Explanation:
In physics, a force is any interaction that, when unopposed, will change the motion of an object. A force can cause an object with mass to change its velocity, i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity.
Formula
Newton's Second Law
F = m * a
F = force
m = mass of an object
a = acceleration
how do forces act when they are not in contact with each other?
Answer:they seperate from eachother but they don't touch eachother at all
Explanation:
what evidence supports the information consolidation theory?
PLEASE HELP ME ITS DUE NEXT PERIOD
Explanation:
Memory loss in retrograde amnesia has long been held to be larger for recent periods than for remote periods, a pattern usually referred to as the Ribot gradient. One explanation for this gradient is consolidation of long-term memories. Several computational models of such a process have shown how consolidation can explain characteristics of amnesia, but they have not elucidated how consolidation must be envisaged. Here findings are reviewed that shed light on how consolidation may be implemented in the brain. Moreover, consolidation is contrasted with alternative theories of the Ribot gradient. Consolidation theory, multiple trace theory, and semantization can all handle some findings well but not others.
A 76.O kg person is being pufed away from a burning building as shown below
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IRLA
If the tension in rope 1.11. As 193.5 N. calculate, without using the component method,
the tension in rope 2.
15a
Answer:
Final Answer
T1=736 NT1=736 N
T2=194
A car is traveling at 40 m/s. The brakes are applied an after 3 seconds the car is traveling 13 m/s. What is the acceleration of the car?
Answer:
- 9m/s²
Explanation:
Given parameters:
Initial velocity = 40m/s
Time taken = 3s
Final velocity = 13m/s
Unknown:
Acceleration of the car = ?
Solution:
Acceleration is the rate of change of velocity with time taken.
Acceleration = [tex]\frac{Final velocity - Initial velocity }{Time taken}[/tex]
Acceleration = [tex]\frac{13 - 40 }{3}[/tex] = - 9m/s²
A proton is given an acceleration of 1.5x109 m/s² when it is placed in an electric field.
What is the strength of the electric field?
Answer:
The strength of the electric field is [tex]E=15.66\: N/C[/tex]
Explanation:
Here the electric force is equal to Newton's second law.
[tex]F_{e}=ma[/tex]
Let's recall that electric force is the electric field times the charge, so we have:
[tex]qE=ma[/tex]
[tex]E=\frac{ma}{q}[/tex] (1)
Where:
m is the proton mass
q is the proton charge
a is the acceleration
Using the equation (1) we have:
[tex]E=\frac{1.67*10^{-27}1.5x10^{9}}{1.6*10^{-19}}[/tex]
Therefore, the strength of the electric field is [tex]E=15.66\: N/C[/tex]
I hope it helps you!
The half-life for U 238 is 4.5x109 years.
a) If five half-lives have gone by how many years have gone by?
b) If you start with 240 grams of U 238 and end up with 60 grams, how many years have gone by?
c) If you start with 240 grams of U 238 and 1.8 x 10^10 years go by, how much U 238 is left?
d) If you start with 562 g and six half lives go by how many grams are left?
Answer:
a. 2.25 × 10^10 yrs
b. 9 × 10^9 yrs
c. 59.5g
d. 8.78g
Explanation:
(a) Original sample(N) = 238g
Half-life = 4.5 × 10^9 yrs
5 half-life 5T½ = 5 × 4.5 × 10^9 yrs
= 2.25 × 10^10 yrs
(b) If N = 240g
N/2 = 120g
N/4 = 60g
Meaning 2T½
= 2 × 4.5 × 10^9 yrs
= 9 × 10^9 yrs
(c) 1.8 × 10^10 ÷ 4.5 × 10^9 = 4
at 4T½ we have N/16 = 238/4 = 59.5
=> the sample left = 59.5g
(d) from the question N = 562g
at 6T½
the amount left will be N/64
= 562÷64 = 8.78g
NB: For questions please you can read more on radioactivity
BIDEN WON MY RIGHTS ARENT GONNA BE TAKEN AWAY ‼️
Scenario
You have recently joined the team at A&L, an engineering firm with a broad portfolio. A&L has recently been hired to help plan a supply drop following a natural disaster. Due to conditions on the ground, the supply drop will be done from the air. Your supervisor has asked you to use that information, as well as your knowledge of kinematics, to create a supply drop plan detailing how far the payload should be from the drop site when it is delivered. Additionally, due to adverse conditions in the area, your supervisor has asked you to prepare for two contingencies.
Directions
The plane carrying the supplies will be cruising at a constant velocity of 250 miles per hour relative to the ground and at a height of 2,650 meters above the target site. Using this information, create a supply drop plan including all required information and calculations outlined below. As you are completing your supply drop plan, remember that correct SI units are a required component of your calculations and descriptions.
Construct a diagram that describes the horizontal and vertical motion of the payload. Your diagram should visually represent the initial velocity and height of the payload as it approaches the drop site. To create the diagrams, you may use drawing tools, or you may photograph or screenshot a drawing of your own. This diagram should include the following elements:
The initial velocity of the payload
The initial height of the payload
Horizontal distance to the drop site
Using your understanding of kinematic equations and the given variables in the scenario, calculate the horizontal and vertical motion of the payload to ensure it arrives at the drop site. In your calculations, account for both the horizontal and vertical motion of the payload. Your calculations should address the following:
Initial velocity of the payload when launched
The velocity of the payload when it hits the ground
Following your calculations, describe the equations used in calculating the vertical and horizontal motion of an object. In your descriptions, address the following:
What is the relationship between vertical and horizontal motion in kinematics equations?
How did you analyze the vertical motion of the payload in your solution?
How did you analyze the horizontal motion of the payload in your solution?
What other kinematics principles did you consider in analyzing the motion of the payload?
Answer:
t = 23.255 s, x = 2298.98 m, v_y = - 227.90 m / s
Explanation:
After reading your extensive writing, we are going to solve the approach.
The initial speed of the plane is 250 miles / h and it is at an altitude of 2650 m; In general, planes fly horizontally for launch, therefore this is the initial horizontal speed.
As there is a mixture of units in different systems we are going to reduce everything to the SI system.
v₀ₓ = 250 miles h (1609.34 m / 1 mile) (1 h / 3600 s) = 111.76 m / s
y₀ = 2650 m
Let's set a reference system with the x-axis parallel to the ground, the y-axis is vertical. As time is a scalar it is the same for vertical and horizontal movement
Y axis
y = y₀ + v₀ t - ½ g t²
the initial vertical velocity when the cargo is dropped is zero and when it reaches the floor the height is zero
0 = y₀ + 0 - ½ g t²
t = [tex]\sqrt{ \frac{2 y_o}{g} }[/tex]
t = √(2 2650/ 9.8)
t = 23.255 s
Therefore, for the cargo to reach the desired point, it must be launched from a distance of
x = v₀ₓ t
x = 111.76 23.255
x = 2298.98 m
at the point and arrival the speed is
vₓ = v₀ₓ = 111.76
vertical speed is
v_y = v_{oy} - gt
v_y = 0 - gt
v_y = - 9.8 23.25 555
v_y = - 227.90 m / s
the negative sign indicates that the speed is down
in the attachment we have a diagram of the movement
Caris parked on road slopes upward at angle θ. The magnitude the force of the road on the is mg cosθ. Is the magnitude of the stalie friction force on the less equal to or greater than μ2mgcosθ? Explain.
Answer:
fr = μ[tex]\miu _{k}[/tex] m g cos θ
the correct answer is less than fr = μ[tex]\miu _{k}[/tex] 2m g cos θ
Explanation:
Let's propose the solution of the problem to be able to answer the question, let's fix a reference system with one axis parallel to the plane (x-axis) and the other perpendicular, let's write Newton's second law
Y Axis
N - W cos θ = 0
N = mg cos θ
X axis.
fr - w sin θ = 0
fr = mg sin θ
the friction force is described by the expression
fr = μ N
fr = μ[tex]\miu _{k}[/tex] m g cos θ
When we analyze this expression we see that the friction force is equal to (μ_k m g cos θ)
If everything is well written in your problem, the correct answer is less, even though I think you have an error when writing the number
3. A train travels 450 m north traveling at a velocity of 50 m/s . How long did it take the train to make this trip?
Answer:
9 seconds
Explanation:
450 divided by 50 equals 9.
Answer:
9s
Explanation:
Given parameters:
Displacement = 450m
Velocity = 50m/s
Unknown:
Time taken for the train to make the trip = ?
Solution:
Since velocity is displacement divided by the time taken,
Time = [tex]\frac{Displacement}{Velocity}[/tex]
Time = [tex]\frac{450}{50}[/tex] = 9s
Two quantities that have the same dimension but have a different physical concepts.
Please help me
Answer:
In general, if the dimensions are same, the quantities do represent the same physical content. Like work and energy have the same dimensions and represent inter-convertible quantity. ... These two quantities represent the same quantity - same meaning and content.
Explanation: i hope this helped.
A racecar reaches 24 m/s in 6 seconds at the start of a race. What is the acceleration of the car?
Answer:
4m/s^2
Explanation:
