Answer:
Explanation:
a) If it flies into a 20 m/s wind heading west
The wind blows against the velocity of plane
resultant velocity = 200 - 20 = 180 m /s
velocity relative to ground = 180 m /s
b ) If it flies into a 20 m/s wind heading east,
wind will support the velocity of plane
resultant velocity = 200 + 20 = 220 m /s .
velocity relative to ground = 220 m /s .
Answer:
so mas o menos
Explanation:
a) If it flies into a 20 m/s wind heading west
The wind blows against the velocity of plane
resultant velocity = 200 - 20 = 180 m /s
velocity relative to ground = 180 m /s
b ) If it flies into a 20 m/s wind heading east,
wind will support the velocity of plane
resultant velocity = 200 + 20 = 220 m /s .
velocity relative to ground = 220 m /s .
THE ONE WHO WILL ANSWER IT CORRECT WILL BE MARKED AS "BRAINLIST"....... If the radius of the orbit of a spaceship from the centre of the earth is twice the radius of the earth . then calculate the acceleration due to the gravity on space ship.
Answer:
Explanation:
19116 being the distance from the centre of the earth to the point you want to check in metres. So after all that you get G = 1.0888m/second squared. Hope that helps you/ you should be able to pop in any height you want.
plz mark as brainliest
A car slows down from 65 km/s to 30 km/s in 5 seconds. What is its acceleration?
A standing wave has a frequency of 471 Hz and a wavelength of 1.9. What is the speed of the
wave? (Round to the 2nd number after the decimal)
1
I REALLY NEED HELP !
Answer:
c = 894.90 m/s
Explanation:
Given data:
Frequency of wave = 471 Hz
Wavelength of wave = 1.9 m
Speed of wave = ?
Solution:
Formula:
Speed of wave = frequency × wavelength
c = f×λ
c = 471 Hz × 1.9 m
Hz = s⁻¹
c = 471s⁻¹ × 1.9 m
c = 894.90 m/s
The speed of wave is 894.90 m/s.
Question 7 (2 points)
Rachel performed an experiment testing the hours students slept with their
performance on a test. In this experiment, the hours that they slept was the_____ variable, while the grade they got on
the test was the
_____ variable.
True or False: A cloud’s only purpose is to create precipitation.
Answer:
I think it's false
Explanation:
clouds also help regulate the Earth's energy balance by reflecting and scattering solar radiation and by absorbing the Earth's infrared radiation.
Answer:
false
Explanation:
HELPP ITS DUE IN 5 MINUTES FREE BODY DIAGRAMS
Answer:
I think it's part c . but sorry if its wrong
Answer the following questions
Answer:
9 - 10N to the left
10 - There is no change on the object
Explanation:
Can I have brainliest answer pls?
A 1500 kg car is parked at the top of a hill 5.2 m high. What is the velocity of the car, in meters per second, when it reaches the bottom of the hill?
Answer:
Explanation:
The car will fall with acceleration due to gravity which is equal to 9.8 m /s²
For downward fall ,
initial velocity u = o
acceleration due to gravity = g = 9.8 m /s
final velocity v = ?
displacement h = 5.2 m
v² = u² + 2 gh
v² = 0 + 2 x 9.8 x 5.2
= 101.92
v = 10.1 m /s
A motorcyclist travelling at 30m/s starts to apply his brake when he is 50m from the traffic light that has just turned red .if he reached the traffic light. what is the deceleration
A 1800 kg car moves along a horizontal road at speed v₀ = 18.4 m/s. The road is wet, so the static friction coefficient between the tires and the road is only µ static = 0.188 and the kinetic friction coefficient is even lower, µ kinetic = 0.1316. The acceleration of gravity is 9.8 m/s². What is the shortest possible stopping distance for the car under such conditions? (neglect the reaction time of the driver and round your answer to 4 decimal places)
Answer:
The shortest possible distance is [tex]|s| = 91.9 \ m[/tex]
Explanation:
From the question we are told that
The mass of the car is [tex]m = 1800 \ kg[/tex]
The speed along the horizontal road is [tex]v_o =u = 18.4 \ m/s[/tex]
The static friction coefficient is [tex]\mu_s = 0.188[/tex]
The kinetic friction coefficient is [tex]\mu_k = 0.1316[/tex]
Generally the static frictional force acting on the car is mathematically represented as
[tex]F_f = m * g * \mu_s[/tex]
Generally the force propelling the car is mathematically represented as
[tex]F = m * a[/tex]
Here a is the maximum acceleration
at the point which the car stops ,
[tex]F = F_f[/tex]
=> [tex]m * g * \mu_s = ma[/tex]
=> [tex]g * \mu_s =a[/tex]
=> [tex]a = 9.8 * 0.188[/tex]
=> [tex]a = 1.8424 \ m/s^2[/tex]
Generally from kinematic equation
[tex]v^2 = u^2 + 2as[/tex]
Here v is the final velocity of the car and the value is zero given that the car comes to rest
So
[tex]0^2 = 18.4^2 + 2* 1.8424 s[/tex]
=> [tex]s = - \frac{18.4^2}{2 * 1.8424}[/tex]
=> [tex]|s| = 91.9 \ m[/tex]
How much air resistance acts on a 100-N bag of nails that falls at its terminal speed?
In order to determine the wavelength, you must know the distance from the (2 points)
Select one:
a. trough of one wave to the crest of the next wave
b. midpoint of a wave to the highest point of the next wave
c. crest of the wave to the equilibrium of the same wave
d. the crest of one wave to the crest of the next wave
Answer:
b
Explanation:
Answer:
a
Explanation:
because the crest and trough are the same length when together and they squeeze together making the length shorter for higher frequencies and are spaced apart when lower frequencies occur
gold has a density of 19300 kg/m3 calculate the mass of 0.02m3 of gold in kilograms
Mass if a substance is the product of its volume and density. The mass of gold of 0.02 m³ with a density of 19300 kg/m³ is 386 kg.
What is density?Density of a substance is the measure of its mass per unit volume. Thus, it says how much denser the object is in a given volume. Density of a substance is dependent on its bond type, temperature and pressure beside the volume and mass.
Volume can be defined as the space occupied by the substance. Larger the volume , less dense the substance is. However as the mass increases volume also increases.
Mass of an object is the product of its volume and density.
Given the volume of gold = 0.02 m³
density = 19300 kg/m³.
mass = volume × density
= 0.02 m³ × 19300 kg/m³
= 386 kg
Therefore, the mass of gold is 386 Kg.
To find more on gold, refer here:
https://brainly.com/question/11405288
#SPJ6
2 sfere electrizate cu sarcini egale,situate in vid la distanta de 10 cm una de alta,interactioneaza cu o forta de 0,1 N.Aflati sarcina electrica a uneia din sfere
Răspuns:
1,054 × 10 ^ -6Combombs
Explicaţie:
Conform legii coloumbs, forța de atracție sau repulsie între sarcină este exprimată ca;
F = kq1q1 / r²
F este forța atracției
q1 și q2 sunt taxele
r este distanța dintre sarcini
Deoarece taxele sunt egale, q1 = q2 = q
Ecuația devine
F = kq² / r²
Dat
F = 0,1N
r = 10cm = 0,1m
Necesar
q
Înlocuiți parametrii dați în formulă;
0,1 = 9 × 10 ^ 9q² / 0,1
0,1 × 0,1 = 9 × 10 ^ 9q²
0,01 = 9 × 10 ^ 9q²
q² = 0,01 / 9 × 10 ^ 9
q² = 0,01 / 9.000.000.000
q² = 1,11 × 10 ^ -12
q = √1.11 × 10 ^ -12
q = 1,054 × 10 ^ -6Combombs
Prin urmare, sarcina electrică pe una dintre sfere este de 1,054 × 10 ^ -6Combombs
What is Newtons third law of motion?
Answer:
His third law states that for every action (force) in nature there is an equal and opposite reaction. Explanation:
Answer:
His third law states that for every action (force) in nature there is an equal and opposite reaction
Explanation:
In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A. Notice that the forces are exerted on different objects.
A car’s velocity changes from 35 m/s to stopped in 13 seconds. Calculate
acceleration.
Answer:
Acceleration = 3m/s^2
Vf= 0 Vi =35m/s t= 13s
Explanation:
[tex]Acceleration = \frac{Change in velocity}{Change in time}\\ = \frac{35m/s}{13s}\\ a = 2.69m/s^2\\ a = 2.7m/s^2\\ a = 3m/s^2[/tex]
A pyrotechnical expert needs to fire a 15 kg projectile from a launching device that has a barrel length of 2 meters. The projectile will need to be launched horizontally 1 km in 5 seconds. Calculate the force needed to launch the projectile.
Answer:
The force needed to launch the projectile is 150000 N.
Explanation:
We can find the force using the following equation:
[tex] F = ma [/tex]
Where:
m: is the mass = 15 kg
a: is the acceleration
First, we need to find the acceleration of the projectile:
[tex] v_{f}^{2} = v_{0}^{2} + 2ax [/tex]
Where:
[tex]v_{f}[/tex]: is the final speed
[tex]v_{0}[/tex]: is the initial speed = 0
x: is the distance = 2 m
The final speed is:
[tex]v_{f} = \frac{1 km}{5 s}*\frac{1000 m}{1 km} = 200 m/s[/tex]
Then, the acceleration is:
[tex]a = \frac{v_{f}^{2}}{2x} = \frac{(200 m/s)^{2}}{2*2 m} = 10000 m/s^{2}[/tex]
Finally, the force is:
[tex]F = ma = 15 kg*10000 m/s^{2} = 150000 N[/tex]
Therefore, the force needed to launch the projectile is 150000 N.
I hope it helps you!
A bowling ball with a mass of 7.0kg strikes a pin that had a mass of 2.0kg the pin flies forward with a velocity of 6.0m/s, and the ball continues forward at 4.0 m/s. What was the original velocity of the ball?
The conservation of momentum P states that the amount of momentum remains constant when there are not external forces.
We don't have external forces, so:
[tex]P_0 = P_1\\m_bv_{0b}+m_pv_{0p}=m_bv_{1b}+m_pv_{1p}\\[/tex]
Where:
mb is the mass of the bowling ball mp the mass of the pin[tex]v_{0b}\quad and\quad v_{0p}[/tex] the initial velocities of the bowling ball and the pin.[tex]v_{1b}\quad and\quad v_{1p}[/tex] the final velocities of the bowling ball and the pin.Solving for v0b:
[tex]v_{0b} =\dfrac{m_bv_{1b}+m_pv_{1p}- m_pv_{0p}}{m_{b}}\\\\v_{0b} =\dfrac{(7\;kg)(4\;m/s)+(2\;kg)(6\;m/s)- (2\;kg)(0 \;m/s)}{7\;kg}\\v_{0b}=\dfrac{40}{7}\;m/s\\\\\boxed{v_{0b}\approx5.71\;m/s}[/tex]
R/ The original velocity of the ball was 5.71 m/s.The original velocity of the ball is 5.71 m/s.
The principle of conservation of momentum: In a closed system, The total momentum before collision is equal to total momentum after collision.
From the principle of conservation of momentum,
MU+mu = MV+mv.................... Equation 1
Where M = mass of the bowling ball, m = mass of the pin, U = initial velocity of the bowling ball, u = initial velocity of the pin, V = final velocity of the bowling ball, v = final velocity of the pin.
From the question,
Given: M = 7 kg, m = 2 kg, u = 0 m/s (at rest), v = 6.0 m/s, V = 4 m/s.
Substitute these values into equation 1 and solve for U
7(U)+2(0) = 7(4)+2(6)
7U = 28+12
7U = 40
U = 40/7
U = 5.71 m/s.
Hence, The original velocity of the ball is 5.71 m/s.
Learn more about velocity here: https://brainly.com/question/6237128
How long does it take for an 8 kg pumpkin to hit the ground if dropped from a height of 55 m.
Answer:
t = 3.35 s
Explanation:
It is given that,
Mass of a pumpkin, m = 8 kg
It is dropped from a height of 55 m
We need to find the time taken by it to hit the ground.
Initial velocity of the pumpkin, u = 0
Using second equation of motion to find it as follows :
[tex]h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2(55)}{9.8}} \\\\t=3.35\ s[/tex]
So, it will take 3.35 seconds to hit the ground.
Suppose a cart of mass 0.5 kg is placed on the table and connected to a mass hanging mass over the edge. If the hanging mass is 0.2 kg, when what would be the acceleration of the cart?
Explanation:
0.10kg
0.5kg×0.2kg=
0.10
A +4.0 uC charge is placed on the x axis at x= +3.0 m, and a -2.0 uC is located on the y-axis at y= -1.0 m. Point A is on the y axis at y= +4.0 m. Determine the electric potential at point A (relative to zero at the origin).
Answer:
The potential is [tex]V_A = 9600 \ V[/tex]
Explanation:
From the question we are told that
The magnitude of the charge is [tex]q_1 = 4 \mu C = 4*10^{-6} \ C[/tex]
The position of the charge is [tex]x = + 3.0 \ m[/tex]
The magnitude of the second charge is [tex]q_2 = -2.0 \mu C = -2.0 *10^{-6} \ C[/tex]
The position is [tex]y_1 = - 1.0 \ m[/tex]
The position of point A is [tex]y_2 = + 4.0 \ m[/tex]
Generally the electric potential at A due to the first charge is mathematically represented as
[tex]V_a = \frac{k * q_1 }{r_1 }[/tex]
Here k is the coulombs constant with value [tex]k = 9*10^{9} \ \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}[/tex]
[tex]r_1[/tex] is the distance between first charge and a which is mathematically represented as
[tex]r_1 = \sqrt{x^2 + y_2 ^2 }[/tex]
=> [tex]r_1 = \sqrt{3^2 + 4 ^2 }[/tex]
=> [tex]r_1 = 5 \ m[/tex]
So
[tex]V_a = \frac{9*10^9 * 4*10^{-6} }{5 }[/tex]
[tex]V_a = 7200 \ V[/tex]
Generally the electric potential at A due to the second charge is mathematically represented as
[tex]V_b = \frac{k * q_2 }{r_2 }[/tex]
Here k is the coulombs constant with value [tex]k = 9*10^{9} \ \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}[/tex]
[tex]r_2[/tex] is the distance between second charge and a which is mathematically represented as
[tex]r_2 = y_2 - y[/tex]
=> [tex]r _2 = 4.0 - (-1.0)[/tex]
=> [tex]r = 5 \ m[/tex]
So
[tex]V_a = \frac{9*10^9 * -2*10^{-6} }{5 }[/tex]
[tex]V_a = -3600 \ V[/tex]
So the net potential difference at point A due to the charges is mathematically represented as
[tex]V_n = V_a + V_b[/tex]
=> [tex]V_n = 7200 - 3600[/tex]
=> [tex]V_n = 3600 V[/tex]
Generally the net potential difference at the origin due to both charges is mathematically represented as
[tex]V_N = V_c + V_d[/tex]
Here
[tex]V_c = \frac{k * q_1 }{x}[/tex]
=> [tex]V_c = \frac{9*10^9 * 4*10^{-6} }{3}[/tex]
=> [tex]V_c = 12000 V[/tex]
and
[tex]V_d= \frac{k * q_2 }{y}[/tex]
=> [tex]V_c = \frac{9*10^9 * -2*10^{-6} }{1}[/tex]
=> [tex]V_c =- 18000 V[/tex]
Generally the net potential difference at the origin is
[tex]V_N = 12000 - 18000[/tex]
=> [tex]V_N = -6000[/tex]
Generally the potential difference at A relative to zero at the origin is mathematically evaluated as
[tex]V_A = V_n - V_N[/tex]
=> [tex]V_A = 3600 - (-6000)[/tex]
=> [tex]V_A = 9600 \ V[/tex]
The chart shows the percentage of different elements in the human body.
Which element is the most prevalent in the human body?
nitrogen
hydrogen
carbon
oxygen
what is energy?
a. the number of atoms in an object
b. a form of sound
c. the ability to do work
d. the size of an object
(ik its not b or d, but i dont know)
Answer:
c
Explanation:
c. the ability to do work or to produce heat
To determine a waves' frequency, you must know the (2 points)
Select one:
a. distance it travels
b. number of oscillations in a given period of time
c. distance between the crest and trough
d. height of the wave
Answer:
b
to determine the frequency you must know oscillations and the given time
How much work do you do when you push a shopping cart with a force of 20 N for a distance of 5m?
A.100J
B.10J
C.1J
D.1000J
Answer:
A.100J
Explanation:
Given parameters:
Force on car = 20N
Distance = 5m
Unknown:
Work done = ?
Solution:
Work done is the product of force and distance;
Work done = force x distance;
Insert given parameters and solve;
Work done = 20 x 5 = 100J
Which evidence did Alfred Wegener’s original theory of continental drift have access to?
Answer:
Evidence for continental drift
Wegener knew that fossil plants and animals such as mesosaurs, a freshwater reptile found only South America and Africa during the Permian period, could be found on many continents. He also matched up rocks on either side of the Atlantic Ocean like puzzle pieces.
Explanation:
1. What mass can be lifted by the hydraulic press if it is known that the are:
of the small piston is 48cm and that of the large piston is 1440cm, and
the force acting on the small piston can be as large as 1000N. (use g =
10m/s)
C. 3000kg
B. 30kg
D. 30,000kg
A. 300kg
Answer:
Answer: C. 3000kg
Explanation:
Hydraulic Jack
It's a device that is used to lift heavy loads by applying force via a hydraulic cylinder.
Since the internal pressure on the liquid of the jack is constant, the following relation applies:
[tex]\displaystyle \frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]
Where
F1 = Force applied in one end of the jack
A1 = Area of the cylinder of the same end
F2 = Force applied in the other end of the jack
A2 = Area of the cylinder of the other end
If we assume the force is being applied on side 1, then the force on side 2 is calculated by solving for F2:
[tex]\displaystyle F_2=\frac{F_1.A_2}{A_1}[/tex]
We have F1=1000N, A1= 48~cm^2, A2= 1440~cm^2, thus:
[tex]\displaystyle F_2=\frac{1000.1440}{48}[/tex]
[tex]F_2 = 30000 ~N[/tex]
Since we need to know the mass that can be lifted by that force, we use the formula:
W = m.g
Where W is the weight of the mass and g= 10~m/s^2. Since the weight is equal to the force exerted by the jack:
[tex]\displaystyle m=\frac{W}{g}=\frac{30000}{10}[/tex]
m = 3000 Kg
Answer: C. 3000kg
In which of the following ways is region B in the diagram different from region D? (1 point)
A
Region B, the corpus callosum, separates the brain into two halves while region D, the cerebellum, communicates between the two halves.
B
Region B, the cerebrum, is one single unit composed of gray matter, while region D, the cerebellum, is divided into two hemispheres.
C
Region B, the corpus callosum, is composed of several lobes while region D, the cerebellum, only has one lobe.
D
Region B, the cerebrum, is divided into two hemispheres while region D, the cerebellum, is one single structure.
Answer:
C
Explanation:
Answer:
C
Region B, the corpus callosum, is composed of several lobes while region D, the cerebellum, only has one lobe.
Explanation:
A scientist is observing a eukarotic cell and a prokaryotic cell. Which structure could she only observe in the eukaryotic cell?
cytoplasm
DNA
ribosomes
a nucleus
Answer:
It's definitely Nucleus
Hope this Helps!
Una larga fanja de pavimento tiene marcas a intervalos de 10m. Los estudiantes usan cronometros para registrar los tiempos en que un automovil pasa por cada marca. Asi han obtenido los datos siguientes: Distancia,m l 0 l 10 l 20 l 30 l 40 l 50 l Tiempo,s l 0 l 2.1 l 4.2 l 6.3 l 8.4 l 10.5 l A) Cual es la rapidez media del vehiculo? B) Al cabo de cuanto tiempo la distancia es igual a 34m? C) Cual es la aceleracion del automovil?
Answer:
men
Explanation: