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
How to Convert 20000N/m2 to N/cm2 ? Explain it Step by Step.
Answer:
2N/cm²
Explanation:
1m=100cm so, 1m²= 100×100 cm² = 10000cm².
therefore , 20000N/m²= 20000/10000 N/cm².
= 2N/cm².
hope this helps you.
is everyone lowkey turning goth?
Answer:
Yeah it's the new trend lol it's kinda weird
Answer:
it is marked by mystery and ambiguity. If to this fascination for creating tension is added a feeling of fear
Explanation:
do I help you
An object with an initial velocity of 2 m/s moves in a straight line under a constant acceleration.
Three seconds later, its velocity is 16 m/s.
(a) How far did the object travel during this time?
(b) What was the acceleration of the object?
Answer:
(a) distance =27m
(b)acceleration =14/3ms-2
Explanation:
s=(u+v)t/2
= (2+16)3/2
=27m
v=u+at
16=2+a*3
a= 14/3ms-2
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
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 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
How far will a bus travel if it averages a speed of 65 km/h for 7 hours?
Answer:
the bus travles 65x7=455 km
Explanation:
hope this helps you
Answer:
455 km/h
Explanation:
65km/h x 7 hours = 455 km/h
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
In a five paragraph essay, the ________ is the main idea you're trying to prove.
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!
Two cars drive on a straight highway. At time t = 0 , car 1 passes ad marker 0 traveling due east with a speed of 20.0 m/s. At the same time, car 2 is 1.0 km east of road marker 0 traveling at 30.0 m/s due west. Car 1 is speeding up, with an acceleration of 2.5 m/s², and car 2 is slowing down, with an acceleration of -3.2 m/s².
(a) Write position-time equations for both cars. Let east be the positive direction.
(b) At what time do the two cars meet?
Answer: (a) Write position-time equations for both cars. Let east be the positive direction.
Explanation:
Hope this helps!
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
Which of the following is true regarding the field of psychology
Answer:
It is a field of scientific inquiry.
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?
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:
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
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
A racket causes a tennis ball to rebound at a speed of 25 m/s The tennis ball is 0.4 kg and is initially moving at a speed of 12 m/s. A high-speed movie film determines that the racket and ball are in contact for 0.05 s. What is the average net force exerted on the ball by the racket
Hi there!
We know that:
I = Δp = mΔv = m(vf - vi)
Let the positive direction be towards the wall. Plug in the given values:
Δp = 0.4(-25 - 12) = -14.8 Ns
Δp = Ft, so:
Δp/t = F
-14.8/0.05 = -296 N (depending on which direction you assign positive/negative, the answer would be positive/negative)
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
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
. 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.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.
srength,weakness,opportunities,threats
Answer:
strength
Explanation:
it is all about strength
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. 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
I NEED HELP WITH THISS
Answer:
its B
Explanation:
pls brainliest
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:
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:
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