Answer: 9.05 s
Explanation: v = u +at
v = 3 m/s
u = 22 m/s
Deceleration = 2.1 m/s^2
∴ 3 = 22 - 2.1 ( t )
∴ t = 9 .05 s
How much would a pair of 0.5 kg shoes weigh on Earth? (Include units in
your answer) *
Answer:
1.1 lbs
Explanation:
To convert kg to lbs you multiply kilograms by 2.2. So 0.5kg × 2.2 equals to 1.1 lbs
NEED AWNSER NOW! WILL MARK BRAINLY! Which term is defined as the ratio of the speed of light in a vacuum to the speed of light in the material it is passing through?
index of reflection
index of refraction
angle of reflection
angle of incidence
Answer:
Index of refraction.
Answer:
index of refraction
Explanation:
I just took the k12 quiz.
A hose on the ground projects a water current upwards at an angle 40 to the horizontal at velocity 20 m/s find height at which water hits a wall at 8 m away from the hose (consider that acceleration due to gravity =9.8 m/s2)
Answer:
The water hits the wall at a height of 5.38 m
Explanation:
Projectile Motion
It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.
The object describes a parabolic path given by the equation:
[tex]{\displaystyle y=\tan(\theta )\cdot x-{\frac {g}{2v_{0}^{2}\cos ^{2}\theta }}\cdot x^{2}}[/tex]
Where:
y = vertical displacement
x = horizontal displacement
θ = Elevation angle
vo = Initial speed
The hose projects a water current upwards at an angle of θ=40° at a speed vo=20 m/s.
The height at which the water hits a wall located at x=8 m from the hose is:
[tex]{\displaystyle y=\tan40^\circ\cdot 8-{\frac {9.8}{2*20^{2}\cos ^{2}40^\circ }}\cdot 8^{2}}[/tex]
Calculating:
y = 5.38 m
The water hits the wall at a height of 5.38 m
A force of 30 N stretches a very light ideal spring 0.73 m from equilibrium. What is the force constant (spring constant) of the spring
The forces constant (spring constant) of the spring will be 41.09 N/m.
What is spring force?The force required to extend or compress a spring by some distance scales linearly concerning that distance is known as the spring force. Its formula is;
F = kx
The given data in the problem is;
F is the spring force = 30 N
K is the spring constant= ?
x is the displacement of spring = 0.73 m
The spring constant is;
K =F/x
K=30/0.73
K=41.09 N/m
Hence the force constant (spring constant) of the spring will be 41.09 N/m.
To learn more about the spring force refer to the link;
https://brainly.com/question/4291098
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Describe what happens to the moving boat when the oars are out of the water and the forward thrust is zero
Answer:
The boat won't be able to move if the oars were out and there was no thruster. If there was a flow of the water then yes there would be a moving boat.
The magnitude of vector vector A is 84.9 m and it points in the +y axis direction. The magnitude of vector vector B is 195.0 m and it points at an angle of 41.0° counterclockwise from +x axis. The magnitude of vector vector C is 126.2 m and it points in the +x axis direction.
Solution:
The magnitude of A vector is 84.9 m in the positive y-axis direction.
So the X component of A =0
the Y component of A = 84.9 m
Now the magnitude of B vector is 195 m and it makes an angle of 41° in the direction from the positive x-axis direction.
So the X component of B = B cos 41°
= 195 x cos 41°
= 195 x 0.75 = 146.25 m
the Y component of B = B sin 41°
= 195 x sin 41°
= 195 x 0.65 = 126.75 m
Now it is given that vector C has a magnitude of 126.2 m and it makes a direction towards the positive x-axis.
So the X component of C =126.2 m
the Y component of C = 0
Comparing all these, we get
1. B vector has the largest X component
2. B vector has the largest Y component
help please asap due 20 minutes please help me
) prove that the acceleration due to gravi
is independent to the mass of the falling body
Explanation:
Let the mass of the body is m. The gravitational force acting on the object is given by :
[tex]F=\dfrac{GMm}{r^2}[/tex] ....(1)
G is universal gravitational constant
M is mass of Earth
r is the distance between the body
The acceleration of falling objects due to the gravitational force of Earth is equal to the acceleration due to gravity (g).
F = mg ...(2)
g is acceleration due to gravity
From equation (1) and (2) :
[tex]\dfrac{GMm}{r^2}=mg\\\\g=\dfrac{GM}{r^2}[/tex]
Here, M is mass of Earth and r is the distance. Hence, the acceleration due to gravi ty is independent to the mass of the falling body
A golf ball is sitting on a tee. The ball is struck with a golf club and flies
through the air. How does the force on the club compare with the force on the
ball when momentum is transferred between the club and ball?
Answer:
c i kn now it is
Explanation:
a squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy
What is the squirrels mass
Answer:
yeet yeet yeet yeet
Explanation:
Kinetic energy (K.E):-
So, the Mass of the Squirrel is 0.51 Kg (or) 510 grams.
A squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy.
What is the squirrel’s mass?
Answer: 0.51 kg
Jerry is pushing a 50-kg box across a moth floor with an acceleration of 0.6 m/s2. What force is he applying to the box? *
83.3 N
0.012 N
0
30 N
Answer:
30 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 50 × 0.6
We have the final answer as
30 NHope this helps you
Can u anser 5,6 on the picture
Answer: Number 6 is Periods
Explanation:
A 250-kg moose stands in the middle of the railroad tracks in Sweden, frozen by the lights of an oncoming 10,000kg train traveling at 20m/s. Even though the engineer attempted in vain to slow the train down in time to avoid hitting the moose, the moose rides down the remaining track sitting on the train’s cowcatcher. What is the final velocity of the train and moose after the collision?
(Momentum & Impulse)
Answer:
The final velocity of the train and the moose after collision is approximately 19.51 m/s
Explanation:
The given mass of the moose, m₁ = 250 kg
The velocity of the moose, v₁ = 0
The mass of the oncoming train, m₂ = 10,000 kg
The velocity of the train, v₂ = 20 m/s
The velocity of the moose and the train after collision = v₃
By the principle of conservation of linear momentum, the total initial momentum before the collision = The total final momentum after collision
m₁·v₁ + m₂·v₂ = (m₁ + m₂)·v₃
Therefore, by substitution, we have;
250×0 + 10,000× 20 = (10,000 + 250) × v₃
200,000 = 10,250 × v₃
v₃ = 200,000/10,250 ≈ 19.51 m/s
The final velocity of the train and the moose after collision = v₃ ≈ 19.51 m/s
Which electron dot diagram shows the bonding between 2 chlorine atoms?2 dots then C l with 2 dots above and 1 dot below then 2 dots then 2 dots then C l with 2 dots above and 1 dot below then 2 dots.2 dots then C l with 2 dots above and 2 dots below then 2 dots then C l with 2 dots above and 2 dot below then 2 dots.2 dots then C l with 2 dots above and 2 dots below then 1 dot then C l with 2 dots above and 2 dots below then 2 dots.2 dots then C l with 2 dots above and 1 dot below then 3 dots then 3 dots then C l with 2 dots above and 1 dot below then 2 dots.
Answer:
It is B
Explanation:
Answer: 2nd answer
Explanation: took exam
Under what circumstances will the projectile have the greatest velocity when it hits the ground? Explain please
Lina holownicza wytrzymuje działanie siły o wartości co najwyżej 1400 N (większa siła powoduje rozerwanie liny). Oblicz maksymalne przyspieszenie, jakie może osiągnąć samochód o masie 2000 kg holowany na takiej linie po poziomej jezdni, jeżeli wiadomo, że łączna wartość sił oporu ruchu tego samochodu w czasie holowania wynosi 400 N.
Odpowiedź:
0,5 m / s²
Wyjaśnienie:
Jeśli się uwzględni:
Masa samochodu (m) = 2000 kg
Wartość siły oporu podczas holowania = 400N
Lina o maksymalnej sile może wytrzymać = 1400N
Maksymalne przyspieszenie = siła / masa samochodu
Siła = maksymalna siła - siła oporu
Siła = 1400 N - 400 N = 1000 N.
Stąd maksymalne przyspieszenie;
1000 N / 2000 kg
= 0,5 m / s²
A friend is coming to Tim’s house to study after school. What directions would Tim give for reaching his house from the entrance of the school?
(I WILL GIVE BRAINLIEST)
Answer:
go up the street as you exit the house and make a right and keep going up for 3 blocks and you should see the school
show your work. john uses a 25N force to push a boulder off a cliff that is 312m tall. What is the work done on the boulder?
Answer:7800
work=force x distance
Force in Newtons
Distance in Meters
Work in Joules
While delivering 125 kg blocks of ice to a local village, Kristoff and his family (and Sven too) come upon a cliff that is 5.7 m above them. To solve their problem, they build a catapult that will launch their blocks of ice with an initial velocity of 15 m/s and at an angle of 45 degrees above the ground.
Prove that the catapult will successfully launch the ice blocks up to the top of the cliff. Quantities you will need to solve for will be initial vertical velocity, horizontal velocity, the distance the catapult is from the cliff, and the amount of time it takes the ice to reach the cliff. Assume that they are experts and have arranged everything so that the ice blocks just barely reach the top of their parabolic path when they land at the top of the cliff and make a smooth landing. Draw a picture of the scene to help you visualize what is happening.
After landing on the flat land above, each block of ice travels 20 meters while slowing to a stop.
What is the rate of acceleration while the blocks slow to a stop?
How long do the blocks take to slow to a stop?
What is the amount of friction between the ice and the snowy ground?
How long do the blocks take to slow to a stop?
What is the amount of friction between the ice and the snowy ground?
Answer:
a. Since h = 5.74 m > 5.7 m, the height of the cliff, the block of ice will successfully launch to the top of the cliff.
b. -2.81 m/s²
c. 3.78 s
d. -351.25 N
Explanation:
a. After landing on the flat land above, each block of ice travels 20 meters while slowing to a stop.
For the block of ice to reach the top of the cliff, its maximum height, h should be greater than or equal to 5.7 m. That is, h ≥ 5.7 m.
The maximum height of a projection h, projected with an initial velocity v at an angle Ф is h = v²sin²Ф/2g where g = acceleration due to gravity = 9.8 m/s².
For the block of ice, v = 15 m/s and Ф = 45°. So,
h = v²sin²Ф/2g
= (15 m/s)²sin²45/(2 × 9.8 m/s²)
= 225 (m/s)²(1/√2)²/19.6 m/s²
= 225 (m/s)²(1/2)/19.6 m/s²
= 112.5 (m/s)²/19.6 m/s²
= 5.74 m
Since h = 5.74 m > 5.7 m, the height of the cliff, the block of ice will successfully launch to the top of the cliff.
The graph is in the attachment.
b. What is the rate of acceleration while the blocks slow to a stop?
Using v² = u² + 2as where u = initial horizontal velocity of block = 15m/scos45° = 10.61 m/s, v = final velocity of block = 0 m/s since it stops, a = acceleration and s = distance block moves = 20 m
So, a = (v² - u²)/2s
substituting the variables into the equation, we have
a = ((0 m/s)² - (10.61 m/s)²)/2(20 m)
= - 112.57 (m/s)²)/40 m
= -2.81 m/s²
c. How long do the blocks take to slow to a stop?
Using v = u + at where u = initial horizontal velocity of block = 10.61 m/s v = final velocity of block = 0 m/s since it stops, a = acceleration = -2.81 m/s² and t = time it takes block of ice to stop
So, making t subject of the formula,
t = (v - u)/a
substituting the values of the variables, we have
t = ( 0 m/s - 10.61 m/s)/-2.81 m/s²
= -10.61 m/s/-2.81 m/s²
= 3.78 s
d. What is the amount of friction between the ice and the snowy ground?
The frictional force, f = net force on block of ice
f = ma where m = mass of bock = 125 kg and a = acceleration of block = -2.81 m/s²
f = ma
= 125 kg(2.81 m/s²)
= -351.25 N
what are the effects of moon rotation and revolution
Answer:
The effects of the Moon's rotation includes;
1) The Moon rotation and revolution gives the appearance of a perfectly still Moon to observers of the Moon from the Earth
2) The Moon has two sides, the near side that continuously faces the Earth and the "back" or far side, which is also known as the dark side of the Moon
The effect of the Moon's revolution
1) The tides in the ocean and water bodies, due to the gravitational forces from the Moon
2) The changes in the observed shape of the Moon due to the amount of Sunlight that is able to be reflected from the Moon as a result of the relative position of the Moon, the Earth and the Sun, at a given point in time
3) Lunar and Solar eclipse, when the Earth and the Moon blocks the light coming from the Sun respectively, due to their combined revolution
Explanation:
The duration of the Moon's orbit round the Earth = 27.322 days
The time it takes the Moon to rotate round its axis = 27 days
The Moon is the closest cosmic body to the Earth.
Es muy común que cuando se viaja hacia un río o lago se juegue "ranita", el cual consiste en lanzar una piedra horizontalmente hacia adelante para que cuando ésta toque la superficie del agua haga varios "saltos" sobre el agua. Durante un juego de estos, un desocupado nota que una de las piedras que arroja se demora 0,4 s en tocar la superficie del agua y la toca a 2,5 m de la orilla del lago, desde donde fue lanzada. Encuentre: a) La altura de la que fue lanzada la piedra. b) La velocidad con la que fue lanzada.
Answer:
a) La piedra es lanzada desde una altura de 0,785 metros.
b) La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.
Explanation:
a) Dado que la piedra es lanzada horizontalmente, tenemos que la piedra experimenta un movimiento horizontal a velocidad constante y uno vertical uniformemente acelerado debido a la gravedad. La altura de la que fue lanzada la piedra se puede determinar mediante la siguiente ecuación cinemática:
[tex]y = y_{o}+v_{o,y}\cdot t +\frac{1}{2}\cdot g\cdot t^{2}[/tex] (1)
Donde:
[tex]y[/tex] - Altura final, medida en metros.
[tex]y_{o}[/tex] - Altura inicial, medida en metros.
[tex]v_{o,y}[/tex] - Componente vertical de la velocidad inicial, medida en metros por segundo.
[tex]t[/tex] - Tiempo, medido en segundos.
[tex]g[/tex] - Aceleración gravitacional, medida en metros por segundo cuadrado.
Si sabemos que [tex]y = 0\,m[/tex], [tex]v_{o,y} = 0\,\frac{m}{s}[/tex], [tex]t = 0,4\,s[/tex] y [tex]g = -9,807\,\frac{m}{s^{2}}[/tex], entonces la altura inicial de la piedra es:
[tex]y_{o} = y-v_{o,y}\cdot t -\frac{1}{2}\cdot g\cdot t^{2}[/tex]
[tex]y_{o} = 0\,m-\left(0\,\frac{m}{s} \right)\cdot (0,4\,s)-\frac{1}{2}\cdot \left(-9,807\,\frac{m}{s^{2}} \right) \cdot (0,4\,s)^{2}[/tex]
[tex]y_{o} = 0,785\,m[/tex]
La piedra es lanzada desde una altura de 0,785 metros.
b) Ahora, obtenemos el componente horizontal de la velocidad inicial a partir de la siguiente ecuación cinemática:
[tex]v_{o,x} = \frac{x-x_{o}}{t}[/tex] (2)
Donde:
[tex]x_{o}[/tex], [tex]x[/tex] - Posiciones horizontales iniciales y finales, medidas en metros.
[tex]t[/tex] - Tiempo, medido en segundos.
Si tenemos que [tex]x_{o} = 0\,m[/tex], [tex]x = 2,5\,m[/tex] y [tex]t = 0,4\,s[/tex], entonces el componente horizontal de la velocidad inicial es:
[tex]v_{o,x} = \frac{2,5\,m-0\,m}{0,4\,s}[/tex]
[tex]v_{o,x} = 6,25\,\frac{m}{s}[/tex]
La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.
In a place covered by shadow of cloud sun cannot be seen . Explain with reasons .
Answer:
Because even though our eyes have a huge dynamic range (ability to pick out details in sharply lit and lesser lit areas simultaneously) than any camera, there's a limit.
When there's strong sunlight, your pupils contract and let less light in, which makes the shadows look darker.
When it's cloudy, your pupils widen and let more light in, which makes the shadows look less dark.
Do some experiments with a camera and you'll soon get the hang of it.
NOTE: Also test HDR (high dynamic range) photography, where the camera takes three or more pictures in quick succession, with different exposure settings, and combines them to get the most detail of both bright and dark areas. The result is more or less what we percieve.
13. Austin rode his bike 10 m/s for two minutes. How far did he travel? A. 200 meters B. 1200 meters C. 1000 meters D. 20 meters
Answer:
B. 1200
Explanation:
60 sec in one min in 2 min there will be 120 sec. 10x120=1200
A Chevy Camaro drives straight off the top level off a parking garage at 13 m/s. If the car landed 90 meters away from the base of the parking garage, how high (height) was the top level?
a200.64
b234.64
c34.64
d134.64
Answer:
234.64 m
Explanation:
Using the formula for calculating range;
Range R = u√2H/g
u is the speed = 13m/s
H is the maximum height
g is the acceleration due to gravity = 9,8m/s²
R = 90m
Get the maximum height;
90 = 13√2H/9.8
90/13 = √2H/9.8
6.923 = √2H/9.8
square both sides
6.923² = 2H/9.8
469.698 = 2H
H = 469.698/2
H = 234.64
Hence the top level was 234.64 m high.
I need help with science homework
Answer:
4. atmosphere and geosphere
5. atmosphere and hydrosphere
6. hydrosphere and geosphere (? not sure about this one sorry)
7. hydrosphere and geosphere
8. biosphere and geosphere
A container with a mass of 5 kg is lifted to a height of 8 m. How much work is done by the gravitational force
Answer:
work done=392J
Explanation:
work done =Mgh
DATA
M=5kg
g= 9.8
h=8m
W=Mgh
W=5*9.8*8
W=392J
The work done by the gravitational force is equal to -400 J.
What is gravitational force?The gravitational force can be described as a force that attracts a body towards the center of the earth or any physical object that has mass. Every object has mass and exerts a gravitational pull on another object with mass.
Given, the mass of the container, m = 5 Kg
The height from the ground level, h = 8 m
The work done can be calculated by using the below-mentioned formula:
[tex]W = m\times g\times h[/tex]
Where g is the gravitational acceleration.
As the container is lifted in the upward direction therefore the value of the g will be negative.
Substitute the values of m, g, and h in the above formula:
W = 5 × (-10) × 8
W = -400 J.
The negative sign shows that the work is done against the gravitational force.
Therefore, the work done by the gravitational force is equal to -400 J.
Learn more about gravitational force, here:
brainly.com/question/24783651
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A 0.15 kg ball is moving with a velocity of
35 m/s. Find the momentum of the ball.
Answer:
5.25 kg.m/sExplanation:
The momentum of an object can be found by using the formula
momentum = mass × velocity
From the question we have
momentum = 0.15 × 35
We have the final answer as
5.25 kg.m/sHope this helps you
draw position time graph when speed is increasing
Explanation:
We need to draw position-time graph when the speed is increasing.
The slope of position-time graph gives the speed of an object.
Position means distance covered.
When the speed of an object is increasing with time. It means it is moving with increasing speed.
The attached figure shows the position -time graph when speed is increasing.
When two ocean plates come together, one ocean plate __________________
under the other, causing a chain of ________________ __________________
to form.
Answer:
A subduction zone is also generated when two oceanic plates collide — the older plate is forced under the younger one — and it leads to the formation of chains of volcanic islands known as island arcs.
Explanation:
The slope of a position-time graph can be used to find the moving obiects
Answer:
Is this a true and false statement?
Explanation: