Answer:
Explanation:
The balloon would require a time of
t = d/v = 13.5/ (23.6cos38) = 0.7259...s
to travel the horizontal distance.
the vertical position relative to the throw point at that time is
h = 0 + (23.6sin38)(0.7259) + ½(-9.8)(0.7259²)
h = 7.9652...
so as long as the adjacent building is at least 8.0 m higher than the student position, the balloon is in the air for 0.726 s.
If the building is shorter than 8.0 m above the student, the balloon will land on the building roof and will be in the air for a longer period of time
Suppose a cannon is sitting on top of a 50.0 m high hill and a 5.00 kg cannon ball is fired with a velocity of 30.0 m/s at some unknown angle. What is the velocity of the cannon ball when it strikes the earth?
The final velocity of the cannon ball when it strikes the earth is 43.36 m/s.
The given parameters:
Height of the hill, h = 50 mMass of the cannon, m = 5 kgVelocity of the ball, v = 30 m/sThe final velocity of the cannon ball when it strikes the earth is calculated by applying the principle of conservation of energy as follows;
[tex]P.E_i + K.E_i = P.E_f + K.E_f\\\\mgh_i + \frac{1}{2} mv_i^2 = mgh_f + \frac{1}{2} mv_f^2\\\\gh_i + \frac{1}{2} v_i^2 = g(0) + \frac{1}{2} v_f^2\\\\2gh_i + v_i^2 = v_f^2\\\\v_f = \sqrt{2gh_i + v_i^2 } \\\\v_f = \sqrt{(2 \times 9.8 \times 50) \ \ + \ \ 30^2} \\\\v_f = 43.36 \ m/s[/tex]
Thus, the final velocity of the cannon ball when it strikes the earth is 43.36 m/s.
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You are on the roof of a building 30 m above the ground. Using hand crank 3 points
you are to lift a 300 kg dresser on to the roof. How much work must to
generated to lift the dresser? *
Answer:
The work will increase the potential energy.
W = PE = mgh = 300(9.8)(30) = 88200 J
A body is thrown up into the air takes a time of 4s to reach the height. What is the velocity with which the body was thrown up.(g=10ms2)
Answer:
40m/s
Explanation:
V= u + at
v= 0
a= -10
t= 4
0= u -40
u= 40m/s
just trial!!!!!!!
A ball is launched from ground level at 30 m/s at an angle of 35° above the horizontal. how far does it go before it is at ground level again
Answer:
Explanation:
Ignoring air resistance
Initial vertical velocity is 30sin35 = 17.2 m/s
Gravity reduces this velocity to zero in a time of
t = v/g =17.2 / 9.8 = 1.755 s
it takes the same time to come back down to ground level for a total flight time of 2(1.755) = 3.51 s
The horizontal velocity is 30cos35 = 24.57 m/s
the distance traveled horizontally is
d = vt = 24.57(3.51) = 86.298... = 86 m
is it true that playing badmenton help you to become a better person?
Answer:
There is no scientific evidence that playing specifically l badminton makes you a better person, but sport and exercise in general release hormones which can make you feel more happy therefore making you nicer to the people around you and 'a better person'.
Answer:
It's true because playing any sport makes a person happy. So a happy person is a better person.
Please Mark as brainliest.
Help me outtttt jejjejejeje
Answer:
do it got a picture
Explanation:
The mass of fifteen washers is _____ kg, which exerts a force of _____ N
Answer:
It could be related with the lesson from which this question belongs as far we did not read the lesson
Sorry
An object following a straight-line path at constant speed
A.) has no forces acting on it.
B.) has a net force acting on it in the direction of motion.
C.) has zero acceleration.
D.) must be moving in a vacuum.
E.) none of the above
An object following a straight-line path at constant speed is option C.) has zero acceleration.
Are there any forces acting on a moving item traveling in a straight line at a constant speed?There are no forces operating on a body if it is travelling straight ahead at a steady speed. There are no forces operating on a body if it is travelling straight ahead at a steady speed.
Note that the physics concept of acceleration measures how quickly an object's motion is changing. An object's speed or velocity is what largely defines its motion.
Therefore, An object is considered to be accelerating when its velocity changes over time and as such since acceleration of the object is said to be zero, one can say that the net force acting on it is also zero.
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Which one of the following statements concerning a collection of gas molecules at a certain temperature is true?A. The lower the temperature, the greater are the molecular speeds. B. Most of the molecules have the same kinetic energy. C. All molecules possess the same momentum. D. The molecules have a range of kinetic energies. E. All molecules move with the same velocity.
Answer:
D Is true - the velocities (and squared) follow the appropriate statistical curve
The molecules have a range of kinetic energies at a certain temperature. As the temperature increases, their kinetic energy and molecular speed increases.
What is kinetic theory of gases ?Kinetic theory of gases describes the nature of ideal gases and their volume, pressure and kinetic energy. As per this theory the gases are made of tiny particles which have negligible mass compared to that of the container.
Kinetic theory states that the kinetic energy of all gases increases with increase in temperature which is independent of the masses and and at certain temperature all the gases are having same range of kinetic energies.
The velocity of all the gaseous particles increases with increasing in temperature which results in the increase in kinetic energy. Hence, option D is correct.
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Check if correct or not:
Directions: Using what you learned about energy describe the energy transfer or transformations for each of the items below.
1. Clapping Your Hands:
Kinetic- sound
2. Dropping Your Pencil:
3. The Toaster:
Electric-Thermal/Heat
4. A Cat Lying in a Sunny Window:
Light-Thermal/heat
5. Lifting a Book Over Your Head:
kinetic-potential
6. The Radio:
Electric-sound
Tell me if correct or not
Answer:
Looks good to me
Explanation:
#2 should probably be turning potential energy to kinetic.
59. (II) The crate shown in Fig. 4-60 lies on a plane tilted at an angle A = 25.0° to the horizontal, with Mk 0.19. (a) Determine the acceleration of the crate as it slides down the plane. (b) If the crate starts from rest 8.15 m up along the plane from its base, what will be the crate's speed when it reaches the bottom of the incline?
Explanation:
a) We need to write down first Newton's 2nd law as applied to the given system. The equations of motion for the x- and y-axes can be written as follows:
[tex]x:\;\;\;\;\;mg\sin 25° - \mu_kN = ma\;\;\;\;\;\;(1)[/tex]
[tex]y:\;\;\;\;\;N - mg\cos 25° = 0\;\;\;\;\;\;\;\;\;(2)[/tex]
From Eqn(2), we see that
[tex]N = mg\cos 25°\;\;\;\;\;\;\;(3)[/tex]
so using Eqn(3) on Eqn(1), we get
[tex]mg\sin 25° - \mu_kmg\cos 25° = ma[/tex]
Solving for the acceleration, we see that
[tex]a = g(\sin 25° - \mu_k\cos 25°)[/tex]
[tex]\;\;\;\;= 2.45\:\text{m/s}^2[/tex]
b) Now that we have the acceleration, we can now solve for the velocity of the crate at the bottom of the plane. Using the equation
[tex]v^2 = v_0^2 + 2ax[/tex]
Since the crate started from rest, [tex]v_0 = 0.[/tex] Thus our equation reduces to
[tex]v^2 = 2ax \Rightarrow v = \sqrt{2ax}[/tex]
[tex]v = \sqrt{2(2.45\:\text{m/s}^2)(8.15\:\text{m})}[/tex]
[tex]\;\;\;\;= 6.32\:\text{m/s}[/tex]
AnswAnswer This!!!!!!
I'll give brainliest to whoever gets it right.
Question 2 of 25
If JKLM is a parallelogram, what is the length of LM?
K
18
8
M
O A. 10
O B. 18
O C. 8
D. 26
SUBMIT
Answer:its 10
Explanation:
Answer:
8
Explanation:
A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. The total mechanical energy is 25 J. The maximum speed of the block is:
Answer:
Explanation:
easy way
when system is all kinetic energy, velocity is at a maximum
E = ½mv²
v = √(2E/m) = √(2(25)/0.5) = √100 = 10 m/s
harder way
ω = √(k/m) = √(80/0.5) = √160 rad/s
When the system is entirely spring potential, the amplitude A is
E = ½kA²
A = √(2E/k) = √(2(25)/80) = 0.790569... = 0.79 m
maximum velocity is ωΑ = 0.79√160 = 10 m/s
Question No. 1 Marks = = 5 +5 +2 = 12
The driver of a 2.0 × 103 kg red car traveling on the highway at 45m/s slams on his brakes to avoid striking a second yellow car in front of him, which had come to rest because of blocking ahead as shown in above Fig. After the brakes are applied, a constant friction force of 7.5 × 103 N acts on the car. Ignore air resistance.
(a) Determine the least distance should the brakes be applied to avoid a collision with the other vehicle?
(b) If the distance between the vehicles is initially only 40.0 m, at what speed would the collision occur?
(c) Write your conclusive observations on the result obtained from this numerical. i.e. the importance of Physics in daily life.
Question No. 2 Marks = 8
Consider an automobile moving at v mph that skids d feet after its brakes lock. Calculate how far it would skid if it was moving at 2v and the brakes were locked.
The kinematics and Newton's second law we can find the results for the questions about the braking movement of the car are;
Question 1.
a) The stopping distance is: x = 270 m
b) The initial velocity is: v₀ = 17.3 m / s
c) Concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.
Question 2.
The stopping distance is: x = 4d
Given parameters
Mass of the red carriage m1 = 2,0 10³ kg Red car speed vo = 45 m / s Friction force fr = 7.5 10³ N.To find
Question 1.
a) Minimum braking distance.
b) If the distance is x = 40.0 m, what speed should the vehicles have?
c) Conclusive importance of physics in daily life.
Question 2.
The distace to stop.
Kinematics studies the movement of bodies, looking for relationships between the position, speed and acceleration of bodies.
v² = v₀² - a2 x
Where v and v₀ are the current and initial velocity, respectively, at acceleration and x the distance traveled.
Newton's second law states that the net force is proportional to the mass and the acceleration of the body.
F = ma
Where F is force, m is mass and acceleration.
In the attachment we see a diagram of the forces in the system. Let's look for the acceleration of the body
fr = m a
a =[tex]\frac{fr}{m}[/tex]
a = [tex]\frac{7.5 \ 10^3}{2.0 \ 10^3 }[/tex]
a = 3.75 m / s²
This acceleration is in the opposite direction to the speed.
Let's find the distance needed to stop, the final speed is zero.
0 = v₀² - 2 ax
x = [tex]\frac{v_o^2 }{ 2a}[/tex]
Let's calculate.
x = [tex]\frac{45^2 }{2 3.75}[/tex]
x = 270 m
This is the minimum distance that the two vehicles must separate to avoid a collision.
b) We look for speed.
v₀ = [tex]\sqrt{2ax}[/tex]
v₀ = [tex]\sqrt{2 \ 3.75 \ 40.0}[/tex]
v₀ = 17.3 m / s
c) The concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.
2) They indicate that the initial velocity is v and the distance traveled to stop is d, let's find the acceleration.
0 = v₀² - 2ax
Let's substitute.
a = [tex]\frac{v^2}{2d}[/tex]
They ask the distance traveled if this car traveled from an initial speed 2v.
0 = v² - 2 a x
x = [tex]\frac{v^2}{2a}[/tex]
We substitute
x = [tex]\frac{(2v)^2 }{2} \ (\frac{2d}{v^2})[/tex]
x = 4 d
In conclusion, using the kinematic relations and Newton's second law we can find the results for the questions about the braking movement of the car are;
Question 1
a) The stopping distance is: x = 270 m
b) The initial velocity is: v₀ = 17.3 m / s
c) concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.
Question 2.
The stopping distance is x = 4d
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A 300 cm rope under a tension of 120 N is set into oscillation. The mass density of the rope is 120 g/cm. What is the frequency of the first harmonic mode (m
Answer:
Explanation:
f = [tex]\sqrt{T/(m/L)} / 2L[/tex]
T = 120 N
L = 3.00 m
(m/L) = 120 g/cm(100 cm/m / 1000 g/kg) = 12 kg/m
(wow that's massive for a "rope")
f = [tex]\sqrt{120/12} /(2(3))[/tex])
f = [tex]\sqrt{10\\}[/tex]/6 = 0.527 Hz
This is a completely silly exercise unless this "rope" is in space somewhere as the weight of the rope (353 N on earth) far exceeds the tension applied.
A much more reasonable linear density would be 120 g/m resulting in a frequency of √1000/6 = 5.27 Hz on a rope that weighs only 3.5 N
why is it so important that you take care of your nervous system?
Answer:
The nervous system handles the stress response, which, if overworked, can eventually lead to diseases ranging from high blood pressure to diabetes.
Explanation:
hope I helped
Why does the earth stay in orbit around the sun instead of drifting away from it into space?
A Electric force between the Sun and Earth
B Magnetic force between the Sun and Earth
Gravitational force between the Sun and Earth
Answer:
b it is b part answer i think so
12) A horizontal force of 200 N is applied to move a 55-kg cart (initially at rest) across a 10 m level surface. What is the final speed of the cart? [hint: use work – energy principle] [3 marks]
Hi there!
We can use the following:
W = ΔKE = F · d
Find the work done on the cart:
W = 200 · 10 = 2000 J
Now, this is equal to the change in kinetic energy of the object. Its initial kinetic energy is 0 J since it starts from rest, so:
2000J = KEf - KEi
KE is given as:
[tex]KE = \frac{1}{2}mv^2[/tex]
2000J = 1/2(55)v²
4000 = 55v²
√(4000/55) = 8.53 m/s
A 2.55 kg piece of lead at 40 degree Celsius is placed in a very large quantity of water at 10 degree Celsius and thermal equilibrium is eventually reached. Calculate the entropy change of the lead that occurs during this process. The specific heat of lead is 130 J/(kg K).
6.6 J/K
1.4 J/K
190 J/K
100 J/K
6.2 J/k
i think this is a answer
Which of the following waves DO NOT require a medium to travel?
Answer:
hey man, think you forgot to put the questions
Word Bank:
Electrical
Mechanical
Chemical
Light
Thermal
Sound
Answer:
Light to
Electrical to
mechanical and sound
Based on the law of conservation of mass, if approximately 20g of water reacts in the following equation,
producing about 15g of oxygen, what mass of hydrogen (H2) was produced in this reaction?
2H20
2H2 + O2
Answer: H2O2
Explanation:
First, calculate the moles of H2O2 reacting. In order to do this, we must evaluate the relative molecular mass ( Mr ) of hydrogen peroxide
4 points
This machine has a sharp edge and helps spread or cut something
The diagram below shows a 5.00-kilogram block
at rest on a horizontal, frictionless table.
5.00-kg
block
Table
Which of the following is the correct name and strength of the force holding the block up?
The name and strength of the force holding the block up is 50 N upward - Normal force.
The given parameters:
Mass of the block, m = 5 kgThe weight of the block acting downwards due to gravity is calculated as follows;
W = mg
where;
g is acceleration due to gravity = 10 m/s²W = 5 x 10
W = 50 N (downwards)
Since the block is at rest, an a force equal to the weight of the block must be acting upwards. This force is known as normal reaction.
Fₙ = 50 N (upwards)
Thus, the name and strength of the force holding the block up is 50 N upward - Normal force.
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The block will remain on the table because the normal force balances with the weight of the block. The correct answer is 50 N upward normal force
From the diagram shown a 5.00-kilogram block at rest on a horizontal, frictionless table. The weight of the block will act downward which will be
Weight W = mg
let g = 10 m/[tex]s^{2}[/tex]
W = 5 x 10
W = 50 N
The block will also produce an equal but in opposite direction of a normal force which is equal to the weight of the block. That is,
Normal force N = 50 N
The block will remain on the table because the normal force balances with the weight of the block.
Therefore, the correct name and strength of the force holding the block up is 50 N upward normal force.
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What is the net force here?
11 N left
6 N right
1 N right
4 N right
answer = 6n to the right
Explanation:
2n plus 4n equals 6n
since 6n is more than 5n it goes 6n to the right
An irregularly shaped object weighs 11.20 N in air. When immersed in water, the object has an apparent weight of 3.83 N. Find its density.
Answer:
Weight of object = 11.2 N
Apparent weight = 3.83 N when immersed
Weight of water displaced = 11.2 - 3.83 = 7.37 N
d (density) W / V weight / volume the weight density
Wo = Vo do weight of object
Ww = Vo dw where Ww is weight of equivalent volume of water = 7.37
Wo / Ww = do / dw dividing previous equations
do = 11.2 / 7.37 dw = 1.52 dw
The density of the object is 1.52 that of water
The density of water is 1000 kg / m^3 * 9.8 m/s^2 = 9800 N/m^3
So the weight density is 14900 N/m^3
An irregularly shaped object weighs 11.20 N in air. When immersed in water, the object has an apparent weight of 3.83 N. It's density can be calculated as 1523 kg/m³.
To find the density, the given values are,
Weight in air = 11.20 N
Weight in water = 3.83 N
density of water = 1000 kg/m³
What is meant by Density?According to the Archimedes principle, when a body is immersed in a liquid partly or wholly, it experiences an upward force which is called buoyant force. The buoyant force is equal to the loss in weight of the body.
Loss in weight of the object = Weight of object in air - weight of object in water
Loss in weight = 11.20 - 3.83 = 7.37 N
Volume of body x density of water x g = 7.37
Let V be the volume of body
V x 1000 x 9.8 =7.37
V = 7.5× 10⁻⁴ m³
Weight in air = Volume of body x density of body x g
11.20 = 7.5× 10⁻⁴ x d x 9.8
d = 1523 kg/m³.
Thus, the density of body is 1523 kg/m³.
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Which sentence states Newton's second law?
Answer:
Force is equal to the change in momentum per change in time.
Explanation:
That situation is described by Newton's Second Law of Motion. According to NASA, this law states, "Force is equal to the change in momentum per change in time. For a constant mass, force equals mass times acceleration." This is written in mathematical form as Force = mass.
Motion Velocity
Reference point Speed
1. An object is in __________ when its distance from a(n) ________ is changing.
2. Speed is given direction is called _______________
3. ____________ can be calculated if you know the distance that an object travels in one unit of time.
Answer:
1. An object is in motion when its distance from another object is changing.
2.Speed is given direction is called velocity.
Speed can be calculated.......
What is activated by the sympathetic system?
1. The digestive system
2. The "rest and digest" response
3. O The "fight or flight" response
4. The "breed and feed" response
Answer:
1 the digestive system thats all