im not sure if im right or not but maybe its "D scientific investigation" but ldk so
Suppose you are riding a bike at a constant velocity by applying a force of 85 N. What is the force of friction acting on the bike?
Hi there!
Force due to friction = -85N.
At a constant velocity, the sum of the forces is equal to 0, so:
Force applied + Force due to friction = 0
85 N + Force of friction = 0
85N = -Force of friction = -85N
A car initially traveling at a speed of 15.0 m/s accelerates uniformly to a speed of 20.0 m/s over a distance of 40.0 meters. What is the magnitude of the car's acceleration?
Answers:
1.1 m/s^2
2.0 m/s^2
2.2 m/s^2
9 m/s^2
Answer:
[tex]\boxed {\boxed {\sf 2.2 \ m/s^2}}[/tex]
Explanation:
We are asked to solve for the magnitude of the car's acceleration.
We are given the initial speed, final speed, and distance, so we will use the following kinematic equation.
[tex]{v_f}^2={v_i}^2+2ad[/tex]
The car is initially traveling at 15.0 meters per second and accelerates to 20.0 meters per second over a distance of 40.0 meters. Therefore,
[tex]v_f[/tex]= 20.0 m/s[tex]v_i[/tex]= 15.0 m/s d= 40.0 mSubstitute the values into the formula.
[tex](20.0 \ m/s)^2= (15.0 \ m/s)^2 + 2 a (40.0 \ m)[/tex]
Solve the exponents.
(20.0 m/s)² = 20.0 m/s * 20.0 m/s = 400.0 m²/s² (15.0 m/s)² = 15.0 m/s * 15.0 m/s = 225.0 m²/s²[tex]400.0 \ m^2/s^2 = 225.0 \ m^2/s^2 + 2 a(40.0 \ m)[/tex]
Subtract 225.0 m²/s² from both sides of the equation.
[tex]400.0 \ m^2/s^2 - 225.0 m^2/s^2 = 225.0 \ m^2/s^2 -225 \ m^2/s^2 +2a(40.0 \ m)[/tex]
[tex]400.0 \ m^2/s^2 - 225.0 m^2/s^2 = 2a(40.0 \ m)[/tex]
[tex]175 \ m^2/s^2 = 2a(40.0 \ m)[/tex]
Multiply on the right side of the equation.
[tex]175 \ m^2/s^2 =80.0 \ m *a[/tex]
Divide both sides by 80.0 meters to isolate the variable a.
[tex]\frac {175 \ m^2/s^2}{80.0 \ m}= \frac{80.0 \ m *a}{80.0 \ m}[/tex]
[tex]\frac {175 \ m^2/s^2}{80.0 \ m}=a[/tex]
[tex]2.1875 \ m/s^2 =a[/tex]
Round to the tenths place. The 8 in the hundredth place tells us to round the 1 up to a 2.
[tex]2.2 \ m/s^2=a[/tex]
The magnitude of the car's acceleration is 2.2 meters per second squared.
a. of water a. of bread a. of soap a. of juice a. of packet a. of sand option_piece packet grain bottle cake slice bale tank pane note tube bar can sack pile sheaf
Answer:
is it a poem or recipe :(
don't know
A student is planning an experiment to find
out how the height from which he drops a ball
affects how high the ball bounces. What is the control variable?
a) The ball used
b) The height the ball bounces
c) Height of the ball being dropped
Answer:
b) the height the ball bounces
Explanation:
the control variable is the variable that you change yourself. since you change the height that the ball bounces from we know this is the answer
Nearly all the cells in an organism have identical ? , which contain cell instructions.
Answer:
Nearly all cells in an organism contain identical DNA, and each DNA strand may contain hun- dreds or thousands of individual genes. ... No, cells only transcribe genes as they are needed to make specific proteins.
You drop a rock down a well and hear a splash 3 s later. As Charlie Brown would say, the well is 3 seconds deep. How fast is the rock moving when it hits the water?
A boy starts from a point A, travels to a point B at a distance of 1.5 km and returns to A. If he takes one hour to do so, his average velocity is ?
Because the displacement is zero(from point A to point A), the average velocity is 0 m/s.
A car slows down uninformly from a speed of 28.0m/s to rest in 8.00s.How far did it travel in that time?
Answer: 112 meters
Explanation:
Use the equation: d = vi+vf/2 x t which turned into: 28/2 times 8 = 112 meters
srength,weakness,opportunities,threats
Answer:
strength
Explanation:
it is all about strength
I NEED HELP WITH THISS
Answer:
its B
Explanation:
pls brainliest
The free body diagram below represents a balloon released from a student's hand so that the air can escape.
Based on this diagram, the balloon must be....
A. Accelerating down
B.moving at a constant speed,not accelerating
C.accelerating left
Stationary, not moving at all
D.accelerating to the right
E.accelerating up
Answer:
answer is D
Explanation:
accelerating to the right
A car traveling at 31 m/s runs out of gas while traveling up a 7.0 ∘ slope. How far will it coast before starting to roll back down?
I tried using a=g in a kinematic equation but my homework hint said my "a" value is incorrect.
Hi there!
On an incline, the acceleration due to gravity is: (g = 9.81 m/s²)
a = gsinФ ≈ 1.196 m/s²
We can use the following kinematic equation to solve:
vf² = vi² + 2ad
Since the car will start to roll back when v = 0 m/s, we can write that vf = 0.
0 = vi² + 2ad
Plug in values:
0 = 31² + 2(-1.196)(d)
Rearrange for d:
31²/(2.392) = d
d = 401.756 m
The car will coast for 7885.49 meters before starting to roll back down.
The car is subject to two forces: the force of gravity, which is pulling it down the hill, and the force of friction, which is trying to slow it down. The force of friction is equal to the coefficient of friction between the tires and the road multiplied by the normal force, which is equal to the weight of the car.
The car will stop coasting when the force of gravity is equal to the force of friction. We can set up an equation to find the distance the car will coast before stopping:
d = ([tex]v^{2}[/tex]) / (2 * μ * g * sin(θ))
where:
d is the distance the car will coast before stopping
v is the initial speed of the car
μ is the coefficient of friction between the tires and the road
g is the acceleration due to gravity
θ is the angle of the slope
Plugging in the values from the problem, we get:
d = [tex](31 m/s)^2[/tex] / (2 * 0.01 * 9.8 m/[tex]s^{2}[/tex] * sin(7.0°))
d = 7885.49 meters
Therefore, the car will coast for 7885.49 meters before starting to roll back down.
To know more about car here
https://brainly.com/question/31448756
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Give TWO conditions in which a force is exerted on an object without doing any work on the object.
Answer:
1)There is no movement. You stand still, gravity pulls you down but the ground prevents you from moving so gravity does no work on you.
2)There is movement but it is not in the direction of the force (at all). The most common example of this is a catr travellling along a horizontal road. gravity neither helps nor hinders the car. It does no work on the car because the car does not move in the direction of the graviational force. Notice that once the road is inclined, the car partly moves in the direction of gravity and it has some effect. It reduces your kinetic energy if you were going uphill and increaes it when coming down.
Answer: ANY time a force is exerted but the object is not displaced in a direction parallel to the force, work is not done. Force of tension on a ball rotating on the end of a string. Normal force and gravitational force on an object at rest or moving across a level floor.
Explanation:
. When trying to observe the frictional force of an object, what formula is needed? Include the units for each variable
The formula given by
[tex]\\ \sf\longmapsto F_f=\mu N[/tex]
Or
[tex]\\ \sf\longmapsto F_f=\mu mg[/tex]
u is coefficient of frictionN is normal reaction.Imagine a 15 kg block moving with a velocity of 20 m/s to the left. Calculate the Kinetic Energy of this block. Formula: Work with Units: Variables with units: Final Answer with units:
Answer:
3000 J (joules)
Explanation:
Use the formula for kinetic energy which is:
[tex] k.e = \frac{1}{2} mv ^{2} [/tex]
Where m is the mass and v is the velocity. If we put the values into the formula we get: 0.5 * 15 * (20)*(20) = 3000J
A bike takes 6 s to accelerate from 10 m/s
to 14 m/s. The mass of the bike and its
rider is 90 kg. Find the net force and the
final momentum of the bicyclist.
Answer:
Explanation:
F = ma = mΔv/Δt = 90(14 - 10) / (6 - 0) = 60 N
p = mv = 90(14) = 1,260 kg•m/s (for bicyclist AND bike together)
a bike accelerates from 0 m/s to 5 m/s in 4.5s.Then continues at a Constant speed for another 2.5s How far does it move?
Answer:
Below
Explanation:
First we will find the distance from when the bike is accelerating
Use this formula to find this :
d1 = ( vf - vi / 2 ) t
d1 = ( 5m/s - 0 m/s / 2 ) 4.5 s
d1 = (2.5)(4.5)
d1 = 11.25 m
Now we can find the distance for another 2.5 seconds
Since the car accelerated to 5 m/s, the constant velocity will be 5 m/s
Use this formula to find this :
d2 = vt
d2 = (5m/s)(2.5s)
d2 = 12.5 m
Finally we can add them up to get the total distance :
dt = 12.5 m + 11.25 m
dt = 23.75 m
*Make sure to round this number to how your teacher taught you
Hope this helps!
What does it mean when an object is magnified?
The object appears darker.
The object appears larger.
The object appears lighter.
The object appears smaller.
Answer:
The object appears larger
Explanation:
B
a car, which has a mass of 2000kg traveled a distance of 200 meters in 5 seconds. After 20 seconds the car was raveling at a speed of 60 m/s what is the force of the car?
Answer:
Explanation:
vi = 200/5 = 40 m/s
a = (vf - vi)/t = (60 - 40)/20 = 1 m/s²
F = ma = 2000(1) = 2000 N
MRUV, pido con procedimiento porfa
Answer:
Explanation:
Sa=Sa0 +Vot+at²/2
Sa = 0 - 2t+2t²/2
Sa = -2t+t²
Sb = Sb0+Vot+at²/2
Sb = 53 + 0 + 4t²/2
Sb = 53 + 2t²
-2t+t² = 53 + 2t²
t² + 2t - 53 = 0
t = -2 ± [tex]\sqrt{216}[/tex] / 2
t = 3[tex]\sqrt{6}[/tex] -1 s
Sb = 53 + 2( 3[tex]\sqrt{6}[/tex] -1) m
Which of the following is true regarding the field of psychology
Answer:
It is a field of scientific inquiry.
A racing car is travelling at 70 m/s and accelerates at -14 m/s2. What would the car’s speed be after 3 s?
A racing car is travelling at 70 m/s and accelerates at -14 m/s^2. What would the car’s speed be after 3 s?
Statement:A racing car is travelling at 70 m/s and accelerates at -14 m/s^2.
Solution:Initial velocity (u) = 70 m/sAcceleration (a) = -14 m/s^2Time (t) = 3 sLet the velocity of the car after 3 s be v m/sBy using the formula,v = u + at, we have
[tex]v = 70 + ( - 14)(3) \\ = > v = 70 - 42 \\ = > v = 28[/tex]
So, the velocity of the car after 3 s is 28 m/s.Answer:The car's speed after 3 s is 28 m/s.
Hope it helps
One strategy in a snowball fight is to throw
a snowball at a high angle over level ground.
While your opponent is watching this first
snowball, you throw a second snowball at a
low angle and time it to arrive at the same
time as the first.
Assume both snowballs are thrown with
the same initial speed 15.1 m/s. The first
snowball is thrown at an angle of 69◦
above
the horizontal. At what angle should you
throw the second snowball to make it hit the
same point as the first? Note the starting and
ending heights are the same. The acceleration
of gravity is 9.8 m/s^2
How many seconds after the first snowball
should you throw the second so that they
arrive on target at the same time?
Answer in units of s.
Answer:
Range formula: R = v^2 sin (2 theta) / g
If theta = 69 deg and v = 15.1
R = 15.1^2 sin 138 / 9,8 = 15.6 m
sin 138 = .669 = sin 42
So a snowball thrown at 21 deg will travel
R = 15.1 * .669^2 / 9.8 = 15.6 m
The second snowball can be thrown at 21 deg to travel the same distance
Vx = V cos theta = 15.1 * cos 69 = 5.41 first snowball
t1 = 15.6 / 5.41 = 2.88 sec
Vx = V cos theta = 15.1 cos 21 = 14.1 m/s
t2 = 15.6 / 14.1 = 1.11 sec
Difference = t1 - t2 = 1.77 sec time delay for second snowball
5word puzzle about sounds waves across
Answer:
amplitude
frequency
wave
transition
transverse waves
Nearby jet airplane
Fast train
Siren
Lawn move
This is an example of what kind of wave?
Answer:
transverse wave. it goes up and down :)
Answer:
Longitudinal wave
Explanation:
This is an example of Longitudinal wave.
Waves produced in spring is Longitudinal wave.
A car drives 16 miles south and then 12 miles west. What is the magnitude of the car's displacement?
Answer:the answer is 20 miles
Explanation:
Rosa uses a spring scale to pull a cart along a track for 2 meters. The spring scale shows the force exerted on the cart while being pulled. She uses metal plates to vary the mass of the cart, and motion sensors to calculate the average speed and acceleration of the cart.
She records the results in a table.
Rosa claims that the force needed to accelerate an object is proportional to the mass of the object. Which statement BEST supports or disproves Rosa’s claim using the data from the table?
A- Rosa’s claim is false. When the mass of the cart increases by 1.0 kg, the force on the cart increases by more than 1.0 kilogram.
B- Rosa’s claim is false. Even though the mass and the force increase in each trial, the acceleration of the object remains the same.
C-Rosa’s claim is true. The acceleration for each trial is the same. Doubling the mass of the cart causes the force to double in order to maintain the same rate of acceleration.
D-Rosa’s claim is true. Each time the mass increases by 1.0 or 2.0 kilograms, the force increases by 1.0 or 2.0 newtons respectively. This causes the rate of acceleration to remain constant.
Rosa’s claim is true. The acceleration for each trial is the same. Doubling the mass of the cart causes the force to double in order to maintain the same rate of acceleration.
According to Newton's second law of motion, the force applied to an object is directly proportional to the product of the mass and acceleration of the object.
F = ma
[tex]a = \frac{F}{m} \\\\\frac{F_1}{m_1} = \frac{F_2}{m_2} \\\\\frac{2}{1} = \frac{F_2}{2} \\\\F_2 = 4 \ N \\\\when \ the \ mass \ increases \ by \ 2\\\\m_1 = 2 \ kg,\ \ m_2 = 2(m_1) = 2(2) = 4 \ kg\\\\F_1 = 4 \ N, \ \ F_2 = 2(F_1) = 2(4) = 8 \ N[/tex]
Thus, we can conclude that from the table, Rosa’s claim is true. The acceleration for each trial is the same. Doubling the mass of the cart causes the force to double in order to maintain the same rate of acceleration.
Learn more here:https://brainly.com/question/19887955
Answer:
is C
Explanation:
A 11.8-kg monkey is hanging by one arm from a branch and swinging on a vertical circle. As an approximation, assume a radial distance of 99.8 cm is between the branch and the point where the monkey's mass is located. As the monkey swings through the lowest point on the circle, it has a speed of 3.22 m/s. Find:
a. the magnitude of the centripetal force acting on the monkey
b. the magnitude of the tension in the monkey's arm.
Answer:
Explanation:
Fc = mv²/R = 11.8(3.22)²/0.998 = 123 N
T = m(g + v²/R) = 11.8( 9.81 + 3.22²/0.998) = 238 N
Use the simulation to complete the activity.
Force and Motion Basics (select Acceleration)
In one or two sentences make a hypothesis about whether or not it is easier to start the box moving than to keep it moving on a rough surface. Then, explain how the simulation can be used to prove the
hypothesis
(2 points)
I'll mark brainliest to whoever gives me the right answer first
Answer:
The acceleration is directly proportional to the net force; the net force equals mass times acceleration; the acceleration in the same direction as the net force; an acceleration is produced by a net force
Force is related to acceleration through the equation F=ma. “F” stands for force, “m” stands for mass and “a” stands for acceleration. Force is a push or pull that an object can exert on other objects. Acceleration is the rate of change of an object's speed.
Explanation:
Una varilla de 5m de longitud y 1.5 cm^2 de sección transversal se alarga 0.10 cm al someterla a una tensión de 700 N. Determinar el Módulo de Young de la varilla