In a game of tug of war, team one pulls to the right with a force of 500 newtons and team two pulls to the left with a force of 600 newtons. What is the net force?
Answer:
Explanation:
There is no set way to do this. All you have to do is define left and right. Left will be minus and right will be the opposite --- plus.
That is completely arbitrary. It could be the other way around. It does not matter.
Left is minus so: - 600 N is the force going left.
Right plus so: + 500 N
Now just add.
Net Force = +500 - 600
Net Force = - 100 N
So the Net Force is - 100 N going to the left.
Which of the following formulas describes the change in momentum of an
object?
A. change in momentum = force x time over which force is applied
B. change in momentum = acceleration distance over which
acceleration is applied
C. change in momentum = force x distance over which force is
applied
O D. change in momentum = acceleration time over which
acceleration is applied
Atoms of which two elements could combine with atoms of carbon (C) to
form covalent bonds?
A. S
B. Na
C. N
D. K
E. Ca
Answer:
Correct answer is letter E.Ca
The volume of an ideal gas is increased from 0.6 m3 to 2.4 m3 while maintaining a constant pressure of 1000 Pa (1000 N/m2). Determine, in J, the amount of work done by the gas in this expansion.
The amount of work done by the gas in the given expansion is 1800 J.
The given parameters;
initial volume of the ideal gas, V₁ = 0.6 m³final volume of the ideal gas, V₂ = 2.4 m³constant pressure of the gas, P = 1000 PaThe amount of work done by the gas in the given expansion is calculated as follows;
W = PΔV
where;
ΔV is the change in volume of the gasSubstitute the given parameters and solve for the work done;
W = 1000(2.4 - 0.6
W = 1800 J
Thus, the amount of work done by the gas in the given expansion is 1800 J.
Learn more here:https://brainly.com/question/14978695
Ex 2) A cannon ball is shot straight up into the air with an initial velocity of 25 m/s[Up).
What is the maximum height of the cannonball?
Explanation:
S=(V^2-U^2)/2a a=g (gravity) a=10
=(0^2-25^2/2*(-10)
=625/20
=31.25m
In hockey activities, a warm hockey puck and a frozen hockey puck has a different coefficient of restitution: 0.5 for a warm hockey puck, and 0.35 for a frozen one. NHL requires the frozen pucks to be used in games. To make sure the puck can be used in the game, the referee drops the puck on its side from a height of 2.5 m. How high should the puck bounce if it is a frozen puck
If its is a frozen hockey puck, it bounce off the ground after collision to a height of 0.3m.
Given the data in the question;
Since the hockey puck was initially in the referee's hands
Initial velocity; [tex]u = 0m/s[/tex]Distance or height from which it was dropped; [tex]h = 2.5m[/tex]Acceleration due to gravity; [tex]g = 9.8 m/s^2[/tex]Coefficient of restitution a frozen puck; [tex]0.35[/tex]First we will find the velocity of the Puck when it hits the ground
From the Third Equation of Motion:
[tex]v^2 = u^2 + 2as[/tex]
Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the distance.
Since the pluck is under gravity, we will have:
[tex]v^2 = u^2 + 2gh[/tex]
We substitute in our value and find "v"
[tex]v^2 = 0 + (2 \ *\ 9.8m/s^2\ *\ 2.5m )\\\\v^2 = 47.04m^2/s^2\\\\v= \sqrt{47.04m^2/s^2}\\\\v = 6.85857m/s[/tex]
Now, Velocity of the hock puck after it hits the ground and bounce back;
We know that; Coefficient of restitution [tex]= \frac{Relative\ velocity\ after\ collision}{Relative\ velocity\ before\ collision}[/tex]
Hence, Relative Velocity after collision = Coefficient of restitution × Relative Velocity before collision
we substitute in our values;
Relative Velocity after collision [tex]= 0.35 \ *\ 6.85857m/s[/tex]
Relative Velocity after collision [tex]= 2.4 m/s[/tex]
Now, to determine how high should the puck bounced back
We use the Third Equation of Motion:
[tex]v^2 = u^2 + 2as[/tex]
Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the distance.
Since the pluck is under gravity, we will have:
[tex]v^2 = u^2 + 2gh[/tex]
Now, since the hockey puck bounces back, it is experiencing a negative acceleration
Hence, the equation becomes
[tex]v^2 = u^2 - 2gh[/tex]
We substitute our values into the equation and find "h"
[tex](0m/s)^2 = (2.4m/s)^2 - ( 2*9.8m/s^2*h)\\\\0 = 5.76m^2/s^2 - (19.6m/s^2*h)\\\\(19.6m/s*h) = 5.76m^2/s^2 \\\\h= \frac{ 5.76m^2/s^2 }{19.6m/s^2}\\\\h = 0.3m[/tex]
Therefore, If its is a frozen hockey puck, it bounce off the ground after collision to a height of 0.3m.
Learn more; https://brainly.com/question/16908019
2. Calculate; The acceleration of any object due to Earth's gravity is -9.81 m/s. For every
second an object falls, its velocity changes by 9.81 meters per second. For several different
times on the table, multiply the time by the acceleration
A. What do you notice:
I
As the time of motion increases, the velocity of the object increases downwards.
The given parameters;
acceleration due to gravity, g = -9.81 m/s²velocity of the object, v₀ = 9.8 m/sThe final velocity of the object at different time is calculated as follows;
when the time = 1 second;
v = v₀ - gt
v = 9.8 - 9.8(1)
v = 0
when the time = 2 second;
v = v₀ - gt
v = 9.8 - 9.8(2)
v = -9.8 m/s
when the time = 3 second;
v = v₀ - gt
v = 9.8 - 9.8(3)
v = -19.6 m/s
Thus, we can conclude that as the time of motion increases, the velocity of the object increases downwards.
Learn more here:https://brainly.com/question/16782759
A man with a mass of 60 kg rides a bike with a mass of 13 kg. What is the force needed by the man to accelerate the bike at 0.90 m/s2?
Um cubo fechado com 1 m3 de um líquido (d=0,8; =21 000 kgf/cm2
) é submetido a um aumento de pressão de
20107 N/m2
. Calcule, no Sistema Internacional, as seguintes grandezas:
a) massa final;
b) peso final;
c) volume final;
d) massa volúmica final;
e) peso volúmico final.
Answer:
a. massa final
Explanation:
eu escolhi isso porque essa é a resposta espero que ajude você pode me dar uma ideia?
can anyone heelp me pls pls
Answer:
Lotion : semisolid
Suspension : liquid
capsule : solid
Explanation:
3)
Car A, moving in a straight line at a constant speed of 20. meters per second, is initially 200
meters behind car B, moving in the same straight line at a constant speed of 15 meters per
second. How far must car A travel from this initial position before it catches up with car B?
A)
200 m
B)
1000 m
C)
800 m
D)
400 m
Answer:
C) 800
Explanation:
How can stretching affect the range of motion of the neck? Hypothesis
Answer:
reduce passive stiffness and increase range of movement during exercise.
Explanation:
stretching performed as part of a warm up prior to exercise is thought to reduce passive stiffness and increase range of movement during exercise. in general it appears that is static stretching is most beneficial for athletes requiring flexibility for their sports.
Please HELP!!!!!!!!!!!!!!!!!!!!!
Using Newton's First Law, explain why, in a frictionless environment, a car that is under motion will not stop moving?
Answer:
According to Newton's first law of motion, an object maintains its state unless a force acts on it. Therefore, a moving car does not change its direction and keeps its speed unless a force acts on it.
In which device is chemical energy transformed into electrical energy?
A.
battery
B.
hair dryer
C.
television
D.
hydroelectric plant
Answer:
A) Battery
Explanation:
A Battery because it holds lithium whatever stuff and we can use to power our electronics (Chemical -> electrical)
Hair dryers (electrical-> kinetic)
Television ( Electrical -> ???)
Hydroelectric plant ( Kinetic -> electrical)
*Sorry for the bad quality picture!*
A frictionless pendulum with a mass of 0.4 kg and a length of 2.1 m starts at point A, at an angle 0 of 60°. As it swings downward, it passes through point B, which is 30 degrees from equilibrium. What is the kinetic energy of the pendulum at point B?
A) 3.9 J
B) 3.0 J
C) 1.1 J
D) 4.1 J
The conservation of mechanical energy allows finding the result for the speed of the pendulum when it is at 30º is:
The speed is: 3.88 m / s
The conservation of mechanical energy is a theorem of greater importance in physics and ordinary life, it states that if there is no friction force the total mechanistic energy remains constant at all points.
Mechanical energy is the sum of kinetic energy plus all potential energies. In the attachment we see a diagram of the pendulum's movement at the two points of interest.
They indicate that the pendulum is released from an initial angle of θ₁ = 60º, let's find the mechanical energy at that point.
Em₀ = U = m g h
Where the height is measured from the lowest point of the movement.
h = L - L cos tea1 = L (1 cos tea1)
The second point of interest occurs for θ₂ = 30º.
At this point part of the energy is indica and part gravitational potential.
[tex]Em_f[/tex] = K + U₂
[tex]Em_f[/tex] = ½ m v² + m g h ’
There is no friction in the system, therefore mechanical energy is conserved.
Em₀ = Em₀_f
mg L (1 - cos θ₁) = ½ m v² + m g L (1 - cos θ₂)
v² = 2g L (cos θ₂ - cos θ₁)
Let's calculate.
v² = 2 9.8 2.1 (cos 30 - cos 60)
v² = 41.16 0.366
v = 3.88 m / s
In conclusion using the conservation of mechanical energy we can find the result for the speed of the pendulum when it is at 30º is:
The speed is: 3.88 m / s
Learn more here: brainly.com/question/14688403
Answer:
3.0 J
Explanation:
Just took the test
can anyone heelp me pls pls
Answer: Liquid - Lotion
Suspension - Semisolid
Capsule - Solid
Explanation:
Give Brainliest if correct :)
Help !!!
I solved Anthor one but didn’t understand this one ??
Answer:
Explanation:
It's a velocity•time chart. As distance = vt, the area under the curve between limits is the distance traveled.
Pic is fuzzy so I will ASSUME the horizontal axis is seconds and vertical is meters per second.
0 s ≤ t ≤ 5 s = ½(5 - 0)(30 - 0) = 75 m
5 s ≤ t ≤ 10 s = (10 - 5)(30 - 0) = 150 m
10 s ≤ t ≤ 15 s = ½(30 + 20)(15 - 10) = 75 m
0 s ≤ t ≤ 25 s = 75 + 150 + 75 + 20(20 - 15) + ½(20)(25 - 20) = 450 m
Most gasoline engines in today's automobiles are belt driven. This means that the crankshaft, a rod which rotates and drives the
pistons, is timed to the camshaft, the mechanism which actuates the valves, by means of a belt. Starting from rest, assume it
takes t = 0.0320 s for a crankshaft with a radius of r = 3.75 cm to reach 1250 rpm. If the belt does not stretch or slip, calculate
the angular acceleration ay of the larger camshaft, which has a radius of r2 = 7.50 cm, during this time period.
The angular acceleration of the larger camshaft is 995.72 rad/s².
The given parameters;
initial angular velocity, [tex]\omega _i[/tex] = 0time of motion, t = 0.032 sradius of the crankshaft, r = 3.75 cm final angular speed, [tex]\omega _f[/tex] = 1250 rpmThe angular acceleration of the 3.75 cm camshaft is calculated as follows;
[tex]\omega _f = \omega _i + \alpha t\\\\\omega _f =0 + \alpha t\\\\\omega _f = \alpha t\\\\(1250 \ \frac{rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1\min}{60 \ s} ) = 0.032 \alpha \\\\130.92 = 0.032\alpha \\\\\alpha = \frac{130.92}{0.032} = 4091.25 \ rad/s^2[/tex]
The angular momentum of the camshaft is calculated as follows;
[tex]I_1 \alpha _1 = I_2 \alpha_2 \\\\\frac{1}{2} mr_1^2 \alpha _1 = \frac{1}{2}m R^2 \alpha_2\\\\r_1^2 \alpha _1 = R^2 \alpha_2\\\\\alpha_2 = \frac{r_1^2 \alpha _1 }{R^2} \\\\\alpha_2 =\frac{(0.037)^2 \times (4091.25)}{(0.075)^2} \\\\\alpha _2 = 995.72 \ rad/s^2[/tex]
Thus, the angular acceleration of the larger camshaft is 995.72 rad/s².
Learn more here:https://brainly.com/question/14695341
a place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. when kicked, the ball leaves the ground with a speed of 23.9 m/s at an angle of 51.5 degrees to the horizontal. (a) by how much does the ball clear or fall short of clearing the crossbar?
(b) does the ball approach the crossbar while still rising or while falling?
Explanation:
Let's calculate the components of the football's velocity:
[tex]v_{0x} = (23.9\:\text{m/s})\cos{51.5°} = 14.9\:\text{m/s}[/tex]
[tex]v_{0y} = (23.9\:\text{m/s})\sin{51.5°} = 18.7\:\text{m/s}[/tex]
a) The time it takes for the football to travel 36.0 m horizontally is
[tex]t = \dfrac{x}{v_{0x}} = \dfrac{36.0\:\text{m}}{14.9\:\text{m/s}} = 2.4\:\text{s}[/tex]
During this time, the y-displacement of the football is
[tex]y = v_{0x}t - \frac{1}{2}gt^2[/tex]
[tex]\:\:\:\:= (18.7\:\text{m/s})(2.4\:\text{s}) - \frac{1}{2}(9.8\:\text{m/s}^2)(2.4\:\text{s})^2[/tex]
[tex]\:\:\:\:= 16.7\:\text{m}[/tex]
This means that the football cleared the crossbar by 16.7 m - 3.05 m = 13.7 m
b) To determine whether the football was rising or falling while clearing the crossbar, let's look at the y-component of its velocity after 2.4 s:
[tex]v_y = v_{0y} - gt = 18.7\:\text{m/s} - (9.8\:\text{m/s}^2)(2.4\:\text{s})[/tex]
[tex]\:\:\:\:\:\:= -4.82\:\text{s}[/tex]
Since its sign is negative, this means that the football was already on its way down.
If you traveled 50m/s for 60 seconds, how far did you travel? Remember speed=distance/time
Question options:
300 m/s
500 m/s
3,000 m/s
300 km/h
for some reason my question got removed-_-
Jonathan wants to separate stones, insects and other unwanted materials in his mixture of grains and corn. What technique of separating mixture is appropriate
A. Winnowing
B. physical manipulation
C. Filtering
D. Magnetism
Answer:
What is B physical manipulation
Explanation:
Physical manipulation means fertilizers that are manufactured, blended, or mixed, or animal manures or compost that have been changed from their initial physical state by manipulations such as drying, cooking, chopping, grinding, shredding, ashing, or pelleting.
Please allow me to know if my answer helped you with a thank you!
Miss Hawaii
A car accelerates at 4 m/s/s from rest. What is the car's velocity after it travels 20 m?
Once you start pulling your object with less force than friction, what should you expect your object to do? What about when your object is pulled with more force than friction?
NO LINK S
#Case -1
If Pulling force is less than frictional force the object won't move .
#Case-2
If Pulling force is greater than frictional force then object will be .
In order to calculate friction force you need Limiting friction first .
[tex]\\ \sf\longmapsto F_L=\mu sN[/tex]
u s is coefficient of static friction and N is normal reaction
Or
[tex]\\ \sf\longmapsto F_L=\mu smg[/tex]
As N=mgWhich of the following is the current best hypothesis for the formation of the solar system?
A. Formed by an exploding super nova star which then collapsed and coalesced into a spinning
disk forming Sun and planets
B. Our solar system has always been here and has never changed
C. Formed from the Sun’s explosion releasing particles into space forming planets and other
objects
D. Our solar system was formed by a great collision of other stars with one another
Answer:
A
Explanation:
all galaxies exploded in order to create the sun/stars
A block slides down an incline plane that makes a 30 degree angle with the
horizontal. If the coefficient of kinetic friction is 0.3. Calculate the acceleration of the block.
Hi there!
We know that:
Force due to gravity = Mgsinθ
Force due to friction = μMgcosθ
Let the positive direction be directed in the direction of the block's acceleration, which is downward.
Thus:
ΣF = Mgsinθ - μMgcosθ
Solving for acceleration requires diving all terms by the mass, so:
a = gsinθ - μgcosθ
Substitute in given values. (g = 9.8 m/s²)
a = 9.8sin(30) - 0.3(9.8)cos(30) = 2.354 m/s²
What force is required to accelerate a 1000-kilogram car at 6 m/s??
Answer:
Hence, Force (F) is 4000 Newtons.
Explanation:
ɢɪᴠᴇɴ :-
Mass (m) = 1000 kg
Acceleration (a) = 4 m/s²
ᴛᴏ Find:-
Force (F)
solution :-
We know that,
Force = mass × acceleration
➮ F = ma
➮ F = 1000 × 4
➮ F = 4000 N
Hence,
Force (F) is 4000 Newtons
(Hope this helps can I pls have brainlist (crown)☺️)
--[50 POINTS]--
1)A block of mass 25 kg is placed on flat ground. The coefficient of static friction and kinetic friction are 0.73 and 0.16
a.If a person pushes the block and the block is moving, what will be the acceleration of the block?
2) A block has a mass of 79 kg. The coefficient of static and kinetic friction between the sled and the ground is 0.87 and 0.37. Person A tries to pull the block with 210N, but fails.
a) Person B successfully pulls the sled with 909N. What is the acceleration of the sled?
Newton's second law allows us to find the results for the acceleration of the blocks are:
1) The acceleration is a = 559 m / s²
2) The acceleration is a = 7.88 m / s²
Newton's second law states that the net force is equal to the product of the mass and the acceleration of the body.
∑ F = m a
Where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the body.
The reference system is a coordinate system with respect to which the decomposition of the forces is carried out, in the attached we have a free body diagram of the system.
1) They indicate that the body mass is 25 kg.
y-axis
N - W = 0
N = W = m g
x-axis
F -fr = ma
The friction force is a macroscopic force that results from the sum of all the microscopic interactions between the two surfaces, it has the formula
fr = μ N
Where fr is the friction force, N the normal and very the friction coefficient.
This friction coefficient has two values:
Static. For when with there is not relative motion between the two surfaces. Dynamic. When there is relative motion between the two surfaces.
We substitute.
F - μ m g = m a
a) The system moves which is the acceleration.
Suppose that the force that star to move the system keeps constant, just before the system begins to move the coefficient of friction is static, let's find the applied force.
F = μ m g
F = 0.73 25 9.8
F = 178.85 N
The block begins to move and the friction coefficient decreases to the dynamic value, we look for the acceleration.
a = [tex]\frac{F - \mu \ m g}{m}[/tex]
a = [tex]\frac{178.85 - 0.16 \ 25 \ 9.8 }{25}[/tex]
a = 5.59 m / s²
2) In this case the mass of the block is 79 kg and the applied force is
F = 909 N
We look for acceleration.
a = [tex]\frac{909 - 0.37 \ 79 \ 9.8 }{79}[/tex]
a = 7.88 m / S²
In conclusion using Newton's second law we can find the results for the acceleration of the blocks are:
1) The acceleration is a = 559 m / s²
2) The acceleration is a = 7.88 m / s²
Learn more here: brainly.com/question/11329269
The model represents a fluorine (F) atom. What is the mass of the atom?
Answer:
19
Explanation:
The mass of an atom is found in the nucleus: number of protons + number of neutrons; 9 + 10 = 19
The mass number of fluorine is 19
A football is kicked with a velocity of 5 m/s at an angle of 53° above the horizontal. What is its speed at the
maximum height?
A) 3 m/s
B) 6 m/s
C) 9 m/s
D) 12 m/s
E) 15 m/s
Un péndulo de 4 m de longitud tiene una frecuencia de 5Hz. Calcular la longitud de otro péndulo que en el mismo lugar tiene una frecuencia de 4Hz
Conociendo la longitud y frecuencia de un péndulo, queremos encontrar la longitud de otro pendulo de tal forma que tenga otra frecuencia.
Veremos que la longitud del nuevo péndulo debe ser 6.25m
Sabemos que un péndulo de 4m de longitud tiene una frecuencia de 5Hz.
La frecuencia de un péndulo está dada por:
[tex]f = \frac{1}{2*\pi} *\frac{g}{l}[/tex]
Donde g es la aceleración gravitatoria y l es la longitud del péndulo, remplazando los datos que tenemos en esa ecuación obtenemos:
[tex]5 Hz = \frac{1}{2*3.14} *\sqrt{\frac{g}{4m} } \\\\(5Hz*2*3.14)^2*4m = g = 3,943.8 m/s^2[/tex]
Ahora debemos encontra la longitud de tal forma que la frecuencia sea 4Hz, entonces debemos resolver:
[tex]4Hz = \frac{1}{2*3.14} *\sqrt{ \frac{3943.8m/s^2}{l} }\\\\4hz*2*3.14 = \sqrt{ \frac{3943.8m/s^2}{l} }\\\\l =\frac{3943.8m/s^2}{ (4hz*2*3.14)^2} = 6.25m[/tex]
La longitud del nuevo péndulo deve ser 6.25m
Sí quieres aprender más, puedes leer:
https://brainly.com/question/23483504