Answer:
When covalent compounds dissolve in water, they break apart into molecules, but not individual atoms. Water is a polar solvent, but covalent compounds are usually nonpolar. This means covalent compounds typically don't dissolve in water, instead, it makes a separate layer on the water's surface.
Explanation:
I don't understand the second part but hope this helps.
You carry a 20. N bag of dog food up a 6.0 m flight of stairs. How much work was done?
Answer:
Solution given:
force=20N
distance travelled=6m
work done=force *displacement=20*6=120joule
Work done=120joule.
Work done=Force×Displacement
[tex]\\ \rm\longmapsto Work\:done=20(6)[/tex]
[tex]\\ \rm\longmapsto Work\:done=120J[/tex]
Who do you simp for in Harry Potter?
Answer:
I don't rlly know anyone in harry Potter so I simp for........MYSELF :)
Answer:
Either Draco Malfoy or Harry Potter
Explanation:
UvU
what is this simple machines name
Answer:
bottle opener and door knob
Explanation:
Answer:
bottle opener
door handle
if a person weighs 140 lb on earth their weight in newtons is
Answer:
623.6N
Explanation:
There are 2.2 pounds in one kilogram, so you must divide 140 lb by 2.2 to convert the units to kilograms.
140/2.2 = 63.6 kg
The gravitational force also known as weight would equal mass times acceleration due to gravity.
Fg = (63.6 kg)(9.8 m/s^2) = 623.6N
help me pls. will give brainliest
Answer:
B: Relay
Explanation:
Out of all the options, the only device that relies upon the magnetic field of an electric current.
The principle relies upon the principle of electro magnetic attraction whereby when the circuit detects a faulty current, it makes the electromagnetic field to become energized which in turn produces the temporary magnetic field. This magnetic field then moves the relay armature which is used in opening or closing the connections.
joe loves joe so yes
Answer:
noice.
Explanation:
thanks for points.... idrk what im supposed to do
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yeah same.
Two dogs are pulling on a chew toy. One dog pulls the chew toy with 64 N [E] and
the other pulls the chew toy with 31 N [W]. Determine the ACCELERATION of the 3
kg chew toy.
Answer:
Eastward, at 11 m/s^2
Explanation:
64N-31N=unbalanced force of 33N
F=ma
33N=(3kg)a
a=11m/s^2 to the East
Sean applies a force of 100N to move a box 5 meters. How much work did he do?
Answer:
500 Joules
Explanation:
W(work)= force * distance
w = 100*5
W= 500
Choose the word that best completes each sentence.
Plants are
. They make their own food.
Animals are
. They consume other organisms for food.
will mark as brainlest pls help
Answer:
Explanation:
one and three seem right soo answer C
What is the kinetic energy of a 25 kg object moving at a velocity of 4 m/s?
Thus Kinetic energy of the object is 200J when mass is 25 kg and velocity is 4 m/s.
What are the five elements of fitness?
Answer:
5 Components of Physical Fitness
Cardiovascular Endurance.
Muscular Strength.
Muscular endurance.
Flexibility.
Body Composition.
Which element is represented by the model of the atom above?
Carbon is the element modeled
What is the kinetic energy of a 2.7kg book
moving at 3.5m/s?
Give your answer rounded to one place after the
decimal, no units.
!!
As the ball moves from point A to point B, kinetic energy...
Answer:
KE=1/2mv*
Explanation:
write the numbers inorder to the symbols
Ans use this formula to solve the problem
5. What is the velocity of a 0.5 kg ball that has a momentum of 3.00 kg m/s?
Answer:
6 m/s
Explanation:
p=mv
p=3 kg.m/s
m=0.5 kg
rearrange and substitute to find v:
(3)=(0.5)v
v=3/0.5
v=6 m/s
Please explain why
True or False
Burning paper is a formation of a precipitate
How would the force of gravity between two objects change if the mass of
each object were doubled?
A. The force between them would be 2 times larger than with the
initial masses.
B. The force between them would be 1/4 the force with the initial
separation.
C. The force between them would be 4 times larger than with the
initial masses.
D. The force between them would be 1/2 the force with the initial
separation
Answer:
Option C. The force between them would be 4 times larger than with the
initial masses.
Explanation:
To know which option is correct, we shall determine the force of attraction between the two masses when their masses are doubled. This can be obtained as follow:
From:
F = GMₐM₆/ r²
Keeping G/r² constant, we have
F₁ = MₐM₆
Let the initial mass of both objects to be m
F₁ = MₐM₆
F₁ = m × m
F₁ = m²
Next, let the masses of both objects doubles i.e 2m
F₂ = MₐM₆
F₂ = 2m × 2m
F₂ = 4m²
Compare the initial and final force
Initial force (F₁) = m²
Final (F₂) = 4m²
F₂ / F₁ = 4m² / m²
F₂ / F₁ = 4
F₂ = 4F₁ = 4m²
From the above illustrations, we can see that when the mass of both objects doubles, the force between them would be 4 times larger than with the
initial masses.
Thus, option C gives the correct answer to the question.
A car covers 400m distance in 50 sec find it's speed?
Answer:
[tex]\boxed {\boxed {\sf 8 \ m/s}}[/tex]
Explanation:
Speed is equal to distance over time.
[tex]s=\frac{d}{t}[/tex]
The distance is 400 meters and the time is 50 seconds.
[tex]d= 400 \ m \\t= 50 \ s[/tex]
Substitute the values into the formula.
[tex]s=\frac { 400 \ m }{ 50 \ s}[/tex]
Divide.
[tex]s= 8 \ m/s[/tex]
The speed of the car is 8 meters per second.
1.
X, Y and Z are the weights of three objects
suspended by pulleys as shown.
Assuming the pulleys in this system are frictionless
and weightless and that Z= 12 N, what are the
values of X and Y?
24°
50°
Х
Y
Z
Answer:
The weight of the object X is approximately 3.262 N (Acting downwards)
The weight of the object Y is approximately 8.733 N (Acting downwards)
Explanation:
The question can be answered based on the principle of equilibrium of forces
The given parameters are;
The weight of Z = 12 N (Acting downwards)
The weight of the pulleys = Negligible
From the diagram;
The tension in the in the string attached to object Z = The weight of object Z = 12 N
The tension in the in the string attached to object X = The weight of the object X
The tension in the in the string attached to object Y = The weight of the object Y
Given that the forces are in equilibrium, we have;
The sum of vertical forces acting at a point, [tex]\Sigma F_y[/tex] = 0
Therefore;
[tex]T_{1y} + T_{2y} + T_{3y} = 0[/tex]
[tex]T_{1y} = -( T_{2y} + T_{3y} )[/tex]
Where;
[tex]T_{1y}[/tex] = The weight of object Z = 12 N
[tex]T_{1y}[/tex] = 12 N
[tex]T_{2y}[/tex] = The vertical component of tension, T₂ = T₂ × sin(24°)
∴ [tex]T_{2y}[/tex] = T₂ × sin(156°)
Similarly;
[tex]T_{3y}[/tex] = T₃ × sin(50°)
From [tex]T_{1y} = -( T_{2y} + T_{3y} )[/tex], and [tex]T_{1y}[/tex] = 12 N, we have;
12 N = -(T₂ × sin(156°) + T₃ × sin(50°))...(1)
Given that the forces are in equilibrium, we also have that the sum of vertical forces acting at a point, ∑Fₓ = 0
Therefore at point B, we have;
T₁ₓ + T₂ₓ + T₃ₓ = 0
The tension force, T₁, only has a vertical component, therefore;
∴ T₁ₓ = 0
∴ T₂ₓ + T₃ₓ = 0
T₂ₓ = -T₃ₓ
T₂ₓ = T₂ × cos(156°)
T₃ₓ = T₃ × cos(50°)
From T₂ₓ = -T₃ₓ, we have;
T₂ × cos(156°) = - T₃ × cos(50°)...(2)
Making T₃ the subject of equation (1) and (2) gives;
Making T₃ the subject of equation in equation (1), we get;
12 = -(T₂ × sin(156°) + T₃ × sin(50°))
∴ T₃ = (-12 - T₂ × sin(156°))/(sin(50°))
Making T₃ the subject of equation in equation (2), we get;
T₂ × cos(156°) = - T₃ × cos(50°)
∴ T₃ = T₂ × cos(156°)/(-cos(50°))
Equating both values of T₃ gives;
(-12 - T₂ × sin(156°))/(sin(50°)) = T₂ × cos(156°)/(-cos(50°))
-12/(sin(50°)) = T₂ × cos(156°)/(-cos(50°)) + T₂ × sin(156°)/(sin(50°))
∴ T₂ = -12/(sin(50°))/((cos(156°)/(-cos(50°)) + sin(156°)/(sin(50°))) ≈ -8.02429905283
∴ T₂ ≈ -8.02 N
From T₃ = T₂ × cos(156°)/(-cos(50°)), we have;
T₃ = -8.02× cos(156°)/(-cos(50°)) = -11.3982199717
∴ T₃ ≈ -11.4 N
The weight of the object X = -T₂ × sin(156°)
∴ The weight of the object X ≈ -(-8.02 × sin(156°)) = 3.262 N
The weight of the object X ≈ 3.262 N (Acting downwards)
The weight of the object Y = -(T₃ × sin(50°))
∴ The weight of the object Y = -(-11.4 × sin(50°)) ≈ 8.733 N
The weight of the object Y ≈ 8.733 N (Acting downwards)
Two objects have a gravitational force of 800 N. If they are moved twice as far apart and the mass of one object doubles, what is the new gravitational force?
Answer:
400 N
Explanation:
From the question,
F = kmm'/r²........................ Equation 1
Where F = gravitation force, m and m' = mass 1 and mass 2 respectively, r = distance between the masses.
Given; F = 800 N
Substitute these values into equation 1
800 = kmm'/r².............. Equation 2
If the distance is doubled (2r), and one of the mass is doubled (2m), The new force is
F' = k(2m)(m')/(2r)²
F' = 2kmm'/4r²
F' = kmm'/2r²................. Equation 3
Comparing equation 2 and equation 3
F' = F/2............................ Equation 4
Substitute the value of F into equation 4
F' = 800/2
F' = 400 N
The parking brake on a 1200kg automobile has broken, and the vehicle has reached a momentum of 7800kg.M/s. What is the velocity of the vehicle?
Explain the changes in energy as a ball is held in the air , and then dropped
Answer:
Potential Energy to Kenetic Energy
Explanation:
When holding a ball in the air, the ball has potential energy. Once you drop the ball, the ball gains Kenetic Energy
In lab, a radiation detector was used to calculate the background radiation. The
readings taken every minute were recorded in the table below. What is the reason
that the readings are not the same every minute?
If a car is traveling 35 m/s for 200 seconds, how far will the car travel?
Answer:
7.14285714286
Explanation:
who invented Pascal??
4. What is the best estimate of the frequency of the wave shown below?
O A. 0.40 Hz
OB. 0.85 Hz
O C. 1.25 Hz
O D. 2.50 Hz
Answer:
The answer is "Option D"
Explanation:
Please find the graph file in the attached file.
Using formula:
[tex]\to v= \rho \lambda \\\\\ \to \lambda= 130 \cm \\\\ \to v= 335 \frac{cm}{s}\\\\ \to \rho= \frac{v}{\lambda} \\\\[/tex]
[tex]=\frac{335}{130} \\\\ = 2.57 \approx 2.50 \ Hz[/tex]
What is the acceleration of a 24 kg mass pushed by an 8 N force?
Answer:
0.25 m/s squared
Explanation:
Mass of the body (m)= 24kg
Applied force on the body(F)= 6N
Acceleration of the body (a)= ?
Solution
F = ma
a = F/m
a = 6N/24kg
a = 0.25m/s^2
Answer: The acceleration of the body is 0.25m/s^2
I hope this helps
Answer:
[tex]\boxed {\boxed {\sf a=\frac {1}{3} \ m/s^2}}[/tex]
Explanation:
Force is the product of mass and acceleration.
[tex]F=ma[/tex]
The mass is 24 kilograms and the force is 8 Newtons. Let's convert the Newtons to make the problem easier.
1 Newton is equal to 1 kilogram meter per square second. The force of 8 N equals 8 kg*m/s²Substitute the known values into the formula.
[tex]8 \ kg*m/s^2= (24 \ kg )*a[/tex]
Since we are solving for a, we must isolate the variable. It is being multiplied and the inverse of multiplication is division. Divide both sides by 24 kilograms.
[tex]\frac {8 \ kg*m/s^2}{24 \ kg}= \frac{(24 \ kg )*a}{24 \ kg}[/tex]
The kilograms will cancel, hence our earlier unit conversion.
[tex]\frac {8 \ m/s^2}{24 }= a[/tex]
[tex]\frac {1}{3} \ m/s^2=a[/tex]
The acceleration is 1/3 meter per square second.
What is the volume of the rock? A. 7 cm3
B. 12 cm3
C. 19 cm3
D. 5 cm3
The pressure was 50pa ,when a perpendicular force was applied on a surface area of 2.5m2.what is the force applied??????
Answer:
Force = 125 Newton
Explanation:
Given the following data;
Pressure = 50 pa
Area = 2.5 m²
To find the force applied;
Force = pressure * area
Substituting into the formula, we have;
Force = 50 * 2.5
Force = 125 Newton
