A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is falling vertically at a constant velocity of 7.0 m/s downward with their parachute open (terminal velocity). A horizontal gust of wind helps push them towards the target. What must be their total speed if they want to just hit their target?
Answer:
Answer:
15.67 seconds
Explanation:
Using first equation of Motion
Final Velocity= Initial Velocity + (Acceleration * Time)
v= u + at
v=3
u=50
a= - 4 (negative acceleration or deceleration)
3= 50 +( -4 * t)
-47/-4 = t
Time = 15.67 seconds
We have that the speed must be at the speed below if they want to just hit their target
From the Question we are told that
Distance [tex]d=50m[/tex]
Height [tex]h=70m[/tex]
Constant Velocity [tex]v= 7.0 m/s[/tex]
Generally the equation for the time is mathematically given as
[tex]T=\frac{h}{v}\\\\T=\frac{70}{7}\\\\T=10s[/tex]
Therefore
The velocity required to make horizontal movement is
[tex]V=\frac{d}{T}\\\\V=\frac{50}{10}\\\\V=5m/s[/tex]
Given that
Velocity on the Vertical axis is
[tex]v_y=7m/s[/tex]
Velocity on the horizontal axis is
[tex]v_x=5m/s[/tex]
Therefore resultant speed
[tex]v_r=\sqrt{v_x^2+V_y^2}\\\\v_r=\sqrt{(5)^2+(7)^2}[/tex]
[tex]v_r=8.6023m/s[/tex]
In conclusion
[tex]v_r=8.6023m/s[/tex] must be their total speed if they want to just hit their target
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The deepest part of the ocean is the Challenger Deep, at 10,900 m. The depth was first measured in 1875 by the HMS Challenger by depth sounding (which does not involve sound waves). If you were to measure the depth by echo sounding (which does involve sound), what would you expect the time for a sound pulse at the surface to return in s, naively assuming a constant sound velocity throughout the ocean
Answer:
t = 14.53 s
Explanation:
The speed of a wave is constant and is given by
v = [tex]\sqrt{ \frac{B}{ \rho} }[/tex]
in this exercise they indicate that we assume the constant velocity, therefore we can use the uniform motion relations
v = x / t
t = x / v
in this case the sound pulse leaves the ship and must return so the distance is
x = 2d
where d is the ocean depth d = 10900m and the speed of sound in seawater is v = 1500 m / s
let's calculate
t = 2 10900/1500
t = 14.53 s
In the 1986 Olympic Games, Abdon Pamich of Italy won the 50 km walk, in
4h, 11 min and 11.2 s. Find his average speed in m/s.
Answer: 4
Explanation: because my pet monkey said it was
A car starts from rest and accelerates uniformly at a rate of 2.0 meter per second squared for 4.0 seconds. During this time interval, the car traveled ________ meters.
Answer:
16
Explanation:
[tex] \frac{1}{2} \times 2 \times {4}^{2} = 16[/tex]
A π meson of rest energy 139.6 MeV moving at a speed of 0.921c collides with and sticks to a proton of rest energy 938.3 MeV that is at rest. (a) Find the total relativistic energy of the resulting composite particle. (b) Find the total linear momentum of the composite particle. (c) Using the results of (a) and (b), find the rest energy of the composite particle.
Answer:
A) 1268 MeV
B) 299MeV/c
C) 1268 MeV
Explanation:
Given :
π meson rest energy = 139.6 MeV
Speed = 0.921c
proton at rest energy = 938.3 MeV
a) Find the total relativistic energy of resulting composite particle
E = E(meson) + E(proton)
= [tex]\frac{(mc^2)_{meson} }{\sqrt{1-\frac{v^2}{c^2} } } + (mc^2)_{proton}[/tex]
= [tex]\frac{139.6MeV}{\sqrt{1-\frac{(0.906c)^2}{c^2} } } + 938.3[/tex]
E = 1268 MeV
B) determine the total linear momentum of the composite particle
= 299MeV/c
attached below is the detailed solution
C) Determine the rest energy of the composite particle
E = 1268 MeV
Pluto has a radius of 1.15 x 10^6 m, and its acceleration due to gravity is 0.61 | m/s^2. What is Pluto's mass?
so we will use Newton's gravitational law :
gravitational acceleration = G*m/r^2
G is the gravitational constant = G = 6.673×10^-11 N m^2 kg^-2
after substitution :
6.673×10^-11 * m / (1.15 x 10^6)^2 = 0.61
= 5.04575*10^-23 * m = 0.61
dividing over 5.04575*10^-23 :
m = 1.20894*10^22 kg
pls give me brainliest
Which of the following is an example of Newton's Third Law?* O A stack of pennies will not move unless you flick them over. O Falling off of a skateboard after you run into a curb A ball hits the ground and the ground pushes up on it with the same force
Answer:
A ball hits the ground and the ground pushes up on it
Explanation:
Newton's third law basically states that for every action, there's a reaction.
a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.
Hope this Helps!!! :)
17. During a game of tug of war, two teams of students pull on opposite sides of a rope. During the
game, the rope begins to accelerate towards the left. What must be true about the forces acting on the
rope at the time of the acceleration towards the left?
A. The team pulling towards the right is pulling with a force greater than the team pulling towards the left.
B. The team pulling towards the right is pulling with a force equal to the team pulling towards the left.
C. The team pulling towards the right is pulling with a force less than the team pulling towards the left.
D. The team pulling towards the right stopped pulling the rope while the team pulling towards the left
continued
Answer:
c
Explanation:
plsss answer this plsss answer this plsss answer this plsss answer this
Answer:
I dont see file
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Answer:
ye ek rod h or electric ⚡ field h P point
In traveling a distance of 2.3 km between points A and D, a car is driven at 83 km/h from A to B for t seconds and 41 km/h from C to D also for t seconds. If the brakes are applied for 4.4 seconds between B and C to give the car a uniform deceleration, calculate t and the distance s between A and B.
Answer:
- time t taken for car to travel is 64.57 s
- distance travelled between A and B is 1.4887 km
Explanation:
Given the data in the question;
[tex]U_{BC}[/tex] = 83 km/h = ( 83×1000 / 60×60) = 23.0555 m/s
[tex]U_{CD}[/tex] = 41 km/h = ( 41×1000 / 60×60) = 11.3888 m/s
now, we calculate the acceleration;
a = ( [tex]U_{BC}[/tex] - [tex]U_{CD}[/tex] ) / t
we substitute
a = ( 23.0555 - 11.3888 ) / 4.4
a = 11.6667 / 4.4
a = 2.6515 m/s²
Now equation for displacement from BC
[tex]S_{BC}[/tex] = [tex]U_{BC}[/tex]t + 1/2.at²
we substitute
[tex]S_{BC}[/tex] = 23.0555×4.4 + 1/2×a×(4.4)²
we substitute -2.6515m/s² for a
[tex]S_{BC}[/tex] = 23.0555×4.4 + 1/2×(-2.6515)×(4.4)²
= 101.4442 - 25.6665
[tex]S_{BC}[/tex] = 75.7792 m
Now, for total distance covered = 2.3km = ( 2.3×1000) = 2300m
so
[tex]S_{AB}[/tex] + [tex]S_{BC}[/tex] + [tex]S_{CD}[/tex] = 2300 m
we substitute substitute
[tex]S_{AB}[/tex] + 75.7792 m + [tex]S_{CD}[/tex] = 2300 m
[tex]S_{AB}[/tex] + [tex]S_{CD}[/tex] = 2300 - 75.7792
[tex]S_{AB}[/tex] + [tex]S_{CD}[/tex] = 2224.2208 m
so we substitute 23.0555t for [tex]S_{AB}[/tex] and 11.3888t for [tex]S_{CD}[/tex]
23.0555t + 11.3888t = 2224.2208
34.4443t = 2224.2208
t = 2224.2208 / 34.4443
t = 64.57 s
Therefore, time t taken for car to travel is 64.57 s
Distance Between A to B
[tex]S_{AB}[/tex] = t × [tex]U_{AB}[/tex]
we substitute
[tex]S_{AB}[/tex] = 64.57 s × 23.0555
[tex]S_{AB}[/tex] = 1488.69 m
[tex]S_{AB}[/tex] = 1.4887 km
Therefore, distance travelled between A and B is 1.4887 km
5. An astronaut has a mass of 65kg where the gravitational field strength is 10N/kg
a. Calculate the weight of the astronaut on earth
[3]
Answer: a) weight on Earth = mass of the object and gravity n the Earth. = 65*10 = 650 kg.
Explanation:
An astronaut has a mass of 65 kg on Earth where the gravitational field strength is 10 N kg A calculate the astronaut's weight on Earth
hope this helps :)
Answer:
650
Explanation:
use the equation
weight = gm
In order for the eye to see an object _____ from the object myst be reflected to your eye.
Light or particle ?
Answer: light from the object
Explanation:
When light is reflected off an object like a lamp or a door is travels in a straight line but in a new direction so if the light enters our eyes we will see the object because our eyes can detect light
PLEASE HELP GIVING BRAINLIEST ANSWER
Explain why your PE and KE are usually not both high at the same time (If PE is high then usually KE is low)
If the car has a mass of 0.2 kg, the ratio of height to width of the ramp is 12/75, the initial displacement is 2.25 m, and the change in momentum is 0.58 kg*m/s, how far will it coast back up the ramp before changing directions
Answer:
l = 0.548 m
Explanation:
For this exercise we compensate by finding the speed of the car
p = m v
v = p / m
v = 0.58 / 0.2
v = 2.9 m / s
this is how fast you get to the ramp, let's use conservation of energy
starting point. Lowest point
Em₀ = K = ½ m v²
final point. Point where it stops on the ramp
[tex]Em_{f}[/tex] = U = m g h
mechanical energy is conserved
Em₀ = Em_{f}
½ m v² = m g h
h = [tex]\frac{m v^2}{2 g}[/tex]
let's calculate
h = [tex]\frac{0.2 \ 2.9^2}{2 \ 9.8}[/tex]
h = 0.0858 m
to find the distance that e travels on the ramp let's use trigonometry, we look for the angle
tan θ = y / x
tan θ = 12/75 = 0.16
θ = tan⁻¹ 0.16
θ = 9º
therefore
sin 9 = h / l
l = h / sin 9
l = 0.0858 / sin 9
l = 0.548 m
A pendulum has a period of 5.14s and a length of 0.25m. What is the acceleration
due to gravity? *
Answer:
Acceleration due to gravity, g = 2.68m/s²
Explanation:
Given the following data;
Period = 5.14s
Length = 0.25m
To find acceleration due to gravity, g;
[tex] Period, T = 2 \pi \sqrt {lg} [/tex]
Substituting into the equation, we have;
[tex] 5.14 = 2*3.142 \sqrt {0.25g} [/tex]
[tex] 5.14 = 6.284 \sqrt {0.25g} [/tex]
[tex] \frac {5.14}{6.284} = \sqrt {0.25g} [/tex]
[tex] 0.8180 = \sqrt {0.25g} [/tex]
Taking the square of both sides
[tex] 0.8180^{2} = 0.25g [/tex]
[tex] 0.6691 = 0.25*g[/tex]
[tex] g = \frac {0.6691}{0.25} [/tex]
Acceleration due to gravity, g = 2.68m/s²
noooooooooooooooooooooooo
Answer:
yes
Explanation:
Answer:yeeeeeeeeeeeeeeeeeeeeeees
Explanation:
A fan is set on a desk next to a stack of paper. The fan is turned on and then turned
on HIGH SPEED. Which of the following would best apply Newton's First Law to this
example.
The papers accelerated due to the force of the fan.
The papers are acted upon by an unbalanced force from the fan.
O The papers exerted an equal force on the air blown by the fan.
O The papers that were the heaviest were blown the closest.
Answer:
The second option - the papers are acted upon by an unbalanced force from the fan.
Explanation:
. Why is it harder to stop an elephant accelerating at 1m/s2 than a rabbit accelerating at 1m/s2
(10 Points)
the elephant has more mass
the rabbit is faster
the rabbit has more mass
the elephant is faster
Answer:
this is about momentum p=mv
A, the elephant has more mass
Pls help!!
1 example of a conductor and 1 example of a insulator in your EVERYDAY world.
Answer:
here
Explanation:
Examples of insulators include plastics, Styrofoam, paper, rubber, glass and dry air.
Examples of conductors include metals, aqueous solutions of salts
Match the measurements with the proper SI unit.
Acceleration:
A. Meters
B. Meters per second
C. Meters per second squared
Velocity:
A. Meters
B. Meters per second
C. Meters per second squared
Distance:
A. Meters
B. Meters per second
C. Meters per second squared
Explanation:
C. meter per second squared
B. meter per second
A. meter
Answer:
b. meters per second
c.meters per second squared
c.meters
Explanation:
ginawa ko na rin KC toh
A 7300 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.20 m/s2 and feels no appreciable air resistance. When it has reached a height of 575 m , its engines suddenly fail so that the only force acting on it is now gravity. Part A What is the maximum height this rocket will reach above the launch pad
Answer:
Explanation:
We shall first calculate the velocity at height h = 575 m .
acceleration a = 2.2 m /s²
v² = u² + 2 a s
u is initial velocity , v is final velocity , s is height achieved
v² = 0 + 2 x 2.2 x 575
v = 50.3 m /s
After 575 m , rocket moves under free fall so g will act on it downwards
If it travels further by height H
from the relation
v² = u² - 2 g H
v = 0 , u = 50.3 m /s
H = ?
0 = 50.3² - 2 x 9.8 H
H = 129.08 m
Total height attained by rocket
= 575 + 129.08
= 704.08 m .
If an ocean wave has a wavelength of 2 m and a frequency of 1 wave/s, then its speed is m/s Enter the answer Check it CRATCHPAD Improve this questic 트
Answer:
2m/s
Explanation:
v=f×wavelength
v=2×1
=2m/s
A football of mass 2.5kg is lifted up to the top of a cliff that is 180m high. How much
potential energy does the football gain?
The potential energy of the football with mass 2.5 kg which is lifted up to the top of a cliff is 4410 Joules.
What is Potential energy?Potential energy is the stored energy which depends upon the relative position of the various parts of a system of objects. Potential energy is the product of mass of the object, acceleration due to gravity, and the height. The SI unit of potential energy is Joule (J).
PE = m × g × h
PE = Potential energy,
m = mass of the object,
g = acceleration due to gravity,
h = height
PE = 2.5 × 9.8 × 180
PE = 4410 Joules
Therefore, the potential energy of the football is 4410 Joules.
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6. The petrol in a petrol can weighs 2000g. The density of petrol is 0.8g/cm3.
What is the volume of the petrol in the can in a) cm3 b)litres (1000cm3=1 litre)
Pls help :((
Answer:
a. 2500 cm³.
b. 2.5 litres.
Explanation:
Given the following data:
Density = 0.8g/cm³
Mass = 2000g
To find the volume of the petrol;
Density can be defined as mass all over the volume of an object.
Simply stated, density is mass per unit volume of an object.
Mathematically, density is given by the equation;
[tex]Density = \frac{mass}{volume}[/tex]
Making volume the subject of formula, we have;
[tex]Volume = \frac{mass}{density}[/tex]
Substituting into the equation, we have:
[tex]Volume = \frac{2000}{0.8}[/tex]
Volume = 2500 cm³
a. The volume of the petrol in the can in cubic centimeters (cm³) is 2000.
b. The volume of the petrol in the can in litres;
1000 cm³ = 1 litre
2500 cm³ = x litres
Cross-multiplying, we have;
1000x = 2500
x = 2500/1000
x = 2.5 litres.
Therefore, the volume of the petrol in the can in litres is 2.5.
A laser positioned on a ship is used to communicate with a small two man research submarine resting on the bottom of a lake. The laser is positioned 12 m above the surface of the water, and it strikes the water 20 m from the side of the ship. The water is 76 m deep and has an index of refraction of 1.33. How far is the submarine from the side of the ship
Answer:
84.1 m
Explanation:
Given :
The distance from the ship to submarine :
20 + y
Using Pythagoras :
Tan θ = opposite / Adjacent
Tan θ = 20 / 12
12 tan θ = 20
θ = tan^-1(20/12)
20
θ = 59.036°
The angle phi;
n1sinθ1 = n2sin θ
Sin 59.036 = 1.33 * sin phi
Sin phi = sinsin(59.04) ÷1.33
0.8574907 = 1.33 * sin phi
Sin phi = 0.8574907 / 1.33
Sin phi = 0.6447298
phi = sin(0.6447298
Phi = 40.15°
From Pythagoras :
y = 76tan40.15°
y = 76 * 0.8435707
y = 64.11
20 + y
20 + 64.11 = 84.11
Starting from rest, a wheel with constant angular acceleration turns through an angle of 25 rad in a time t. What will its angular velocity be after 3t?
Answer:
θ = 225 rad
Explanation:
given data
angle = 25 rad
to find out
angular velocity after 3t?
solution
let angular acceleration α in t
θ = ω × t + 0.5 × α × t² ........................1
here ω = 0 (initial velocity )
so put this value here
25 = 0 + 0.5 × α × t² ..........................2
α = 25 ÷ (0.5 t²)
α = 50 ÷ t² .........................3
now here we take in 3t
θ = ω × 3t + 0.5 × α × (3t)²
for ω = 0
θ = 0 + 0.5 × α × 9t²
now put value in eq 2
so
θ = (0.5) × (50 ÷ t²) × (3t)²
θ = 25 × 9
θ = 225 rad
A toy car accelerates uniformly from rest at a constant rate. The car travels 1.0 meters in 1.0 seconds. The acceleration of the car is ________ meters per second squared.
Answer:
a=1m/s^2
Explanation:
1÷1÷1=1m/s^2
50 points Two waves combine with constructive interference. What must be true of the
combined wave that forms?
A. It has a lower frequency than that of the original waves.
B. It has a higher amplitude than that of the original waves.
C. It has a higher frequency than that of the original waves.
D. It has a lower amplitude than that of the original waves.
Answer:
it has a higher amplitude than that of the original waves
Explanation:
trust me its right
The two waves combined with constructive interference have a higher amplitude than the original wave.
Which wave has the highest amplitude?High energy waves are characterized by high amplitude. Low energy waves are characterized by their small amplitude. As explained in Lesson 2, wave amplitude is the maximum amount of particle movement on a medium from a resting position.
What are superposition and interference?Superposition is a combination of two waves in the same place. Constructive interference occurs when two identical waves interfere in phase. Destructive interference occurs when two identical waves are exactly out of phase and overlap.
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I love buying physics toys. I recently broke out my new electromagnetic field meter and started playing with it. After turning it on, I noticed the device kept showing an electric field value of 200 N/C towards the ground, without being near anything obvious (e.g., an electronic device) that would be producing the electric field. I even took a long walk to check if the reading was somehow localized to my house, but I got the same result. How might you explain the reading (assuming the device is working properly)
Answer:
ionized particles from the sun.
* interactions in radiation belts.
* the friction of the planet in the solar wind
q = +9 10⁵ C
Explanation:
Due to being made up of matter, the planet Earth has a series of positive and negative charges, in general these charges should be balanced and the net charge of the planet should be zero, but there are several phenomena that introduce unbalanced charges, for example:
* ionized particles from the sun.
* interactions in radiation belts.
* the friction of the planet in the solar wind
This creates that the planet has a net electrical load
We can roughly calculate the charge of the planet
E = k q / r²
q = E r² / k
let's calculate
q = 200 (6.37 10⁶)²/9 10⁹
q = +9 10⁵ C
If a solid metal sphere and a hollow metal sphere of equal diameters are each given the same charge, the electric field (E) midway between the center and the surface is...A. greater for the solid sphere than for the hollow sphere.B. greater for the hollow sphere than for the solid sphere.C. zero for bothD. equal in magnitude for both, but one is opposite in direction from the other.
Answer:
C. zero for both
Explanation:
In case of solid metal sphere , when it is given any charge , all the charges are transferred on the surface and within surface no charge exists . In case of hollow metal sphere , all charges reside on surface . In this way , in both solid and hollow sphere , all charge reside on the surface and no charge resides inside it . Hence due to absence of any charge inside , there is no electric field inside the sphere in both the cases .
Hence in both the case electric field is zero .
option C is correct .