Explanation:
this is actually not as simple as it sounds here.
quite some energy is lost in the deformation of the bodies of car and truck, and it also needs more energy to get a standing object going than to accelerate an already moving object.
but assuming the simple described circumstances, then the energy and impulse of the moving truck of 5000 kg is transferred to a new combined system of car and truck of now 5000 + 3000 = 8000 kg.
so, the 20m/s inertia energy of the truck is now distributed to the truck/car combination.
since the same energy has to move now more mass, it is clear that the combined speed will be lower.
20×5000 = x×8000
20×5 = x×8
x = 100/8 = 12.5 m/s
that is the resulting speed of the combined truck/car object.
can anyone heelp me pls pls
Answer: Liquid - Lotion
Suspension - Semisolid
Capsule - Solid
Explanation:
Give Brainliest if correct :)
What is the magnitude of the electric field at a point 0.0075 m from a 0.0035
C charge?
Use E-
kq
and k - 9.00 x 10° N.m²/C2.
A. 1.9 x 1010 N
B. 4.2 x 10' N
O c. 5.6 x 1011 N
D. 5.5 x 1012 N
Explanation:
E = 9×10^9 × 0.0035 /(0.0075)²
= 9×10^9 × 3.5×10`³/(7.5×10`³)²
= 31.5×10^6 /(56.25 ×10^-6)
= 0.56 × 10^12
= 5.6 × 10^11 N/C (option C)
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.
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--[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
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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)☺️)
define potential difference as used in electricity
Answer:
This ability of charged particles to do work is called an electric potential.
Explanation:
When two negative charges are brought close to each other, they also repel. But when a positive and a negative charge are brought close together, they attract each other. When these two opposite charges are combined, they can be used to work. This is why we need a positive (+) and a negative (–) to light a bulb or run any electrical tool, equipment, mobile phone or home appliance.
Potential difference is the difference in the amount of energy that charge carriers have between two points in a circuit. ... The energy is transferred to the electrical components in a circuit when the charge carriers pass through them. We use a voltmeter to measure potential difference (or voltage).
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-_-
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.
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
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?
A 12.0-g bullet is fired horizontally into a 112-g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 149 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 83.5 cm, what was the speed of the bullet at impact with the block
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.
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A car accelerates at 4 m/s/s from rest. What is the car's velocity after it travels 20 m?
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
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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:
*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
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Answer:
3.0 J
Explanation:
Just took the test
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
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
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.
help with this question please.
Answer: Yes Because it matches with the mass and the amount of force Hope this helps :>
Explanation:
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.
What force causes the passengers in a car to move forward when the car comes to a sudden stop?
Answer:
You and the car have inertia. If the car comes to a sudden stop, your body tends to keep moving forward. When the car starts moving again, your body tends to stay at rest. You move forward because the car seat exerts an unbalanced force on your body.
Explanation:
This is due to inertia. When the speeding bus stops suddenly, lower part of the body comes to rest while the upper part of the body tends to maintain uniform motion. Hence, the passenger's are thrown forward.
can anyone heelp me pls pls
Answer:
Lotion : semisolid
Suspension : liquid
capsule : solid
Explanation:
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)
Which material(s) listed below is an example of a persistent organic pollutant?
(*select all that apply)
Select 4 correct answer(s)
Ricin
PCBs
DDT
Mercury
Arsenic
Answer:
mercury
arsenic
ricin
ddt
Explanation:
pls mark me as brainliest
Among the materials listed, examples of persistent organic pollutants (POPs) are PCBs (Polychlorinated Biphenyls) DDT (Dichlorodiphenyltrichloroethane). The correct option is (2) and (3).
Persistent Organic Pollutants (POPs) are a class of chemicals that are long-lasting, bioaccumulative, and harmful to humans and wildlife. These pollutants are resistant to degradation and can persist in the environment for long periods of time, resulting in long-distance transport and bioaccumulation in the food chain.
PCBs (Polychlorinated Biphenyls): Once utilized in a variety of industrial applications such as electrical transformers and capacitors, PCBs are now prohibited in many countries due to their toxicity.
DDT (Dichlorodiphenyltrichloroethane): An pesticide that was widely used for agricultural and public health objectives in the past but is now restricted or banned due to its environmental and health effects.
Hence, the correct answer is PCBs and DDT. The correct option is (2) and (3).
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#SPJ3
Explain what is meant by the elastic limit of a spring.
Answer:
The elastic limit of a material is the furthest
point it can be stretched or deformed while
being able to return to its previous shape.
Explanation:
Once a material has gone past its elastic
limit, its deformation is said to be inelastic.
The higher the spring constant, the stiffer
the spring.
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.
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
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.
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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