Answer:
The centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s²
Explanation:
Given;
distance traveled in the given time = 200 m
time to cover the distance, t = 29.6 s
speed of the runner, v = d / t
v = 200 / 29.6
v = 6.757 m/s
The centripetal acceleration of the runner is given by;
[tex]a_c = \frac{V^2}{r}[/tex]
where;
r is the radius of the circular arc, given as 50 m
Substitute the givens;
[tex]a_c = \frac{V^2}{r}\\\\a_c = \frac{(6.757)^2}{50}\\\\a_c = 0.91 \ m/s^2[/tex]
Therefore, the centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s².
For satellite travelling on circular orbit if radius of the orbit increased 4 times then the period of the satellite increased *
2
4
8
none of the above
Answer:
8
Explanation:
A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?
Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.
In a simulation on earth, an astronaut in his space suit climbs up a vertical ladder. On the moon, the same astronaut makes the same climb. In which case does the gravitational potential energy of the astronaut change by a greater amount?
Answer:
Gravitational potential energy of the astronaut will change by a greater amount on the earth
Explanation:
Gravitational potential energy is expressed by the formula;
GPE = mgh
This means that the gravitational potential energy is directly proportional to the gravity(g)
Now, from constant values, gravity of moon is 1.62 m/s² while gravity of the earth is 9.81 m/s².
This means that if we plug in the values of g on the earth and g on the moon, the potential energy on the earth would be greater than that of the moon
Thus, gravitational potential energy of the astronaut will change by a greater amount on the earth
Which environment is least likely to support protists
A soil
B open ocean
C shallow pod
D organisms blood
Answer:
A: Soil
Explanation:
Protists need a moist environment to survive, and shallow ponds, oceans, and blood is all moist. So, the answer would be the soil, because that is the least moist environment out of these options.
What is the volume of a box if he has Length=7 cm Width=5cm , Height=10cm ?
Answer:
Volume of Cuboid = Height*Width*Length
Explanation:
Volume of Cuboid = 10*5*7
= 350 cu² cm
Answer:
Diagram:-[tex]\setlength{\unitlength}{0.74 cm}\begin{picture}\thicklines\put(5.6,5.4){\bf A}\put(11.1,5.4){\bf B}\put(11.2,9){\bf C}\put(5.3,8.6){\bf D}\put(3.3,10.2){\bf E}\put(3.3,7){\bf F}\put(9.25,10.35){\bf H}\put(9.35,7.35){\bf G}\put(3.5,6.1){\sf 5cm}\put(7.7,6.3){\sf 7cm}\put(11.3,7.45){\sf 10cm}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}[/tex]
Required Answer:-It is a cuboid
where
length =l=7cmwidth=b=5cmheight =h=10cmAs we know that in a cuboid
[tex]{\boxed{\sf Volume=lbh}}[/tex]
Substitute the values[tex]{:}\longrightarrow[/tex][tex]\sf Volume =7×5×10 [/tex]
[tex]{:}\longrightarrow[/tex][tex]\sf Volume=35×10 [/tex]
[tex]{:}\longrightarrow[/tex][tex]\sf Volume=350cm^3 [/tex]
1000 kg car takes travels on a circular track having radius 100 m with speed 10 m/s. What is the direction of the acceleration of the car? *
A- outside track, and normal to track
B- towards the center and normal to the track
C- up
D- down
Answer:
B.
Explanation:
A locomotive creates a
59,400 N force, which creates
an acceleration of 0.145 m/s.
What is the mass of the
locomotive?
Unit=kg
Answer:
410,000 kg
Explanation:
Use Newton's second law
F = ma
m = F/a
m = 59,400 N/(.145 m/s) = 410,000 kg
The spring constant, k, for a 22cm spring is 50N/m. A force is used to stretch the spring and when it is measured again it is 32cm long. Work out the size of this force
Answer:
5N
Explanation:
Given parameters:
Original length = 22cm
Spring constant, K = 50N/m
New length = 32cm
Unknown
Force applied = ?
Solution:
The force applied on a spring can be derived using the expression below;
Force = KE
k is the spring constant
E is the extension
extension = new length - original length
extension = 32cm - 22cm = 10cm
convert the extension from cm to m;
100cm = 1m;
10cm will give 0.1m
So;
Force = 50N/m x 0.1m = 5N
3. Sarah drops a 10.0 kg objed from a 200 m high bridge over a river. a) What work is done by the objec as a result of its fall?
1. When asteroids collided some of the broken materials fall into Earth's orbit. What do
astronomers call the debris when it hits planet Earth?
Answer:
meteoroids
Explanation:
when an asteroid (or really anything else) falls to earth, it is called a meteoroid
When a mass of a cart is 10 kg, and an applied force is 5 N, The acceleration of the cart is
5.0 m/s2
2.0 m/s2
0.5 m/s2
0.2 m/s2
Answer:
0.5 m/s2
Explanation:
F = ma
5 = 10a
a = 5/10
a =0.5
A medicine ball has a mass of 5kg and is thrown with a speed of 3 m/sec what is it's kinetic energy
A 6.0-kilogram cart initially traveling at 4.0 meters per second east accelerates uniformly at 0.50 meter per second squared east for 3.0 seconds. What is the speed of the cart at the end of this 3.0 second interval? A. 1.5 m/s B. 5.5 m/s G. 3.0 m/s D. 7.0 m/s
Answer: 5.5m/s
Explanation:
vf=vi+at
vf= 4.0m/s + (0.50m/s^2)(3.0s)
The speed of the cart at the end of this 3.0 second interval is 5.5 meter per seconds.
Given the following data:
Initial velocity = 4 m/sMass of cart = 6 KgAcceleration = 0.5 [tex]m/s^2[/tex]Time = 3 secondsTo find the speed of the cart at the end of this 3.0 second interval, we would use the first equation of motion;
[tex]V = U + at[/tex]
Where:
U is the initial velocity.V is the final velocity. a is the acceleration. t is the time measured in seconds.Substituting the given parameters into the formula, we have;
[tex]V = 4 + 0.5(3)\\\\V = 4 + 1.5[/tex]
Final velocity, V = 5.5 m/s.
Therefore, the speed of the cart at the end of this 3.0 second interval is 5.5 meter per seconds.
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Why are the coral reefs suffering? (site 2) explain
Answer:
bcuz ov yo fat mamma
Explanation:
Answer:
Water pollution
Explanation:
You are lifting a 10 kg block straight up at a constant speed of 10 m/s. How much force are you exerting on the block?
Answer:
The force exerted is [tex]F = 100 \ N[/tex]
Explanation:
From the question we are told that
The mass of the block is [tex]m_b = 10 \ kg[/tex]
The speed is [tex]v = 10 \ m/s[/tex]
Generally the force exerted to lift the object at constant speed is equivalent to the wight of the ball, this is mathematically represented as
[tex]F = m * g[/tex] Here [tex]g = 10 \ m/s^2[/tex]
=> [tex]F = 10 * 10[/tex]
=> [tex]F = 100 \ N[/tex]
The force are you exerting on the block when the block is lifting straight up with constant speed is 98 N and this can be determined by using the given data.
Given :
You are lifting a 10 kg block straight up at a constant speed of 10 m/s.
The following steps can be used in order to determine the force are you exerting on the block:
Step 1 - According to the given data, the block is lifting straight up at a constant speed. So, the acceleration is zero.
Step 2 - So, the only force exerted on the block is the weight of the block.
Step 3 - So, the force are you exerting on the block is given by:
F = mg
F = 10 [tex]\times[/tex] 9.8
F = 98 N
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Which type of mass movement is likely to result in considerable property damage, but rarely causes loss of life? a. debris avalanche b. rock fall c. mudflow d. creep
Answer:
The correct answer is D. Creep.
Explanation:
Ground creep is a slow downward movement of a hill or mountain slope without the formation of demolition forms. The decisive factor for this is the continuous flow of movement of the soil.
The main driver of collapse is the movement of the surface layer particles during expansion in a direction perpendicular to the slope, followed by vertical collapse on contraction. The visible effect of the collapsing is the inclination of fences and poles, as well as trees that grow out of the ground towards the slope and have trunks curved vertically, in more extreme cases it may be cracks on the walls of buildings.
A 7.50 kg bowling ball has 70.4
kg•m/s of momentum. What is its
velocity?
Answer:
9.39 m/sExplanation:
The velocity of the bowling ball can be found by using the formula
[tex]v = \frac{p}{m} \\ [/tex]
p is the momentum
m is the mass
From the question we have
[tex]v = \frac{70.4}{7.5} \\ = 9.38666666..[/tex]
We have the final answer as
9.39 m/sHope this helps you
a vertical solid steel post 29cm in diameter and 2.0m long is required to support a load of 8200kg, ignore the weight of the post. determine the stress in the post
Answer:
The stress is [tex]\sigma = 1.218*10^{6} \ N/m^2[/tex]
Explanation:
From the question we are told that
The diameter of the post is [tex]d = 29 \ cm = 0.29 \ m[/tex]
The length is [tex]L = 2.0 \ m[/tex]
The weight of the loading mass
Generally the radius of the post is mathematically represented as
[tex]r = \frac{0.29}{2}[/tex]
=> [tex]r = 0.145 \ m[/tex]
Generally the area of the post is
[tex]A = \pi r^2[/tex]
=> [tex]A = 3.14 * 0.145 ^2[/tex]
=> [tex]A = 0.066 \ m^2[/tex]
Generally the weight exerted by the load is mathematically represented as
[tex]F = m * g[/tex]
=> [tex]F = 8200 * 9.8[/tex]
=> [tex]F = 80360 \ N[/tex]
Generally the stress is mathematically represented as
[tex]\sigma = \frac{F}{A}[/tex]
=> [tex]\sigma = \frac{80360 }{0.066}[/tex]
=> [tex]\sigma = 1.218*10^{6} \ N/m^2[/tex]
A spaceship of mass mm circles a planet of mass M in an orbit of radius R. How much energy is required to transfer the spaceship to a circular orbit of radius 3R?
Answer:
ΔE = GmM/3R
Explanation:
The absolute potential energy of an object in a planet's field is given as:
E = -GmM/2r
where,
E = Potential Energy
G = Universal Gravitational Constant
m = mass of spaceship
M = Mass of Planet
r = distance from surface of planet
Therefore, for initial state:
E = E₁ and r = R
E₁ = - GmM/2R
and for final state:
E = E₂ and r = 3R
E₂ = - GmM/6R
So, the required energy will be:
ΔE = E₂ - E₁ = - GmM/6R + GmM/2R
ΔE = GmM(- 1/6R + 1/2R)
ΔE = GmM/3R
find the fundamental units involved in derived units
newton
watt
joule
pascal
cubic meter
9ycy8c8t 7f fixfuozofuxt8lsrupsurpaurae6pUeoUe6eoUeFipzuroz6d0, 7d0z6e0z7e0zurpz6e0z
Explanation:
force newton N - m·kg·s-2
pressure, stress pascal Pa N/m2 m-1·kg·s-2
energy, work, quantity of heat joule J N·m m2·kg·s-2
power, radiant flux watt W J/s m2·kg·s-3
volume cubic meter m3
when is thermal equilibrium achived between two identical objects
need help ASAP
Answer: When two objects in contact with each other are at different temperatures, they are said to be in thermal equilibrium.
Explanation: . When two objects not in contact with each other are at the same pressure, they are said to be in thermal equilibrium.
Question 1 of 5
In which way are electromagnetic waves different from mechanical waves?
Answer:
Electromagnetic waves differ from mechanical waves in that they do not require a medium to propagate. This means that electromagnetic waves can travel not only through air and solid materials, but also through the vacuum of space.
Explanation:
Answer:
In which way are electromagnetic waves different from mechanical waves?Electromagnetic waves can travel through empty space
Explanation:
I took A P E X Quiz.
The time constant for RC circuit with the values of R1 and C1 is 5ms. What will be the time constant for a new RC circuit with the values: R2=10R1 and C2 = 0.5C1.
a. 2.5ms
b. 15.5ms
c. 50ms
d. 25ms
e. 15ms
Answer:
d. 25 ms
Explanation:
In a RC circuit we call time constant to the product of the resistance times the capacitance, which represents the time when the charge reaches to the 63% of the final value, as follows:[tex]\tau_{1} = R_{1} *C_{1} = 5 ms (1)[/tex]
If we have a new circuit with new values for R and C, the time constant will be defined in the same way, as follows:[tex]\tau_{2} =10* R_{1} *0.5*C_{1} = 5*(R_{1}* C_{1}) = 5* \tau_{1} = 5* 5 ms = 25 ms (2)[/tex]
An electromagnet needs a magnetic metal core. To produce a magnetic field,
what else is required?
A second metal core?
A solenoid with current or no current running through it? Or a permanent magnet?
Answer:solenoid with current running through it
Explanation:just took the test
A 0.12-m-radius grinding wheel takes 5.5 s to speed up from 2.0 rad/s to 11.0 rad/s. What is the wheel's average angular acceleration?
Answer:
0.56rad/s²
Explanation:
Using the equation of motion
wf = wi + αt
wf is the final angular velocity
wi is the initial angular velocity
α is the angular acceleration
t is the time
Given
wf = 11.0rad/s
wi =2.0rad/s
t = 5.5secs
Substitute into the formula and get α
11.0 = 2.0+5α
11.0-2.0 = 5α
9.0 = 5α
α = 5/9.0
α ≈ 0.56rad/s²
Hence the wheel's average angular acceleration is 0.56rad/s²
The wheel's average angular acceleration is equal to 1.64 [tex]rad/s^2[/tex].
Given the following data:
Radius = 0.12 meterTime = 5.5 secondsInitial angular velocity = 2.0 rad/sFinal angular velocity = 11.0 rad/sTo determine the wheel's average angular acceleration, we would apply the first equation of kinematics:
Mathematically, the angular acceleration of an object is given by the formula:
[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
Where:
[tex]\omega_i[/tex] is the initial angular velocity.[tex]\omega_f[/tex] is the final angular velocity.t is the time.Substituting the given parameters into the formula, we have;
[tex]\alpha =\frac{11.0\;-\;2.0}{5.5} \\\\\alpha =\frac{9.0}{5.5}[/tex]
Angular acceleration = 1.64 [tex]rad/s^2[/tex]
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An object that moves in uniform circular motion has a centripetal acceleration of 13 m/s^2 . If the radius of the motion is 0.02m, what is the frequency of motion?
Answer:
f = 3.97 Hz
Explanation:
Given that,
Centripetal acceleration, [tex]a=13\ m/s^2[/tex]
The radius of motion is 0.02 m
The formula for the centripetal acceleration is given by :
[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{13\times 0.02} \\\\v=0.5\ m/s[/tex]
The speed of an object in a circular path is given by :
[tex]v=\dfrac{2\pi r}{t}[/tex]
t is time period
Also, f=1/t (f is frequency)
[tex]f=\dfrac{v}{2\pi r}\\\\f=\dfrac{0.5}{2\pi \times 0.02}\\\\f=3.97\ Hz[/tex]
Hence, the frequency of motion s 3.97 Hz.
The frequency of the motion is 4.1 Hz.
Linear velocity?The linear velocity of the of the object is calculated as follows;
[tex]a = \frac{v^2}{r} \\\\v^2 = ar\\\\v = \sqrt{ar} \\\\v = \sqrt{13 \times 0.02} \\\\v = 0.51 \ m/s[/tex]
Angular speed of the objectThe angular speed of the object is calculated as follows;
[tex]\omega =\frac{v}{r} \\\\\omega = \frac{0.51}{0.02} \\\\\omega = 25.5 \ rad/s[/tex]
Frequency of motionThe frequency of the motion is calculated as follows;
[tex]\omega = 2\pi f\\\\f = \frac{\omega }{2\pi} \\\\f = \frac{25.5}{2\pi } \\\\f = 4.1 \ Hz[/tex]
Thus, the frequency of the motion is 4.1 Hz.
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A 1150 kg car is on a 8.70° hill.
Using X-Y axes tilted down the
plane, what is the y-component
of the weight?
Answer:
Explanation:
y-component = - mgcos(8.7)
= - (1150)(9.81)cos(8.7)
= - 11151.69378
= - 11151.69 N
The weight of the y-component is 11140.33N.
How to find the weight of the y-component?
To find the weight of the y-component:
Given,
Car weight = 1150 kg
Anfle = 8.70 degree
weight = mg = 1150 * 9.8
= 11270 N
Y-component = mg cos∅
= 11270 * cos(8.70)
= 11140.33N
The aspect that pushes proper or left is referred to as the x-factor, and the element that pushes up or down is known as the y-component.
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What specific changes in two climate variables are expected to lead to major decreases in soil moisture southern Africa and the Mediterranean region?
Answer:
Less precipitation, droughts9: How might agriculture in southern Europe change by the end of the century if conditions follow the RCP8.
Explanation:
Precipitation and droughts are the specific changes in two climate variables that are expected to lead to major decreases in soil moisture.
What is drought?
Drought is defined as a period of protracted water scarcity, whether it is due to atmospheric surface water, or groundwater constraints.
Droughts can last months or years, although they can be proclaimed in as little as 15 days.
It has the potential to have a significant influence on the afflicted region's ecology and agriculture as well as harm the local economy.
Precipitation and droughts are the specific changes in two climate variables that are expected to lead to major decreases in soil moisture in southern Africa and the Mediterranean region.
Hence Precipitation and droughts are the specific changes in two climate variables.
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The following is the longitudinal characteristic equation for an F-89 flying at 20,000 feet at Mach 0.638. The Short Period natural frequency is _____.
Answer:
hello your question is incomplete attached below is the missing part
answer : short period oscillations frequency = 0.063 rad / sec
phugoid oscillations natural frequency ( [tex]w_{np}[/tex] ) = 4.27 rad/sec
Explanation:
first we have to state the general form of the equation
= [tex]( S^2 + 2\alpha _{p} w_{np} S + w^{2} _{np} ) (S^{2} + 2\alpha _{s} w_{ns}S + w^{2} _{ns} ) = 0[/tex]
where :
[tex]w_{np} = Natural frequency of plugiod oscillation[/tex]
[tex]\alpha _{p} = damping ratio of plugiod oscilations[/tex]
comparing the general form with the given equation
[tex]w^{2} _{np}[/tex] = 18.2329
[tex]w^{2} _{ns} = 0.003969[/tex]
hence the short period oscillation frequency ( [tex]w_{ns}[/tex] ) = 0.063 rad/sec
phugoid oscillations natural frequency ( [tex]w_{np}[/tex] ) = 4.27 rad/sec
A motorcycle moving at a constant velcoity suddenly accelerates at a rate of 4.0 m/s/s to a speed of 35 m/s in 5.0 s. What was the initial speed of the motorcycle?
Answer:
15 m/s
Explanation:
v = u+ at
35 = u + 20
35-20 = u
u= 15 m/s