Answer:
F = 2250 [N]
Explanation:
In order to solve this problem, we must first use the following equation of kinematics.
[tex]v_{f}^{2} =v_{o}^{2}-2*a*x[/tex]
where:
Vf = final velocity = 0 (come to rest)
Vo = initial velocity = 30 [m/s]
a = acceleration or desaceleration [m/s²]
x = distance = 100 [m]
[tex](0)=30^{2} -2*a*100\\900 = 200*a\\a = 4.5 [m/s^{2}][/tex]
Now we must use the following equation of kinetics, which is based on Newton's second law that explains that the sum of forces on a body is equal to the product of mass by acceleration.
∑F = m*a
where:
F = force [N]
m = mass = 500 [kg]
a = acceleration = 4.5 [m/s²]
[tex]F = 500*4.5\\F = 2250 [N][/tex]
(1-dimension) A fish has a mass of 6 kg and is moving at a speed of 4m/s to the right. What is its momentum?
Answer:
24 kg m/sExplanation:
The momentum of an object can be found by using the formula
momentum = mass × velocity
From the question we have
momentum = 6 × 4
We have the final answer as
24 kg m/sHope this helps you
A sled is pulled with a force of 540 N at an angle of 40° with the horizontal. What are the horizontal and vertical components of this force?
Answer:
Fx = 467.65N
Fy = 270N
Explanation:
Given
Force = 540N
angle of inclination = 40 degree
Horizontal component Fx = Fcos 30
Fx = 540cos30
Fx = 540(0.8660)
Fx = 467.65N
Hence the horizontal component is 467.65N
Vertical component Fy = Fsin 30
Fy = 540sin30
Fy = 540(0.5)
Fy = 270N
Hence the vertical component is 270N
what is the force of an egg that is thrown at a brick wall if the egg has a mass of 0.3 kg and an acceleration of 50 m/s/s
Answer:
15N
Explanation:
F=ma so F=.3*50 therefore F=15N
The force of an egg that is thrown at a brick wall is equal to 15 N.
What is force?Force can be defined as the influence or effect that changes the state of the body of from motion to rest or vice versa. The S.I. unit of force is Newton (N) as well as force is a vector quantity. Force can change the direction or the speed of the moving object.
The force acting on an object can be calculated from the multiplication of the mass(m) and acceleration(a). The mathematical form of the second law of motion for force can be written as follows:
F = ma
Given, the mass of the egg, m = 0.3 Kg
The acceleration of the egg with which it is thrown on the wall, a = 50 m/s²
The force of an egg that is thrown at a brick wall can be calculated as:
F = ma = 50 ×0.3 = 15 N
Learn more about force, here:
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what is the initial position of the object?
a. 2m b. 4m c. 6m d. 8m e. 10 m
what is the velocity of the object?
a. -10 m/s b. -5 m/s c. 0 m/s d. 5 m/s e. 10 m/s
which of the following is true?
a. the object increase its velocity.
b. the object decrease its velocity.
c. the objects velocity stays unchanged.
d. the object stays at rest.
e. more information is required.
Answer:
e:10 m b: -5 m/s b.The object decreases its velocity.
Explanation:
3) A 10kg object rests on a frictionless surface when it is struck by a 300N force. At what rate will it accelerate?
3m/s/s
30m/s/s
0.3m/s/s
300m/s/s
Answer: 0.3m/s/s
(i'm really sorry if i'm wrong)
:(
d. If a dog has a mass of 12 kg, what is its weight on Neptune?
11.7N/kg
Answer:
133.8 N
Explanation:
Recall that the acceleration of gravity in Neptune is estimated as 11.15 m/s^2
Therefore, the weight of the dog on this planet would be:
Weight = mass x acceleration of gravity = 12 kg x 11.15 m/s^2 = 133.8 N
Which scenario is an example of the transfer of thermal energy by radiation?
A. Water boils in a pan.
B. Hot air circulates in an oven.
C. An ice cube melts in a person's hand.
D. A frozen lake melts under the Sun.
Correct answer is D
Answer:
its D: A frozen lake melts under the sun.
Explanation:
Radiation is the transfer of heat energy through space by electromagnetic radiation. Most of the electromagnetic radiation that comes to the earth from the sun is invisible. Only a small portion comes as visible light. Light is made of waves of different frequencies.
a mass of 2.00 kg rest on a rough horizontal table. The coefficient of static friction between the block and the table is 0.60. The block is attached to a hanging mass by a string that goes over a smooth pulley,as shown in the diagram. Determine the largest mass that can hang in this way without forcing the block to slide.
Answer:
1.2 kg
__________________________________________________________
We are given:
Mass of the block = 2 kg
Coefficient of Static Friction = 0.6
__________________________________________________________
Friction Force on the Block:
Finding the Normal Force:
We know that the normal force will be equal and opposite to the weight of the 2 kg block
So, Normal Force = mg
replacing the variables with the given values
Normal Force = (2)(9.8) [Taking g = 9.8]
Normal Force = 19.6 N
Friction force on the Block:
We know that:
Coefficient of Static Friction = Static Friction Force/Normal Force
replacing the variables
0.6 = Static Friction force / 19.6
Static Friction force = 0.6*19.6 N [Multiplying both sides by 19.6]
Static Friction force = 11.76 N
__________________________________________________________
Largest Mass that can Hang:
We know that the Static Friction force is 11.76 N, this means that a force of 11.76 N will be applied to keep the object at rest
So, if the weight of the second block is less than the static friction force, it will hang
Weight of the second block ≤ 11.76
We know that weight = mg
mg ≤ 11.76
m(9.8) ≤ 11.76 [since g = 9.8]
m ≤ 1.2 kg [dividing both sides by 9.8]
From this, we can say that the maximum mass of the second block is 1.2 Kg
Carter's favorite ride at playland amusement park is the rollercoaster. The roller coaster
car and passengers have a combined mass of 1620kg and they descend the first hill at
an angle of 45.0 degrees to the horizontal. With what force is the rollercoaster pulled
down the hill?
Answer:
F = 11226.02 N
Explanation:
The roller coaster car and passengers have a combined mass of 1620kg.
They descend the first hill at an angle of 45.0 degrees to the horizontal.
We need to find force is the rollercoaster pulled down the hill.
We firstly find the rectangular component of the downward. The force acting in the downward direction is mgsinθ such that,
F = mgsinθ
= 1620 × 9.8 × sin(45)
= 11226.02 N
So, the roller coaster is pulled down the hill with a force of 11226.02 N.
How high does a rocket have to go above the earth's surface to be subject to a gravitational field from the earth that is 50.0 percent of its value at the earth's surface?
A) 2.650 km
B) 3,190 km
C) 9.020 km
D) 12.700 km
Answer:
A) 2.650 km
Explanation:
The relationship between acceleration of gravity and gravitational constant is:
[tex]g = \frac{Gm}{R^2}[/tex] ---- (1)
Where
[tex]R = 6,400 km[/tex] -- Radius of the earth.
From the question, we understand that the gravitational field of the rocket is 50% of its original value.
This means that:
[tex]g_{rocket} = 50\% * g[/tex]
[tex]g_{rocket} = 0.50 * g[/tex]
[tex]g_{rocket} = 0.5g[/tex]
For the rocket, we have:
[tex]g_{rocket} = \frac{Gm}{r^2}[/tex]
Where r represent the distance between the rocket and the center of the earth.
Substitute 0.5g for g rocket
[tex]0.5g = \frac{Gm}{r^2}[/tex] --- (2)
Divide (1) by (2)
[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}/\frac{Gm}{r^2}[/tex]
[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}*\frac{r^2}{Gm}[/tex]
[tex]\frac{1}{0.5} = \frac{1}{R^2}*\frac{r^2}{1}[/tex]
[tex]2 = \frac{r^2}{R^2}[/tex]
Take square root of both sides
[tex]\sqrt 2 = \frac{r}{R}[/tex]
Make r the subject
[tex]r = R * \sqrt 2[/tex]
Substitute [tex]R = 6,400 km[/tex]
[tex]r = 6400km * \sqrt 2[/tex]
[tex]r = 6400km * 1.414[/tex]
[tex]r = 9 049.6\ km[/tex]
The distance (d) from the earth surface is calculated as thus;
[tex]d = r - R[/tex]
[tex]d = 9049.6\ km - 6400\ km[/tex]
[tex]d = 2649.6\ km[/tex]
[tex]d = 2650\ km[/tex] --- approximated
Allure of the seas is one of the most expensive cruise ships around the world with a length of 362 meters(1,187 ft) and a height of 72 meters(236 ft) above water line. On her first day of operation she moves with a uniform acceleration of 83.5 km/hr2 from rest has gone 10 nautical miles. How many seconds she is in motion? Note: 1 nautical mile = 1.852 km (help 3 mins left
Answer:
about 2398 seconds
Explanation:
The relation between time, distance, and acceleration is ...
d = (1/2)at²
t = √(2d/a) = √(2·10·1.852 km/(83.5 km/h²)) ≈ √0.4436 h ≈ 0.6660 h
That is about ...
(0.6660 h)(3600 s/h) ≈ 2397.7 s
The cruise ship takes about 2397.7 seconds to cruise 10 nautical miles, accelerating all the way.
Describe the motion of an object as it accelerates. IN YOUR OWN WORD!! ASAP
Answer:
The aceleration of an object is in the direction of the net force. If you push or pull an object in a particular direction, it accelerates in that direction. The aceleration has a magnitude directly proportional to the magnitude of the net force.
Explanation:
Hope this helps Plz mark brainliest
A 100 kg. football player and a 20 kg. child sit on rolling carts and push off
from each other. Which person applies more force?*20 kg. child applies more force
100 kg. football player applies more force
Same force is applied
Answer:
the forces are the same
Explanation:
8x = -6. What does x equal?
Answer:
x=-3/4
Explanation:
Find the state of Georgia in the southeastern U.S. What can you say about the weather in this state?
A 100kg couch is being pushed with 196N of force. As it slides along the ground it experiences a coefficient of friction of 0.1. What is the net force in this situation?
A 300N
B 202N
C 398N
D 98N
Answer:98
Explanation:hope this helps!
Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular acceleration of 0.311 rad/s2 for 4.13 s. What is the drill's angular displacement during that time interval?
Answer:
The drill's angular displacement during that time interval is 24.17 rad.
Explanation:
Given;
initial angular velocity of the electric drill, [tex]\omega _i[/tex] = 5.21 rad/s
angular acceleration of the electric drill, α = 0.311 rad/s²
time of motion of the electric drill, t = 4.13 s
The angular displacement of the electric drill at the given time interval is calculated as;
[tex]\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad[/tex]
Therefore, the drill's angular displacement during that time interval is 24.17 rad.
HELPP physics final will give brainliest
A vertical tube one meter long is open at the top. It is filled with 50 cm of water. If the velocity of sound is 344 m/s, what will the fundamental resonant frequency be (in Hz)?
Answer:
The fundamental resonance frequency is 172 Hz.
Explanation:
Given;
velocity of sound, v = 344 m/s
total length of tube, Lt = 1 m = 100 cm
height of water, hw = 50 cm
length of air column, L = Lt - hw = 100 cm - 50 cm = 50 cm
For a tube open at the top (closed pipe), the fundamental wavelength is given as;
Node to anti-node (N ---- A) : L = λ / 4
λ = 4L
λ = 4 (50 cm)
λ = 200 cm = 2 m
The fundamental resonance frequency is given by;
[tex]f_0 = \frac{v}{\lambda}\\\\f_0 = \frac{344}{2}\\\\f_0 = 172 \ Hz\\\\[/tex]
Therefore, the fundamental resonance frequency is 172 Hz.
Hey guys this is Ap physics please help I need this to pass i will mark brainliest for a good attempt
Split up the forces into components acting parallel to and perpendicular to the slope. See the attached picture for the reference axes.
The box stays on the surface of the plane, so that the net force acting perpendicular to it is 0, and the only acceleration is applied in the parallel direction.
Let m be the mass of the box, θ the angle the plane makes with the ground, and a the acceleration of the box. By Newton's second law, we have
• net parallel force
∑ Force (//) = W (//) - F = m a
(that is, the net force in the parallel direction is the sum of the parallel component of the weight W and the friction F which acts in the negative direction)
• net perpendicular force
∑ Force (⟂) = W (⟂) + N = 0
Notice that
W (//) = W sin(θ) … … … which is positive since it points down the plane
W (⟂) = -W cos(θ) … … … which is negative since it points opposite the normal force N
So the equations become
W sin(θ) - F = m a
-W cos(θ) + N = 0
Solving for a gives
a = (W sin(θ) - F ) / m
which is good enough if you know the magnitude of the friction force.
If you don't, you can write F in terms of the coefficient of kinetic friction between the box and plane, µ, as
F = µ N
so that
a = (W sin(θ) - µ N ) / m
and the normal force itself has a magnitude of
N = W cos(θ)
so that
a = (W sin(θ) - µ W cos(θ) ) / m
The weight W has magnitude m g, where g is the magnitude of the acceleration due to gravity, so
a = (m g sin(θ) - µ m g cos(θ) ) / m
a = g (sin(θ) - µ cos(θ))
There is a bell at the top of a tower that is 45 m high. The bell weighs 190 N. The bell has ____________
energy.
Answer:
The bell has a potential energy of 8550 [J]
Explanation:
Since the belt is 45 [m] above ground level, only potential energy is available. And this energy can be calculated by means of the following equation.
[tex]E_{p}= W*h\\E_{p} = 190*45\\E_{p}=8550[J][/tex]
3.00 kg block moving 2.09 m/s right hits a 2.22 kg block moving 3.92 m/s left. afterwards, the 3.00 kg block moves 1.71 m/s left. find the velocity of the 2.22 kg block afterwards
Momentum is conserved, so the total momentum before collision is equal to the total momentum after collision. Take the right direction to be positive. Then
(3.00 kg) (2.09 m/s) + (2.22 kg) (-3.92 m/s) = (3.00 kg) (-1.71 m/s) + (2.22 kg) v
where v is the velocity of the 2.22 kg block after collision. Solve for v :
6.27 kg•m/s - 8.70 kg•m/s = -5.13 kg•m/s + (2.22 kg) v
(2.22 kg) v = 2.70 kg•m/s
v = (2.70 kg•m/s) / (2.22 kg)
v ≈ 1.22 m/s
i.e. a velocity of about 1.22 m/s to the right.
What is the wavelength of a wave whose velocity is 12m/s and has a frequency of .75 Hz?
Answer:
16 meters
Explanation:
Use the formula that relates frequency velocity and wavelength:
velocity = wave-length x frequency
in our case:
12 m/s = wave-length * 0.75 Hz
wave-length= 12/0.75 m = 16 meters
in the case shown below, the 1 kg rock rides on a horizontal disk that rotates at constant speed 5m/s about its vertical axis. the radius of the disk is 1 meter. What is the magnitude of the friction?
Answer:
25
Explanation:
Calculate the RMS speed of helium atoms near the surface of the Sun at a temperature of about 5300 K.
Answer:
RMS velocity, [tex]v_{rms}=5748.75\ m/s[/tex]
Explanation:
We need to find the RMS speed of helium atoms near the surface of the Sun at a temperature of about 5300 K.
The formula for RMS speed of a gas is given by :
[tex]v_{rms}=\sqrt{\dfrac{3RT}{m}}[/tex]
Where
R is ideal gas constant, R = 8.314 J /mol K
T = 5300 K
m is molar mass of Helium, [tex]m = 4\times 10^{-3}\ Kg/mol[/tex]
Substituting all the values in above formula :
[tex]v_{rms}=\sqrt{\dfrac{3\times 8.314\times 5300}{4\times 10^{-3}}}\\\\=5748.75\ m/s[/tex]
So, the RMS speed Helium atoms 5748.75 m/s.
A skier is moving 8.33 m/s when
he starts to slide UP a 8.44°
frictionless slope. How much time
does it take him to come to a stop?
(Unit = s)
Answer:
5.79
Explanation:
Vf=Vi+at
0=8.33+(-9.8)sin8.44t
t=8.33/(9.8sin8.44)=0.83/sin 8.44
=5.79
If two exactly the same cars are driving down a road, which one would have the most kinetic energy. The one that is moving faster, the one that is moving downhill, the one that is moving uphill, or the one that is moving slower.
Answer: the car that is moving downhill
Explanation:
A rocket will move upward as long as which condition applies?
A. A piece of paper near a magnet
B. An aluminum nail near a magnet
C. An iron nail, not near a magnet
D. An iron nail near a magnet
Answer:
it’s c not d
Explanation:
took the test
Answer: D!!!
Explanation: jus got it wrong from the other answer.
A hare can run at a rate of 15 m/s, while a turbocharged tortoise can now crawl at a rate of 3 m/s, how much of a head-start (time-wise) does the tortoise need in order to tie the hare in a 250 meter race?
A.
16.7 seconds
B.
66.7 seconds
C.
83.3 seconds
D.
100 seconds
Answer:
t = 66.7 s
Explanation:
Given that,
Speed of a hare, v = 15 m/s
Speed of a turbocharged tortoise, v' = 3 m/s
The hare in a 250 meter race
Let the Hare takes time t. It can be calculated as follows :
[tex]t=\dfrac{250}{15}=16.67\ s[/tex]
Let a turbocharged tortoise takes t'. It can be calulated as follows :
[tex]t'=\dfrac{250}{3}= 83.33\ s[/tex]
To tie the race, required time is given by :
[tex]\Delta t = t'-t\\\\=83.33-16.67\\\\=66.66\ s\\\\\approx 66.7\ s[/tex]
Hence, the correct option is (b) i.e. 66.7 seconds.