On a slope where does a marble have to most kinetic energy?
a) it is always the same
b) at the initial position
c) at the final position
d) somewhere between the initial and the final position
Answer:
C
Explanation:
Kinetic energy is the energy of motion. It has the most potential energy at the top but the most kinetic at the bottom after it's accelerated fully down the slope.
•What is the gravitational potential energy of a girl
who has a mass of 40 kg and is standing on the
edge of a diving board that is 5 m above the water?
Answer:
1960 joule
Explanation:
PLEASE HELP IM HEING TIMED
Answer:
Uh, I could be wrong but doesn’t it mean that the wave and particle are reacting together to make light? I think it’s something like that... I hope this helps!
Which of these objects are constantly in motion? Select all that apply.
A.
Earth
B.
Planes
C.
Trains
D.
Blood
E.
Sun
F.
Cars
The object Earth, Sun, and Blood are constantly in motion. The correct option is A, D, and E.
What is motion?if a body changes its position with respect to its surroundings in a given interval of time, Then the body is said to be in motion
.
Motion is generally classified as follows.:
i) Rectilinear motion.
ii) Circular motion.
iii) Rotational motion.
iv) Periodic motion.
The Earth is continuously in motion because it continuously revolves around The Sun in an elliptical orbit, due to which a year is 365 days. and also The earth rotates about its own axis once a day.
The Sun also revolves around the galactic center of our Milkyway galaxy. and it also rotates about its own axis continuously. so that the sun is also continuously in motion.
The Human blood is continuously in motion Because our blood is continuously circulating whole over the body with the help of our heart. The heart continuously pumps our blood and circulates it inside the human body.
Hence the Earth, Sun, and blood are continuously in motion.
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If an object has applied force of 20 N and a frictional force of 5 N what is the net force?
Answer:
Net force = 15 N
Explanation:
Given that,
Applied force on an object = 20 N
Frictional force = 5 N
We need to find the net force acting on the object.
Friction is an opposing force. It acts in the opposite direction.
Net force = Applied force - Frictional force
= 20 N - 5 N
= 15 N
Hence, the net force acting on the object is 15 N.
A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 34.5 N is applied tangent to the rim of the disk.
a) What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through .200 revolution?
b) What is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through .200 evolution?
Answer:
a) v = 1.01 m/s
b) a = 5.6 m/s²
Explanation:
a)
If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:[tex]\tau = I * \alpha (1)[/tex]
Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:τ = F*r (2)For a solid uniform disk, the rotational inertia regarding an axle passing through its center is just I = m*r²/2 (3).Replacing (2) and (3) in (1), we can solve for α, as follows:[tex]\alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)[/tex]
Since the angular acceleration is constant, we can use the following kinematic equation:[tex]\omega_{f}^{2} - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)[/tex]
Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:[tex]0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)[/tex]
Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:[tex]\omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)[/tex]
Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:[tex]v = \omega * r (8)[/tex]
where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of the disk, r becomes the radius of the disk, 0.200 m.Replacing this value and (7) in (8), we get:[tex]v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)[/tex]
b)
There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:[tex]a_{t} = \alpha * r (9)[/tex]
where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.Replacing this value and (4), in (9), we get:[tex]a_{t} = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)[/tex]
Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.The radial acceleration is just the centripetal acceleration, that can be expressed as follows:[tex]a_{c} = \omega^{2} * r (11)[/tex]
Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.Replacing this value and (7) in (11) we get:[tex]a_{c} = \omega^{2} * r = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)[/tex]
The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:[tex]a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)[/tex]
Planet Earth is held in orbit by the sun's gravitational force pulling on the Earth.
Which one of the following statements are true about that gravitational force?
The sun does not have a gravitational force because it is made of gas.
The earth pulls with more force since it is less massive than the sun.
The sun and the earth pull on each other with equal and opposite forces.
The sun pulls with more force since the sun is more massive than the earth.
Answer:
The sun pulls with more force since the sun is more massive than the earth.
Explanation:
That your answer because the Earth is getting pulled by the gravitational pull from the Sun
I will give brainly
Defend Democritus' work on the atom and its contribution to the modern atomic model.
What is the torque on a bolt applied with a wrench that has a
lever arm of 30 cm with a force of 30 N?
Answer:
9 Nm
Explanation:
The formula for torque is;
τ = lever arm * force applied
τ = 30/100 * 30
τ = 9 Nm
The torque on a bolt will be "9 Nm".
Given values are:
Force applied,
30 NLever arm,
30 cmAs we know the formula,
→ [tex]\tau = Lever \ arm\times force \ applied[/tex]
By substituting the values, we get
→ [tex]= \frac{30}{100}\times 30[/tex]
→ [tex]= 9 \ Nm[/tex]
Thus the above response is correct.
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Suppose it takes a constant force a time of 6.0 seconds to slow a 2500 kg truck
from 26.0 m/sec to 18.0 m/sec. What is the magnitude of the force? Give
your answer in scientific notation rounded correctly.
Answer:
[tex]3.3\cdot 10^3\:\mathrm{N}[/tex]
Explanation:
Impulse on an object is given by [tex]\mathrm{[impulse]}=F\Delta t[/tex].
However, it's also given as change in momentum (impulse-momentum theorem).
Therefore, we can set the change in momentum equal to the former formula for impulse:
[tex]\Delta p=F\Delta t[/tex].
Momentum is given by [tex]p=mv[/tex]. Because the truck's mass is maintained, only it's velocity is changing. Since the truck is being slowed from 26.0 m/s to 18.0 m/s, it's change in velocity is 8.0 m/s. Therefore, it's change in momentum is:
[tex]p=2500\cdot 8.0=20,000\:\mathrm{kg\cdot m/s}[/tex].
Now we plug in our values and solve:
[tex]\Delta p=F\Delta t,\\F=\frac{\Delta p}{\Delta t},\\F=\frac{20,000}{6}=\fbox{$3.3\cdot 10^3\:\mathrm{N}$}[/tex](two significant figures).
A racecar makes 24 revolutions around a circular track of radius 2 meters in
162 seconds. Find the racecar's frequency
Answer:
[tex]0.15\: \mathrm{Hz}[/tex]
Explanation:
The frequency is of an object is given by [tex]f=\frac{1}{T}[/tex], where [tex]T[/tex] is the orbital period of the object.
Since the racecar makes 24 revolutions around a circular track in 162 seconds, it will take the racecar [tex]\frac{162}{24}=6.75\:\mathrm{s}[/tex] per revolution.
Therefore, the frequency of the racecar is [tex]\frac{1}{6.75}=\fbox{$0.15\:\mathrm{Hz}$}[/tex] (two significant figures).
The radius of the track is irrelevant in this problem.
Suppose Isaac Newton can swim with a velocity of 35 m/sec. If Isaac Newton swims straight
across a river with a current of 22 m/sec, then what is the magnitude of Newton's resulting
velocity as he swims across the river. Note: Sir Isaac Newton swims like Aquaman.
Answer:
Let Vx = 35 m/s speed of swimmer across river
Vy = 22 m/s speed of river dowstream
V = (Vx^2 + Vy^2)^1/2 = 41.3 m/s net speed of swimmer
tan theta = Vy / Vx = .629 theta = 32.2 deg
(Theta would be zero if speed of river was zero)\
A frictionless cart of mass M is attached to a spring with spring constant k. When the cart is displaced 6 cm from its rest position and released, it oscillates with a period of 2 seconds.Four English majors are discussing what would happen to the period of oscillation if the cart was displaced 12 cm from its rest position instead of 6 cm and again released.With which, if any, of these students do you agree?
Answer:
Time period of horizontal Oscillation = T = 2[tex]\pi[/tex][tex]\sqrt{\frac{m}{k} }[/tex]
As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.
Explanation:
Solution:
As we know that:
F = Kx
Here,
K = Spring constant
x = displacement.
First, they are displacing it with 6 cm from its rest position for which Time period of the oscillation is T = 2 seconds.
But next, they want to know the effect on the time period of the oscillation if the displacement x is doubled from 6cm to 12 cm.
First of all, let us see the equation of the time period of the oscillation.
We need to check, if time period does depend on the displacement or not.
As we know,
Time period of horizontal Oscillation = T = 2[tex]\pi[/tex][tex]\sqrt{\frac{m}{k} }[/tex]
As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.
Since, K is the constant for a particular spring, we need to change the mass of the cart to change the time period.
Hence the Time period will remain same.
The time period will remain the same in both conditions. The time period of horizontal Oscillation will be [tex]2 \pi \sqrt{\frac{m}{k} }[/tex].
What is the time period of oscillation?The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.
From Hooke's law;
F = Kx
Where,
K is the spring constant
x is the displacement
First, they move it 6 cm from its rest position, with a T = 2 second oscillation period.
However, they want to know what influence doubling the displacement x from 6 cm to 12 cm has on the oscillation's time period.
Let's start by looking at the oscillation's time period equation. We need to see if the time period is affected by the shift.
The time period of the horizontal oscillation is given by;
[tex]\rm T = 2 \pi \sqrt{\frac{m}{k} }[/tex]
As the equation shows, the period of oscillation is determined by the mass and the spring constant. On the displacement, no.
We must modify the mass of the cart to change the time period since K is the constant for each spring.
Hence the time period will not change.
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A beaker with water resting on a scale weighs 40 N. A block
suspended on a hanging spring weighs 20 N. The spring scale
reads 15 N when a block is fully submerged in the water. What is
the reading of a scale on which the beaker with water rests, while
the block is submerged in the water after detached from the
hanging spring?
A. 25 N B. 60 N C. 55 N D. 45 N
Answer:
D. 45 N
Explanation:
The weight of the block is 20 N, when the block is fully immerged in water, it weighs 15 N. Hence the loss of weight = 20 N - 15 N = 5 N.
The loss of weight is as a result of the buoyant force. The buoyant force is the upward force exerted by a fluid when an object is fully or partially immersed in a fluid.
The buoyant force of 5 N acts in the upward direction, the weight of the beaker that would be read by the scale when the beaker is immersed in water = 40 N + 5 N = 45 N
A person's speed around the Earth is faster at the poles than it is at the equator.
True or False?
Answer:
no
Explanation:
it is faster at the equator
When the sum of all the forces acting on a block on an inclined plane is zero, the block
A) must be at rest
B) must be accelerating
C) may be slowing down
D) may be moving at constant speed
Answer:
hmmm thats too hard for me.
Explanation:
A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequencies is 195 Hz. The next higher resonance frequency is 260 Hz.
Required:
a. What is the fundamental frequency of this string?
b. Which harmonics have the given frequencies?
c. What is the length of the string?
Answer:
(a) the fundamental frequency of this string is 65 Hz
(b) the harmonics of the given frequencies are third and fourth respectively.
(c) the length of the string is 2.74 m
Explanation:
Given;
mass density of the string, μ = 3 x 10⁻³ kg/m
tension of the string, T = 380 N
resonating frequencies, 195 Hz and 260 N
For the given resonant frequencies;
[tex]195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3[/tex]
(c) From any of the equations, solve for Length of the string (L);
[tex]195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m[/tex]
(a) the fundamental frequency is calculated as;
[tex]f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o = 65 \ Hz[/tex]
(b) harmonics of the given frequencies;
the first harmonic (n = 1) = f₀ = 65 Hz
the second harmonic (n = 2) = 2f₀ = 130 Hz
the third harmonic (n = 3) = 3f₀ = 195 Hz
the fourth harmonic (n = 4) = 4f₀ = 260 Hz
Thus, the harmonics of the given frequencies are third and fourth respectively.
Of the charge Q initially on a tiny sphere, a portion q is to be transferred to a second, nearby sphere. Both spheres can be treated as particles. For what value of q/Q>0.5 will the electrostatic force between the two parts have 1/3 of the maximum possible value?
Answer:
Explanation:
Maximum value of force will be possible when both the sphere will have same charge . In that case charge on each sphere = Q / 2 =.5Q
F( max ) = k .5Q x .5Q / R²
=.25kQ² /R²
For the second case
F = k q ( Q-q)/ R²
F = .25kQ² /3R²
.25kQ² /3R² = k q ( Q-q)/ R²
.25 Q² = 3qQ - 3q²
3q² - 3qQ + .25 Q² = 0
q =
What is the gravitational force between two students, John and Mike, if John has a mass of 81.0 kg, Mike has a mass of 93.0 kg, and their centers are separated by a distance of .620 m?
Answer:
1.31×10¯⁶ N
Explanation:
From the question given above, the following data were obtained:
Mass of John (M₁) = 81 Kg
Mass of Mike (M₂) = 93 Kg
Distance apart (r) = 0.620 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Force (F) =?
The gravitational force between the two students, John and Mike, can be obtained as follow:
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × 81 × 93 / 0.62²
F = 6.67×10¯¹¹ × 7533 / 0.3844
F = 1.31×10¯⁶ N
Therefore, the gravitational force between the two students, John and Mike, is 1.31×10¯⁶ N
Surface water is replaced by the blank cycle.
Answer: Surface water is replaced by the water cycle.
Explanation: The water cycle is a cycle that describes the movement of water.
Answer:
it is water cycle when it's one of the process occurs and process is precipation this replaces surface water
Heat traveling through a pan to warm food in the pan is an example of what kind of heat transfer?
A. Convection
B. Radiation
C. Insulation
D. conduction
Answer:
A
Explanation:
Answer:
D. Conduction
Explanation:
Convection heat transfer occurs in fluids, while conduction heat transfer occurs in solids.
You are holding two balloons of the same shape and size. One is filled with helium, and the other is filled with ordinary air. On which balloon the buoyant force is greater?
a. The helium filled balloon experiences the greater buoyant force.
b. The air filled balloon experiences the greater buoyant force.
c. Both balloons experiences same buoyant force.
Answer:
Both balloons experiences same buoyant force.
Explanation:
Buoyant force is defined as the upward force that is exerted on an object which is wholly or partly immersed in a fluid. We can also define it as the upward force exerted by any fluid upon a body immersed in it. We may also refer to this buoyant force as the upthrust.
This buoyant force is the push of air on the balloon and it is independent of the contents of the balloons. Hence, both balloons experiences same buoyant force.
A car of weight 6,400 N has four wheels. Each wheel has an area of
80 cm-touching the road. Find the pressure the car puts on the
ground. I
Answer:
80N/cm^2
Explanation:
Given data
A car of weight 6,400N
Area= 80cm^2
The weight is distributed evenly on the 4 wheels
hence 6400/4
=1600 N
We know that
Pressure = force/Area
Pressure= 1600/80
Pressure= 20 N/cm^2
For the four wheels, the pressure is
=20*4
=80N/cm^2
Various amplifier and load combinations are measured as listed below using rms values. For each, find the voltage, current, and power gains ( A v , Ai , and Ap, respectively) both as ratios and in dB:
(a) vI= 100 mV, iI = 100 μA, vO = 10 V, RL = 100Ω
(b) vI = 10 μV, iI = 100 nA, vO =1 V, RL= 10 kΩ
(c) vI =1 V, iI = 1 mA, vO =5 V, RL = 10Ω
Answer:
The solution to this question can be defined as follows:
Explanation:
In point (a):
[tex]v_i= 100 \ mV\\\\ i_I = 100 \mu \ A\\\\ v_O = 10 \ V\\\\ R_L = 100 \ \Omega \\\\i_L = \frac{V_0}{R_L} = \frac{10}{100} = 100 \ MA \\\\A_v = \frac{V_0}{V_i} = \frac{10}{100 \times 10^{-3}} =100 \\\\A_v(db) = 20 \lag (100) =40 \ db \\\\ A_i= \frac{i_L}{i_i} = \frac{100 \times 10^{-3}}{100 \times 10^{-6}} =1000 \\\\A_i(db) = 20 \lag (100) =60 \ db \\\\[/tex]
[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_L}{v_i i_i} = \frac{ 10(100 \times 10^{-3})}{100 \times 10^{-6} \times 100 \times 10^{-3}} =100000\\\\ A_p(db) =10 \log (100000) =50 \ db \\\\[/tex]
In point (b):
[tex]v_i = 10 \mu V\\\\ i_i = 100 \ nA \\\\ v_O =1 \ V \\\\ R_L= 10 \ k \Omega \\\\i_0 = \frac{V_0}{R_L} = \frac{1}{10 \ K} = 100 \ \muA \\\\A_v = \frac{V_0}{V_i} = \frac{10}{10 \times 10^{-6}} =100000 \\\\A_v(db) = 20 \lag (100000) =100 \ db \\\\ A_i= \frac{i_0}{i_i} = \frac{100 \times 10^{-6}}{100 \times 10^{-9}} =1000 \\\\A_i(db) = 20 \lag (1000) =60 \ db \\\\[/tex]
[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_0}{v_i i_i} = \frac{ 1 \times 100 \times 10^{-6})}{10 \times 10^{-6} \times 100 \times 10^{-9}} =100000000\\\\A_p(db) =10 \log (100000000) =80 \ db \\\\[/tex]
In point (C):
[tex]v_i =1\ V\\\\ i_I = 1 \ mA\\\\ v_O =5\ V \\\\ R_L = 10 \ \Omega \\\\i_0 = \frac{V_0}{R_L} = \frac{5}{10 } = 0.5 \ A \\\\A_v = \frac{V_0}{V_i} = \frac{5}{1} =5 \\\\A_v(db) = 20 \log 5 =13.97 \ db = 14 \db \\\\ A_i= \frac{i_0}{i_i} = \frac{0.5}{1\times 10^{-3}} =500 \\\\A_i(db) = 20 \log (500) =53.97 \ db = 54 \db \\\\[/tex]
[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_0}{v_i i_i} = \frac{ 5 \times 0.5 }{1 \times 1 \times 10^{-3}} =2500\\\\A_p(db) =10 \log (2500) = 33.97 \ db = 34 \db\\\\[/tex]
Calculate the moment on the following seesaws
5Kg
4m
The earth rotates through one complete revolution every 1,440 minutes. Since the axis of rotation is perpendicular to the equator, you can think of a person standing on the equator as standing on the edge of a disc that is rotating through one complete revolution every 1440 minutes. Find the angular velocity of a person standing on the equator.
Answer:
The angular velocity of a person standing on the equator is approximately [tex]7.272\times 10^{-5}[/tex] radians per second.
Explanation:
The Earth rotates at constant speed. From Rotational Physics, the angular velocity ([tex]\omega[/tex]), measured in radians per second, is defined by the following formula:
[tex]\omega = \frac{2\pi}{T}[/tex] (1)
Where [tex]T[/tex] is the period of rotation of the Earth, measured in seconds.
If we know that [tex]T = 86400\,s[/tex], then the angular velocity of a person standing on the equator is:
[tex]\omega = \frac{2\pi}{86400\,s}[/tex]
[tex]\omega \approx 7.272\times 10^{-5}\,\frac{rad}{s}[/tex]
The angular velocity of a person standing on the equator is approximately [tex]7.272\times 10^{-5}[/tex] radians per second.
What is one characteristic of an electron?
Answer:
A is the best character that defines an electron.
a boy of mass 40kg sits at point 2m from the pivot of a see-saw . find the weight if a girl who can balance the see-saw by sitting at a distance of 3•2m from the pivot.( take g=10NKg)
Answer:
The weight of the girl is 250 N
Explanation:
Static Equilibrium
Static equilibrium occurs when an object is at rest, i.e., neither rotating nor translating.
In the static rotational equilibrium, the total torque is zero with respect to any rotational axis.
The torque applied by a force F perpendicular to a displacement X with respect to a reference rotating point is:
T = F*X
The seesaw will be in rotational equilibrium if the torque applied by the boy of mass m1=40 Kg at x1=2 m from the pivot is equal to the torque applied by the girl of unknown mass m2 at x2=3.2 m from the pivot.
The force applied by both children is their weight:
[tex]F_1 = W_1 = m_1g[/tex]
[tex]F_2 = W_2 = m_2g[/tex]
It must be satisfied:
[tex]m_1gx_1=m_2gx_2[/tex]
Simplifying:
[tex]m_1x_1=m_2x_2[/tex]
Solving for m2:
[tex]\displaystyle m_2=\frac{m_1x_1}{x_2}[/tex]
[tex]\displaystyle m_2=\frac{40*2}{3.2}[/tex]
[tex]m_2=25\ kg[/tex]
Her weight is:
[tex]\mathbf{W_2=25*10 = 250\ N}[/tex]
The weight of the girl is 250 N
Let w(x)=3x-7.If w(x)=14, find x
Answer:
7
Explanation:
We are given:
w(x) = 3x - 7
w(x) = 14
The problem here entails us to solve for x;
To solve for x; equate the two expressions:
3x - 7 = 14
3x = 14 + 7
3x = 21
x = 7
So the value of x = 7
The speed limit on some segments of interstate 4 is 70 mph. What is this in km/h?
Answer:
112.63km/hr
Explanation:
The given dimension is :
70mph
We are to convert this to km/hr
1 mile = 1.609km
so;
70mph x 1.609 = 112.63km/hr
So,
The solution is 112.63km/hr