Answer:
d
Explanation:
i just know (pls check out my question pls i will give nayone who answers brainiest)
In which type of collision is the kinetic energy of the system unchanged?
A. Perfectly inelastic
O B. Elastic
C. No type of collision
D. Inelastic
B.) Elastic
Took the test and this is the answer
What's the name for how you see the world through your perception
in cognitive theory.
Answer:
Perception: Perception is a cognitive process that allows people to take in information through their senses (sensation) and then utilize this information to respond and interact with the world. Thought: Thought is an essential part of every cognitive process.
I hope it helped
Two car's are racing! The winner of the race will always be be the vehicle with the greater...
When Royce was 10 years old, he had a mass of
30 kg. By the time he was 16 years old, his mass
increased to 60 kg. How much larger is the
gravitational force between Royce and Earth at
age 16 compared to age 10?
Answer:
The gravitational force between Royce and Earth would be doubled at 16 years.
Explanation:
"Newton's law of universal gravitation states that gravitation force between two masses is proportional to the magnitude of their masses and inverse-squared of their distance".
Royce Scenario
At the age of 10 Royce's mass = 30kg
At the age of 16 Royce's mass = 60kg
From Newton's law of universal gravitation, an Increase in the mass of a body would amount to a corresponding increase in the gravitational force.
In the case of Royce, the mass double between the age of 10 and 16, so there would be an increase of the gravitation force by double.
The gravitational force at the age of 60 is doubled of the gravitational force at the age of 10. A complete solution is below.
Given:
At 10 years old,
Mass,
m = 30 kgnow,
→ [tex]Gravitational \ force (F_1) = \frac{GMm}{r^2}[/tex]
[tex]= mg[/tex]
or,
→ [tex]g = \frac{GM}{r^2}[/tex]
then,
Gravitational force = 30 g
Now,
At 16 years old,
Mass,
m = 60 kgNow,
→ [tex]\frac{F_1}{F_2} = \frac{30}{60}[/tex]
[tex]=\frac{1}{2}[/tex]
[tex]F_2 = 2F_1[/tex]
Thus the above response is right.
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A book on a 2-meter high shelf has a mass of 0.4 kg. What is its potential energy?
Answer:
[tex]\boxed {\boxed {\sf 7.84 \ Joules}}[/tex]
Explanation:
The formula for potential energy is:
[tex]PE=m*g*h[/tex]
where m is the mass, g is the gravitational acceleration, and h is the height.
The mass of the book is 0.4 kilograms. The gravitational acceleration on Earth is 9.8 m/s². The height of the book is 2 meters.
[tex]m=0.4 \ kg \\g=9.8 \ m/s^2 \\h=2\ m[/tex]
Substitute the values into the formula.
[tex]PE=(0.4 \ kg)(9.8 \ m/s^2)(2 \ m)[/tex]
Multiply the first two numbers.
0.4 kg*9.8 m/s²= 3.92 kg*m/s²If we convert the units now, the problem will be much easier later on. 1 kg*m/s² is equal to 1 Newton. So, our answer of 3.92 kg*m/s² is equal to 3.92 N[tex]PE=(3.92 \ N )(2 \ m)[/tex]
Multiply.
3.92 N* 2 m=7.84 N*m1 Newton meter is equal to 1 Joule (this is why we converted the units). Our answer is equal to 7.84 Joules.[tex]PE=7.84 \ J[/tex]
Answer:
7.84 is your answer ok bye
Problem (8):
A projectile is fired at an angle of 53° above the horizontal with initial speed of 10.0 m/s. After
one second, what is X and Y components of the position of that projectile?
A60m 20 m
B. 6.0 m. 8.0 m
C. 3.0 m, 6.0 m
D.4.0 m, 3.0 m
Answer:
x=6 m, y= 3 m
Explanation:
Projectile Motion
It's the type of motion that experiences an object launched near the Earth's surface and moves along a curved path under the action of gravity.
Being vo the initial speed of the object, θ the initial launch angle, and g the acceleration of gravity, then the components of the velocity at a given time t are:
[tex]v_x=v_o\cos\theta[/tex]
[tex]v_y=v_o\sin\theta-g.t[/tex]
And the components of the displacement are:
[tex]x=v_o\cos\theta .t[/tex]
[tex]\displaystyle y=v_o\sin\theta.t-\frac{g.t^2}{2}[/tex]
The projectile of the problem is fired at θ=53° with an initial speed of vo=10 m/s.
Thus, after 1 second, the components of the displacement are:
[tex]x=10\cos 53^\circ \cdot 1[/tex]
[tex]x\approx 6~m[/tex]
[tex]\displaystyle y=10\sin 53^\circ\cdot 1-\frac{9.8\cdot 1^2}{2}[/tex]
[tex]y\approx3~m[/tex]
Answer: x=6 m, y= 3 m
None of the options match this answer.
A projectile is fired horizontally from a height of 78.4 m at a speed of 300 m/sec. How far did it travel horizontally before hitting the ground?
Answer:
Explanation:
Using the formula for calculating range expressed as;
R = U√2H/g
U is the speed = 300m/s
H is the maximum height = 78.4m
g is the acceleration due to gravity = 9.8m/s²
Substitute into the fromula;
R = 300√2(78.4)/9.8
R = 300 √(16)
R = 300*4
R = 1200m
Hence the projectile travelled 1200m before hitting the ground
A group of big city tenants were upset about rising rent for their apartments, and decided to play a prank on the Building Manager. They went up to the top floor of the
building, which was 176.4 m above the sidewalk, and waited. They wanted to drop the balloon at just the right time, so they needed to make a quick calculation. How many
seconds before the Building Manager walks to the spot should they drop the balloon?
Answer:
t = 6 s
Explanation:
This is a free fall exercise
y = y₀ + v₀ t - ½ g t²
If the balloon is released its initial velocity is zero, when it reaches the floor its height is also zero, we substitute
0 = y₀ + 0 - ½ g t2
t = √(2yo / g)
let's calculate
t = √ (2 176.4 / 9.8)
t = 6 s
the balloon must be released 6 s before the person reaches the building
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Comentarios
As you take the stoppered part of the tube up the staircase you begin to see the water level drop around the 4th floor. As you continue up it does not continue up with you but stays at a constant level. What does that mean?
a. The pressure in the tubing is equal to the barometric pressure.
b. The tubing was unable to supply any more water to the tube for use.
c. The pressure outside the tube is higher that the water pressure inside the tube.
Answer:
a. The pressure in the tubing is equal to the barometric pressure.
Explanation:
Since in the question it is mentioned that the if you take the stoppert part of the tube than the level of warer would be fall approx 4th floor and if it is continued than it wont be continue but remains constant.
Now here first we do that the tube i.e. connected to the bucket should be taken up. In the first instance, the bucket supplies the water to the tube but it would not increased far away to the level of the barometric pressure
Hence, the correct option is a.
On an alien planet, a simple pendulum of length 0.300 m has oscillation frequency 0.709 Hz. Find the period of the pendulum. Find the acceleration due to gravity on this planet's surface. What must be the length of a simple pendulum if its oscillation frequency is to be equal to that of an air-track glider of mass 0.450 kg attached to a spring of force constant 9.75 N/m?
Answer:
1.41s
5.95m/s
0.2746m
Explanation:
The time period
T = 1/f
= 1/0.709s
= 1.41 seconds
We have
T = 2π√l/g
T² = 4π²l/g
g = 4π²l/T²
g = 4x3.14²x0.300/1.41²
g = 5.95m/s² this is the acceleration due to gravity.
Then the time period of the glide
T2 = 2π√m/k
Length of pendulum = l
Time period T
T2 = 2π√l/g
Then T1 = T2
2π√m/k = 2π√l/g
M/k = l/g
L = g.m/k
L = 5.95x0.450/9.75
L = 0.2746
This must be the length of the simple pendulum
A student rides a bicycle in a circle at a constant speed and constant radius. A force diagram for the student-bicycle system is
shown in the figure above. The value for each force is shown in the figure. What is the acceleration of the student-bicycle system?
m
А
0
2
B
0.2"
C) 5
D
25"
Solution:
The gravitational force that acts on the bicycle system is
[tex]$F_{g} = 500 \ N$[/tex]
Now the force, that is the gravitational force is related to mass of the system and the acceleration due to gravity of the system, 'm' and 'g' respectively.
Therefore, we can write
[tex]$F_g=mg$[/tex]
500 = m x 10 (since , g = 10 m/s-s)
∴ m = 50 kg
Now the net vertical force acting on the student bicycle system is 0. And the vertical acceleration of system is also 0. The total horizontal force acts to the right of the system. So by Newton's 2nd law of motion, we can write
[tex]$F_f = ma$[/tex]
[tex]$a=\frac{F_f}{m}$[/tex]
[tex]$=\frac{250}{50}$[/tex]
Therefore [tex]$a= 5 \ m/s^2$[/tex]
Hence (C) is correct option.
A student walks 300 meters due east and then 100 meters due north,
then 500 meters due west. The total distance travelled by the student
-
is
meters.
Answer:
900 meter hhgffhffhhfcbjhggjdswexaazcgh
Answer:
900 meters
Explanation:
If you are talking about how many meters student walked then its 900 metes
But if you are talking about how many meters student travelled from the start point then its 200 meter
Witch is not true about longitudinal waves?
Answer:
light is a longitudinal wave.
Explanation:
lol
Help please an observer at the North Pole sees the moon phase shown below. What Moon phase will be observed approximately one week later?
Answer:
it should be a new moon
Explanation:
(30points pls help ASAP) How many grams of sodium would there be in 70 g of NaCl?
27.52g grams of sodium would there be in 70 g of NaCl.
What is molecular mass ?The molecular mass gives the total mass of the molecule, it is the sum of masses of all the atoms present in a molecule. for example in Nacl, there is one atom of Na and one atom of Cl both makes NaCl. the sum of masses of Na and Cl gives the mass of the NaCL. Molecular mass is expressed in amu. 1 amu = Avogadro's number of atom or molecule.
The Molar mass is defined as the mass of a given substance divided by the amount of a substance. its unit is g/mol.
Here I have direct method for the calculation of mass of particular atom in the molecule.
Molecular weight of the NaCl is,
(23+35.5)g=58.5g where 23= mass of Na and 35.5 mass Cl.
in this we can see that 58.5 g of Nacl contains 23 g of Na.
we can write 58.5 NaCl = 23g Na
therefore 70g of NaCl = (23÷58.5)×70 Na
70g of NaCl = 27.52g of Na
Hence 70g of NaCl contains 27.52g of sodium.
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Does lighting striking the earth considered the speed of light?
Answer:
i would tell you but im a hobbit
Explanation:
biden 2020
A metal block of density 900kgm^-3 weighs 60N in air, find its weight when it is immersed in paraffin wax of density 800kgm^-3
Nicki rides her bike at a constant speed for 6 km. That part of her ride takes her 1 h. She then rides her bike at a constant speed for another 9
km. That part of her trip takes her 2 h. What is her average speed?
Answer:
24km/h
Explanation:
go it right on ingenuity 2020
How much heat does 56 L of water need to absorb to raise its temperature from 350C to 740C?
Answer:
300 L of water
Explanation:
A wire 50.0 m long and 2.00 mm in diameter is connected to a source with a potential difference of 9.11 V, and the current is found to be 36.0 A. Find the resistivity of the material.
Answer:
ρ = 1.6*10⁻⁸ Ω/m.
Explanation:
Applying Ohm's Law to the wire, assuming that it can be treated as a pure resistance, the resistance of the wire can be obtained as follows:[tex]R = \frac{V}{I} = \frac{9.11V}{36.0A} = 0.253 \Omega (1)[/tex]
At the same time, we know that there exists a relationship between the resistance, the resistivity ρ, the length L and the area A of the wire, that is given for the following expression:[tex]R = \rho* \frac{L}{A} (2)[/tex]
The area of the circular section of the wire, can be expressed as a function of the diameter d, as follows:[tex]A = \frac{\pi*d^{2} }{4} = \frac{\pi*(0.002m)^{2}}{4} = \pi*10e-6 (3)[/tex]
Replacing the left side of (2) by (1), and (3) on the right side, we can solve for the resistivity ρ as follows:[tex]\rho = \frac{R*A}{L} = \frac{0.253\Omega*\pi*10e-6}{50.0m} = 1.6e-8 \Omega/m[/tex]
ρ = 1.6*10⁻⁸ Ω/mWhat name is given to the rate at which velocity changes?
Answer mere
Answer:
Acceleration
Explanation:
A segment of wire of total length 3.0 m carries a 15-A current and is formed into a semicircle. Determine the magnitude of the magnetic field at the center of the circle along which the wire is placed. a. 1.6 mu T b. 4.9 mu T c. 1.0 mu T d. 9.8 mu T e. 15 mu T
Answer:
c. 1.0 mu T
Explanation:
From, the ampere's circuit law:
∫B.dl = μ₀ I
B.(2πr) = μ₀ I
where,
B = Magnetic Field = ?
r = radius of the circle = length of wire = 3 m
μ₀ = Permeability of free space = 4π x 10⁻⁷ Tm/A
I = Current = 15 A
Therefore,
B(2π)(3 m) = (4π x 10⁻⁷ Tm/A)(15 A)
B = 10 x 10⁻⁷ T
B = 1 x 10⁻⁶ T = 1 μT
Hence, the correct option is:
c. 1.0 mu T
can a body have mass but no weight give reasion for your answer
Answer:
Hey buddy, here is your answer. Hope it helps you.
Explanation:
Yes, weight of a body is not constant, it varies with the value of acceleration due to gravity, g. Weight of a body is zero, when it is taken to the centre of the earth or in the interplanetary space, where g=0. Mass is the total matter in a body, while weight is the force by which the body is attracted.
Weight is never constant while mass is. So we can have weight zero but with some mass. This will happen when an object is at the centre of earth, as g=0 at centre of earth , so weight will be 0 while mass will always be constant everywhere
Answer:
yes a body can have mass but no wt at a centre of the earth
Explanation:
this is because acceleration due to gravity decreases as we go deep into earth and becomes 0 at the centre
A car with a mass of 1.5 × 103 kilograms is traveling west at a velocity of 22 meters/second. It hits a stationary car with a mass of 9.0 × 102 kilograms. If the collision is inelastic, what is the final direction and approximate velocity of the two cars?
Answer:
A. 14 meters/second to the west
Explanation:
The collision is inelastic, which means that:
- only the total momentum is conserved (not the kinetic energy)
Answer:
14 meters/second to the west
Explanation:
That's the answer!
Picture question please answer do in 5 mins
Answer:
Can't see anything, please share clearly
According to the Law of the Conservation of Matter, if you dissolve 25 grams of sugar into 150 grams of water, the mixture should weigh how much?
In the picture below, sphere X swings down and collides with sphere Y. Identify the action and reaction forces.
Action force:
Reaction force:
Answer:
Action force is force acting in one direction. Reaction force is force acting in the opposite direction.
An unknown mass is hung from a very light spring with a spring constant of 13.6 N/m. Neglect the mass of the spring. If the period of oscillation of the mass-spring system is 3.35 s, then the unknown mass is:__________
Answer:
The unknown mass is 3.87 kg.
Explanation:
Given;
spring constant of the spring, k = 13.6 N/m
period of oscillation, T = 3.35 s
The period of oscillation of the mass-spring system is given by;
[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]
where;
m is the mass attached to the spring
[tex]T = 2\pi \sqrt{\frac{m}{k} }\\\\\frac{T}{2\pi} = \sqrt{\frac{m}{k} }\\\\\frac{T^2}{4\pi^2}= \frac{m}{k}\\\\m = \frac{kT^2}{4\pi^2}\\\\m = \frac{(13.6)(3.35)^2}{4\pi^2}\\\\m = 3.87 \ kg[/tex]
Therefore, the unknown mass is 3.87 kg.
the nitrogen group has 5 electrons true or false
Answer:
True
Nitrogen family elements consists of atoms having 5 electrons in their outer energy level.
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Hope it may helps you
The Sputnik 1 satellite orbited Earth
(mass = 5.98 x 1024 kg) in a circle of
radius 6.96 x 106 m. What was its
period of orbit in seconds?
[?] s
Answer:
5778.1931
Explanation:
Given that:
Mass (M) = 5.98 x 10^24 kg
Radius (r) = 6.96 x 10^6 m
Gravitational constant (G) = 6.67 * 10^-11
Using the relation:
T² / R³ = 4π² / GM
Where T = period of orbit
T = 2π√(R³ /GM)
T = 2π√((6.96 x 10^6)³) /(6.67 * 10^-11) * (5.98 x 10^24)
T = 2π√((3.37153 * 10^20) / (39.866 * 10^13))
T = 2π √845715.64
T = 2π * 919.62799
T = 5778.1931
