Hello!
To solve, we can begin by using the kinematic equation:
[tex]d = v_it + \frac{1}{2}at^2[/tex]
Where:
vi = initial velocity (m/s)
t = time (s)
a = acceleration (in this case, due to gravity. g = 9.8 m/s²)
Since the object falls from rest, the initial velocity is 0 m/s.
[tex]d = \frac{1}{2}at^2[/tex]
Plug in the given values:
[tex]d = \frac{1}{2}(9.8)(3.5^2) = \boxed{60.025 m}[/tex]
What causes an astigmatism?
A. damaged lens
B. retina not focusing the image
C. cornea being wavy or not spherical
D. sclera not refracting light properly
Answer:
c) cornea being wavy or not spherical
PLEASE HELP I DONT GET THISS
Answer:
I feel like its the second one but I'm not completely sure..
Explanation:
how do all organisms begin life
Answer:
All organisms begin their lives as single cells.Overtime,these organisms grow and take on the characteristics of their species...All organisms grow,and different parts of organisms may grow at different rates.Organisims made out of only one cell
may change little during their lives, but they do grow
Explanation:
brainlest me please
CAN SOMEONE PLEASE HELP ME
Answer:
she will eventually slow down and come to a stop
which of the following statements might be used to defend the Act of 1848
1. Explain who is doing more work and why: a bricklayer carrying bricks and placing them on the wallof a building being
constructed, or a project supervisor observing and recording the progress of the
workers from an observation booth.
A man applies a force of 540 N to the barrow in a direction 75 from the horizontal. He moves the barrow 30 m along the level ground. Calculate the work he does against friction?
The work done by the man against friction is 4,192.86 J.
The given parameters;
force applied, F = 540 Nangle of inclination, θ = 75⁰horizontal distance, x = 30 mThe work done by the man against friction is calculated as follows;
[tex]W = F \times d \times cos(75)\\\\W = 540 \times 30 \times cos(75)\\\\W = 4,192.86 \ J[/tex]
Thus, the work done by the man against friction is 4,192.86 J.
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At the molecular level, as the kinetic energy increases, what happens to the temperature?
decreases
increases
stays the same
Answer: increases
Explanation:
Temperature is a measure of the average velocity of the molecular particles. The faster they go, the higher the temperature.
MCQ
A body of mass 5kg is pushed for distance x with accleration a. Then workdone against static friction is
1.ma*X cosB
2.ma*X sinB
3.zero
4.ma/X
Answer:
ma*XsinB
option 2 is correct
b) A satellite is in a circular orbit around the Earth at an altitude of 1600 km above the Earth's surface. Determine the orbital period of the satellite in hours. [3]
Explanation:
The orbiting period of a satellite at a height h from earth' surface is
T=2πr32gR2
where r=R+h.
Then, T=2π(R+h)R(R+hg)−−−−−−−−√
Here, R=6400km,h=1600km=R/4
T=2πR+R4−−−−−−√R(R+R4g)−−−−−−−−−⎷=2π(1.25)32Rg−−√
Putting the given values,
T=2×3.14×(6.4×106m9.8ms−2)−−−−−−−−−−−−√(1.25)32=7092s=1.97h
Now, a satellite will appear stationary in the sky over a point on the earth's equator if its period of revolution around the earthh is equal to the period of revolution of the earth up around its own axis whichh is 24h. Let us find the height h of such a satellite above the earth's suface in terms of the earth,'s radius.
Let it be nR.Then
T=2π(R+nR)R(R+nRg)−−−−−−−−−−√
=2π(Rg)−−−−−√(1+n)32
=2×3.14(6.4×106m/s9.8m/s2)−−−−−−−−−−−−−−−⎷(1+n)32
(5075s)(1+n)32=(1.41h)(1+n)32
For T=24h, we have (24h)=(1.41h)(1+n)32
or (1+n)32=241.41=17
or 1+n(17)23=6.61
or n=5.61
The height of the geostationary satellite above the earth's surface is nR=5.61×6400km=6.59×104km.
d what is
7 A rocket of mass 10000 kg uses 5.0kg of fuel and oxygen
to produce exhaust gases ejected at 5000 m/s. Calculate the
increase in its velocity
Answer:
Approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex], assuming that no external force (e.g., gravitational pull) was acting on this rocket.
Explanation:
Assume that no external force is acting on this rocket. The system of the rocket and the fuel on the rocket would be isolated (an isolated system.) The momentum within this system would be conserved.
Let [tex]v_{0}\; \rm m\cdot s^{-1}[/tex] be the initial velocity of the rocket.
The velocity of the exhaust gas would be [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] since the gas is ejected away from the rocket.
Let [tex]\Delta v\; \rm m\cdot s^{-1}[/tex] denote the increase in the velocity of the rocket. The velocity of the rocket after ejecting the gas would be [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex].
The momentum [tex]p[/tex] of an object of velocity [tex]v[/tex] and mass [tex]m[/tex] is [tex]p = m \cdot v[/tex].
The combined mass of the rocket and the fuel was [tex]10000\; \rm kg[/tex]. The initial momentum of this rocket-fuel system would be:
[tex]\begin{aligned}p_{0} &= m \cdot v\\ &= 10000\; {\rm kg} \times v_{0}\; {\rm m \cdot s^{-1}} \\ &= (10000\; v_{0})\; \rm {kg \cdot m\cdot s^{-1}}\end{aligned}[/tex].
The momentum of the [tex]5.0\; \rm kg[/tex] of fuel ejected at [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] would be:
[tex]\begin{aligned} & 5.0 \; {\rm kg} \times (v_{0} - 5000)\; {\rm m\cdot s^{-1}}\\ =\; & (5.0\, v_{0} - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].
After ejecting the [tex]5.0\; \rm kg[/tex] of the fuel, the mass of the rocket would be [tex]10000\; \rm kg - 5.0\; \rm kg = 9995\; \rm kg[/tex]. At a velocity of [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex], the momentum of the rocket would be:
[tex]\begin{aligned} & 9995 \; {\rm kg} \times (v_{0} + \Delta v)\; {\rm m\cdot s^{-1}}\\ =\; & (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].
Take the sum of these two quantities to find the momentum of the rocket-fuel system after the fuel was ejected:
[tex]\begin{aligned}p_{1} &= (5.0\, v_{0} - 25000)\; {\rm kg \cdot m\cdot s^{-1}\\ &\quad\quad + (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}} \\ &= (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].
The momentum of the rocket-fuel system would be conserved. Thus [tex]p_{0} = p_{1}[/tex].
[tex](10000\, v_{0})\; {\rm kg \cdot m\cdot s^{-1}} = (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}[/tex].
Solve this equation for [tex]\Delta v[/tex], the increase in the velocity of the rocket.
[tex]10000\, v_{0} = 10000\, v_{0} + 9995\, \Delta v - 25000[/tex].
[tex]9995\, \Delta v = 25000[/tex].
[tex]\begin{aligned}\Delta v &= \frac{25000}{9995} \approx 2.5\end{aligned}[/tex].
Thus, the velocity of the rocket would increase by approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex] after ejecting the [tex]5.0\; \rm kg[/tex] of fuel.
A solid sphere rolls without slipping down an incline, starting from rest. At the same time, a box starts from rest at the same altitude and slides down the same incline, with negligible friction. Which object arrives at the bottom first
Answer:
The box arrives first.
Explanation:
Hope this helps!! :))
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According to the information, a solid sphere is an object that arrives at the bottom first. This is because it occupies less friction as compared to the box.
What is Friction?Friction may be defined as the resistance that is offered by the surfaces that are in contact when they move past each other. It is a type of force that opposes the motion of a solid object over another.
There are mainly four types of friction: static friction, sliding friction, rolling friction, and fluid friction. According to the context of this question, the sphere possesses less friction as compared to the box. This is because the box has an irregular surface that possesses high friction over the inclined surface.
Therefore, according to the information, a solid sphere is an object that arrives at the bottom first. This is because it occupies less friction as compared to the box.
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A metal wire of length 1.2 m and cross-sectional area 2.0 x 10 raised to power-7 m 2 is stretched by a force of 50 N. assuming the force constant of the metal is 6000 Nm-1. Calculate the tensile stress
L=1m
=2mm²=(2/1000²)=2(10^-6)m
y=4x10¹¹N/m²
∆l=2mm=2/10.00=0.002m
=(4x10¹¹x2x10^-6x2x10^-3)÷1m
=16x10¹¹-⁶-³
=16x10¹¹-⁹
=16x10²
=1600N
where a=cross sectional area=2x10^-6m
C=2mm= 2x10^-3m
L=1m
hope it helps
Objects 1 and 2 attract each other with a gravitational force of 34 units. If the distance separating objects 1 and 2 is changed to one-third the original value, then the new gravitational force will be ____ units.
Answer:
F12 = G M1 M2 / R12^2
F12' = G M1 M2 / R12'^2
F12' / F12 = R12'^2 / R12^2 = (1/3)^2
F12' = 1/9 F12
The new force is 1/9 the of the old force
Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall?
Answer:
d = -33.1 m and Vf = -25.5 m/s
Explanation:
Given:a = -9.8 m
t = 2.6 s
Vᵢ = 0 m/s
To Find:d = ?
Vf = ?
Now,
d = Vᵢ × t + 0.5 × a × t²
d = (0 m/s) × (2.60 s) + 0.5 × (-9.8 m/s²) × (2.60 s)²
d = -33.1 m (- indicates direction)
Vf = Vᵢ + a × t
Vf = 0 + (-9.8 m/s²) × (2.60 s)
Vf = -25.5 m/s (- indicates direction)
Thus, d = -33.1 m and Vf = -25.5 m/s
-TheUnknownScientist 72
A 2457 kg car moves with initial speed of 18 ms-l. It is stopped in 62 m by its brakes.
What is the force applied by the brakes?
Answer:
Explanation:
The work of the brakes will equal the initial kinetic energy of the car
Fd = ½mv²
F = mv²/2d
F = 2457(18²) / (2(62))
F = 6,419.903...
F = 6.4 kN
Why is it important for the community to take action
Answer:
Why is community action important? ... Involving communities in the design and delivery of services can help to achieve a number of objectives, including: Building community and social capacity – helping the community to share knowledge, skills and ideas. Community resilience – helping the community to support itself
in rotational motion the angular frequency of all the particles is
Answer:
L=mvr
α=0
v=ωr
L=mωr
2
2
1
mv
2
=k
L=Iω
k=
2
1
Iω
2
ω
1
=2ω
k
1
=1/2k
∴I
1
=
8
1
I
L
1
=
8
1
I×2ω
=
4
1
Iω
=L/4
Physical motion that occurs when an item rotates or spins on an axis is known as “rotatory motion,” sometimes known as “rotational motion” or “circular motion.”
What angular frequency of particles in rotational motion?Angular displacement per unit of time is measured by angular frequency, also referred to as radial or circular frequency (). Therefore, it has degrees (or radians) per second as its units. 1 Hz = 6.28 rad/sec, thus. 1 radian equals 57.3° because 2 radians equal 360°.
The frequency of rotation is the rate at which an item rotates around an axis, sometimes referred to as rotational speed or rate of rotation.
However, the linear velocity also depends on how far the particles are from the axis of rotation. Because every particle distances are different, so too will the linear velocities for each particle.
Therefore, All the particles in a rigid body's rotational motion move in the same direction in a given interval. Therefore, all the particles' angular velocities will be the same.
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A gold doubloon 6.1 cm in diameter and 2.0mm thick is dropped over the side of a Pirate Ship. When it comes to rest on the ocean floor at a depth of 770m how much has its volume changed
The volume of a material is the total amount of matter that it can contain. The volume of the given coin has been determined to be 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex]. Since the gold doubloon do not absorb water, then its volume remains constant at the ocean floor.
The volume of the gold doubloon can be determined by;
volume = [tex]\pi r^{2}[/tex] + h
where r is the radius of the coin and h is its thickness.
Such that; diameter = 6.1 cm (61 mm) and h = 2.0 mm
r = [tex]\frac{diameter}{2}[/tex]
= [tex]\frac{61}{2}[/tex]
r = 30.5 mm
Thus,
volume of the coin = [tex]\frac{22}{7}[/tex] x [tex](30.5)^{2}[/tex] x 2
= 5847.2857
Therefore, the volume of the gold doubloon is 5847.3 [tex]mm^{3}[/tex]. This can also be expressed as 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex].
Since the gold doubloon is not miscible with water, thus its volume at a depth of 770 m at the ocean floor is the same as its initial volume.
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How would you best define the word drug?
A: Something that makes you tired
B: Something that can kill you
C: Something that effects your body and mind
D: Stored for energy
someone help
Answer:
C
Explanation:
Definition of drug: a medicine or other substance which has a physiological effect when ingested or otherwise introduced into the body
According to the table what was the hikers total displacement the graph has 4km 6km 4km 6km
Answer:
0
Explanation:
0 is the answer
A vessel with an unknown volume is filled with 10 kg of water at 90oC. Inspection of the vessel at equilibrium shows that 8 kg of the wateris in the liquid state. What is the pressure in the vessel, and what is the volume of the vessel
In this case, it is possible to solve this problem by using the widely-known steam tables which show that at 90 °C, the pressure that produces a vapor-liquid mixture at equilibrium is about 70.183 kPa (Cengel, Thermodynamics 5th edition).
Moreover, for the calculation of the volume, it is necessary to calculate the volume of the vapor-liquid mixture, given the quality (x) it has:
[tex]x=\frac{m_{steam}}{m_{total}}[/tex]
Thus, since 8 kg correspond to liquid water, 2 kg must correspond to steam, so that the quality turns out:
[tex]x=\frac{2kg}{10kg} =0.20[/tex]
Now, at this temperature and pressure, the volume of a saturated vapor is 2.3593 m³/kg whereas that of the saturated liquid is 0.001036 m³/kg and therefore, the volume of the mixture is:
[tex]v=0.001036m^3/kg+0.2(2.3593-0.001036 )m^3/kg=0.4727m^3/kg[/tex]
This means that the volume of the container will be:
[tex]V=10kg*0.4727m^3/kg\\\\V=4.73m^3[/tex]
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https://brainly.com/question/23339302https://brainly.com/question/15081976What is the birth rate of a population of 3000 chipmunks if 200 chipmunks are born each year?
o 0.022 births per chipmunk per year.
0.067 births per chipmunk per year.
O 0.82 births per chipmunk per year.
o 15 births per chipmunk per year.
Answer:
0.067 births per chipmunk per year
Explanation:
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