Answer:
b) During the turn, the door exerts a leftward force on you.
Explanation:
Although at the curve the car makes a turn without deceleration or accelerating, we must consider the fact that the direction of the instantaneous velocity of the car, which is always pointing tangentially away from the curve, is changing, producing a centripetal acceleration (acceleration is the rate of change of velocity) towards the center of the curve. The result is that a force inwards towards the center of the curve from the door, is exerted on you in the leftwards direction when your body collides with the door.
In separate experiments, a large number of particles (all with the same charge, but with a wide variety of masses and speeds) are fired into a region containing a uniform magnetic field. The velocities of the particles are all perpendicular to the field. What do the particles that move in orbits of the same radius have in common
Answer:
they have the same mass
Explanation:
The force applied by the field is a function of the charge and velocity, so the acceleration experienced by a particle will be dependent upon its mass. Particles in orbits with the same radius are exhibiting the same acceleration, so must have the same mass.
If 2 balls had the same volume but ball a has twice as much mass as babil which one will have the greater density
Water is boiled at 1 atm pressure in a 20-cm-internal diameter polished copper pan on an electric range. If it is observed that the water level in the pan drops by 8.00 cm in 15 minutes, determine the inner surface temperature of the pan.
Answer:
11.3298W
Explanation:
The rate of heat transfer is determined from the enthalpy of vaporization at the give pressure obtained and the mass flow rate. The mass flow rate is determined from the volume of the boiled water, the given time interval and the specific volume of the saturated liquid.
Given that
1atm as the atmospheric pressure
Internal diameter = 20cm = 0.2m
Time = 15mins = (15×60)secs
Latent heat of vaporization (hevap) = 2256.6
Q = mh(evap)
= m/∆t . hevap
= V/αliq∆t ×h(evap)
D^2π∆h/4αliq ∆t × hevap
= 0.2^2 ×π×0.8×2256.5/4×0.001043×15×60
=0.04×3.142×0.08×2256.6/2.00256
= 22.68876/2.00256
Q = 11.3298W
According to the model in which active galactic nuclei are powered by supermassive black holes, the high luminosity of an active galactic nucleus primarily consists of
Answer:
the high luminosity of an active galactic nucleus primarily consists of light emitted by hot gas in an accretion disk that swirls around the black hole
A 2.0-kg stone is tied to a 0.50 m long string and swung around a circle at a constant angular velocity of 12 rad/s. The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin. The magnitude of the torque about the origin is
Answer:
108 NmExplanation:
Given data
mass m= 2 kg
radius r= 0.5 m
angular velocity ω= 12 rad/s
distance d= 0.75 m
we know that
[tex]Force= mass * acceleration\\\ acceleration= w^2r\\\\ Force= m*w^2r\\\\Force =2*12^2*0.5= 144 N[/tex]
we know that torque = F*d= 144*0.75= 108 Nm
The ball tends to come back to the centerline of the flow when it is pushed by an external disturbance. Explain this phenomenon using the curvature of streamlines.
Answer is given below
Explanation:
given data
we will consider here
Ping-Pong ball weighs = 3.1 g
diameter = 4.2 cm
solution
Whenever the ball is pushed, the length of the airflow along the outer edge increases and it accelerates. According to Bernoulli's equation. As the speed increases, the pressure decreases, so the pressure at the outer end is reduced. As the pressure at the outer edge is low, the extra air jet pushes it back to the center line.
A solid wooden door, 90 cm wide by 2.0 m tall, has a mass of 35 kg. It is open and at rest. A small 500-g ball is thrown perpendicular to the door with a speed of 20 m/s and hits the door 60 cm from the hinged side, causing it to begin turning. The ball rebounds along the same line with a speed of 16.0 m/s relative to the ground.
Required:
How much energy is lost during this collision?
a. 15J
b. 16J
c. 13J
d. 4.8J
e. 30J
Answer:
the kinetic energy lost in the collison is a) 30 J
Explanation:
given data
mass of door m1 = 35 kg
width a = 90 cm = 0.9 m
the mass of ball m2 = 500 g = 0.5 kg
initial speed of ball u = 20 m/s
final speed of ball v = 16 m/s
r = 60 cm = 0.6 m
soluion
we will consider here final angular speed of the door = w
so now we use conservation of angular momentum that is
Li = Lf ........................1
that is express as
m2 × u × r = I × w + m2 × v × r
put here value and we get
0.5 × 20 × 0.6 = [tex](m1 \times \frac{a^2}{12})[/tex] × w + 0.5 × 16 × 0.6
solve it we get
w = 0.508 rad/s
so that here
the kinetic energy lost in the collison,
KE = KE initial - KE final ..................2
put here value
KE = 0.5 × m2 × u² - (0.5 × I × w² + 0.5 × m2 × v²)
KE = 0.5 × (0.5 × 20² - (35 × 0.9² ÷ 12) × 0.508² - 0.5 × 16²) J
KE = 30 J
the kinetic energy lost in the collison is a) 30 J
The air is partially removed at the upper part of the tube and the valve is closed. The height of the water above its pool surface is 0.7 m.The height of the liquid above its pool surface is 0.33 m.The difference in the heights of the pool surfaces is 0.22 m.
Required:
Find the density of the unknown liquid.
Answer:
2121.2kg/m^3 is the density of the test liquid on the left
Explanation:
See attached file
Fermat's principle is of least time rather than of least distance. Would least distance apply for reflection? For refraction? Why are your answers different?
Answer:
Fermat's principle states that the path taken by a ray between two given points is the path that can be traversed in the least time.
Thus this least distance being the same as least time will only apply to reflection alone because in reflection light travels In the same direction so least distance will also mean least time. But for refraction light travels in different mediums at different speeds so least distance and least time paths will definately not be the same
Two people play tug of war. The 100-kg person on the left pulls with 1,000 N, and the 70-kg person on the right pulls with 830 N. Assume that neither person releases their grip on the rope with either hand at any time, assume that the rope is always taut, and assume that the rope does not stretch. What is the magnitude of the tension in the rope in Newtons
Answer:
The tension on the rope is T = 900 N
Explanation:
From the question we are told that
The mass of the person on the left is [tex]m_l = 100 \ kg[/tex]
The force of the person on the left is [tex]F_l = 1000 \ N[/tex]
The mass of the person on the right is [tex]m_r = 70 \ kg[/tex]
The force of the person on the right is [tex]F_r = 830 \ N[/tex]
Generally the net force is mathematically represented as
[tex]F_{Net} = F_l - F_r[/tex]
substituting values
[tex]F_{Net} = 1000-830[/tex]
[tex]F_{Net} = 170 \ N[/tex]
Now the acceleration net acceleration of the rope is mathematically evaluated as
[tex]a = \frac{F_{net}}{m_I + m_r }[/tex]
substituting values
[tex]a = \frac{170}{100 + 70 }[/tex]
[tex]a = 1 \ m/s ^2[/tex]
The force [tex]m_i * a[/tex]) of the person on the left that caused the rope to accelerate by a is mathematically represented as
[tex]m_l * a = F_r -T[/tex]
Where T is the tension on the rope
substituting values
[tex]100 * 1 = 1000 - T[/tex]
=> T = 900 N
An inquisitive physics student and mountain climber climbs a 47.0-m-high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.12 m/s.
(a) How long after release of the first stone do the two stones hit the water?
(b) What initial velocity must the second stone have if the two stones are to hit the water simultaneously?
magnitude =
(c) What is the speed of each stone at the instant the two stones hit the water?
first stone =
second stone =
Answer:
a) Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds, b) The initial velocity of the second stone is -16.038 meters per second, c) The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.
Explanation:
a) The time after the release after the release of the first stone can be get from the following kinematic formula for the first rock:
[tex]y_{1} = y_{1,o} + v_{1,o} \cdot t +\frac{1}{2}\cdot g \cdot t^{2}[/tex]
Where:
[tex]y_{1}[/tex] - Final height of the first stone, measured in meters.
[tex]y_{1,o}[/tex] - Initial height of the first stone, measured in meters.
[tex]v_{1,o}[/tex] - Initial speed of the first stone, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]g[/tex] - Gravity constant, measured in meters per square second.
Given that [tex]y_{1,o} = 47\,m[/tex], [tex]y_{1} = 0\,m[/tex], [tex]v_{1,o} = -2.12\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the following second-order polynomial is built:
[tex]-4.984\cdot t^{2} - 2.12\cdot t + 47 = 0[/tex]
Roots of the polynomial are, respectively:
[tex]t_{1} \approx 2.866\,s[/tex] and [tex]t_{2}\approx -3.291\,s[/tex]
Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds.
b) As the second stone is thrown a second later than first one, its height is represented by the following kinematic expression:
[tex]y_{2} = y_{2,o} + v_{2,o}\cdot (t-t_{o}) + \frac{1}{2}\cdot g \cdot (t-t_{o})^{2}[/tex]
[tex]y_{2}[/tex] - Final height of the second stone, measured in meters.
[tex]y_{2,o}[/tex] - Initial height of the second stone, measured in meters.
[tex]v_{2,o}[/tex] - Initial speed of the second stone, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]t_{o}[/tex] - Initial absolute time, measured in seconds.
[tex]g[/tex] - Gravity constant, measured in meters per square second.
Given that [tex]y_{2,o} = 47\,m[/tex], [tex]y_{2} = 0\,m[/tex], [tex]t_{o} = 1\,s[/tex], [tex]t = 2.866\,s[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the following expression is constructed and the initial speed of the second stone is:
[tex]1.866\cdot v_{2,o}+29.926 = 0[/tex]
[tex]v_{2,o} = -16.038\,\frac{m}{s}[/tex]
The initial velocity of the second stone is -16.038 meters per second.
c) The final speed of each stone is determined by the following expressions:
First stone
[tex]v_{1} = v_{1,o} + g \cdot t[/tex]
Second stone
[tex]v_{2} = v_{2,o} + g\cdot (t-t_{o})[/tex]
Where:
[tex]v_{1,o}, v_{1}[/tex] - Initial and final velocities of the first stone, measured in meters per second.
[tex]v_{2,o}, v_{2}[/tex] - Initial and final velocities of the second stone, measured in meters per second.
If [tex]v_{1,o} = -2.12\,\frac{m}{s}[/tex] and [tex]v_{2,o} = -16.038\,\frac{m}{s}[/tex], the final speeds of both stones are:
First stone
[tex]v_{1} = -2.12\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (2.866\,s)[/tex]
[tex]v_{1} = -30.227\,\frac{m}{s}[/tex]
Second stone
[tex]v_{2} = -16.038\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (2.866\,s-1\,s)[/tex]
[tex]v_{2} = -34.338\,\frac{m}{s}[/tex]
The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.
what are all the rays that come from the sun called List all?
Explanation:
what are all the rays that come from the sun called List all?
cosmetic rays, gamma rays, x-rays, ultraviolet rays, infrared rays, microwaves, short radio waves and long radio waves.
1) A net force of 75.5 N is applied horizontally to slide a 225 kg crate across the floor.
a. Compute the acceleration of the crate?
Answer:
The acceleration of the crate is [tex]0.3356\,\frac{m}{s^2}[/tex]
Explanation:
Recall the formula that relates force,mass and acceleration from newton's second law;
[tex]F=m\,a[/tex]
Then in our case, we know the force applied and we know the mass of the crate, so we can solve for the acceleration as shown below:
[tex]F=m\,a\\75.5\,N=225\,\,kg\,\,a\\a=\frac{75.5}{225} \,\frac{m}{s^2} \\a=0.3356\,\,\frac{m}{s^2}[/tex]
A particle with charge q is to be brought from far away to a point near an electric dipole. Net nonzero work is done if the final position of the particle is on:__________
A) any point on the line through the charges of the dipole, excluding the midpoint between the two charges.
B) any point on a line that is a perpendicular bisector to the line that separates the two charges.
C) a line that makes an angle of 30 ∘ with the dipole moment.
D) a line that makes an angle of 45 ∘with the dipole moment.
Answer:
Net nonzero work is done if the final position of the particle is on options A, C and D
Explanation:
non zero work is done if following will be the final position of the charges :
A) Any point on the line through the charges of the dipole , excluding the midpoint between the two charges.
C) A line that makes an angle 30° with the dipole moment.
D) A line that makes an angle 45° with the dipole moment.
Light in vacuum is incident on the surface of a glass slab. In the vacuum the beam makes an angle of 38.0° with the normal to the surface, while in the glass it makes an angle of 26.0° with the normal. What is the index of refraction of the glass?
Answer:
n_glass = 1.404
Explanation:
In order to calculate the index of refraction of the light you use the Snell's law, which is given by the following formula:
[tex]n_1sin\theta_1=n_2sin\theta_2[/tex] (1)
n1: index of refraction of vacuum = 1.00
θ1: angle of the incident light respect to normal of the surface = 38.0°
n2: index of refraction of glass = ?
θ2: angle of the refracted light in the glass respect to normal = 26.0°
You solve the equation (1) for n2 and replace the values of all parameters:
[tex]n_2=n_1\frac{sin\theta_1}{sin\theta_2}=(1.00)\frac{sin(38.0\°)}{sin(26.0\°)}\\\\n_2=1.404[/tex]
The index of refraction of the glass is 1.404
A charging bull elephant with a mass of 5500 kg comes directly toward you with a speed of 4.70 m/s . You toss a 0.160-kg rubber ball at the elephant with a speed of 7.50 m/s(a) When the ball bounces back toward you, what is its speed? (b) How do you account for the fact that the ball's kinetic energy has increased?
Answer:
v2 = - 16.899 m/s
velocity of ball increases so that the kinetic energy of the ball increases.
Explanation:
given data
mass of elephant, m1 = 5500 kg
mass of ball, m2 = 0.160 kg
initial velocity of elephant, u1 = - 4.70 m/s
initial velocity of ball, u2 = 7.50 m/s
solution
we consider here final velocity of ball = v2
so collision formula is express as for v2
[tex]v_{2}=\left ( \frac{2m_{1}}{m_{1}+m_{2}} \right )u_{1}+\left ( \frac{m_{2}-m_{1}}{m_{1}+m_{2}} \right )u_{2}[/tex] .................1
put here value and we get
[tex]v_{2}=\left ( \frac{2\times 5500}{5500+0.160} \right )(-4.70)+\left ( \frac{0.16-5500}{5500+0.160} \right )(7.50)[/tex]
solve it we get
v2 = - 16.899 m/s
here negative sign shows that the ball bounces back towards you
and
here we know the velocity of ball increases so that the kinetic energy of the ball increases.
and due to this effect, it will gain in energy is due to the energy from the elephant mass
a certain volume of dry air at NTP is allowed to expand five times of it original volume under adiabatic condition.calculate the final pressure.(air=1.4)
Answer:
Final pressure 0.105atm
Explanation:
Let V1 represent the initial volume of dry air at NTP.
under adiabatic condition: no heat is lost or gained by the system. This does not implies that the constant temperature throughout the system , but rather that no heat gained or loss by the system.
Adiabatic expansion:
[tex]\frac{T_1}{T_2} =(\frac{V_1}{V_2} )^{\gamma -1}[/tex]
273/T2=(5V1/V1)^(1.4−1)
273/T2=5^0.4
Final temperature T2=143.41 K
Also
P1/P2=(V2/V1)^γ
1/P2=(5V1/V1)^1.4
Final pressure P2=0.105atm
Suppose your friend claims to have discovered a mysterious force in nature that acts on all particles in some region of space. He tells you that the force is always pointed away a definite point in space, which we can call the force center. The magnitude of the force turns out to be proportional to B/r3, where r is the distance from the force center to any other point. Your friend says that it has been determined that the constant of proportionality has been determined to be B= 2 (in units to be determined later), so that the magnitude of the force on a particle (in newtons) can be written as 2r^3, when the particle is at a distance r from the force center.
Required:
Write an expression of potential energy.
Answer:
U = 1 / r²
Explanation:
In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related
F = - dU / dr
this derivative is a gradient, that is, a directional derivative, so we must have
dU = - F. dr
the esxresion for strength is
F = B / r³
let's replace
∫ dU = - ∫ B / r³ dr
in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product
let's evaluate the integrals
U - Uo = -B (- / 2r² + 1 / 2r₀²)
To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)
U = B / 2r²
we substitute the value of B = 2
U = 1 / r²
A woman living in a third-story apartment is moving out. Rather than carrying everything down the stairs, she decides to pack her belongings into crates, attach a frictionless pulley to her balcony railing, and lower the crates by rope.
Required:
How hard must she pull on the horizontal end of the rope to lower a 49 kg crate at steady speed?
Answer:
T = 480.2N
Explanation:
In order to find the required force, you take into account that the sum of forces must be equal to zero if the object has a constant speed.
The forces on the boxes are:
[tex]T-Mg=0[/tex] (1)
T: tension of the rope
M: mass of the boxes 0= 49kg
g: gravitational acceleration = 9.8m/s^2
The pulley is frictionless, then, you can assume that the tension of the rope T, is equal to the force that the woman makes.
By using the equation (1) you obtain:
[tex]T=Mg=(49kg)(9.8m/s^2)=480.2N[/tex]
The woman needs to pull the rope at 480.2N
An experiment is set up to test the angular resolution of an optical device when red light (wavelength ????r ) shines on an aperture of diameter D . Which aperture diameter gives the best resolution? D=(1/2)????r D=????r D=2????r
Explanation:
As per Rayleigh criterion, the angular resolution is given as follows:
[tex]\theta=\frac{1.22 \lambda}{D}[/tex]
From this expression larger the size of aperture, smaller will be the value of angular resolution and hence, better will be the device i.e. precision for distinguishing two points at very high angular difference is higher.
The first Leyden jar was probably discovered by a German clerk named E. Georg von Kleist. Because von Kleist was not a scientist and did not keep good records, the credit for the discovery of the Leyden jar usually goes to physicist Pieter Musschenbroek from Leyden, Holland. Musschenbroek accidentally discovered the Leyden jar when he tried to charge a jar of water and shocked himself by touching the wire on the inside of the jar while holding the jar on the outside. He said that the shock was no ordinary shock and his body shook violently as though he had been hit by lightning. The energy from the jar that passed through his body was probably around 1 J, and his jar probably had a capacitance of about 1 nF.A) Estimate the charge that passed through Musschenbroek's body.
B) What was the potential difference between the inside and outside of the Leyden jar before Musschenbroek discharged it?
Answer:
a) q = 4.47 10⁻⁵ C
b) ΔV = 4.47 10⁴ V
Explanation:
A Leyden bottle works as a condenser that accumulates electrical charge, so we can use the formula of the energy stored in a capacitor
U = Q² / 2C
Q = √ (2UC)
let's reduce the magnitudes to the SI system
c = 1 nF = 1 10⁻⁹ F
let's calculate
q = √ (2 1 10⁻⁹-9)
q = 0.447 10⁻⁴ C
q = 4.47 10⁻⁵ C
b) for the potential difference we use
C = Q / ΔV
ΔV = Q / C
ΔV = 4.47 10⁻⁵ / 1 10⁻⁹
ΔV = 4.47 10⁴ V
Four point charges have the same magnitude of 2.4×10^−12C and are fixed to the corners of a square that is 4.0 cm on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.
Answer:
7.2N/C
Explanation:
Pls see attached file
Find the potential energy associated with a 79-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level. Take the zero of potential energy at sea level.
Answer:
P = 1470980 J
Explanation:
We have,
Mass of the hiker is 79 kg
It is required to find the potential energy associated with a 79-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level.
It is given by :
[tex]P=mgh\\\\P=79\times 9.8\times 1900\\\\P=1470980\ J[/tex]
So, the potential energy of 1470980 J is associated with a hiker.
The normal force equals the magnitude of the gravitational force as a roller coaster car crosses the top of a 58-m-diameter loop-the-loop. What is the car's speed at the top?
Answer:
16.87 m/s
Explanation:
To find the speed of the car at the top, when the normal force is equal the gravitational force, we just need to equate both forces:
[tex]N = P[/tex]
[tex]m*a_c = mg[/tex]
[tex]a_c[/tex] is the centripetal acceleration in the loop:
[tex]a_c = v^2/r[/tex]
So we have that:
[tex]mv^2/r = mg[/tex]
[tex]v^2/r = g[/tex]
[tex]v^2 = gr[/tex]
[tex]v = \sqrt{gr}[/tex]
So, using the gravity = 9.81 m/s^2 and the radius = 29 meters, we have:
[tex]v = \sqrt{9.81 * 29}[/tex]
[tex]v = \sqrt{284.49} = 16.87\ m/s[/tex]
The speed of the car is 16.87 m/s at the top.
Rope BCA passes through a pulley at point C and supports a crate at point A. Rope segment CD supports the pulley and is attached to an eye anchor embedded in a wall. Rope segment BC creates an angle of ϕ = 51.0 ∘ with the floor and rope segment CD creates an angle θ with the horizontal. If both ropes BCA and CD can support a maximum tensile force Tmax = 120 lb , what is the maximum weight Wmax of the crate that the system can support? What is the
Answer:
Wmax = 63.65 ≈ 64 lb
Explanation:
How much work will it take to lift a 2-kg pair of hiking boots 2 meters off the
ground and onto a shelf in your closet?
O A. 2.45 J
OB. 4J
C. 39.2 J
D. 20 J
Answer:
Option C - 39.2 J
Explanation:
We are given that;
Mass; m = 2 kg.
Distance moved off the floor;d = 10 m.
Acceleration due to gravity;g = 9.8 m/s².
We want to find the work done.
Now, the Formula for work done is given by;
Work = Force × displacement.
In this case, it's force of gravity to lift up the boots, thus;
Formula for this force is;
Force = mass x acceleration due to gravity
Force = 2 × 9.8 = 19.2 N
∴ Work done = 19.6 × 2
Work done = 39.2 J.
Hence, the Work done to life the boot of 2 kg to a height of 2 m is 39.2 J.
Answer:39.2J
Explanation: I just answered this question and this was the correct answer. 4J is the wrong answer.
If there is a potential difference V between the metal and the detector, what is the minimum energy Emin that an electron must have so that it will reach the detector
Answer:
Emin= eV
Explanation:
This minimum amount of energy is the work function and so for an electron to reach the detector the energy must b more than work function and the stopping potential so
Emin=eV
5. (10 points) Which of the following statements is(are) correct: A. Resistivity purely depends on internal properties of the conductor; B. Resistance purely depends on internal properties of the conductor; C. Resistivity depends on the size and shape of the conductor; D. Resistance depends on the size and shape of the conductor; E. A and D; F. B and C.
Answer:
B and D
Explanation:
Because
R= resistivity xlenght/ Area
Where R= resistance
which of the following has the lowest density? A. Water B. Air C. Mineral Water D. Salt Water
Air has lower density than water, mineral Water, or salt water. (B)
A goalie kicks a soccer ball straight vertically into the air. It takes 5.00 s for the ball to reach its maximum height and come back down to the level of the crossbar. Assume the crossbar of a soccer goal is 2.44 m above the ground. (a) How fast was the ball originally moving when it was kicked. (b) How much longer would it take the ball to reach the ground?
Answer:
(a) vo = 24.98m/s
(b) t = 5.09 s
Explanation:
(a) In order to calculate the the initial speed of the ball, you use the following formula:
[tex]y=y_o+v_ot-\frac{1}{2}gt^2[/tex] (1)
y: vertical position of the ball = 2.44m
yo: initial vertical position = 0m
vo: initial speed of the ball = ?
g: gravitational acceleration = 9.8m/s²
t: time on which the ball is at 2.44m above the ground = 5.00s
You solve the equation (1) for vo and replace the values of the other parameters:
[tex]v_o=\frac{y-y_o+1/2gt^2}{t}[/tex]
[tex]v_o=\frac{2.44m-0.00m+1/2(9.8m/s^2)(5.00s)^2}{5.00s}\\\\v_o=24.98\frac{m}{s}[/tex]
The initial speed of the ball is 24.98m/s
(b) To find the time the ball takes to arrive to the ground you use the equation (1) for y = 0m (ground) and solve for t:
[tex]0=24.98t-\frac{1}{2}(9.8)t^2\\\\t=5.09s[/tex]
The time that the ball takes to arrive to the ground is 5.09s
We have that for the Question, it can be said that the speed of ball and How much longer would it take the ball to reach the ground is
u=25.13m/sX=0.095sec
From the question we are told
A goalie kicks a soccer ball straight vertically into the air. It takes 5.00 s for the ball to reach its maximum height and come back down to the level of the crossbar. Assume the crossbar of a soccer goal is 2.44 m above the ground.
(a) How fast was the ball originally moving when it was kicked.
(b) How much longer would it take the ball to reach the ground?
a)
Generally the Newton equation for the Motion is mathematically given as
[tex]S=ut+1/2at^2\\\\Therefore\\\\2.44=ut+1/2(9.8)(5)^2\\\\u=25.13m/s\\\\[/tex]
b)
Generally the Newton equation for the Motion is mathematically given as
[tex]S=ut+1/2at^2\\\\Therefore\\\\t=\frac{-24}{a}\\\\t=\frac{-2*25.013}{9.81}\\\\t=5.095sec\\\\[/tex]
Therefore
[tex]X=5.095-5[/tex]
X=0.095sec
For more information on this visit
https://brainly.com/question/23379286