We know, force is given by :
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
Here, k in a constant.
[tex]Net \ Force = F_1+F_2\\\\Net \ Force = k(\dfrac{Qq_1}{r_1^2}+\dfrac{Qq_2}{r_2^2})\\\\Net \ Force = k( \dfrac{-3}{50}+\dfrac{2}{50})\\\\Net \ Force = 9\times 10^9\times \dfrac{-1}{50}\ N\\\\Net\ Force = -1.8\times 10^8\ N[/tex]
Hence, this is the required solution.
According to the kinetic theory, what happens when the temperature of water
increases?
The water molecules increase their speed.
The water molecules do not change speed.
The water molecules decrease their speed
PLEASE HELP
8. A dog runs with an initial speed of 7.5 m/s on a waxed floor. It slides to a stop in 15 sec. What
is the acceleration?
(10 Points)
O-7.5 m/s
-0.5 m/s2
O -7.5 m/s2
7.5 mi/hr
Answer:
-0.5 m/s²
Explanation:
Step 1: Given data
Initial velocity of the dog (u): 7.5 m/sFinal velocity of the dog (v): 0 m/s (Stop)Time elapsed (t): 15 sStep 2: Calculate the acceleration of the dog
The acceleration (a) is equal to the change in the velocity over the time elapsed.
a = v - u / t
a = 0 m/s - 7.5 m/s / 15 s
a = -0.5 m/s²
a single frictionless roller coaster car of mass m=750 kg tops the first hill with speed v=15 m/s ar height h =40m as shown
1) Find the speed of the car at B and C
2) if mass m were doubled, would the speed at B increase, decrease, or remain the same?
1. The speed of the roller coaster car at point B and C: are 19.8 m/s and 0 m/s respectively.
2. If mass (m) of the roller coaster car were doubled, the speed at B would increase because energy is dependent on the mass of an object.
Given the following data:
Mass of roller coaster car = 750 kgSpeed of roller coaster car = 15 m/sHeight = 40 meters1. To find the speed of the roller coaster car at point B and C:
First of all, we would determine the potential energy of the roller coaster car by using the formula:
[tex]P.E = mgh[/tex]
Where:
m is the mass of object.g is the acceleration due to gravity ([tex]9.8\;m/s^2[/tex]).h is the height of an object.From the diagram, height at point B = [tex]\frac{h}{2} = \frac{40}{2} = 20[/tex] meters
[tex]P.E = 750(9.8)(20)[/tex]
P.E = 147000 Joules.
Next, we would determine the speed by applying the law of conservation of energy.
[tex]Kinetic\;energy = Potential\;energy\\\\\frac{1}{2} mv^2 = mgh[/tex]
Substituting the values, we have;
[tex]147000 = \frac{1}{2} (750)v^2\\\\147000 = 375v^2\\\\v^2 = \frac{147000}{375} \\\\v^2 = 392\\\\v = \sqrt{392}[/tex]
Speed, v = 19.8 m/s
From the diagram, at point C, the speed is equal to zero (0) meters per seconds.
2. If mass (m) of the roller coaster car were doubled, the speed at B would increase because energy is dependent on the mass of an object.
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How are gas giants similar to one another?
Answer:
they are all made of gass and they are all giants.
Explanation:
Answer:
How are the gas giants similar to one another? dont have solid surfaces and are much larger than earth. Why do all of the gas giants have thick atmospheres? Because they are so massive, the gas giants exert a much stronger gravitational force than the terrestrial planets
Explanation:
If a wave is traveling at a constant speed, and the frequency increases, what would happen to the wavelength?
It would increase
It would decrease.
It would remain constant.
It would drop to zero.
If a wave is traveling at a constant speed, and the frequency increases, the wavelength would decrease.
What is wavelength?The wavelength is the distance between the adjacent crest or trough of the sinusoidal wave. The wavelength is the reciprocal of the frequency of the wave.
Wavelength λ = 1 / frequency
The wavelength and frequency are inversely related to each other. So, when frequency is increased, the wavelength increases.
Thus, the wavelength of would decrease if the frequency increases.
Learn more about wavelength.
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Answer:
B. It would decrease
Explanation:
Write the vectors labeled A, B and C with rectangular coordinates.
Focus on vector A. Start at the base of the vector which is at (1,1)
Then notice how we have to move 3 units to the right and 2 units up to get to the tip of the vector where the arrow is located. So this is at (4,3)
The fact that we moved 3 units to the right and 2 units up means that vector A is <3, 2>
So we'll type 3 in the first box and 2 in the second box.
======================================================
We use the same idea for vector B
Start at the base and move to the tip. Doing so has us move 1 unit left and 3 units down.
Vector B = <-1, -3>
We type -1 in the first box, and -3 in the second box.
======================================================
Finally onto vector C
Start at the base (0,0). To move to the tip, we need to go 3 units left and 3 units up
Vector C = <-3, 3>
-3 goes in the first box and 3 goes in the second box
======================================================
In general, vectors are useful to tell us a specific direction in which to go. They're useful in physics to determine how exactly a force is being applied, or how a particle is moving.
The two main types of weathering are (4 points)
A. mechanical and physical
B. physical and kinetic
C. chemical and physical
D. chemical and acidic
Answer:
b
Explanation:
Answer:
its acually c
Explanation:
Suppose Tom Harmon17 is standing at the exact center of the Ohio State football field18 on the 50 yard line. The field is 300 feet long and 160 feet wide. Suppose the ball leaves Harmon’s hand at a height of six feet and has initial velocity v(0) = 40i + 39j + 32k feet per second where j points toward the sideline, i points toward the end zone, and k = i × j points straight up. The acceleration due to gravity is −32k feet per second squared and the mass of the football is .45 kilograms.
(a) Determine a parameterization for the position of the football at time t seconds after it is thrown.
(b) Suppose David Nelson19 catches the ball when it is six feet above the ground. Is Nelson in bounds or out of bounds when he catches the ball? (Assume Nelson’s toes are touching the ground directly below the ball when he catches the ball.)'
Answer:
a) x = 40 t , y = 39 t , z = 6 + 32 t - 16 t ², b) x = 80 feet , y = 78 feet , the ball came into the field
Explanation:
a) This is a projectile launch exercise, where in the x and y axes there is no acceleration and in the z axis the acceleration of the acceleration of gravity, let's write the equations of motion in each axis
Since the cast is in the center of the field, let's place the coordinate system
x₀ = 0
y₀ = 0
z₀ = 6 feet
x-axis (towards end zone, GOAL zone)
x = xo + v₀ₓ t
x = 40 t
y-axis (field width)
y = y₀ + [tex]v_{oy}[/tex] t
y = 39 t
z axis (vertical)
z = z₀ + v_{oz} t - ½ g t²
z = 6 + 32 t - ½ 32 t²
z = 6 + 32 t - 16 t ²
b) The player catches the ball at the same height as it came out, so we can find the time it takes to arrive
z = 6
6 = 6 + 32 t - 16 t²
(t - 2)t = 0
t=0 s
t= 2 s
The ball position
x = 40 2
x = 80 feet
y = 39 2
y = 78 feet
the dimensions of the field from the coordinate system (center of the field) are
x_total = 150 feet
y _total = 80 feet
so we can see that the ball came into the field