The amount of time the ball stay in the air is (t)=3.04 Seconds
What is time?The best example that would help us understand and know what time are the clock and the calendar. This clock gives us the exact hour, minutes and seconds. The calendar tells us the exact day, month and year.
How can we calculate the time?To calculate the amount of time the ball stay in the air, we are using the formula,
T=2Vsinθ/g
Here we are given,
V= The velocity of the football.=26 m/s.
θ= The angle that the football makes in the air= 35°
g= The acceleration due to gravity = 9.8 m/s².
We have to calculate, the amount of time the ball stay in the air = T
Now, we put the known values in the above equation,
T=2Vsinθ/g
Or, T= 2*26*sin (35°)/9.80
Or, T= 3.04 Seconds.
Thus, from the above calculation we can conclude that the amount of time the ball stay in the air is (t)=3.04 Seconds
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3. Two balls are rolling toward each other. One has a momentum of 85kg*m/s, and the other has a momentum of -85kg*m/s. What will be the total momentum of the system after they collide? In what direction will they move after they collide if it is an inelastic collision? Answer:
Answer:
The total momentum is zero.
Explanation:
This problem can be solved by applying the momentum conservation theorem and the amount of motion. This theorem tells us that the amount of motion is conserved before and after a collision.
In the next equation, we will write to the left of the equal sign the amount of motion before the collision and to the right the amount of motion after the collision.
[tex](P_{1})-(P_{2})=P_{3}[/tex]
where:
P₁ = momentum of the ball moving to the right, before the collision = 85 [kg*m/s]
P₂ = momentum of the ball moving to the left, before the collision = - 85 [kg*m/s]
P₃ = Final momentum after the collision [kg*m/s]
[tex](85) - 85 = P_{3}\\P_{3}= 0[/tex]
There is no movement of any of the balls, they remain at rest after the impact.
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
Exponential population growth is a major issue that threatens the stability of on Earth. Which of the following is not a factor used to measure human population growth for a given country?
are active centers of galaxies that radiate powerful x-rays visible across the universe
Answer:
no
Explanation:
PLS HELP. What is the result of two waves meeting each other?
A. diffraction
B. reflection
C. refraction
D. interference
Answer:
D- Interference
Explanation:
When two waves meet they interfere with one another resulting in Interference
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.
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
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
Which situation illustrates conservation of mechanical energy?
a. When friction and air resistance are balanced.c. Only external forces are acting on the object.
b.
A pendulum swinging back and forth.
Both a and b.
Please select the best answer from the choices provided
A
OB
O C
OD
d.
Mark this and return
Save and Exit
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Submit
Situations that illustrate the conservation of mechanical energy is (d) Both a and b.
According to the principle of mechanical energy conservation, a system's entire mechanical energy is conserved, meaning that energy cannot be generated or destroyed and can only be internally transformed from one form to another provided the forces acting on the system are conservative in nature.
(a) Sometimes referred to as a dissipative force, air resistance contributes to a loss in the total amount of mechanical energy. So, if friction and air resistance are balanced, the total mechanical energy will be conserved. Hence, friction and air resistance balanced is an example of conservation of mechanical energy.
(b) The pendulum bob's height above the tabletop changes continuously as it swings back and forth, which affects its speed. Potential energy is lost as the height drops, whereas kinetic energy is gained simultaneously. However, the bob's total potential and kinetic energy are always constant. There is a conservation of all mechanical energy. There is only a change from kinetic to potential energy; there is neither a gain nor a loss of mechanical energy (and vice versa).
(c) Let's assume that an external force exerted on an object over a certain distance causes the object's total mechanical energy to change. The item gains mechanical energy if the external force (or nonconservative force) exerts positive work. Energy is gained in a quantity equal to the work done on the object. The item loses mechanical energy if the external force (or nonconservative force) exerts negative work. Hence, energy is not conserved.
Both (a) and (b) illustrate the conservation of energy.
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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.
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 0.5 kg ball was kicked and was going 15 m/s.What is the kinetic energy of the ball?
Answer:
[tex]\boxed {\boxed {\sf 56.25 \ Joules}}[/tex]
Explanation:
Kinetic energy can be found using the following formula:
[tex]E_k=\frac{1}{2}mv^2[/tex]
where m is the mass and v is the velocity.
The mass of the ball is 0.5 kilograms and the velocity is 15 meters per second
[tex]m= 0.5 \ kg \\v= 15 \ m/s \\[/tex]
Substitute the values into the formula.
[tex]E_k=\frac{1}{2} (0.5 \ kg ) * (15 \ m/s)^2[/tex]
First, evaluate the exponent.
(15 m/s)²= (15 m/s) * (15 m/s) = 225 m²/s²[tex]E_k=\frac{1}{2} (0.5 \ kg ) *(225 \ m^2/s^2)[/tex]
Multiply 0.5 kg by 225 m²/s²
[tex]E_k=\frac{1}{2}(112.5 \ kg* m^2/s^2)[/tex]
Multiply 112 kg*m²/s² by 1/2, or divide by 2.
[tex]E_k=56.25 \ kg*m^2/s^2[/tex]
1 kg*m²/s² is equal to 1 JouleTherefore, our answer of 56.25 kg*m²/s² is equal to 56.25 Joules.[tex]E_k= 56.25 \ J[/tex]
The kinetic energy of the ball is 56.25 Joules
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.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
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
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.
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:
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 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
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
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
Push a box across a table with a force of 5 newtons and the box moves
0.5 meters. How much work has been accomplished?
If you do
Answer:
2.5J
Explanation:
Work done = Force x Distance
= 0.5 x 5
= 2.5J
The work done that should be accomplished should be 2.5 J.
Calculation of the work done:Since the force is 5 newtons
The box moves 0.5 meters
So here the work done should be
= Force x Distance
= 0.5 x 5
= 2.5J
here we basically multiply the force with the distance so that the work done should be determined.
hence, The work done that should be accomplished should be 2.5 J.
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How long does it take to accelerate from 8.0 m/s to 20.0 m/s at a rate of acceleration of 3.0 m/s2?
Question 17 options:
4.0 s
0.25 s
36 s
9.33 s
Answer:
4.0s
Explanation:
v = v_0 + a*t -> t = (v - v_0) / a
v = 20.0m/s, v_0 = 8.0 m/s, a = 3.0 m/s^2
t = (20.0 - 8.0) / 3.0 = 4.0s
Answer:
t = 4.0 s
Explanation:
Given:
V₁ = 8.0 m/s
V₂ = 20.0 m/s
a =3.0 m/c²
______________
t - ?
a = ( V₂ - V₁ ) / t
t = ( V₂ - V₁ ) / a
t = ( 20.0 - 8.0 )/ 3.0 = 12.0 /3.0 = 4.0 s
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
what is transferred by a radio wave?
A. air
B. energy
C. matter
D. space
(for science)
Answer: B. energy
Explanation:
One type of gas mixture used in anesthesiology is a 50%/50% mixture (by volume) of nitrous oxide (N2O) and oxygen (O2), which can be premixed and kept in a cylinder for later use. Because these two gases don't react chemically at or below 2000 psi, at typical room temperatures they form a homogeneous single gas phase, which can be considered an ideal gas. If the temperature drops below -6∘C, however, N2O may begin to condense out of the gas phase. Then any gas removed from the cylinder will initially be nearly pure O2; as the cylinder empties, the proportion of O2 will decrease until the gas coming from the cylinder is nearly pure N2O. In a test of the effects of low temperatures on the gas mixture, a cylinder filled at 24.0 ∘C to 2000 psi (gauge pressure) is cooled slowly and the pressure is monitored.
What is the expected pressure at -5.00∘C if the gas remains a homogeneous mixture?
Answer:
1850 psi
Explanation:
Which element is usually listed first in the formula of a binary
molecular compound
a the element with the smallest atomic radius
b the element with the lowest electronegativity
C the element with the smallest atomic mass
d the element that is first in alphabetical order help!
Answer: the element with the lowest electronegativity
Explanation:
You throw a rock from the upper edge of a 88.0 m vertical dam with a speed of 21.0 m/s at 58.0∘ above the horizon. Neglect any effects due to air resistance.
How much time 1 after throwing the rock will it return to its initial height?
how much time t2 after throwing the rock will it hit the water flowing out at the base of the dam
The time taken for the rock to return to its initial height is 6.4 seconds.
The time taken for the rock to hit the water flowing out at the base of the dam is 2.8 seconds.
Time of motion of the rockThe time taken for the rock to return to its initial height is calculated as follows;
h = Vy(t) + ¹/₂gt²
88 = (21 x sin58)(t) + 4.9t²
88 = 17.8t + 4.9t²
4.9t² + 17.8t - 88 = 0
solve the quadratic equation using formula method;
a = 4.9, b = 17.8, c = -88
t₁ = -6.4 seconds
t₂ = 2.8 seconds
The negative time indicates the time taken for the rock to travel back to its initial position.
Also, the positive time indicates the time taken for the rock to fall from the given height.
Thus, the time taken for the rock to return to its initial height is 6.4 seconds.
The time taken for the rock to hit the water flowing out at the base of the dam is 2.8 seconds.
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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