Answer:
one sec let me think
Explanation:
(a)The average velocity of the ball over the following time intervals will be [3,4] ft/sec.
(b)The instantaneous velocity at time t = 3 will be32 ft/sec.
(c)The instantaneous velocity at time t = 6 will be -64 ft/sec.
(d)The ball will hit the ground at 13.4 sec.
What is velocity?The change of displacement with respect to time is defined as the velocity. velocity is a vector quantity. it is a time-based component.
The given data in the question will be ,
u is the initial velocity by which ball thrown=128 ft/sec.
V₃ is the instantaneous velocity at time t=3 sec.
V₆ is the instantaneous velocity at time t=6 sec.
t is the time when ball hits the ground=?
(a) Given equation for the displacement
s(t) =128t-16t² (on differenting got the velocity )
v(t) = 128-32t
Time when velocity is zero will be
[tex]\rm{ t=\frac{128}{32}[/tex]
[tex]\rm{ t=4 sec[/tex]
If the velocity got in the equation is 128 and 32 ft /sec. it can be only when the average velocity is [3,4] ft/sec .
Hence the average velocity obtained from the problem will be [3,4] ft/sec
(b)
s(t) =128t-16t² (on differenting got the velocity )
v(t) = 128-32t
At time( t=3 sec)
v(3) = 128-32×3
v(3) =32 m/sec.
Hence the instantaneous velocity at time t = 3 will be32 ft/sec.
(c)
s(t) =128t-16t² (on differenting got the velocity )
v(t) = 128-32t
At time( t=6 sec)
v(6) = 128-32×6
v(6) = -64 m/sec.
Hence the instantaneous velocity at time t = 6 will be -64 ft/sec.
(d)
According to Newtons third equation of motion we got
v=u+gt
If the body returens from a certain height at max height its velocity must be zero; ( u=0)
[tex]\rm t=\frac{(v-u)}{g} \\\\\ \rm t=\frac{(128-0)}{9.81}\\\\\rm t=13.04 sec.[/tex]
Hence the ball will hit the ground at 13.4 sec.
To learn more about the velocity refer to the link ;
https://brainly.com/question/862972
What happens when a moving object experiences no net force?
Answer:
An object with no net forces acting on it which is initially at rest will remain at rest. If it is moving, it will continue to move in a straight line with constant velocity. Forces are "pushes" or "pulls" on the object, and forces, like velocity and acceleration are vector quantities.
Which of the following best describes the location of the
mantle?
A
Above the crust
B
Between the crust and the lithosphere
С
Between the crust and the core
D
Beneath the core
Answer:
The mantle exists above the crust of the earth
A scientist plans to release a weather balloon from ground level, to be used for high-altitude atmospheric measurements. The balloon is spherical, with a radius of 2.10 m, and filled with hydrogen. The total mass of the balloon (including the hydrogen within it) and the instruments it carries is 24.0 kg. The density of air at ground level is 1.29 kg/m3. (a) What is the magnitude of the buoyant force (in N) acting on the balloon, just after it is released from ground level
Answer:
Fb = 490.4 N
Explanation:
According to Archimedes' principle, any object submerged in a fluid, receives a push upward (which we call buoyant force) equal to the weight of the volume of the fluid removed by the object.We can express this force (Fb), in terms of the density and the volume of the fluid, as follows:[tex]F_{b} = \rho * V * g (1)[/tex]
The volume removed from the fluid by the balloon is just the volume of the balloon, assuming that it is a perfect sphere, as follows:[tex]V = \frac{4}{3}*\pi * R^{3} = \frac{4}{3} *\pi *(2.1m)^{3} = 38.8 m3 (2)[/tex]
Replacing by the givens and (2) in (1), we get:[tex]F_{b} = \rho * V * g = 1.29 kg/m3* 38.8m3*9.8m/s2 = 490.4 N (3)[/tex]
In a mass spectrometer, a specific velocity can be selected from a distribution by injecting charged particles between a set of plates with a constant electric field between them and a magnetic field across them (perpendicular to the direction of particle travel). If the fields are tuned exactly right, only particles of a specific velocity will pass through this region undeflected. Consider such a velocity selector in a mass spectrometer with a 0.105 T magnetic field.
Part (a) What electric field strength, in volts per meter, is needed to select a speed of 3.8 × 106 m/s?
Part (b) What is the voltage, in kilovolts, between the plates if they are separated by 0.75 cm?
Answer:
a).[tex]$3.99 \times 10^5 \ v/m$[/tex]
b). 2.9925 kV
Explanation:
Given :
For mass spectrometer
The magnetic field = B
B = 0.105 T
a). Given speed, v = [tex]$3.8 \times 10^6 \ m/s$[/tex]
We known
[tex]$\frac{E}{B}=v$[/tex]
∴ [tex]$E= 3.8 \times 10^6 \times 0.105$[/tex]
[tex]$=3.99 \times 10^5 \ v/m$[/tex]
b). Now spectrometer, d = 0.75 cm
[tex]$d=0.75 \times 10^{-2} \ m$[/tex]
We known
[tex]$E=\frac{V}{d}$[/tex]
[tex]$V = E\times d$[/tex]
[tex]$V = 3.99 \times 10^5 \times 0.75 \times 10^{-2}$[/tex]
[tex]$V = 2.9925 \times 10^3 \ V$[/tex]
= 2.9925 kV
When you are driving on the freeway and following the car in front of you, how close is too close? Let's do an estimation.
1. Pick a car model (preferably the one you drive, but can also be any car of your dream), and find its stopping distance at highway speeds (you can usually find this type of data online).
2. Assuming that the car in front of you suddenly does a hard brake. For simplicity, assume that its braking performance is about the same as yours. Then also assume a reasonable amount of reaction time on your part (the time delay between seeing the brake lights lit up and applying your own brake). In order for you not to run into the car your are following, what's the closest distance you need to keep between the two cars?
3. Redo the same calculation if the vehicle in front of you is a typical big-rig truck. Find its braking data online.
4. There is a rule of thumb which says that you must stay one car length behind the car in front of you for every 10 mi/h of driving speed. From your calculation, does this rule make sense?
Answer:
1) v= 90km/h d = 70 m, 2) x₁ = v t_r, x₁ = 6.25 m, 3) x₁=6.25 no change
4) x = 22 m
Explanation:
1) for the first part, you are asked to find the minimum safety distance with the vehicle in front
The internet is searched for the stopping distance for two typical speeds on the highway
v (km/ h) v (m/s) d (m)
90 25 70
100 27.78 84
the safe distance is this distance plus the distance traveled during the person's reaction time, which can be calculated with infirm movement
v = x / t_r
x₁ = v t_r
the average reaction time is t_r = 0.25s for a visual stimulus and t_r 0.17 for an auditory stimulus
therefore the safe distance is
x_total = x₁ + d
2) The distance is the sum of the distance traveled in the reaction
x₁ = v t_r
for v = 90 km / h
x₁ = 25 0.25
x₁ = 6.25 m
for v = 100 km / h
x₁ = 27.78 0.25
x₁ = 6.95 m
the total distance is
x_total = x₁ + d
for v = 90 km / h
x_total = 25 0.25 + 70
x_total = 76.25 m
this is the distance until the cars stop and do not collide
3) the stopping distance of a truck is
v = 90 km / h d = 100 m
in this case we see that the braking distance is much higher,
the safe distance is given by the distance traveled during the reaction, as the truck brakes slower than the car this distance does not change
4) let's analyze the empirical rule: maintain the length of a car for each increase in speed of v = 10 m / h = 4.47 m / s
for the car case at v = 90km / h = 25 m / s
according to this rule we must this to
x = 25 / 4.47 = 5.6 cars
each modern car is about 4 m long so the distance is
x = 22 m
we see that this distance is much greater than the reaction distance so it does not make much sense
What is The first stage in cellular respiration, what cycle?
ANSWER:
What is The first stage in cellular respiration, what cycle?- GLYCOLYSIS
In glycolysis, the beginning process of all types of cellular respiration, two molecules of ATP are used to attach 2 phosphate groups to a glucose molecule, which is broken down into 2 separate 3-carbon PGAL molecules. PGAL releases electrons and hydrogen ions to the electron carrier molecule NADP+.What are the 4 stages of cellular respiration?It has four stages known as glycolysis, Link reaction, the Krebs cycle, and the electron transport chain. This produces ATP which supplies the energy that cells need to do work.Identify two technologies that can be used to find the locations of objects. Identify the type or types of waves each
technology uses.
what are three physical properties of gases
Answer: Gases have three characteristic properties: (1) they are easy to compress, (2) they expand to fill their containers, and (3) they occupy far more space than the liquids or solids from which they form. An internal combustion engine provides a good example of the ease with which gases can be compressed.
Explanation:
I'm asking for a quick favor. I'm trying to understand an equation that has to do with Projectile motion, Bernoulli's principle, and Magnus Effect. Basically focused on understanding air resistance on a projectile. I would like to discuss this privately rather than have it on this public forum. I'll give you 100 of my points if you help.
Explanation:
Projectile motion, Bernoulli's principle, and Magnus Effect.
Sure I would be happy to discuss projectile motion!
I'll do it if you mark brainliest :) I need the points thanks
A 12.0 V storage battery is connected to three
resistors, 4.75, 14.0 and 21.0, respectively. The resistors are joined in series.
(a)Calculate the equivalent resistance.
(b)What is the current in the circuit?
Answer in units of A.
(a). The equivalent resistance of a series circuit is the SUM of the individual resistances in the circuit.
(4.75Ω + 15Ω + 21Ω ) = 39.75Ω
(b). Current = (voltage) / (resistance)
Current = (12 v) / (39.75Ω)
Current = 0.302 Amperes (rounded to the nearest MilliAmpere))
What us a magnetic domain?
magnetic domain is a region within a magnetic material in which the magnetization is in a uniform direction. This means that the individual magnetic moments of the atoms are aligned with one another and they point in the same direction.
Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for slinging supplies to affected areas, similar to the electromagnetic lift catapults used to launch planes from aircraft carriers. This catapult is located at a fixed point 400 meters away and 50 meters below the target site. The catapult is capable of launching the payload at 67 meters per second and an initial launch angle of 50 degrees. Using your knowledge of kinematics equations, determine whether this would be sufficient to deliver the payload to the drop site.
Answer:
Please see below as the answer is self-explanatory.
Explanation:
We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).Both movements are independent each other, due to they are perpendicular.In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:[tex]v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)[/tex]
Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:[tex]t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s (2)[/tex]
In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:[tex]\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (3)[/tex]
In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:[tex]v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)[/tex]
Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:[tex]\Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)[/tex]
Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.A flat loop of wire consisting of a single turn of cross-sectional area 8.80 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.80 T in 1.10 s. What is the resulting induced current if the loop has a resistance of 2.20
Answer:
The magnitude of the induced current is 4.73 x 10⁻³ A.
Explanation:
Given;
number of turns, N = 1
cross sectional area of the loop, A = 8.8 cm² = 8.8 x 10⁻⁴ m²
change in magnetic field strength, ΔB = 1.8 T - 0.5 T = 1.3 T
change in time, Δt = 1.10 s
resistance of the loop, R = 2.2 ohm
The magnitude of the induced emf is calculated as;
[tex]emf = \frac{NA \Delta B}{\Delta t} \\\\emf = \frac{1 \times 8.8\times 10^{-4} \times 1.3}{1.10} \\\\emf = 1.04 \times 10^{-3} \ V[/tex]
The induced current in the loop is calculated as;
[tex]I = \frac{emf}{R} \\\\I = \frac{1.04 \times 10^{-3}}{2.2} \\\\I= 4.73 \times 10^{-4} \ A[/tex]
Therefore, the magnitude of the induced current is 4.73 x 10⁻³ A
Why are dominant alleles always shown as capital letters?
Answer:
When writing a genotype, the dominant allele is usually represented by a capital letter, while the recessive allele has a lowercase letter.
Explanation:
Please mark me brainlest.
Answer:
Because in a punnett square, dominant alleles need to be differentiated from the recessive alleles, to do this, we write the dominant alleles as capital letters.
Explanation:
If a person walks 3 m north and 5 meters east, how would you find the displacement for that person? what would the displacement be?
Answer:
AC)=(AB)2+(BC)2−−−−−−−−−−−−√=42+32−−−−−−√
⇒displacement=16+9−−−−−√=25−−√=5m
How fast is a ball going when it hits the ground after being dropped from a
height of 16 m? The acceleration of gravity is 9.8 m/s2
A. 22.5 m/s
B. 28.1 m/s
O C. 17.7 m/s
D. 19.3 m/s
Hi there!
[tex]\large\boxed{\text{C. 17.7m/s}}[/tex]
Use the following kinematic equation to solve:
vf² = vi² + 2(ad)
Since the initial velocity is 0 m/s because it started at rest, we can eliminate this part of the equation:
vf² = 2ad
Plug in the given acceleration and distance:
vf² = 2(9.8)(16)
vf ≈ 17.7. The correct answer is C.
An object is moving with constant non-zero velocity. Which of thw following statements about it must be true
Answer:
The net force on the object is zero.
Explanation:
An object is moving with constant non-zero velocity. If velocity is constant, it means that the change in velocity is equal to 0. As a result, acceleration of the object is equal to 0. Net force is the product of mass and acceleration. Hence, the correct option is (d) "The net force on the object is zero".
Plsss I want answer???
Answer:
s=136.89/2g meter
s=6.98 meter (correct to 3 sig.fig. taking g=9.81ms^-2)
Explanation:
u= + 11.7 ms^-1
a= - g ms^-2
At highest point: v=0ms^-1
v^2=u^2+2as
0=11.7^2+2(-g)s
s=136.89/2g meters
Which of the following is a contact force?
Magnetic
Friction
Electric
Gravity
Answer:
magnetic
Explanation:
because magnetic attracted other magnetic by pulling them contact forcely that creates a strong force when they are nearly close to one another
What factors determine electric potential?
A. Mass and distance
B. Charge and density
C. Charge and distance
O D. Mass and charge
Answer: C. Charge and distance
Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of m A = 17.0 kg and an initial velocity of v 0A = 8.00 m/s, due east. Object B, however has a mass of m B = 29.0 kg and an initial velocity of v 0B = 5.00 m/s, due north. Find the magnitude and direction of the total momentum of the two-object system after the collision.
Answer:
pf = 198.8 kg*m/s
θ = 46.8º N of E.
Explanation:
Since total momentum is conserved, and momentum is a vector, the components of the momentum along two axes perpendicular each other must be conserved too.If we call the positive x- axis to the W-E direction, and the positive y-axis to the S-N direction, we can write the following equation for the initial momentum along the x-axis:[tex]p_{ox} = p_{oAx} + p_{oBx} (1)[/tex]
We can do exactly the same for the initial momentum along the y-axis:[tex]p_{oy} = p_{oAy} + p_{oBy} (2)[/tex]
The final momentum along the x-axis, since the collision is inelastic and both objects stick together after the collision, can be written as follows:[tex]p_{fx} = (m_{A} + m_{B} ) * v_{fx} (3)[/tex]
We can repeat the process for the y-axis, as follows:[tex]p_{fy} = (m_{A} + m_{B} ) * v_{fy} (4)[/tex]
Since (1) is equal to (3), replacing for the givens, and since p₀Bₓ = 0, we can solve for vfₓ as follows:[tex]v_{fx} = \frac{p_{oAx}}{(m_{A}+ m_{B)}} = \frac{m_{A}*v_{oAx} }{(m_{A}+ m_{B)}} =\frac{17.0kg*8.00m/s}{46.0kg} = 2.96 m/s (5)[/tex]
In the same way, we can find the component of the final momentum along the y-axis, as follows:[tex]v_{fy} = \frac{p_{oBy}}{(m_{A}+ m_{B)}} = \frac{m_{B}*v_{oBy} }{(m_{A}+ m_{B)}} =\frac{29.0kg*5.00m/s}{46.0kg} = 3.15 m/s (6)[/tex]
With the values of vfx and vfy, we can find the magnitude of the final speed of the two-object system, applying the Pythagorean Theorem, as follows:[tex]v_{f} = \sqrt{v_{fx} ^{2} + v_{fy} ^{2}} = \sqrt{(2.96m/s)^{2} + (3.15m/s)^{2}} = 4.32 m/s (7)[/tex]
The magnitude of the final total momentum is just the product of the combined mass of both objects times the magnitude of the final speed:[tex]p_{f} = (m_{A} + m_{B})* v_{f} = 46 kg * 4.32 m/s = 198.8 kg*m/s (8)[/tex]
Finally, the angle that the final momentum vector makes with the positive x-axis, is the same that the final velocity vector makes with it.We can find this angle applying the definition of tangent of an angle, as follows:[tex]tg \theta = \frac{v_{fy}}{v_{fx}} = \frac{3.15 m/s}{2.96m/s} = 1.06 (9)[/tex]
⇒ θ = tg⁻¹ (1.06) = 46.8º N of E
As dancers twirl faster and faster around their partners, they are demonstrating what type of energy?
A. Binding energy,
B. Nuclear energy,
C. Kinetic energy,
When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how fast is the angle of elevation of the rocket increasing, as seen by an observer on the ground 5 kilometers from the launching pad
Answer:
The angle of elevation of the rocket is increasing at a rate of 48.780º per second.
Explanation:
Geometrically speaking, the distance between the rocket and the observer ([tex]r[/tex]), measured in kilometers, can be represented by a right triangle:
[tex]r = \sqrt{x^{2}+y^{2}}[/tex] (1)
Where:
[tex]x[/tex] - Horizontal distance between the rocket and the observer, measured in kilometers.
[tex]y[/tex] - Vertical distance between the rocket and the observer, measured in kilometers.
The angle of elevation of the rocket ([tex]\theta[/tex]), measured in sexagesimal degrees, is defined by the following trigonometric relation:
[tex]\tan \theta = \frac{y}{x}[/tex] (2)
If we know that [tex]x = 5\,km[/tex], then the expression is:
[tex]\tan \theta = \frac{y}{5}[/tex]
And the rate of change of this angle is determined by derivatives:
[tex]\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y[/tex]
[tex]\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}[/tex]
[tex]\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}[/tex]
[tex]\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}[/tex]
Where:
[tex]\dot \theta[/tex] - Rate of change of the angle of elevation, measured in sexagesimal degrees.
[tex]\dot y[/tex] - Vertical speed of the rocket, measured in kilometers per hour.
If we know that [tex]y = 4\,km[/tex] and [tex]\dot y = 400\,\frac{km}{h}[/tex], then the rate of change of the angle of elevation is:
[tex]\dot \theta = 48.780\,\frac{\circ}{s}[/tex]
The angle of elevation of the rocket is increasing at a rate of 48.780º per second.
In 1666 at the age of 23, what scientist
developed the theories of gravitation inspired
by an apple falling from a tree?
A. Galileo Galilei
B. Issac Newton
C. Albert Einstein
D. Nicolaus Copernicus
Answer:
B - Isaac NewtonExplanation:
He first thought of his system of gravitation which he hit upon by observing an apple fall from a tree,
The incident occurring in the late summer of 1666.
A transverse wave is represented below. 1.5 m 0.20 m What is the approximate amplitude and wavelength of the wave? amplitude = 0.20 m, wavelength = 0.60 m B. amplitude = 0.20 m, wavelength = 0.30 m C. amplitude = 0.10 m, wavelength = 0.60 m OO amplitude = 0.10 m, wavelength = 0.30 m
Answer:
C. amplitude = 0.10 m, wavelength = 0.60 m
Explanation:
The diagram shows an oscillating progressive wave, with its amplitude and wavelength.
Amplitude of a wave is the maximum distance covered either upward or downward.
So that,
amplitude of the wave, A = [tex]\frac{0.2}{2}[/tex]
= 0.1
Amplitude of the wave = 0.1 m
Wavelength in this case is the distance from crest to crest, or trough to trough of the wave.
So that,
wavelength = [tex]\frac{1.5}{2.5}[/tex]
= 0.6
wavelength of the wave = 0.6 m
Therefore, the amplitude of the wave is 0.10 m, while the wavelength is 0.60 m.
The course an object travels along is called DIRECTION.
A
TRUE
B
FALSE
DO NOT WRITE DOWN NONSENSE OR YOU WILL BE REPORTED!!!!!!!!
Physics gravity question, Please help
Answer: 2.7 x10^-4 N
Explanation: 6.674 ×10^-11 × 1000 x 1000 divided by 0.5 squared.
An object is spun in a circle of radius 1.5m with a frequency of 6.0Hz. what is it’s velocity ?
Answer:
56.55 m/s
Explanation:
Frequency=1/T
1/6 = 0.1666666
V=(2*pi*1.5)/0.166666
An object is experiencing an acceleration of 0.4 m/s^2 while traveling in a circle of 35 m. What is it’s velocity?
Answer:
v = 3.74 m/s
Explanation:
Given that,
The acceleration of the object in circular path, a = 0.4 m/s²
The radius of the circle, r = 35 m
We need to find the velocity of the object. The acceleration of an object on the circular path is given by :
[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{0.4\times35}\\\\v=3.74\ m/s[/tex]
So, the velocity of the object is equal to 3.74 m/s.
why type of volcano is built almost entirely from ejected lava fragments
Answer:
Shield volcanoes
Explanation: