The head of a rattlesnake can accelerate at 50 m/s2 in striking a victim. If a car could do as well, how long would it take to reach a speed of 100 km/h from rest
Answer:
the time for the car to reach the final velocity is 0.56 s.
Explanation:
Given;
acceleration of the car, a = 50 m/s²
final velocity of the car, v = 100 km/h = 27.778 m/s
the initial velocity of the car, u = 0
The time for the car to reach the final velocity is calculated as;
v = u + at
27.778 = 0 + 50t
27.778 = 50t
t = 27.778 / 50
t = 0.56 s
Therefore, the time for the car to reach the final velocity is 0.56 s.
16. An object has a gravitational potential energy 41,772.5 Jof and has a mass of 1550 kg. How high is it
above the ground?
Plz help
Answer:
2.75 m.
Explanation:
From the question given above, the following data were obtained:
Potential energy (PE) = 41772.5 J
Mass (m) of object = 1550 kg
Height (h) =?
Potential energy is the energy possess by an object due to its location. Mathematically, potential energy is expressed as shown below:
PE = mgh
Where
PE => potential energy
m => mass of the object
g => acceleration due to gravity
h => height to which the object is located.
With the above formula, we can obtain the height to which the object is located as follow:
Potential energy (PE) = 41772.5 J
Mass (m) of object = 1550 kg
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) =?
PE = mgh
41772.5 = 1550 × 9.8 × h
41772.5 = 15190 × h
Divide both side by 15190
h = 41772.5 / 15190
h = 2.75 m
Thus, the object is located at 2.75 m above the ground.
Can someone please help me get this right pleaseee I’ll mark brainless .
Answer:i think it is c
Explanation:
Answer:
Explanation: i think its c to try it
20 points!!!! A 2,00ON steel rod that is 5 meters long is placed in a corner between the floor and a wall, and balanced at an angle using a cord attached to the wall The rod is balanced such that its top end is 2.38 meters away from the wall, The cord is 40 cm long, and it is attached to the wall at a height of 75 cm above the floor. The diagram to the right shows the situation If the lower end of the rod does not slip from the corner, what is the tension in the cord?
Answer:
WE NEED TO ADD ALL 40+2.38 +75+5
Explanation:
PLSE GIVE SOME POINTS DUDE
Two loudspeakers are about 10 mm apart in the front of a large classroom. If either speaker plays a pure tone at a single frequency of 400 HzHz, the loudness seems pretty even as you wander around the room, and gradually decreases in volume as you move farther from the speaker. If both speakers then play the same tone together, what do you hear as you wander around the room
Answer:
I hear points of low volume sound and points of high volume of sound.
Explanation:
This is because, since the two sources of sound have the same frequency and are separated by a distance, d = 10 mm, there would be successive points of constructive and destructive interference.
Since their frequencies are similar, we should have beats of high and low frequency.
So, at points of low frequency, the amplitude of the wave is smallest and there is destructive interference. The frequency at this point is the difference between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, f - f' = 400 Hz - 400 Hz = 0 Hz. So, the volume of the sound is low(zero) at these points.
Also, at points of high frequency, the amplitude of the wave is highest and there is constructive interference. The frequency at this point is the sum between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, (f + f') = 400 Hz + 400 Hz = 800 Hz. So, the volume of the sound is high at these points.
So, as you wander around the room, I should hear points of high and low sound across the room.
How high above the ground would a 2 kg object need to be in order to have 180 J
of gravitational potential energy?
Answer:
energy= MGH
2*9.8*h=180
h=180/19.6
h=9.32 m
The height of the object above the ground would be equal to 9.18 m.
What is gravitational potential energy?When an object of mass (m) is moved from infinity to a certain point inside the gravitational influence, the amount of work done in displacing it is stored in the form of potential energy and is known as gravitational potential energy.
The mathematical equation for gravitational potential energy can be written as:
Gravitational potential energy = m⋅g⋅h
Where m is the mass, g is the gravitational acceleration and h is the height above the ground.
Given, the mass of the given object, m = 2 Kg
The gravitational potential energy = 180 J
[tex]GPE = m\times g\times h[/tex]
180 = 2 × 9.8 × h
h = 9.18 m
Therefore, the object should be at a height of 9.18 meters in order to have 180 J of gravitational potential energy.
Learn more about gravitational potential energy, here:
https://brainly.com/question/3884855
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how do positive and negative acceleration differ?
1. positive acceleration represents an object speeding up; negative acceleration represents an object slowing down
2. positive acceleration moves North or east; negative acceleration moves south or west
3. positive acceleration occurs when there is more velocity than speed; negative acceleration occurs when there is less velocity than speed.
4. positive acceleration occurs when an object changes its speed but not its direction; negative acceleration occurs when an object changes both its speed and direction
Answer:
1. positive acceleration represents an object speeding up; negative acceleration represents an object slowing down
Explanation:
Acceleration is clearly defined as the rate of change of velocity with time. When are body is speeding up as we say, it is accelerating. When a body is coming to rest, it is decelerating.
Positive acceleration occurs when the speed of a moving continues to increase.
Negative acceleration is when the speed of a moving body reduces drastically.
Violet pulls a rake horizontally on a frictionless driveway with a net force of 2.0 N for 5.0 m.
How much kinetic energy does the rake gain?
Answer:
10 J.
Explanation:
Given that,
Net force acting on the rake, F = 2 N
Distance moved by the rake, d = 5 m
We need to find the kinetic energy gained by the rake. We know that,
Kinetic energy = work done
So,
K = F×d
K = 2 N × 5 m
K = 10 J
So, 10 J of kinetic energy is gained by the rake.
Violet pulls a rake horizontally on a frictionless driveway with a net force of 2.0 N for 5.0 m.
How much kinetic energy does the rake gain?
Answer: 10 J
a point charge q1 = 2.40 uC is held stationary at the origin. A second point charge q2 = -4.30uC moves from the point x= .150 m, y= 0.0 m, to the point x = .250 m, y= 0.0m
a) what is the charge in potential energy of the pair of charges?
b) How much work is done by the electric force on q2
Answer:150M
Explanation:
3. Do you think Lynn’s (the protagonist)actions were justifiable by her motives? Why or why not? Please help me Bad Genius the movie
Answer:
I do believe her actions were justified.
Explanation:
Due to the school charging extra fee from her father who makes a modest amount as a teacher. There was sum of money involved that could change how he lived and her.
I do not believe her actions where justified
She had a lot going for her. She could have skipped the hardship of helping grace and pass. She could have easily have gotten a good job with a degree and paid back all the debts owed. Alot of troubles could have been avoided just by doing her own thing.
15 points!!:-) Need help ASAP!
When does a compass NOT point towards magnetic north?
A.during a solar eclipse, which changes
Earth's magnetic field.
B. When there is another magnet close by.
C.When there is an unused battery close by.
D. When there is a coil of copper wire close by.
Answer:
b i think so because it makes senes
Answer: When there is another magnet close by.
Explanation: The needle of a compass is itself a magnet, and thus the north pole of the magnet always points north, except when it is near a strong magnet. ... When you take the compass away from the bar magnet, it again points north. So, we can conclude that the north end of a compass is attracted to the south end of a magnet.
To understand the behavior of the electric field at the surface of a conductor, and its relationship to surface charge on the conductor. A conductor is placed in an external electrostatic field. The external field is uniform before the conductor is placed within it. The conductor is completely isolated from any source of current or charge.
Answer:
Explanation:
The electric field inside of a conductor is 0 because the conduction electrons are pushed to the outer edges of the conductor. The surface of the conductor still has charge.
You are on the Pirates of the Caribbean attraction in the Magic Kingdom at Disney World. Your boat rides through a pirate battle, in which cannons on a ship and in a fort are firing at each other. While you are aware that the splashes in the water do not represent actual cannonballs, you begin to wonder about such battles in the days of the pirates. Sup-pose the fort and the ship are separated by 75.0 m. You see that the cannons in the fort are aimed so that their cannon-balls would be fired horizontally from a height of 7.00 m above the water.
(a) You wonder at what speed they must be fired in order to hit the ship before falling in the water.
(b) Then, you think about the sludge that must build up inside the barrel of a cannon. This sludge should slow down the cannonballs. A question occurs in your mind: if the can-nonballs can be fired at only 50.0% of the speed found ear-lier, is it possible to fire them upward at some angle to the horizontal so that they would reach the ship?
Answer:
a) v₀ₓ = 62.76 m / s, b) θ₁ = 17.6º, θ₂ = 67.0º
Explanation:
We can solve this exercise using the projectile launch ratios
a) Let's find the time it takes for the bullet to reach the water level
y = y₀ + v_{oy} t - ½ g t²
when it reaches the water its height is zero y = 0, as the bullet is fired horizontally its initial vertical velocity is zero
0 = y₀ + 0 - ½ g t²
t =[tex]\sqrt{2y_o/g}[/tex]
t = [tex]\sqrt{2 \ 7 /9.8}[/tex]
t = 1,195 s
now we can calculate the speed with the horizontal movement
x = v₀ₓ t
v₀ₓ = x / t
v₀ₓ = 75.0 / 1.195
v₀ₓ = 62.76 m / s
b) if the speed of the bullets is half of that found
v₀ = 62.76 / 2 = 31.38 m / s
let's write the expressions for the distance
x = v₀ cos θ t
y = y₀ + v_{oy} sin θ t - ½ g t²
t = [tex]\frac{x}{v_o \ cos \theta}[/tex]
we substitute
[tex]0 = y_o + v_o sin \theta \ \frac{x}{v_o \cos \thetay} - 1/2 g \ (\frac{x}{v_o \ cos \theta})^2[/tex]
[tex]0 = y_o + x tan \theta - \frac{1}{2} g \ \frac{x^2}{ v_o^2 \ cos^2 \theta}[/tex]
let's use the identified trigonometry
sec² θ = 1 + tan² θ
sec θ = 1 / cos θ
we substitute
[tex]0 = y_o + x tan \theta - \frac{g x^2}{2 v_o^2} ( 1 + tan^2 \theta)[/tex]
[tex]\frac{g x^2}{2v_o^2} tan^2 \theta - x tan \theta + \frac{gx^2}{2v_o^2} - y_o = 0[/tex]
we change variable
tan θ = H
[tex]\frac{gx^2}{2 v_o^2 } H^2 - x H + \frac{gx^2}{2v_o^2}-y_o =0[/tex]
we subtitle the values
[tex]\frac{9.8 \ 75^2}{2 \ 31.38^2} H^2 - 75 H + \frac{9.8 \ 75^2}{2 \ 31.38^2}-7 =0[/tex]
27.99 H² - 75 H + 20.99 = 0
H² - 2.679 H + 0.75 = 0
we solve the quadratic equation
H = [2.679 ± [tex]\sqrt{2.679^2 - 4 0.75}[/tex]] / 2
H = [2,679 ± 2,044] / 2
H₁ = 0.3175
H₂ = 2.3615
now we can find the angles
H₁ = tan θ₁
θ₁ = tan⁻¹ H₁
θ₁ = tan⁻¹ 0.3175
θ₁ = 17.6º
θ₂ = 67.0º
for these two angles the bullet hits the boat
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
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
A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.4 cm. Assuming heat to be transferred uniformly from all surfaces, determine (a) the amount of heat this resistor dissipates during a 24-hour period, (b) the heat flux, and (c) the fraction of heat dissipated from the top and bottom surfaces.
Answer:
(a) The resistor disspates 103680 joules during a 24-hour period.
(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.
(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.
Explanation:
(a) The amount of heat dissipated ([tex]Q[/tex]), measured in joules, by the cylindrical resistor is the power multiplied by operation time ([tex]\Delta t[/tex]), measured in hours. That is:
[tex]Q = \dot Q \cdot \Delta t[/tex] (1)
If we know that [tex]\dot Q = 1.2\,W[/tex] and [tex]\Delta t = 86400\,s[/tex], then the amount of heat dissipated by the resistor is:
[tex]Q = (1.2\,W)\cdot (86400\,s)[/tex]
[tex]Q = 103680\,J[/tex]
The resistor disspates 103680 joules during a 24-hour period.
(b) The heat flux ([tex]Q'[/tex]), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ([tex]A[/tex]), measured in square meters:
[tex]Q' = \frac{\dot Q}{A}[/tex] (2)
[tex]Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }[/tex] (3)
Where:
[tex]D[/tex] - Diameter, measured in meters.
[tex]h[/tex] - Length, measured in meters.
If we know that [tex]\dot Q = 1.2\,W[/tex], [tex]D = 4\times 10^{-3}\,m[/tex] and [tex]h = 2\times 10^{-2}\,m[/tex], the heat flux of the resistor is:
[tex]Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }[/tex]
[tex]Q' \approx 4340.589\,\frac{W}{m^{2}}[/tex]
The heat flux of the resistor is approximately 4340.589 watts per square meter.
(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ([tex]r[/tex]), no unit, is the ratio of the top and bottom surfaces to total surface:
[tex]r = \frac{\frac{\pi}{2}\cdot D^{2}}{A}[/tex] (3)
If we know that [tex]A \approx 2.765\times 10^{-4}\,m^{2}[/tex] and [tex]D = 4\times 10^{-3}\,m[/tex], then the fraction is:
[tex]r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}[/tex]
[tex]r = 0.045[/tex]
The fraction of heat dissipated from the top and bottom surfaces is 0.045.
In traveling a distance of 2.3 km between points A and D, a car is driven at 83 km/h from A to B for t seconds and 41 km/h from C to D also for t seconds. If the brakes are applied for 4.4 seconds between B and C to give the car a uniform deceleration, calculate t and the distance s between A and B.
Answer:
- time t taken for car to travel is 64.57 s
- distance travelled between A and B is 1.4887 km
Explanation:
Given the data in the question;
[tex]U_{BC}[/tex] = 83 km/h = ( 83×1000 / 60×60) = 23.0555 m/s
[tex]U_{CD}[/tex] = 41 km/h = ( 41×1000 / 60×60) = 11.3888 m/s
now, we calculate the acceleration;
a = ( [tex]U_{BC}[/tex] - [tex]U_{CD}[/tex] ) / t
we substitute
a = ( 23.0555 - 11.3888 ) / 4.4
a = 11.6667 / 4.4
a = 2.6515 m/s²
Now equation for displacement from BC
[tex]S_{BC}[/tex] = [tex]U_{BC}[/tex]t + 1/2.at²
we substitute
[tex]S_{BC}[/tex] = 23.0555×4.4 + 1/2×a×(4.4)²
we substitute -2.6515m/s² for a
[tex]S_{BC}[/tex] = 23.0555×4.4 + 1/2×(-2.6515)×(4.4)²
= 101.4442 - 25.6665
[tex]S_{BC}[/tex] = 75.7792 m
Now, for total distance covered = 2.3km = ( 2.3×1000) = 2300m
so
[tex]S_{AB}[/tex] + [tex]S_{BC}[/tex] + [tex]S_{CD}[/tex] = 2300 m
we substitute substitute
[tex]S_{AB}[/tex] + 75.7792 m + [tex]S_{CD}[/tex] = 2300 m
[tex]S_{AB}[/tex] + [tex]S_{CD}[/tex] = 2300 - 75.7792
[tex]S_{AB}[/tex] + [tex]S_{CD}[/tex] = 2224.2208 m
so we substitute 23.0555t for [tex]S_{AB}[/tex] and 11.3888t for [tex]S_{CD}[/tex]
23.0555t + 11.3888t = 2224.2208
34.4443t = 2224.2208
t = 2224.2208 / 34.4443
t = 64.57 s
Therefore, time t taken for car to travel is 64.57 s
Distance Between A to B
[tex]S_{AB}[/tex] = t × [tex]U_{AB}[/tex]
we substitute
[tex]S_{AB}[/tex] = 64.57 s × 23.0555
[tex]S_{AB}[/tex] = 1488.69 m
[tex]S_{AB}[/tex] = 1.4887 km
Therefore, distance travelled between A and B is 1.4887 km
Which of the following is NOT a characteristic of noble gases?
unreactive
odorless
solid at room temperature
colorless
Based on your average reaction time, how much time would it take to react to a traffic situation and stop a car traveling at 60 mph (1 mph equals 0.45 m/s) if you could decelerate the car at a rate of -3.4m/s2?
What distance would you travel (in meters) as the car came to a stop in the above situation?
Avg Reaction time: 0.218 ms
Answer:
d = 106.41 m
Explanation:
Given that,
Initial speed of the car, u = 60 mph = 26.9 m/s
The deceleration in the car, a = -3.4 m/s²
The average reaction time, t = 0.218 m/s
It finally stops, final velocity, v = 0
We need to find the distance covered by the car as it come to a stop.
Using third equation of motion to find.
[tex]v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-26.9^2}{2\times (-3.4)}\\\\d=106.41 m[/tex]
So, the car will cover 106.41 m as it comes to a stop.
kinetic energy portfolio in part 2 the independent changes to----?
The nucleus of an atom can be modeled as several protons and neutrons closely packed together.Each particle has a mass of 1.67 3 10227 kg and radius on the order of 10215 m.
(a) Use this model and the data provided to estimate the density of the nucleus of an atom.
(b) Compare your result with the density of a material such as iron. What do your result and comparison suggest about the structure of matter?
Answer:
Explanation:
Let n be number of total number of nucleons ( protons + neutrons )
Total mass inside nucleus = n x 1.67 x 10⁻²⁷ Kg
volume of nucleus = 4/3 π r³
= 1.33 x 3.14 x (10⁻¹⁵)³ m
= 4.17 x 10⁻⁴⁵ m³
Density = mass / volume
= n x 1.67 x 10⁻²⁷ / 4.17 x 10⁻⁴⁵
= .4 n x 10¹⁸ kg / m³
or of the order of 10¹⁸ kg/m³
b )
Density of iron = 7900 kg / m³
or of the order of 10⁴ kg / m³
So nucleus of a matter is about 10¹⁴ times denser than iron .
Could I get help on this question please
Answer:
124.51 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 49.4 m/s
Final velocity (v) = 0 m/s (at maximum height)
Maximum height (h) =?
NOTE: Acceleration due to gravity (g) = 9.8 m/s²
The maximum height to which the cannon ball attained before falling back can be obtained as illustrated below:
v² = u² – 2gh ( since the ball is going against gravity)
0² = 49.4² – (2 × 9.8 × h)
0 = 2440.36 – 19.6h
Collect like terms
0 – 2440.36 = –19.6h
–2440.36 = –19.6h
Divide both side by –19.6
h = –2440.36 / –19.6
h = 124.51 m
Therefore, maximum height to which the cannon ball attained before falling back is 124.51 m
A constant torque of 3 Nm is applied to an unloaded motor at rest at time t = 0. The motor reaches a speed of 1,393 rpm in 4 s. Assuming the damping to be negligible, calculate the motor inertia in Nm·s2.
Answer:
The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.
Explanation:
From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system ([tex]\tau[/tex]), measured in Newton-meters, is:
[tex]\tau = I\cdot \alpha[/tex] (1)
Where:
[tex]I[/tex] - Moment of inertia, measured in Newton-meter-square seconds.
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
If motor have an uniform acceleration, then we can calculate acceleration by this formula:
[tex]\alpha = \frac{\omega - \omega_{o}}{t}[/tex] (2)
Where:
[tex]\omega_{o}[/tex] - Initial angular speed, measured in radians per second.
[tex]\omega[/tex] - Final angular speed, measured in radians per second.
[tex]t[/tex] - Time, measured in seconds.
If we know that [tex]\tau = 3\,N\cdot m[/tex], [tex]\omega_{o} = 0\,\frac{rad}{s }[/tex], [tex]\omega = 145.875\,\frac{rad}{s}[/tex] and [tex]t = 4\,s[/tex], then the moment of inertia of the motor is:
[tex]\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}[/tex]
[tex]\alpha = 36.469\,\frac{rad}{s^{2}}[/tex]
[tex]I = \frac{\tau}{\alpha}[/tex]
[tex]I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }[/tex]
[tex]I = 0.0823\,N\cdot m\cdot s^{2}[/tex]
The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.
8) A train enters a curved horizontal section of the track at a speed of 100 km/h and slows down with constant deceleration to 50 km/h in 12 seconds. If the total horizontal acceleration of the train is 2 m/s2 when the train is 6 seconds into the curve, calculate the radius of curvature of the track for this instant.
Answer:
the radius of curvature of the track for this instant is 266 m
Explanation:
Given that;
The Initial Velocity u = 100 km/h = 100 × [tex]\frac{5}{18}[/tex] = 27.77 m/s
velocity of the train at t=12 s is;
[tex]V_{t=12}[/tex] = 50 km/h = 50 × [tex]\frac{5}{18}[/tex] = 13.89 m/s
now, we calculate the deceleration of the train
[tex]V_{t=12}[/tex] = u + at
13.89 = 27.77 + [tex]a_{t}[/tex]12
[tex]a_{t}[/tex] = (13.89 - 27.77) / 12
[tex]a_{t}[/tex] = -13.88 / 12
[tex]a_{t}[/tex] = - 1.1566 m/s²
Now, the velocity of the train at 6 seconds is;
[tex]V_{t=6}[/tex] = u + at
[tex]V_{t=6}[/tex] = 27.77 + ( - 1.1566 m/s²)6
[tex]V_{t=6}[/tex] = 27.77 - 6.9396
[tex]V_{t=6}[/tex] = 20.83 m/s
The acceleration at t=6 s is;
a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]
a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]
we substitute
2m/s² = √[ (- 1.15 )² + ([tex]a_{n}[/tex])²]
4 = (- 1.1566 )² + ([tex]a_{n}[/tex])²
4 = 1.3377 + ([tex]a_{n}[/tex])²
([tex]a_{n}[/tex])² = 4 - 1.3377
([tex]a_{n}[/tex])² = 2.6623
[tex]a_{n}[/tex] = √2.6623
[tex]a_{n}[/tex] = 1.6316 m/s²
Now the radius of curve is;
a = V² / p
[tex]p_{t=6}[/tex] = ( [tex]V_{t=6}[/tex] )² / [tex]a_{n}[/tex]
[tex]p_{t=6}[/tex] = ( 20.83 m/s )² / 1.6316 m/s²
[tex]p_{t=6}[/tex] = 433.8889 / 1.6316
[tex]p_{t=6}[/tex] = 265.9 m ≈ 266 m
Therefore; the radius of curvature of the track for this instant is 266 m
What school did Ronald McNair go to and what kind of science did he work in
Answer:
McNair graduated as valedictorian of Carver High School in 1967. In 1971, he received a Bachelor of Science degree in engineering physics, magna cu.m laude, from the North Carolina Agricultural and Technical State University in Greensboro, North Carolina.
What quantity measures the amount of space an object occupies?
A. Volume B.Temperature C. Mass D. Density
Answer:
mas
Explanation:
mass is the amount of space something occupies.
The deepest part of the ocean is the Challenger Deep, at 10,900 m. The depth was first measured in 1875 by the HMS Challenger by depth sounding (which does not involve sound waves). If you were to measure the depth by echo sounding (which does involve sound), what would you expect the time for a sound pulse at the surface to return in s, naively assuming a constant sound velocity throughout the ocean
Answer:
t = 14.53 s
Explanation:
The speed of a wave is constant and is given by
v = [tex]\sqrt{ \frac{B}{ \rho} }[/tex]
in this exercise they indicate that we assume the constant velocity, therefore we can use the uniform motion relations
v = x / t
t = x / v
in this case the sound pulse leaves the ship and must return so the distance is
x = 2d
where d is the ocean depth d = 10900m and the speed of sound in seawater is v = 1500 m / s
let's calculate
t = 2 10900/1500
t = 14.53 s
how is friction involved in the movement of space
Answer:
Friction can stop or slow down the motion of an object.
Explanation:
The slowing force of friction always acts in the direction opposite to the force causing the motion.
What makes electromagnets useful for sorting metals in recycling centers?
O A. The current can be turned on to pick up items containing all
metals and turned off to drop them.
O B. The current can be turned off to pick up items containing all
metals and turned off to drop them.
O C. The current can be turned on to pick up items containing iron and
turned off to drop them.
D. The current can be turned off to pick up items containing iron and
turned on to drop them.
C
It is right because I took this and I got this answer correct
A skydiver is using wind to land on a target that is 120 m away horizontally. The skydiver starts from a height of 70 m and is falling vertically at a constant velocity of 7.0 m/s downward with their parachute open (terminal velocity). A horizontal gust of wind helps push them towards the target. What must be their total speed if they want to just hit their target
Answer:
13.9 m/s.
Explanation:
Since the vertical velocity of the skydiver is constant at v = 7.0 m/s, we find the time, t it takes him to drop from a height of h = 70 m.
So, distance = velocity time
h = vt
t = h/v = 70 m/7 m/s = 10 s
This is also the time it takes him to move horizontally a distance of d = 120 m to the target.
So, his horizontal velocity is v' = distance/time = d/t = 120m/10 s = 12 m/s.
Since both vertical and horizontal velocities are perpendicular, we add them vectorially to obtain the skydivers total speed, V.
So, V = √(v² + v'²)
= √((7.0 m/s)² + (12.0 m/s)'²)
= √(49 m²/s² + 144 m²/s²)
= √(193 m²/s²)
= 13.9 m/s.
The direction of this velocity is Ф = tan⁻¹(v/v')
= tan⁻¹(7 m/s/12 m/s)
= tan⁻¹(0.5833)
= 30.3°
Which cell line is pointing to the body?
Answer:
The answer is B .........number 2
Explanation:
