Answer:
[tex]5.241\ \text{m/s}[/tex]
Explanation:
m = Masa del coche
g = Aceleración debida a la gravedad = [tex]9.81\ \text{m/s}^2[/tex]
h = Altura = [tex]1.4\ \text{m}[/tex]
v = Velocidad del automóvil en la parte inferior de la pista
Aquí asumimos que el automóvil desciende verticalmente. La energía potencial del automóvil se completará convertida en energía cinética en la parte inferior de la pista ya que no hay pérdida de energía.
[tex]mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.4}\\\Rightarrow v=5.241\ \text{m/s}[/tex]
La velocidad máxima que puede alcanzar el coche es [tex]5.241\ \text{m/s}[/tex].
help plz any one
1. In each of the following questions find the density. State the units of your answer.
Mass 45g, volume 5cm3
Answer:
Explanation:
Density=mass÷volume
=45÷3
=15
A standard 1 kilogram weight is a cylinder 40.0 mm in height and 53.5 mm in diameter. What is the density of the material?
Answer:
1.15*10^-5 kg/m^3
Explanation:
Given data
mass= 1kg
hieght= 40 mm
diameter= 52.5mm
radius= 53.5/2= 26.25mm
The volume of the cylinder
V=πr^2h
V=3.142*26.25^2*40
V=3.142*689.0625*40
V=86601.375 mm^3
Density= mass/volume
Density= 1/86601.375
Density=0.00001154716
Density= 1.15*10^-5 kg/m^3
Hence the density is 1.15*10^-5 kg/m^3
Mr. Rudman drives his race car for 4 hrs at 150
miles/hr. How far will he travel?
A 25 kg object is at rest. Calculate the momentum of the object.
Answer:
there is no momentum
Explanation:
if an object is at rest there is no momentum to build, no speed, no acceleration it does not move unless force is acted upon it.
Given that the mass of Earth is 5.98x1024 kg, what is the orbital radius of a satellite that has an orbital period of
exactly one day (assume that a day is exactly 24 hours in length)?
Answer:
The orbital radius is approximately 42,259 kilometers.
Explanation:
From Newton's Law of Gravitation we find that acceleration experimented by the satellite ([tex]a[/tex]), measured in meters per square second, is defined by:
[tex]a = \frac{G\cdot M}{r^{2}}[/tex] (1)
Where:
[tex]G[/tex] - Gravitational constant, measured in cubic meters per kilogram-square second.
[tex]M[/tex] - Mass of Earth, measured in kilograms.
[tex]r[/tex] - Orbital radius, measured in meters.
By supposing the satellite rotates at constant speed and in a circular path, we find that acceleration is entirely centripetal and can be defined in terms of period, that is:
[tex]\frac{4\pi^{2}\cdot r}{T^{2}} = \frac{G\cdot M}{r^{2}}[/tex]
[tex]4\pi^{2}\cdot r^{3} = G\cdot M\cdot T^{2}[/tex]
[tex]r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}[/tex]
[tex]r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }[/tex]
Where [tex]T[/tex] is period, measured in seconds.
If we know that [tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]M = 5.98\times 10^{24}\,kg[/tex] and [tex]T = 86400\,s[/tex], then orbital radius of the satellite is:
[tex]r = \sqrt[3]{\frac{\left(6.674\cdot 10^{-11}\,\frac{m^{2}}{kg\cdot s^{2}} \right)\cdot (5.98\times 10^{24}\,kg)\cdot (86400\,s)^{2}}{4\pi^{2}} }[/tex]
[tex]r \approx 42.259\times 10^{6}\,m[/tex]
[tex]r \approx 42.259\times 10^{3}\,km[/tex]
The orbital radius is approximately 42,259 kilometers.
- An object in equilibrium has three forces exerted on it. A 33-N force act at 90° from the x-axis and a 44-N force act at 60°. What are the magnitude and direction of the third force
Answer:
The magnitude of third force is 74.4 N and direction of third force is 72.8 degrees South.
Explanation:
Let F1, F2 and F3 are three forces exerted on an object.
[tex]\theta_1=90^{\circ}[/tex]
[tex]\theta_2=60^{\circ}[/tex]
[tex]|F_1|=33 N[/tex]
[tex]|F_2|=44 N[/tex]
We have to find the direction and magnitude of third force i.e F3.
[tex]F_{1x}=33cos(90^{\circ})=0 N[/tex]
[tex]F_{1y}=33sin(90^{\circ})=33 N[/tex]
[tex]F_{2x}=44cos(60^{\circ})=22 N[/tex]
[tex]F_{2y}=44 sin(60^{\circ})=22\sqrt{3}=38.11 N[/tex]
Now,
x-component of resultant
[tex]R_x=F_{1x}+F_{2x}=0+22=22 N[/tex]
y-component of resultant
[tex]R_y=F_{1y}+F_{2y}=33+38.11=71.11 N[/tex]
[tex]|R|=\sqrt{R^2_x+R^2_y}[/tex]
[tex]|R|=\sqrt{(22)^2+(71.11)^2}=74.4 N[/tex]
[tex]\theta=tan^{-1}(\frac{R_y}{R_x})[/tex]
[tex]\theta=tan^{-1}(\frac{71.11}{22})=72.8^{\circ}[/tex] South
Hence, the magnitude of third force is 74.4 N and direction of third force is 72.8 degrees South.
The magnitude and direction of the third force is;
F3 = 74.44 N
θ3 = 72.81° in the south direction
We are given;
F1 = 33 N
F2 = 44 N
θ1 = 90°
θ2 = 60°
Let the third force be F3 which will serve as the resultant
Let's first find the x and y component of the forces.
F1x = F1 cos θ1
F1x = 33 × cos 90
F1x = 0 N
F1y = F1 sin θ1
F1y = 33 × sin 90
F1y = 33 N
F2x = F2 cos θ2
F2x = 44 × cos 60
F2x = 22 N
F2y = F2 sin θ2
F2y = 44 × sin 60
F2y = 38.11 N
Thus, the resultant of the x component is;
F3x = F1x + F2x
F3x = 0 + 22
F3x = 22 N
The resultant of the y component is;
F3y = F1y + F2y
F3y = 33 + 38.11
F3y = 71.11 N
Thus, magnitude of resultant of the F3 force is;
F3 = √((F3x)² + (F3y)²)
F3 = √(22² + 71.11²)
F3 = 74.44 N
The direction of the resultant of F3 is;
θ3 = tan^(-1) F3y/F3x
θ3 = tan^(-1) 71.11/22
θ3 = 72.81° in the south direction
Read more about resultant of forces at;
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A cube icebox of side 3cm has a thickness of 5.0cm. If 4.0 kg of ice is put in the box estimate the amount of ice remaining after 6hrs. The outside T° is 45°c and coefficient of thermal conductivity of thermacole is 0.01J /s/m/k. The heat of fusion of water is 3.35× 10^5 J/k/hr/kg
Answer:
The amount of solid ice remaining after 6 hours is approximately 3.68664 kg
Explanation:
The given parameters are;
The side length of the cube box, s = 3(0) cm = 0.3 m
The thickness of the cube box, d = 5.0 cm = 0.05 m
The mass of ice in the box, m = 4.0 kg
The outside temperature of the cube box, T₁ = 45°C
The temperature of the melting ice inside the box, T₂ = 0°C
The latent heat of fusion of ice, [tex]L_f[/tex] = 3.35 × 10⁵ J/K/hr/kg
The surface area of the box, A = 6·s² 6 × (0.3 m)² = 0.54 m²
The coefficient of thermal conductivity, K = 0.01 J/s·m⁻¹·K⁻¹
For thermal equilibrium, we have;
The heat supplied by the surrounding = The heat gained by the ice
The heat supplied by the surrounding, Q = K·A·ΔT·t/d
Where;
ΔT = T₁ - T₂ = 45° C - 0° C = 45° C
ΔT = 45° C
Q = K·A·ΔT·t/d = 0.01 × 0.54 × 45 × 6× 60×60/0.05 = 104976
∴ The heat supplied by the surrounding, Q = 104976 J
The heat gained by the ice = [tex]L_f[/tex] × [tex]m_{melted \ ice}[/tex] =3.35 × 10⁵ J/kg × [tex]m_{melted \ ice}[/tex]
Therefore, from Q = [tex]L_f[/tex] × [tex]m_{melted \ ice}[/tex], we have;
Q = 104976 J = [tex]L_f[/tex] × [tex]m_{melted \ ice}[/tex] = 3.35 × 10⁵ J/kg × [tex]m_{melted \ ice}[/tex]
104976 J = 3.35 × 10⁵ J/kg × [tex]m_{melted \ ice}[/tex]
[tex]m_{melted \ ice}[/tex] = 104976 J/(3.35 × 10⁵ J/kg) ≈ 0.31336 kg
The mass of melted ice, [tex]m_{melted \ ice}[/tex] ≈ 0.31336 kg
∴ The amount of solid ice remaining after 6 hours, [tex]m_{ice}[/tex] = m - [tex]m_{melted \ ice}[/tex]
Which gives;
[tex]m_{ice}[/tex] = m - [tex]m_{melted \ ice}[/tex] = 4.0 kg - 0.31336 kg ≈ 3.68664 kg
The amount of solid ice remaining after 6 hours, [tex]m_{ice}[/tex] ≈ 3.68664 kg.
A ball with a mass of 0.8 kg is thrown straight upward, flies up to its maximum height, and
then falls back down. If the ball reaches a maximum height of 10.7 meters, how fast was the
ball thrown initially? Round your answer to the tenths place.
Answer:
v = 14.5 m/s
Explanation:
The Principle Of Conservation Of Mechanical Energy
In the absence of friction, the total mechanical energy is conserved. That means that
Em=U+K is constant, being U the potential energy and K the kinetic energy
U=mgh
[tex]\displaystyle K=\frac{mv^2}{2}[/tex]
The ball with a mass of m=0.8 kg is thrown straight upward from the zero height reference (h=0) and with some speed (v). The potential energy is zero, but the kinetic speed is given by the equation above.
When the ball reaches its maximum height of h=10.7 m, the speed is zero and all the initial kinetic energy was transformed into potential energy, thus:
[tex]\displaystyle \frac{mv^2}{2}=mgh[/tex]
Simplifying by m:
[tex]\displaystyle \frac{v^2}{2}=gh[/tex]
Solving for v:
[tex]\displaystyle v=\sqrt{2gh}[/tex]
Substituting:
[tex]\displaystyle v=\sqrt{2*9.8*10.7}[/tex]
Calculating:
v = 14.5 m/s
The place below earth's surface where the earthquake begins.
а
focus
fault
en oo
epicenter
P-wave
Answer:
The position where an earthquake begins below the earths surface is called the hypo center. The point directly above the hypo center is called the epicenter.
Explanation:
www.usga.gov | United States of America Department of Geological Surveys.
The energy of a given wave in the electromagnetic spectrum is 2.64 × 10-21 joules, and the value of Planck's constant is 6.6 × 10-34 joule·seconds. What is the value of the frequency of the wave?
Answer:
the answer is b
wjejehebe ehshw es
Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m each. If Margy is walking at 1.4 m/s and accelerates at 0.20 m/s2 during one of the running portions, what is her final velocity at the end of the 100.0 m? Round your answer to the nearest tenth. m/s
Answer:
6.5 m/s
Explanation:
We are given that
Distance, s=100 m
Initial speed, u=1.4 m/s
Acceleration, [tex]a=0.20 m/s^2[/tex]
We have to find the final velocity at the end of the 100.0 m.
We know that
[tex]v^2-u^2=2as[/tex]
Using the formula
[tex]v^2-(1.4)^2=2\times 0.20\times 100[/tex]
[tex]v^2-1.96=40[/tex]
[tex]v^2=40+1.96[/tex]
[tex]v^2=41.96[/tex]
[tex]v=\sqrt{41.96}[/tex]
[tex]v=6.5 m/s[/tex]
Hence, her final velocity at the end of the 100.0 m=6.5 m/s
Answer:
6.5
Explanation:
When applying a force to a book on a desk, the force must be greater than the force of
the book to begin moving.
A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. Its initial position is x0 = 1.3 m at t0 = 0 s
Answer:
Follows are the solution to this question:
Explanation:
In point a:
Place of particles
[tex]X(t)=\int V_{x}(t)dt[/tex]
[tex]=\int 2t^{2}dt\\\\=\frac{2}{3}t^{3}+C[/tex]
[tex]\to t=0\\\\ \to X(0)=2.3 \ m[/tex]
[tex]\to X(0)=0+C\\\\ \to C=2.3\ m[/tex]
[tex]\to X(t)=( \frac{2}{3})t^3 + 2.3\\\\ \to t=2.2\\\\\to X=( \frac{2}{3})\times 2.2^3 +2.3 \\\\[/tex]
[tex]= \frac{2}{3}\times 10.648 +2.3\\\\= \frac{21.296}{3}+2.3\\\\ = 7.09+2.3\\\\ =9.39\\\\ =9.4\ m[/tex]
In point b:
when [tex]t=2.2 \ s[/tex]
the Particle velocity [tex](V)=2 \times 2.22 =9.68\ \frac{m}{s}[/tex]
In point c:
Calculating the Particle acceleration:
[tex]\to a=\frac{dV}{dt} =4\ t\\\\\to t=2.2 \ s\\\\\to a=4\times 2.2 =8.8 \ \frac{m}{s^2}[/tex]
A 0.144-kg baseball is moving toward home plate with a speed of 43 m/s when
it is bunted. The bat exerts an average force of 6,500-N on the ball for 0.00135.
The pitcher throws in the positive x direction. (The Force will act in the -
direction).
Twice a year the Sun’s rays strike Earth north or south of the....
Answer:
Tropic of Cancer in the north and the Tropic of Capricorn in the south.
Explanation:
does it take more force to slow something down than to speed it up.
If an object must be slowed quickly, the force applied to it must be greater than that required for gradual slowing. For example, the greater the force applied to a bicycle's brakes, the faster it will slow or stop.
What is force?A force is an influence in physics that can change the motion of an object. A force can cause a mass object to change its velocity, or accelerate.
Intuitively, force can be described as a push or a pull. A force is a vector quantity because it has both magnitude and direction.
Force is defined as the tendency of a body to modify or change its state as a result of an external cause. When applied force, the body can also change shape, size, and direction.
The same force that was used to accelerate something can be used to slow it down. It will stop faster if more force is applied.
Thus, it can be concluded that it take more force to slow something down than to speed it up.
For more details regarding force, visit:
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Which statement best describes how scientists and engineers work together
in the research and development cycle?
O A. Engineers make a new discovery about nature, and then scientists
do experiments to verify it.
O B. Scientists develop new technology, and then engineers use it to
solve design problems.
C. Scientists make a discovery, and then engineers use it to help
them solve problems with their designs.
D. Engineers develop a new technology, and then scientists use it to
solve design problems.
Answer:
C
Explanation:
what is a physical benefit of stretching
How long does it take for a car to change its velocity from 10 m/s to 25 m/s if the acceleration is 5 m/s^2?
Answer:five times five is twenty five divded by 10 is 2.5 seconds of acceleration
How much pressure is being exerted by blood cells that are applying 375N of force over 25.5m2 of your body’s blood vessels?
PLS HELP MEEE
Answer:
Pressure = 14.71N/m²
Explanation:
Given the following data;
Force = 375N
Area = 25.5m²
To find the pressure;
Pressure = force/area
Pressure = 375/25.5
Pressure = 14.71N/m²
Therefore, the amount of pressure that is being exerted by blood cells is 14.71 Newton per meter square.
What is the ƒ if v = 50 m/s and λ = 10 m?
Answer:
5Hz
Explanation:
I hope that by stating f you mean frequency
so I think people in this app just like to give answers without explaining so let me try to explain to you why
first of all, you must know the formula
which is velocity = frequency x wavelengthjust insert the values in
50 = 10 x frequency
50/10 = frequency
5 = frequency
heyy i don't understand this question may u help me pls?
Answer:
What question?
Explanation:
A boy finds an abandoned mine shaft in the woods, and wants to know how deep the hole is. He drops in a stone, and counts 3 seconds before he hears the "plunk" of the stone hitting the bottom of the shaft. Approximately how deep is the shaft? (Assume a gravitational acceleration of 9.8 m/s2 and no significant air resistance.)
Answer: 9.81 times 3s= 29.43m/s
so 29.43 m/s times 3= 88.29m
Explanation:
PLZ Help!! Answer the Question Below:
Answer:
18.78 m/s.
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 50 kg
Initial velocity (u) = 0 m/s
Height (h) of cliff = 18 m
Acceleration due to gravity (g) = 9.8 m/s²
Final velocity (v) =?
The velocity with which the diver reach the water can be obtained as follow:
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 18
v² = 0 + 352.8
v² = 352.8
Take the square root of both side
v = √352.8
v = 18.78 m/s
Thus, the diver will reach the water with a velocity of 18.78 m/s.
An object is continuously changing its velocity by the same rate. What is true about this object?
А The object has decreasing acceleration
B
The object has increasing acceleration.
С
The object has zero acceleration.
D
The object has constant acceleration.
Answer:
DThe object has constant acceleration.Explanation:
I hope it helps
A 4500 kg Honda Civic, a 0.5 kg Matchbox car, and a 45 kg Power Wheel
are each pushed with a 30 N force. Which car will have the GREATEST
acceleration?
Describing the Changes in the Ator Based on the article "Will the real atomic model please stand up?,"describe one major change that occurred in the development of the modern atomic model.
Answer: the last period. i think has the largest energy level
Explanation:
Answer:
honestly?
Explanation:
the atoms changed
At a swim meet, swimmer A swims 50 m in 12 s, swimmer B in 13 s, swimmer C in 11 s, and
swimmer D in 14 s. The race involved swimming back and forth one time. Which swimmer had
the greatest displacement?
Answer:
Swimmer B had the greatest displacement:)
Explanation:
Hope you do great! :)
brainliest would be nice:)
10 m/s^2 is an example of which vocabulary word?
Answer:
Acceleration.
Explanation:
In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.
This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.
Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.
Mathematically, acceleration is given by the equation;
[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]
[tex]a = \frac{v - u}{t}[/tex]
Where,
a is acceleration measured in [tex]ms^{-2}[/tex]
v and u is final and initial velocity respectively, measured in [tex]ms^{-1}[/tex]
t is time measured in seconds.
The S.I unit for measuring acceleration is meters per seconds square (m/s²).
Hence, 10 m/s^2 is an example of acceleration.
How does momentum play in tackling?
Answer:
When two players are running full speed at each other on a football field they build up their momentum
