(a) The ball's height y at time t is given by
y = (20 m/s) sin(40º) t - 1/2 g t ²
where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve y = 0 for t :
0 = (20 m/s) sin(40º) t - 1/2 g t ²
0 = t ((20 m/s) sin(40º) - 1/2 g t )
t = 0 or (20 m/s) sin(40º) - 1/2 g t = 0
The first time refers to where the ball is initially launched, so we omit that solution.
(20 m/s) sin(40º) = 1/2 g t
t = (40 m/s) sin(40º) / g
t ≈ 2.6 s
(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration g. So
0² - ((20 m/s) sin(40º))² = 2 (-g) y
where y in this equation refers to the maximum height of the ball. Solve for y :
y = ((20 m/s) sin(40º))² / (2g)
y ≈ 8.4 m
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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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
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 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
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
A displacement vector with a magnitude of 20. meters could have perpendicular components with magnitudes of A. 10. m and 10. m B. 12 m and 8.0 m 12 m and 16 m D. 16 m and 8.0 m
Answer:10.m and 10. M
Explanation:
A displacement vector with a magnitude of 20. m could have perpendicular components with magnitudes of C. 12 m and 16 m.
A displacement vector with a magnitude of 20. meters can be decomposed in 2 perpendicular components.
They would form a right triangle, in which the displacement vector would be the hypotenuse (a) and the components would be the legs (b, c).
Given the magnitude of the legs, we can calculate the magnitude of the hypotenuse using the Pythagorean theorem.
[tex]c = \sqrt{a^{2} + b^{2} }[/tex]
Let's use this formula to calculate the displacement vector for each pair of legs.
A. 10. m and 10. m[tex]c = \sqrt{a^{2} + b^{2} } = \sqrt{(10.m)^{2} + (10.m)^{2} }= 14.1m[/tex]
B. 12 m and 8.0 m[tex]c = \sqrt{a^{2} + b^{2} } = \sqrt{(12m)^{2} + (8.0m)^{2} }= 14.4m[/tex]
C. 12 m and 16 m[tex]c = \sqrt{a^{2} + b^{2} } = \sqrt{(12m)^{2} + (16m)^{2} }= 20m[/tex]
D. 16 m and 8.0 m[tex]c = \sqrt{a^{2} + b^{2} } = \sqrt{(16m)^{2} + (8.0m)^{2} }= 17.9m[/tex]
A displacement vector with a magnitude of 20. m could have perpendicular components with magnitudes of C. 12 m and 16 m.
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The speed of a space shuttle is 10 / express this in /�
Answer:
268.22m/s
Explanation:
Given;
10mile/min to m/s
We need to convert between the two units;
1 mile = 1609.34m
60s = 1min
Now;
10 x [tex]\frac{mile}{min}[/tex] x [tex]\frac{1min}{60s}[/tex] x [tex]\frac{1609.34m}{1mile}[/tex]
= 268.22m/s
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.
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?
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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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 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]
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
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
True or False. A projectile is an object that once set in motion, continues in motion by its own inertia.
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Question
An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of
32.7 m/s. It then flies a further distance of 44500 m, and afterwards, its velocity is 50.3 m/s. Find the
airplane's acceleration
acceleration:
.016m/s2
Calculate how much time clapses while the airplane covers those 44500 m
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49 ENG
1:45 PM
9/17/2020
Answer:
Explanation:
initial velocity u = 32.7 m /s
final velocity v = 50.3 m /s
displacement s = 44500 m
acceleration a = ?
v² = u² + 2 a s
50.3² = 32.7² + 2 x a x 44500
2530.09 = 1069.29 + 89000a
a .016 m /s²
time taken t = ?
v = u + at
50.3 = 32.7 + .016 t
t = 1100 s
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 wire carries a current of 11.4 A in a direction that makes an angle of 11.4° with a magnetic field of magnitude 11.4 à 10-3 T. The magnitude of the force on 11.4 cm of this wire is:____.a) 11.4 * 10^-3 N
b) 0.130 N
c) 1.48 * 10^-2 N
d) 2.93 * 10^-3 N
Answer:
(d) 2.93 x 10⁻³ N
Explanation:
Given;
current in the wire, I = 11.4 A
angle of inclination, θ = 11.4⁰
magnetic field on the wire, B = 11. 4 x 10⁻³
length of the wire, L = 11.4 cm = 0.114 m
The magnitude of magnetic force on the wire is given by;
F = BILsinθ
F = (11.4 x 10⁻³)(11.4)(0.114)(sin 11.4°)
F = 0.00293 N
F = 2.93 x 10⁻³ N
Therefore, the correct option is "D"
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
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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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.
Two tiny conducting spheres are identical and carry charges of -20μC and +50μC. They are separeted by a distance of 2.50cm. (a) what is the magnitude of the force that each sphere each sphere experience, and is the force attractive or repulsive ? (b) The spheres are brought into contact and then separated toa distance of 2.50cm. Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive or repulsive.
Answer:
[tex]14400\ \text{N}[/tex], Attractive
[tex]3240\ \text{N}[/tex], Repulsive
Explanation:
[tex]q_1[/tex] = -20 μC
[tex]q_2[/tex] = 50 μC
r = Distance between the charges = 2.5 cm
k = Coulomb constant = [tex]9\times 10^9\ \text{Nm}^2/\text{C}^2[/tex]
Electrical force is given by
[tex]F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{9\times 10^9\times (-20\times 10^{-6})\times (50\times 10^{-6})}{(2.5\times10^{-2})^2}\\\Rightarrow F=-14400\ \text{N}[/tex]
The magnitude of force each sphere will experience is [tex]14400\ \text{N}[/tex]
Since the charges have opposite charges they will attract each other.
Now the charges are brought into contact with each other so the resultant charge will be
[tex]q=\dfrac{q_1+q_2}{2}\\\Rightarrow q=\dfrac{-20+50}{2}\\\Rightarrow q=15\ \mu\text{C}[/tex]
[tex]F=\dfrac{kq^2}{r^2}\\\Rightarrow F=\dfrac{9\times 10^9\times (15\times 10^{-6})^2}{(2.5\times 10^{-2})^2}\\\Rightarrow F=3240\ \text{N}[/tex]
The magntude of the force the spheres experience will be [tex]3240\ \text{N}[/tex]
The spheres have the same charge now so they will repel each other.
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.
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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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
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.
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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Why are the coral reefs suffering? (site 2) explain
Answer:
bcuz ov yo fat mamma
Explanation:
Answer:
Water pollution
Explanation:
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
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