Answer:
The normal force exerted on the ant is 0.75 N.
Explanation:
Given;
diameter of the ball, D = 40 cm = 0.4m
radius of the ball, r = 0.2m
mass of the beach ball, m₁ = 300 g = 0.3 kg
mass of the ant, m₂ = 4 x 10⁻⁶ kg
speed of the ball, v = 4 m/s
The area of the ball, assuming spherical ball is given by;
A = 4πr²
A = 4π(0.2)² = 0.5027 m²
The drag force (resistance) experienced by the spherical ball is given as;
[tex]F_D = \frac{1}{2}C\rho Av^2[/tex]
where;
C is the drag coefficient of the spherical ball = 0.45
ρ is density of air = 1.21 kg/m³
[tex]F_D = \frac{1}{2}C\rho Av^2\\\\F_D = \frac{1}{2}(0.45)(1.21) (0.5027)(4)^2\\\\F_D = 2.19 \ N[/tex]
The downward force of the ball due to its weight and that of the ant is given by;
[tex]F_g = mg\\\\F_g =g(m_{ant} + m_{ball})\\\\F_g = g(4*10^{-6} \ kg\ + \ 0.3\ kg)\\\\F_g = g(0.300004 \ kg) \ \ \ (mass \ of \ the \ ant \ is \ insignificant)\\\\F_g = 9.8(0.3)\\\\F_g = 2.94 \ N[/tex]
The net downward force experienced by the ball is given by;
[tex]F_{net} = F_g - F_D\\\\F_{net} = 2.94 \ N - 2.19 \ N\\\\F_{net} = 0.75 \ N[/tex]
This downward force experienced by the ball is equal to the normal reaction it exerts on the ant.
Thus, the normal force exerted on the ant is 0.75 N.
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.
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.
Can someone help me with this question
Answer:
Net force: 20 N to the right
mass of the bag: 20.489 kg
acceleration: 0.976 m/s^2
Explanation:
Since the normal force and the weight are equal in magnitude but opposite in direction, they add up to zero in the vertical direction. In the horizontal direction, the 195 N tension to the right minus the 175 force of friction to the left render a net force towards the right of magnitude:
195 N - 175 N = 20 N
So net force on the bag is 20 N to the right.
The mass of the bag can be found using the value of the weight force: 201 N:
mass = Weight/g = 201 / 9.81 = 20.489 kg
and the acceleration of the bag can be found as the net force divided by the mass we just found:
acceleration = 20 N / 20.489 kg = 0.976 m/s^2
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:
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
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 football player runs down the field at a speed of 8 m/s how long will it take him to run 20 m?
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
A 7.50 kg bowling ball has 70.4
kg•m/s of momentum. What is its
velocity?
Answer:
9.39 m/sExplanation:
The velocity of the bowling ball can be found by using the formula
[tex]v = \frac{p}{m} \\ [/tex]
p is the momentum
m is the mass
From the question we have
[tex]v = \frac{70.4}{7.5} \\ = 9.38666666..[/tex]
We have the final answer as
9.39 m/sHope this helps you
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
15 points.
An object of mass 100 kg is observed to accelerate at a rate of 15
m/s/s. Calculate the force required to produce this acceleration.
Answer:
its 0.5 for all i beleive
Explanation:
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
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.
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
Net force causes motion
Answer:
yes
Explanation:
What two methods are the best choices to factor this expression 18x2-8
Answer:
Please check the explanation
Explanation:
The best two methods will be:
Factor by groupingFactor out the GCFFactor by grouping
Factor by grouping deals with establishing a smaller groups from each term.
[tex]18x^2=\:\left(2\cdot 3\cdot 3\cdot x^2\right)[/tex]
[tex]8\:=\:\:2\cdot 2\cdot 2[/tex]
Therefore, the expression becomes
[tex]18x^2=\:\left(2\cdot 3\cdot 3\cdot x^2\right)-\left(2\cdot \:2\cdot \:2\right)[/tex]
Now factor out the greatest common factor (GCF) which is 2
[tex]=\:2\left(3\cdot \:\:3x^2-\left(2\right)\left(2\right)\right)[/tex]
[tex]=2\left(9x^2-2\cdot \:2\right)[/tex]
[tex]=2\left(9x^2-4\right)[/tex]
Factor out the GCF
Given the expression
[tex]18x^2-8\:\:\:[/tex]
factor out common term 2
[tex]=2\left(9x^2-4\right)[/tex]
[tex]=2\left(3x+2\right)\left(3x-2\right)[/tex] ∵ [tex]Factors\:\:\left(9x^2-4\right)=\left(3x+2\right)\left(3x-2\right)[/tex]
A crate of books rests on a level floor. To move it along the floor at a constant velocity, why do you exert less force if you pull it at an angle Ï above the horizontal than if you push it at the same angle below the horizontal?
Answer:should be a matter of vector analysis.
Pulling above the horizontal has less surface area for the opposing friction
Explanation:
Why do stretching exercises increase flexibility more than cardio exercises?
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 :)
is 2/2 1 or 0? please help lol
Answer:
1.
Explanation:
Hello!
In this case, for such mathematical operations, we can wee that the slash represents a fraction or a division, say 8 ÷ 4 = 2, 6 ÷ 3 = 2, 20 ÷ 4 = 5, etc. In such a way, since the operation 2/2, represents 2 ÷ 2, it is clear that two is once in 2, therefore, the result is:
2 ÷ 2 = 1.
Best regards!
A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of 50 m. If he completes the 200 m dash in 29.6 s and runs at constant speed throughout the race, what is the magnitude of his centripetal acceleration (in m/s2) as he runs the curved portion of the track?
Answer:
The centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s²
Explanation:
Given;
distance traveled in the given time = 200 m
time to cover the distance, t = 29.6 s
speed of the runner, v = d / t
v = 200 / 29.6
v = 6.757 m/s
The centripetal acceleration of the runner is given by;
[tex]a_c = \frac{V^2}{r}[/tex]
where;
r is the radius of the circular arc, given as 50 m
Substitute the givens;
[tex]a_c = \frac{V^2}{r}\\\\a_c = \frac{(6.757)^2}{50}\\\\a_c = 0.91 \ m/s^2[/tex]
Therefore, the centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s².
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.