Answers
Answer:
The answer is a 4.2m
Explanation:
Given data
Please see attached the rough drawing for your reference.
From the drawing, you ran 18m west and 2.4m south
The displacement is
= 1.8+2.4
=4.2m
Related Questions
What is the period, in seconds, that corresponds to each of
the following frequencies: (a) 10 Hz, (b) 0.2 Hz, (c) 60 Hz?
Answers
Answer:
0.1s,5s,0.017s
Explanation:
T=1÷frequency
Answer:
a =
✔ 6
The period is
✔ 2 seconds.
b =
✔ pi
Explanation:
Graph the function using the graphing calculator. Find the least positive value of t at which the pendulum is in the center.
t =
✔0.5 sec
To the nearest thousandth, find the position of the pendulum when t = 4.25 sec.
d =
✔ 4.243 in.
A 200-N force acts on a 10-kg object. The acceleration of the object is
Answers
If an ocean wave has a wavelength of 2 m and a frequency of 1 wave/s, then its speed is m/s Enter the answer Check it CRATCHPAD Improve this questic 트
Answers
Answer:
2m/s
Explanation:
v=f×wavelength
v=2×1
=2m/s
a sensor light installed on the edge of a home can detect motion for a distance of 50 feet in front and with a range of motion of 200 degrees. what is the arc length of the area covered
Answers
Answer:
4363.3231 feets²
Explanation:
Given that :
Distance, r = 50 ft
θ = 200°
The arc length of area covered :
Arc length = θ/360° * πr²
Arc length = (200/360) * 50 ft ^2 * π
Arc length = 0.5555555 * 2500 * π
Arc length = 4363.3231 feets²
The Moon has a smaller mass than the Earth. If you were to travel to the moon your weight would....
A Increase
B Decrease
C Stay the Same
D Vary with day and night
Answers
Answer:The Moon has a smaller mass than the Earth. If you were to travel to the moon your weight would..
Explanation: It would decrease.
The Moon has a smaller mass than the Earth. If you were to travel to the moon your weight would decrease because the acceleration due by gravity on the moon is less than the acceleration due to gravity on the earth, therefore the correct answer is option C.
What is gravity?It can be defined as the force by which a body attracts another body towards its center as the result of the gravitational pull of one body and another.
In comparison to the Earth, the Moon is less massive. Your weight would drop if you traveled to the moon because the acceleration caused by gravity there is lower than that caused by gravity here on Earth.
As a result of the less gravity on the moon, the weight would decrease.
Thus, the correct answer is option C.
To learn more about gravity from here, refer to the link;
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a 90 kilogram dog runs across the dog park at a speed of 6.5 meters per second. what is the magnitude and direction of the average force required to stop the dog in .85 seconds?
Answers
Answer:
am not sure about the answer
Explanation:
you need to find out the amount of force it's going in for example 10n or 100n then you need to times it the distance then devide by the time
a. Use the graph and the element made in question 2 to determine the mass of the star.
Answers
You are inside the Great Hall, 15 m from the north wall with the doors to the RMC, and centered between two open doors that are 3 m apart. Someone is blairing a 200 Hz tone outside the Great Hall so that it enters the doors as a plane wave. You hear a maximum intensity in your current position. As you walk along the direction of the wall with the doors (but maintain a distance 15 m from the wall), how far will you walk (in m) to hear a minimum in the sound intensity
Answers
Answer:
Δr = 0.425 m
Explanation:
This is a sound interference exercise, the expression for destructive interference is
Δr = (2n + 1) λ / 2
in this case the movement is in the same direction as the sound, therefore the movement is one-dimensional
let's use the relationship between the speed of sound and its frequency and wavelength
v = λ f
λ = v / f
the first minium occurs for n = 0
Δr = λ / 2
Δr = v / 2f
Δr = [tex]\frac{340}{2 \ 400}[/tex]
Δr = 0.425 m
this is the distance from the current position that we assume in the center of the room
An 80 N rightward force is applied to a 10 kg object to accelerate it to the right.
The object encounters a friction force of 50 N.
Answers
net force = 30 N
mass = 8.16 kg
acceleration = 3.68 m/s²
Further explanationGiven
80 N force applied
mass of object = 10 kg
Friction force = 50 N
Required
Net force
mass
acceleration
Solution
net forceNet force = force applied(to the right) - friction force(to the left)
Net force = 80 - 50 = 30 N
massGravitational force(downward) : F = mg
m = F : g
m = 80 : 9.8
m = 8.16 kg
accelerationa = F net / m
a = 30 / 8.16
a = 3.68 m/s²
Rank the four fundamental forces from strongest to weakest. Use 1 to indicate the strongest force and 4 to indicate the weakest force. The gravitational force: The electromagnetic force: The strong nuclear force: The weak nuclear force:
Answers
Answer:
4
2
1
3
Explanation:
Be safe, lovelies <3
how can you rewrite the force formula (f=ma) to solve the acceleration?
Answers
The force formula can be rewritten to solve the acceleration as:
acceleration = force/mass.
What is acceleration?Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.
From Newton's 2nd law of motion, we can write that:
Force = mass × acceleration
⇒ acceleration = force/mass.
Hence, the force formula can be rewritten to solve the acceleration as:
acceleration = force/mass.
Learn more about acceleration here:
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There is a very long straw of charge that is uniformly charged in electro static equilibrium. It has a charge per unit length of 4.0E-9 C/m (4.0 nC/m) and a radius of 0.5 m. What is the strength of the electric field a distance of 10.0 m from its center outside the straw
Answers
Answer:
2880 N/c
Explanation:
Given that:
Charge per unit length ; λ = 4 * 10^-9
radius, r = 10
Radius, R = 0.5m
Using the relation :
2λr / 4πE0R²
Columb's constant, k = 1/4πE0 =. 9* 10^9Nm²/C²
Hence, we have :
2λrk/ R²
(2 * 4 * 10^-9 * 10 * 9 * 10^9) / 0.5^2
(720 ÷ 0.25)
= 2880 N/c
Explain two reasons why astronomers are continually building larger and larger telescopes. Explain two reasons why astronomers are continually building larger and larger telescopes. Larger telescope mirrors have a larger surface area and can therefore collect more light, which makes faint objects bright enough to detect. Also, larger telescope mirrors produce more scattering of light due to diffraction, which contributes to better angular resolution. Larger telescope mirrors have a larger surface area and can therefore collect more light, which contributes to better angular resolution. Also, larger telescope mirrors produce more scattering of light due to diffraction, which makes faint objects bright enough to detect. Larger telescope mirrors have a larger surface area and can therefore collect more light, which contributes to better angular resolution. Also, larger telescope mirrors produce less scattering of light due to diffraction, which makes faint objects bright enough to detect. Larger telescope mirrors have a larger surface area and can therefore collect more light, which makes faint objects bright enough to detect. Also, larger telescope mirrors produce less scattering of light due to diffraction, which contributes to better angular resolution.
Answers
Answer:
* Larger mirrors collect more light and therefore fainter and more distant objects can have enough intensity to be detected
* arger mirrors decreases the angle of dispersion giving a better resolution of the bodies
Explanation:
Refracting telescopes get bigger every day for two main reasons.
* Larger mirrors collect more light and therefore fainter and more distant objects can have enough intensity to be detected
* the diffraction process for circular apertures is given by
θ = 1.22 λ / D
where d is the diameter of the mirror, therefore having larger mirrors decreases the angle of dispersion giving a better resolution of the bodies