Hi there!
We can begin by using the derived kinematic equation:
[tex]t = \sqrt{\frac{2h}{g}[/tex]
Plug in the given values and let g = 9.8 m/s²:
[tex]t = \sqrt{\frac{2(20)}{g}} = \boxed{2.02 s}[/tex]
Now, we can solve for its final velocity using the equation:
[tex]v_f = v_i + at[/tex]
It is dropped from rest, so vi = 0 m/s.
[tex]v_f = at[/tex]
[tex]v_f = 9.8(2.02) = \boxed{19.796 m/s}[/tex]
Answer:
a. The ball would reach the ground in approximately [tex]2.02\; \rm s[/tex].
b. The velocity of the ball right before landing would be approximately [tex]19.8\; \rm m\cdot s^{-1}[/tex].
(Assumptions: the ball was dropped with no initial velocity; air resistance on the ball is negligible; [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex].)
Explanation:
Under these assumptions, the acceleration of this ball would be constantly [tex]a = g = 9.81\; \rm m\cdot s^{-2}[/tex] (same as the gravitational field strength) during the descent.
Displacement of the ball: [tex]x = 20.0\; \rm m[/tex].
Initial velocity of the ball: [tex]v_{0} = 0\; \rm m\cdot s^{-1}[/tex].
Let [tex]t[/tex] denote the duration of this descent.
The SUVAT equation [tex]x = (1/2)\, a\, t^{2} + v_{0}\, t[/tex] relates the known quantities [tex]a[/tex], [tex]x[/tex], and [tex]v_{0}[/tex] to the unknown [tex]t[/tex].
Substitute the known quantities into this equation and solve to find the value of [tex]t\![/tex]:
[tex]\displaystyle 20.0\; {\rm m} = \frac{1}{2}\times 9.81\; {\rm m\cdot s^{-2}} \times t^{2} + 0\; {\rm m \cdot s^{-1}} \times t[/tex].
[tex]\displaystyle 20.0\; {\rm m} = \frac{1}{2}\times 9.81\; {\rm m\cdot s^{-2}} \times t^{2}[/tex].
[tex]\displaystyle t^{2} = \frac{20.0\; \rm m}{(1/2) \times 9.81\; \rm m\cdot s^{-2}}[/tex].
Since [tex]t > 0[/tex]:
[tex]\begin{aligned}t &= \sqrt{\frac{20.0\; \rm m}{(1/2) \times 9.81\; \rm m\cdot s^{-2}}} \\ &\approx 2.01928\; \rm s\\ &\approx 2.02\; \rm s \\ & (\text{Rounded to 2 sig. fig.})\end{aligned}[/tex].
Since the acceleration of this ball is constant, the velocity of the ball right before landing would be:
[tex]\begin{aligned}v_{1} &= a\, t \\ &\approx 9.81\; \rm m\cdot s^{-2} \times 2.01928\; \rm s \\ &\approx 19.8\; \rm m \cdot s^{-1} \end{aligned}[/tex].
Terri is excited about the baby she is carrying in her womb. One day, she tells her
cousin that the child is now the size of a grape. Weeks later, she describes the child as
the size of an apple. A few weeks after that, she describes the child as the size of a
banana. "Seriously, Terri, are you having a baby or shopping for groceries?" asks her
cousin. Terri explains that this is just the easiest way to describe the size of the baby
now that it is a(n)
Zygote
O Fetus
O Embryo
Infant
Answer:
Fetus
Explanation:
the outsiders I got nuggets out of the elder and I get to make it in and I got I got
A car is moving at a speed of 72 km / h on a horizontal section. What is the braking distance of the car if the friction force is equal to the size of the weight of the car? (20 m) (+ formula)
Answer:
20.4m
Explanation:
72kmhr(1000mkm)(hr3600sec)=20ms
Use principles of conservation of energy.
The initial kinetic energy of the car is equal to the work done by friction during braking:
KE=Wfric
12mv2=(Ffric)d
but the force of friction = weight of the car = mg
12mv2=mgd
mass cancels in both terms, which tells us the answer is independent of mass
12v2=gd
12202=(9.81)d
d=20.4m
Note that the correct answer should actually be 20. m with only 2 significant figure accuracy since 72 km/hr only has 2 sig fig accuracy.
A taxi is travelling at 20 m/s. Its driver accelerates with acceleration 5 m/s2 for 4 s. What is its new velocity?
Please find attached photograph for your answer.
Hope it helps.
Do comment if you have any query.
what is the average speed of the toy car during the two trials to the nearest tenth of m/s
wait nvm the answer is D
Answer:
what is the average speed of the toy car during the two trials to the nearest tenth of m/s
wait nvm the answer is B
Explanation:
CARRY ON LEARNING
What is Space debris ?
Answer:
Space junk, or space debris, is any piece of machinery or debris left by humans in space. It can refer to big objects such as dead satellites that have failed or been left in orbit at the end of their mission. It can also refer to smaller things, like bits of debris or paint flecks that have fallen off a rocket.
Answer:
In addition to derelict man-made objects left in orbit, other examples of space debris include fragments from their disintegration, erosion and collisions or even paint flecks, solidified liquids expelled from spacecraft, and unburned particles from solid rocket motors. Space debris represents a risk to spacecraft.
Explanation:
write the name of simple machine used to changed the direction of force?
Answer:
The variation of the wheel and axle is the pulley. In a pulley, a cord wraps around a wheel. It is a simple machine that is used to change the direction of the force. When the wheel rotates, the cord can move in either of the directions.
Explanation:
A fan operating at 260V with a current being supplied at 13A. What is the effective resistance at which it is being operated?
Answer:
20 ohms
Explanation:
Resistance=Potential difference÷Electric current
X=260÷13
X=20
If the ball has a positive electric charge, what should the charge of the coilgun be to push the ball away?
A)positive
B) negative
C)neutral
Answer:
A
Explanation:if the things have the same charge they will repel makeing the ball move away form its positively charged guy = that the coil gun works.
Who is the author "talking” to in the third stanza of Flanders field
Answer:
One of Canada's best-known poems, “In Flanders Fields” was written on May 2, 1915, when Canadian serviceman John McCrae was stationed at an army hospital in Flanders, Belgium, during World War I.
Explanation:
A small ball is attached to one end of a spring that has an unstrained length of 0.201 m. The spring is held by the other end, and the ball is whirled around in a horizontal circle at a speed of 3.41 m/s. The spring remains nearly parallel to the ground during the motion and is observed to stretch by 0.0176 m.
Required:
By how much would the spring stretch if it were attached to the ceiling and the ball allowed to hang straight down, motionless?
The extension of the spring when the ball is allowed to hang straight down, motionless is 0.0032 m.
The given parameters;
unstrained length, l₁ = 0.201 mextension of the string, x = 0.0176 mspeed of the ball, v = 3.41 m/sThe radius of the circular path when spring is stretched is calculated as;
R = l₁ + x
R = 0.201 + 0.0176
R = 0.2186 m
The spring constant is calculated as follows;
[tex]F = ma\\\\Kx = \frac{mv^2}{R} \\\\K = \frac{mv^2}{Rx} \\\\K = \frac{(3.41)^2 m}{0.2186 \times 0.0176} \\\\K = 3,022.4 m \ N/m[/tex]
The extension of the spring when the ball is allowed to hang straight down, motionless;
[tex]F = mg\\\\Kx=mg\\\\x = \frac{mg}{K} \\\\x = \frac{9.8 m}{3022.4 m} \\\\x = 0.0032 \ m[/tex]
Thus, the extension of the spring when the ball is allowed to hang straight down, motionless is 0.0032 m.
Learn more here:https://brainly.com/question/15589287
A boy standing on a bridge 150m above the canyon floor throws a stone downward with a speed of 20.0m/s.
A) with what velocity will it hit the ground?
B) How long will it take to descend?
Answer:
57.97m/s
12.1s
Explanation:
Use one of the key equations of accelerated motion and sub in the values
vf^2=vi^2+2aΔd
vf^2=(20)^2+(2*9.8)(150)
vf^2 = 3340
vf = 57.79m/s (square rooted both sides)
now use another equation
Δd = (vi+vf/2)Δt
and rearrange for Δt
Δt = (2Δd-vf)/vi
sub in values
Δt = (2*150-57.79)/20
Δt = 12.1s
mark me as brainliest if this helped!
A galaxy spectrum has a redshift of 70,000 km per second. If the Hubble constant is 70 km per second per Mpc (megaparsec), how far away from us in this galaxy
Using the Hubble law v = H₀d where v = recessional speed = 70,000 km per second H₀ = hubble constant = 70 km/s/Mpc and d = distance of galaxy.
Making d subject of the formula, we have
d = v/H₀
Substituting the values of the variables into the equation, we have
d = v/H₀
d = 70000 km/s/70 km/s/Mpc
d = 1000 Mpc
So, the galaxy is 1000 Mpc away from us.
Learn more about hubble law here:
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On some strange planet you find that a 17.4 kg object falls downward 28.8 meters in 5.5 seconds. What is the magnitude of the acceleration due to gravity on this planet
Hi there!
[tex]\large\boxed{g = 1.904 m/s^2}[/tex]
Acceleration of gravity is unaffected by an object's mass.
In the absence of air resistance, we can use the following kinematic equation:
d = vit + 1/2at²
The object falls from rest, so:
d = 1/2at²
The acceleration in this instance is gravity, our unknown. We can rearrange the equation to solve for 'g':
2d/t² = g
2(28.8)/(5.5²) = 1.904 m/s²
Using public transports.
Turn off the lights when not in use
Recycle and Reuse
No to plastic bags
Reduction of forest fires and smoking
Use of fans instead of Air Conditioner
Use filters for chimneys
Avoid usage of crackers
how do I put this in a paragraph.
1. A rocket is launched from a 300 cm rail. The upper Launch Lug is placed 1 point 150 cm from the bottom of the rocket. What is the effective distance of the Launch Rail? 300 cm 150 cm 3 m 300 m O 1m O 1.5 m
Answer:
i think its going to be 150 because its half of 300
Explanation:
A small sphere 0.70 times as dense as water is dropped from a height of 8 m above the surface of a smooth lake. Determine the maximum depth to which the sphere will sink. Neglect any energy transferred to the water during impact and sinking.
Answer:
18.66m
Explanation:
This was actually fun! Let set up the problem first: As you drop the sphere, it will accelerate till it hits the water with a given speed [tex]\dot z[/tex]. Once sinking the ball is subjected to two forces: its own weight [tex]m_sg = \rho_sVg[/tex] directed towards the bottom of the lake, and buoyancy (Archimedes law), ie a force upward equal to the volume of displaced water, [tex]m_wg=\rho_wVg[/tex]. At this point it's the classical "how high can I toss a ball before it falls down" with a water twist. Please note that I'm using z as the height of the sphere, and [tex]\dot z; \ddot z[/tex] represent the velocity (with one dot) and the acceleration (with two dots) in the z direction .Let's assume the sign of all quantities being positive if directed upwards, and negative if towards the bottom of the lake.
Let's first determine how fast the ball hit the water. For me the easiest way is saying "at 8m it has a given potential energy, and 0 kinetic energy. When it hits the water it loses all potential energy and has only kinetic energy". In numbers:
[tex]m_sgz=\frac12m_s\dot z_0^2[/tex] Let's divide by the masses, g is a known value ([tex]9.81 ms^{-2}[/tex]), z is 8 meters, and we get that [tex]\dot z_0 = 4\sqrt g =12.52 m/s[/tex]
At this point, let's determine the force acting on the sphere, courtesy of Newton second law and the debate we had earlier:
[tex]m_s \ddot z = -m_sg + m_wg\\\rho_sV \ddot z = -\rho_sVg + \rho_wV g[/tex]
At this point we can divide by the volume of the sphere, and make use of the fact that [tex]\rho_s = 0.7 \rho_w[/tex]
[tex]0.7\rho_w \ddot z = -.7\rho_w g+ \rho_wg \rightarrow 0.7 \ddot z = 0.3 g\\\ddot z = \frac {0.3}{0.7} g= \frac 37g = 4.2 ms^-2[/tex]
We're almost there. At this point we can write the law of motion for the sphere.
[tex]\dot z = \dot z_0 + \ddot z t[/tex] that we will use to find the time for the ball to stop sinking - ie reaches 0 velocity.
[tex]0 = -12.52 +4.2t \rightarrow t= 12.52/4.2 \approx 3s[/tex]
At this point we can use the other law of motion to find out the distance traveled
[tex]z= z_0 + \dot z_ot + \frac12 \ddot z t^2\\z= 0 -12.52 (3) + \frac12 (4.2)(3)^2 = -37.56+18.90 = -18.66 m[/tex]
as usual, please double and triple check all the calculations, it's almost 1.30 am here and I am not the most confident number cruncher even when fully awake.
URGENT NEED GIVING POINTS question is how do i find R and the angle THANK YOU
Answer:
Theta = 26.93°
R = 0.38m
Explanation:
Theta = cos^-1(0.74m/0.83m)
Theta = 26.93°
R = sqrt(0.83²-0.74²)
R = 0.38m
What kind of feature results from the convergence of two oceanic plates?
A.
island arcs
B.
massive continental volcanoes
C
rift valleys
D.
tall mountain ranges
Answer:
A
Explanation:
Convergence is pushing them together so C is out of the question.
B and D are wrong because it is an oceanic plate. B and D are for continental plates.
How much work is required for a 74 kg sprinter to accelerate from rest to 2.2 m/s?
Answer:
179.1 kg
Explanation:
Answer:
Should be 179.1 kg
Explanation:
Gas sample A has a mass of 32.00 grams. Gas sample B has a mass of 28.00
grams. If the two gas samples are both at 25°C, which sample has the greatest
average kinetic energy?
Answer:
I think 32.00
Explanation:
Because the equation is mass times velocity so is you mutiply 32 by 25 and 28 times 25 sample A would be greater.
The sample with the greatest average kinetic energy is sample A
The average kinetic energy of a sample can be estimated by the formula one-half of the mass sample multiplied by the speed (RMS) squared.
Mathematically, we have:
[tex]\mathbf{K.E_{avg} = \dfrac{1}{2}mv_{rms}^2}[/tex]
If the mass of sample A = 32.00 grams, andThe mass of sample B = 28.00Provided that the speed (RMS) is constant, the average kinetic energy of sample A will be greater than the average kinetic energy of sample B because sample A has a greater mass than sample B.
Learn more about average kinetic energy here:
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1+1 =?
2+2=?
3+3=?
1+1 =?
2+2=?
3+3=?
Answer:
here is your answer
1+1=2
2+2=4
4+4=8
The following diagram shows a person exerting a
300.-newton force on the handle of a shovel that
makes an angle of 60. with the horizontal ground.
Determine the magnitude of the component of the
force perpendicular to the ground.
Answer:
260 Newtons
Explanation:
Okay, so what you have to do is know what the question is asking. It wants to know the force perpendicular to the ground. We know perpendicular makes a right angle, so we use the diagram to make it into a right triangle with a degree of 60, and hypotunese of 300 newtons. Now, we use basic trigonometry, we place x on the horizontal side since it perpendicular to the ground. We know it´s sin because x is opposite to 300 which is the hyp.
Sin= opp/hyp
x/300
sin(60)= x/300
300*sin(60)= 259.8, which rounds to
260 Newtons
There is your answer :D. Happy to help.
The component of the force which is perpendicular to the ground is → F [y] = 261 a[y].
What are the components of a Vector?A vector V making an angle θ with the x - axis, can be resolved into two two components along the x and y - axes respectively as →
V = Vcosθ a[x] + Vsinθ a[y]
where -
a[x] is the unit vector along the +x axis
a[y] is the unit vector along the +y axis.
Given is a force of magnitude 300 N acting on the handle that makes an angle of 60° with the horizontal ground.
The component normal to the ground will be the component along the y - axis. The component along the y - axis is the sine component. Suppose that vector F represents the force of magnitude (|F|) 300 N. The component perpendicular to the ground will be -
F [y] = |F| sinθ a[y]
F [y] = 300 x sin(60°) a[y]
F [y] = 300 x 0.87
F [y] = 261 a[y]
Therefore, the component of the force which is perpendicular to the ground is → F [y] = 261 a[y].
To solve more questions on Vector Resolution, visit the link below-
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Please help thank you!
Answer:
mass, weight
Explanation:
mass is constant, and weight is dependant on your location
Answer:
I think d
Explanation:
because distance j
has to be the last one that's the only one with distance
Kichelle enjoys balloon animals from the carnival. She just received a balloon giraffe that has an initial temperature of 39.0°C and a volume of 1.28 L. If Preston plays a trick on Kichelle, and puts her balloon giraffe into the freezer, what would be the new volume of the balloon if the temperature drops down to 8.0°C?
39°C=1.28L
8°C=? Less
=8°C/39°C×1.28L
=8/39×1.28L
=10.24/39
=512/195
=2.6L
Find the velocity of a body of mass 100 g having kinetic energy of 20 J.
Answer:
Explanation:
Let v be the velocity of object ,
Given that mass of object , m=100 g=
1000
100
kg=0.1 kg
& kinetic energy of object =20 J
K.E =
2
1
mv
2
20=
2
1
×0.1v
2
⟹v
2
=400
⟹v=20 m/s
A car of mass 1500 kg is moving with a velocity of 30 m/s. If the velocity-time graph for this car
is a horizontal line parallel to the time axis, then the velocity of the car at the end of 25 s will be
Answer:
v=30m/s
Explanation:
m = 1500kg
v = 30m/s
t = 25s
The velocity time graph is parallel to the time axis,
that means The Velocity is constant. ('constant' means remains same).
Therefore, after 25s, v=30m/s
how much of the light that reaches the earth from the sun is used for photosynthesis
Answer:
Less than one percent of the total energy that reaches Earth is used by plants for photosynthesis
1 Loudness and pitch are two important properties of a musical sound.
a If a musician plays a softer note, which property has been changed,
loudness or pitch?
b If the musician makes the note lower, which property has been
changed, loudness or pitch?
A piece of wood has a mass of 8g and a volume of 10cm³. Work out the density of the wood in /3 .
no links please just explain how you got the answer.
Answer:
8×10^-3÷10×10^-6=8×10^2=800 kg/m^3
Answer:
2.4/3
Explanation:
first we make 8g/10cm then 0.8 so the only number that can fit for /3 is 2.4
hope helped!
Energy a substance or system has because of its motion Is what?
light energy
chemical energy
sound energy
electrical energy
thermal energy
mechanical energy
Answer:
Mechanical energy is the energy a substance or system has because of its motion. For example machines use mechanical energy to do work.