Which of the above beak depth vs time graphs represent the change in finch beak depth that would occur due to successive years of above average rainfall?
Answer:
A
Explanation:
I just did it :)
Answer:
It is A
Explanation:
I did it and got a 100
A jet airliner moving initially at 538 mph (with respect to the ground) to the east moves into a region where the wind is blowing at 937 mph in a direction 23 degrees north of east. What is the new speed of the aircraft with respect to the ground. Answer in units of mph
Answer:
The speed is "1447.576 mph".
Explanation:
The given values are:
Direction (north to east)
= 23°
Wind's speed
= 937 mph
Jet moving at a speed
= 538 mph
Now,
The velocity of jet airliner with respect to the ground will be:
⇒ [tex]v=v_{1}+v_{2}[/tex]
⇒ [tex]v=\sqrt{(538)^2+(987)^2+2(538)(937)\times Cos23^{\circ}}[/tex]
⇒ [tex]=1447.576 \ mph[/tex]
explain the relationship between physics and biology
Answer:
Biology is the study of living organisms. Physics is the study of matter and the laws of nature to understand the behavior of matter and the universe.
find weight of 45 kg at height 800 km from earth's surface?
Answer:
The weight is 348.4 N
Explanation:
Effect of the Altitude on the Acceleration of Gravity
The effect of gravity decreases with altitude because greater altitudes mean greater distances from the Earth's center.
The practical formula to calculate the gravity as a function of the altitude h is:
[tex]{\displaystyle g_{h}=g_{0}\left({\frac {R_{\mathrm {e} }}{R_{\mathrm {e} }+h}}\right)^{2}}[/tex]
Where go is the conventional gravitational acceleration = [tex]9.80665\ m/s^2[/tex]
R is the Earth's mean radius at the equator = 6,378 Km
h is the altitude.
At a height of h=800 Km, the acceleration of gravity is:
[tex]{\displaystyle g_{h}=9.80665 m/s^2\left(\frac {6,378 Km }{6,378 Km+800Km}}\right)^{2}}[/tex]
Calculating:
[tex]g_h=7.74\ m/s^2[/tex]
The weight is the product of the acceleration of gravity by the mass of the object, thus:
[tex]W=45\ Kg\cdot 7.74\ m/s^2=348.4\ N[/tex]
The weight is 348.4 N
Which of the following is NOT an undefined term?
A. Line
B. Point
C. Line Segment
D. Plane
Cody’s car accelerates from 0M/S 245 in/at Northwoods and 15 seconds what is the acceleration of the car
Answer:
3m/s^2
Explanation:
Acceleration is change in velocity / time = 45/15 = 3m/s^2
How long will it take to travel 150 km traveling at 50 km/hr?
Answer: 3 hours
Explanation: 150 km / 50 km = 3 hrs.
On the Moon, astronauts can “bounce” from location to location, but on Earth they cannot move in this way. What physical property changes based on the gravitational force of a planet or moon?
The maximum tension that a 0.80 m string can tolerate is 15 N. A 0.35-kg ball attached to this string is being whirled in a vertical circle. What is the maximum speed the ball can have at the top of the circle? 3.74 m/s 9.67 m/s 5.86 m/s None of the above
Answer:
v=5.86 m/s
Explanation:
Given that,
Length of the string, l = 0.8 m
Maximum tension tolerated by the string, F = 15 N
Mass of the ball, m = 0.35 kg
We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :
[tex]F=\dfrac{mv^2}{r}[/tex]
v is the maximum speed
[tex]v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{15\times 0.8}{0.35}} \\\\v=5.86\ m/s[/tex]
Hence, the maximum speed of the ball is 5.86 m/s.
You throw a ball straight up into the air at a velocity of 15 m/s. You want to know how high above your hand the ball will be at exactly 2.5 sec after you released it. You must SHOW ALL WORK to receive credit!!! Of the 5 motion variables - Vi, Vf, a, ΔX, and T, identify which one you are solving for and the magnitudes and directions of all of the others that are known: Which of the motion equations is best to use to determine the height of the ball? How high is the ball above your hand at 2.5 sec after you throw it?
Answer:
6.875mExplanation:
Given
Velocity = 15m/s
Time = 2.5secs
Required
Distance above your hand the ball will be at exactly 2.5secs
using the equation of motion:
ΔX = ViT + 1/2aT²
ΔX is the distance
Vi is the velocity of the ball in air
a is the acceleration due to gravity
T is the time taken
Since the ball is thrown straight up, a = -g
ΔX = ViT - 1/2gT²
Substitute
ΔX = 15(2.5) - 1/2(9.8)(2.5)²
ΔX = 37.5-30.625
ΔX = 6.875m
Hence the ball will be 6.875m above your hand at 2.5secs after you throw it.
According to Newton’s First Law of Motion, what will happen to a baseball thrown in outer space
Answer:
The baseball will stay in motion until another force act upon it.
Explanation:
Answer:
Newton's First Law of Motion is that an object in motion will stay in motion unless acted upon by an outside force. Due to the lack of gravity in space, until a different outside force acts upon the object, the baseball will continue to move at a constant speed in the same direction for all time (until acted upon by something else).
Problem #2: An apple is thrown upward with an initial velocity of +24.0 m/s. a. Sketch the apple's trip and label what you know. b. How high will the apple go? (Ans: +29.4 m) c. What is the total time needed for the apple to return to its original position? (Ans: 4.90 s)
Answer:
The answer is below
Explanation:
a) The initial velocity (u) = 24 m/s
We can solve this problem using the formula:
v² = u² - 2gh
where v = final velocity, g= acceleration due to gravity = 9.8 m/s², h = height.
At maximum height, the final velocity = 0 m/s
v² = u² - 2gh
0² = 24² - 2(9.8)h
2(9.8)h = 24²
2(9.8)h = 576
19.6h = 576
h = 29.4 m
b) The time taken to reach the maximum height is given as:
v = u - gt
0 = 24 - 9.8t
9.8t = 24
t = 2.45 s
The total time needed for the apple to return to its original position = 2t = 2 * 2.45 = 4.9 s
difference between simple machine and complex machine?
Answer:
a simple machine has way less parts than a complex machine like just one pulley or lever. A complex machine has multiple different parts that lead up to one thing being done in the long run.
Explanation:
Why does an object slow down when going uphill
Please explain properly
Answer:
The change in speed on slopes is due to gravity. When going downhill, objects will accelerate (go faster), and when going uphill they will decelerate (slow down). On a flat surface, assuming that there is little friction, they will then maintain a constant speed.
Mark me as BRAINLIEST if this answer helped you!
draw a diagram of a seismometer showing how the movement of earth is lated into a siesmogram
idk but it seems hard.
POSSIBLE POINTS: 5
A student throws a ball upward with a force of 50.0 Newtons. The ball reaches a maximum height of 15.0 meters in 2 seconds and then returns back to
his hands. What is the direction of the acceleration at the top?
There is no acceleration at the top
O The acceleration is directed upwards away from the Earth
O The acceleration is directed downwards towards the Earth
O None of the above
Answer:
A
Explanation:
Given that a student throws a ball upward with a force of 50.0 Newtons. The ball reaches a maximum height of 15.0 meters in 2 seconds and then returns back to
his hands.
At maximum height which is the top, the velocity is equal to zero.
According to definition of acceleration, which says; the rate of change of velocity.
Since velocity equal zero, acceleration also will be equal to zero.
Therefore, there is no acceleration at the top.
how does shape of object affect friction
Answer:
There are three types of friction.
1. Slanting friction
2. Sliding friction
3. Rolling friction
In the first two types of friction, the body of the object will come in contact with the irregularities on the surface more than in the case of rolling friction, thus a heavy box in square shape will have more friction than a heavy-duty vehicle.
This is also the reason why all vehicles are having wheels as they reduce friction and makes transportation easier and many people suggest to use a small log of wood to be placed on the bottom of a heavy box with flat bottom so it can moved easily due to reduction of friction
You row a boat perpendicular to the shore of a river that flows at 3 m/s. The velocity of your boat is 4 m/s relative to the water. What is the velocity of your boat relative to the shore?
Answer:
The velocity vBW of the boat relative to the water is 4.0 m/s, directed perpendicular to the current, as in Figure 3.18.
Explanation:
The boat is perpendicular to the shore. Then the net velocity of the boat with respect to river is 5 m/s. The observer feels like the boat is moving slower than they appear actually.
What is relative velocity ?Relative velocity is velocity of an object measured with respect to other object which can be of stationary or moving with a speed. Thus speed relative to another object is called the relative velocity.
The relative velocity is what's the speed for an observer but it differs from actual velocity. Sometimes it may slower or faster depending on the object with which a we are relating the speed.
Given that the speed of the boat = 4 m/s
speed of river = 3 m/s.
Both are perpendicular and the net velocity of the boat is :
r = √(3² + 4²) = 5 m/s.
Hence, the velocity of the boat relative to the shore will be 5 m/s.
Find more on relative velocity :
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state the various use of a periscope
Answer:
the periscope is an instrument for observation over, around or through the object, obstacle or condition that prevents direct line-of-sight observation from the observer's current position.
Explanation:
the periscope is used to monitor the dangerous chemical reactions in the laboratory and they are used in observing the nuclear.
Answer:
A periscope is used to see objects over the heads of a crowd.It is used by soldiers sitting in a bunker to observe and view the opponents' activities over the ground.It is used by a navy officer sitting in a submarine to see ships over the surface of the water in the sea.It is also useful in submarines to determine the distance of the torpedo and decide the right time for an attack.how many seconds are there in 3 days and 36 hours
The answer is 388 800 sec
Answer:
388800 secs
Explanation:
1 day = 24 hrs
so 24 times 3 = 72
and 1 hour is equal to 60 min
108 x 60 = 6480
and 60 secs in a minute
6480 x 60= 388800
a pulling force carried by a rope is called what
Answer:
The pulling force that acts along a stretched flexible connector, such as a rope or cable, is called tension, T. When a rope supports the weight of an object that is at rest, the tension in the rope is equal to the weight of the object: T = mg.
Explanation:
How are the frequencies of the color (visible light) spectrum similar to the Electromagnetic spectrum?
Answer:
[tex]this \: is \: due \: to \: both \: spectrum \: having \\ \: the \: same \: or \: almost \: the \: same \: wave \: speed.[/tex]
What is the weight of a body at a centre of Earth
Estimate the total mass of ocean on earth
why can't you run from momentum ?
A car is traveling 30m/s and slams on the breaks, coming to a complete stop in 5 seconds.
Calculate its acceleration (assuming it was constant). Draw a graph of x-t, v-t and a-t.
Answer:
-6
Explanation:
a=vf-vi/t
vf=0
so that we you subtract vf to vi it will be -30/5=-6
What charge does Chlorine have after bonding with magnesium
Answer:
A magnesium atom will lose 2 electrons to form a stable 2+ ion. Chlorine is in group 7 of the periodic table. A chlorine atom will gain 1 electron to form a stable 1- ion.
Explanation:
4. Which type of compound is sugar?
true for false; "Acceleration is inversely proportional to mass."
Answer:
True
Explanation:
F = ma
We can rearrange to a = F/m
So if we increase a and F is constant, then m has to decrease because it is in the denominator.
A driver of a car traveling at -15m/s applies the brakes, causing a uniform acceleration of +2.0m/s2. If the brakes are applied for 2.5s, what is the velocity of the car at the end of the braking period? How far has the car moved during the braking period?
Given :
Initial velocity, u = -15 m/s.
Acceleration , a = 2 m/s².
Time taken to applied brake, t = 2.5 s.
To Find :
The velocity of the car at the end of the braking period.
How far has the car moved during the braking period.
Solution :
By equation :
[tex]v = u+at\\\\v=-15 + 2\times 2.5\\\\v=-10 \ m/s[/tex]
Now, distance covered by car is :
[tex]s=ut+\dfrac{at^2}{2}\\\\s=(-15)(2.5)+\dfrac{2(2.5)^2}{2}\\\\s=-31.25\ m[/tex]
Hence, this is the required solution.