The term which is described as a seamount with a flattened top is called as guyots.
Guyot, commonly known as Tablemount, is an uninhabited submarine volcano with a flat peak that is more than 200 meters (660 feet) above sea level. These flat summits could be more than 10 kilometres in circumference (6 miles).
Seamounts called guyots have grown above sea level. The seamount's top was eroded by erosion by waves, leaving it with a flattened shape. The sea bottom gradually falls as a result of the ocean floor moving away from oceanic ridges, submerging the flattened guyots to form undersea flat-topped peaks. Once the guyot is formed, on it's top a coral colony will start developing and little by little make a circle shaped island, which represents the formation of the atoll.
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In a car, how does an air bag minimize the force acting on a person during a collision?
Answer:
It increases the time it takes for the person to stop.
Explanation:
Answer:
C: It increases the time it takes for the person to stop.
Explanation:
on edge! hope this helps!!~ o(〃＾▽＾〃)o
Two train whistles have identical frequencies of 220 Hz. When one train is at rest in the station sounding its whistle, a beat frequency of 10.0 Hz is heard from the other train that is approaching the station. What is the speed of the approaching train? Assume that the speed of sound is 340 m/s.
Answer:
Speed of the approaching train = 15.45 m/s
Explanation:
Given:
Frequency F0 = 220 Hz
Beat frequency F1 = 10.0 Hz
Find:
Speed of the approaching train
Computation:
Approaching frequency F2 = 220 + 10.0 Hz
Approaching frequency F2 = 230 Hz
Doppler shift;
F = [(v+v0)/(v-vS)]F0
230 = [(340+v0)/(340-0)]220
V0 = 15.45 m/s
Speed of the approaching train = 15.45 m/s
Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. Suppose the mass of a certain spherical neutron star is twice the mass of the Sun and its radius is 10.0 km. Determine the greatest possible angular speed the neutron star can have so that the matter at its surface on the equator is just held in orbit by the gravitational force.
Answer:
16294 rad/s
Explanation:
Given that
M(ns) = 2M(s), where
M(s) = 1.99*10^30 kg, so that
M(ns) = 3.98*10^30 kg
Again, R(ns) = 10 km
Using the law of gravitation, the force between the Neutron star and the sun is..
F = G.M(ns).M(s) / R²(ns), where
G = 6.67*10^-11, gravitational constant
Again, centripetal force of the neutron star is given as
F = M(ns).v² / R(ns)
Recall that v = wR(ns), so that
F = M(s).w².R(ns)
For a circular motion, it's been established that the centripetal force is equal to the gravitational force, hence
F = F
G.M(ns).M(s) / R²(ns) = M(s).w².R(ns)
Making W subject of formula, we have
w = √[{G.M(ns).M(s) / R²(ns)} / {M(s).R(ns)}]
w = √[{G.M(ns)} / {R³(ns)}]
w = √[(6.67*10^-11 * 3.98*10^30) / 10000³]
w = √[2.655*10^20 / 1*10^12]
w = √(2.655*10^8)
w = 16294 rad/s
Farah is doing an experiment that involves calculating the speed of a longitudinal wave. What will most likely happen if she increases the temperature in the room? The amplitude of the wave will increase. The amplitude of the wave will decrease. The speed of the wave will increase. The speed of the wave will decrease.
Answer:
c
Explanation:
Answer:
C
Explanation:
got it right on edge
A man fires a silver bullet of mass 2g with a velocity of200m/sec into wall. What is the temperature changeof the bullet? Note: Csilver = 234J/kg.°CL
F=nmv
where;
n=no. of bullets = 1
m=mass of bullets=2g *10^-3
V=velocity of bullets200m/sec
F=1
loss in Kinetic energy=gain in heat energy
1/2MV^2=MS∆t
let M council M
=1/2V^2=S∆t
M=2g
K.E=MV^2/2
=(2*10^-3)(200)^2/2
2 councils 2
2*10^-3*4*10/2
K.E=40Js
H=mv∆t
(40/4.2)
40Js=40/4.2=mc∆t
40/4.2=2*0.03*∆t
=158.73°C
A 1.1 kg hammer strikes a nail. Before the impact, the hammer is moving at 4.5 m/s; after the impact it is moving at 1.5 m/s in the opposite direction. If the hammer is in contact with the nail for 0.025 s, what is the magnitude of the average force exerted by the hammer on the nail
Answer:
132 N
Explanation:
Given that a 1.1 kg hammer strikes a nail. Before the impact, the hammer is moving at 4.5 m/s; after the impact it is moving at 1.5 m/s in the opposite direction. If the hammer is in contact with the nail for 0.025 s, what is the magnitude of the average force exerted by the hammer on the nail
From Newton 2nd law of motion,
Change in momentum = impulse.
Change in momentum = m( V - U )
Substitute all the parameters into the formula
Change in momentum = 1.1 ( 4.5 - 1.5 )
Change in momentum = 1.1 × 3
Change in momentum = 3.3 kgm/s
Impulse = Ft
That is,
Ft = 3.3
Substitute time t into the formula above
F × 0.025 = 3.3
F = 3.3 / 0.025
F = 132 N
Therefore, the magnitude of the average force exerted by the hammer on the nail is 132 N.
an athlete whirls an 8.71 kg hammer tied to the end of a 1.5 m chain in a simple horizontal circle where you should ignore any vertical deviations the hammer moves at the rate of 1.16 rev/s what is the tension in the chain
Answer:
T = 692.42 N
Explanation:
Given that,
Mass of hammer, m = 8.71 kg
Length of the chain to which an athlete whirls the hammer, r = 1.5 m
The angular sped of the hammer, [tex]\omega=1.16\ rev/s=7.28\ rad/s[/tex]
We need to find the tension in the chain. The tension acting in the chain is balanced by the required centripetal force. It is given by the formula as follows :
[tex]F=m\omega^2r\\\\=8.71\times (7.28)^2\times 1.5\\\\=692.42\ N[/tex]
So, the tension in the chain is 692.42 N.
Two forces of 50 N and 30N respectively, are acting on an object. Find the net force (in N) on the object if
A) the forces are acting in the same direction.. B) Together, forces are acting in opposite directions
Answer:
A) 80 N
B) 20 N
Explanation:
A) If the forces acting are in the same direction, then the net force will be a sum of both so many faces..
Thus;
ΣF = 50 + 30
ΣF = 80 N
B) If the forces are acting in the in opposite directions with the larger force pointing in the positive y-axis then, the net force is;
ΣF = 50 - 30
ΣF = 20 N
A rocket with total mass 3.00 3 105 kg leaves a launch pad at Cape Kennedy, moving vertically with an acceleration of 36.0 m/s2. If the speed of the exhausted gases is 4.50 3 103 m/s, at what rate is the rocket initially burning fuel?
(a) 3.05 3 103 kg/s
(b) 2.40 3 103 kg/s
(c) 7.50 3 102 kg/s
(d) 1.50 3 103 kg/s
(e) None of these
Answer:
(b) 2.40 x [tex]10^{3}[/tex] kg/s
Explanation:
Given that: Total mass of the rocket = 3.003 x [tex]10^{5}[/tex] kg
acceleration of the rocket = 36.0 m[tex]s^{-2}[/tex]
speed of the exhausted gases = 4.503 x [tex]10^{3}[/tex] m/s
Rate at which rocket was initially burning fuel = [tex]\frac{mass}{time}[/tex]
But,
time = [tex]\frac{velocity}{acceleration}[/tex]
= [tex]\frac{4.503*10^{3} }{36.0}[/tex]
= 125.0833 s
So that;
Rate at which rocket was initially burning fuel = [tex]\frac{3.003*10^{5} }{125.0833}[/tex]
= 2400.8001
= 2.40 x [tex]10^{3}[/tex] kg/s
Therefore, the initial rate at which the rocket burn fuel is 2.40 x [tex]10^{3}[/tex] kg/s.
A water rocket is shot from the ground at a 40 degree angle with an initial
velocity of 25 m/s. What is the velocity after 1.5 seconds?
Answer:
5.14m/s
Explanation:
First we need to get the Maximum height reached by the rocket;
H = u²sin² theta/2g
H = 25²sin² 40/2(9.8)
H = 625(0.5878)/19.6
H = 18.74m
Get the velocity after 1.5secs
Using the equation of motion
H = ut + 1/2gt²
18.74 = u(1.5)+1/2(9.8)(1.5)²
18.74 = 1.5u + 4.9(2.25)
18.74 = 1.5u + 11.025
1.5u = 18.74 - 11.025
1.5u = 7.715
u = 7.715/1.5
u = 5.14m/s
Hence the velocity after 1.5secs is 5.14m/s
Throughout the problem, take the speed of sound in air to be 343 m/s . Part A Consider a pipe of length 80.0 cm open at both ends. What is the lowest frequency f of the sound wave produced when you blow into the pipe
Answer:
the lowest frequency f of the sound wave is 214.375 Hz
Explanation:
The computation of the lowest frequency f of the sound wave is shown below;
Length = L= 80 cm
= 0.8 m
V = 343 m/s (sound speed in air )
Now
V1 = n V ÷ 2 L
= 1 × 343 ÷ 2 × 0.8
V1 = 214.375 Hz
Hence, the lowest frequency f of the sound wave is 214.375 Hz
We simply applied the above formula so that the correct value could come
And, the same is to be considered
A horizontal wire PQ is perpendicular to a uniform horizontal magnetic field. A length of 0.25 m of the wire is subject to a magnetic field strength of 40 mT. A downward magnetic force of 60 mN acts on the wire. What is the magnitude and direction of the current in the wire
Complete question:
Please find the image uploaded for the diagram.
Answer:
The magnitude of the current is 6 A, in the direction of Q to P.
Explanation:
Given;
length of the wire, l = 0.25 m
magnetic field strength, B = 40 mT = 0.04 T
Magnitude of the magnetic force, F = 60 mN = 0.06 N
The magnitude of the current in the conductor is given as;
F = BIL
Where;
I is the current induced in the conductor
I = F / BL
I = (0.06) / (0.04 x 0.25)
I = 6 A
Th current will move in direction of Q to P
Two wave pulses pass each other on a string. The pulse traveling toward the right has positive amplitude, whereas the pulse traveling toward the left has equal amplitude in the negative direction. What happens when they occupy the same region of space at the same time?
a. constructive interference occurs
b. destructive interference occurs.
c. a standing wave is produced.
d. a traveling wave is produced.
e. a wave pulse is produced.
Answer:
destructive interference occurs
If a ball leaves the ground with a velocity of 4.67 m/s,
how high does the ball travel?
Answer:
[tex]Vf^2=Vo^2+2aS\\(0m/s)^2=(4.67m/s)^2+(2*-10m/s^2)S\\-(4.67)^2 m^2/s^2=-20m/s^2*S\\S=(21.8089/20) m\\S=1.090445 m\\[/tex]
3)
What causes the Moon to orbit around the Earth?
0.)
A)
the moon's mass
B)
Earth's gravity
9
the Sun's gravity
the vacuum of
space
Answer:
Explanation:
The path of the Earth–Moon system in its solar orbit is defined as the movement of this mutual centre of gravity around the Sun. Consequently, Earth's centre veers inside and outside the solar orbital path during each synodic month as the Moon moves in its orbit around the common centre of gravity.
Answer:
B
Explanation
the moon is constantly falling but moving fast enough to continue falling around the earth.
A cannon can make a maximum angle of 30 degrees with the horizon. What is the minimum speed of a cannon ball if it must clear a 10-m-high obstacle 30 m away?
1. 28.3 m/s
2. 30.5 m/s
3. 32.8 m/s
4. 34.1 m/s
5. 36.8 m/s
Answer:
28.3 m/s
Explanation:
From the question given above, the following data were obtained:
Angle of projection (θ) = 30°
Maximum height (H) = 10 m
Acceleration due to gravity (g) = 10 m/s²
Initial velocity (u) =?
Thus, we can obtain the minimum velocity cannon ball by using the following formula:
H = u²Sine² θ / 2g
10 = u² × (Sine 30)² / 2× 10
10 = u² × (0.5)² / 20
10 = u² × 0.25 / 20
10 = u² × 0.0125
Divide both side by 0.0125
u² = 10/ 0.0125
u² = 800
Take the square root of both side
u = √800
u = 28.3 m/s
Therefore, the minimum speed of the cannon ball is 28.3 m/s
19 "Made mostly of ice", fits best under which heading in the table below?
Asteroid
Meteor
Comet
Planet
Description
A Asteroid
B Meteor
C Comet
D Planet
Two particles, each of mass 7.0 kg, are a distance 3.0 m apart. To bring a third particle, with mass 21 kg, from far away to a resting point midway between the two particles, an external agent must do work equal to
Answer: the external agent must do work equal to -1.3 × 10⁻⁸ J
Explanation:
Given that;
Mass M1 = 7.0 kg
r = 3.0/2 m = 1.5 m
Mass M2 = 21 kg
we know that G = 6.67 × 10⁻¹¹ N.m²/kg²
work done by an external agent W = -2GM2M1 / r
so we substitute
W = (-2 × 6.67 × 10⁻¹¹ × 21 × 7) / 1.5
W = -1.96098 × 10⁻⁸ / 1.5
W = -1.3 × 10⁻⁸ J
Therefore the external agent must do work equal to -1.3 × 10⁻⁸ J
HELP WILL GIVE BRAINLIEST IF CORRECT
Answer:
B the slope of car postion time graph
Explanation:
Answer:
The slope of the car's position time graph.
Explanation:
Can someone please help me with some physics
Answer:
sure I will helpy you iru
A 100 A current circulates around a 2.00-mm-diameter superconducting ring.
A. What is the ring's magnetic dipole moment?
B. What is the on-axis magnetic field strength 4.70 cm from the ring?
Answer:
0.000314 Am²
6.049*10^-7 T
Explanation:
A
From the definitions of magnetic dipole moment, we can establish that
= , where
= the magnetic dipole moment in itself
= Current, 100 A
= Area, πr² (r = diameter divided by 2). Converting to m², we have 0.000001 m²
On solving, we have
= ,
= 100 * 3.14 * 0.000001
= 0.000314 Am²
B
= (0)/4 * 2 / ³, where
(0) = constant of permeability = 1.256*10^-6
z = 4.7 cm = 0.047 m
B = 1.256*10^-6 / 4*3.142 * [2 * 0.000314/0.047³]
B = 1*10^-7 * 0.000628/1.038*10^-4
B = 1*10^-7 * 6.049
B = 6.049*10^-7 T
are active centers of galaxies that radiate powerful x-rays visible across the universe
Answer:
no
Explanation:
Thomas the Tank Engine (a train) is going 80 m/s and slows down to 30 m/s over a period of 30s. What is his deceleration? Acceleration= (final velocity-initial velocity)/ time A. -1.67 m/s/s B. 0.67 m/s/s C. -50 m/s/s D. 50 m/s/s
Answer: D
Explanation:
A force of 3.40 N is exerted on a 6.30 g rifle bullet. What is the bullet's acceleration?
The correct answer is a = 539.68 m/[tex]s^2[/tex]
A push or pull that one thing applies to another is known as force. An object is considered to exert a force when it accelerates through space because acceleration is the rate at which an item's speed changes. The second law of motion of Newton explains this concept.
The equation F=ma, where "F" stands for force, "m" for mass, and "a" for acceleration, illustrates the link between force and acceleration.
Force applied = F = 3.40 N
The bullet's weight = m = 6.30 g = 0.0063 kg
The bullet's rate of acceleration can be expressed as = a = F/m
a = 3.40 / 0.0063
a = 539.68 m/[tex]s^2[/tex]
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What is the power output in watts and horsepower of a 70.0-kg sprinter who accelerates from rest to 10.0 m/s in 3.00 s
Given that,
Mass of a sprinter = 70 kg
Initial velocity, u = 0
Final velocity, v = 10 m/s
Time, t = 3 s
To find,
Power output.
Solution,
The work done by the sprinter is equal to its kinetic energy.
[tex]W=\dfrac{1}{2}m(v^2-u^2)\\\\=\dfrac{1}{2}\times 70\times 10^2\\\\=3500\ J[/tex]
Let P is power output. Power is equal to work done per unit time. So,
[tex]P=\dfrac{3500\ J}{3\ s}\\\\=1166.67\ W[/tex]
So, the power output is 1166.66 W.
A whale comes to the surface to breathe and then dives at an angle 24 degrees to the horizontal surface of the water. The whale continues in a straight line 145 m. What are the horizontal and vertical components of the displacement of the whale?
Given that,
A whale dives at an angle of 24 degrees to the horizontal surface of the water.
The whale continues in a straight line 145 m.
To find,
The horizontal and vertical components of the displacement of the whale.
Solution,
The horizontal component of displacement is :
[tex]d_x=d\cos\theta\\\\=145\times \cos(24)\\\\=132.46\ m[/tex]
The vertical component of displacement is :
[tex]d_y=d\sin\theta\\\\=145\times \sin(24)\\\\=58.97\ m[/tex]
Hence, the horizontal and vertical components of the displacement of the whale are 132.46 m and 58.97 m.The object has a speed of 15 m/s. It took the object 5 second to travel, how far did the object go?
Answer:
Seventy-Five Metres
Explanation:
You can solve this simply by multiplying 15 m/s by 5 seconds
[tex]15 \frac{m}{s} * 5s\\=75\frac{ms}{s}\\= 75m[/tex]
A crate of oranges on a horizontal floor has a mass of 30 kg. The coefficient of static friction is 0.62. The coefficient of kinetic friction is 0.52. The worker pulls the crate with a force of 200 N.
How do you know if the crate is moving or not? Explain it by comparing the magnitude of the friction with the magnitude of the applied force.
Answer:
Explanation:
The Net Force
The net force is defined as the sum of all the forces acting on a body at a certain moment.
Recall some basic concepts and equations to solve the problem.
If no external forces are applied in the vertical direction, the weight of the object and the normal force have the same magnitude and point to opposite directions.
The friction force is defined as:
[tex]Fr_k=\mu_k N[/tex]
[tex]Fr_s=\mu_s N[/tex]
Where the subindices k and s are referred as to the kinetic and static friction forces respectively.
The condition for the object to move is that the applied force is greater than the friction force.
The crate of oranges has a mass of 30 Kg, thus its weight is:
W = m.g = 30 * 9.8 = 294 N
The normal force is:
N = W = 294 N
The kinetic friction is calculated as:
[tex]Fr_k=0.52* 294[/tex]
[tex]Fr_k=152.88\ N[/tex]
The static friction is calculated as:
[tex]Fr_s=0.62* 294[/tex]
[tex]Fr_s=182.28\ N[/tex]
The applied force is 200 N and it's greater than both the kinetic and the static friction forces, thus the crate is definitely moving at a certain positive acceleration.
In a tug of war, team A pulls the rope with a force of 50 N to the right and team B pulls it with a force of 110 N to the left. What is the resultant force on the rope and which direction does the rope move? (N represents newton, the unit of force)
Given that,
Force in right side = 50 N
Force in left side = 110 N
To find,
The resultant and direction of force.
Solution,
Let the net force F is acting on the rope. Also, we can assume that right side is positive and left side is negative.
F = 50+(-110)
= -60 N
So, the magnitude of the resultant force acting on the rope is 60 N and it acts in left side.
Energy stored because of an object's height above the Earth's surface is_____energy.
nuclear
gravitational
electrical or chemical