Answer:
pagtalon?
Explanation:
because that had to be the answer
Given equal acceleration, which has
most force?
Mosquito
Dog
Horse
Elephant
A small mirror is attached to a vertical wall, and it hangs a distance of 1.87 m above the floor. The mirror is facing due east, and a ray of sunlight strikes the mirror early in the morning and then again later in the morning. The incident and reflected rays lie in a plane that is perpendicular to both the wall and the floor. Early in the morning, the reflected ray strikes the floor at a distance of 3.56 m from the base of the wall. Later on in the morning, the ray is observed to strike the floor at a distance of 1.46 m from the wall. The earth rotates at a rate of 15.0o per hour. How much time (in hours) has elapsed between the two observations
Answer:
t = 1.62 h
Explanation:
A flat mirror fulfills the law of reflection where the incident angle is equal to the reflected angle.
θ_i = θ_r
If we use trigonometry to find the angles, the mirror is at a height of L = 1.87 m, and the reflected rays reach a distance x1 = 3.56 m
tan θ₁ = x₁ / L
tan θ₁ = [tex]\frac{3.56}{1.87}[/tex]
θ₁ = tan⁻¹ 1.90
θ₁ = 62.29º
for the second case x₂ = 1.46 m
tan θ₂ = x₂ / L
θ₂ = tan⁻¹ [tex]\frac{1.46}{1.87}[/tex]
θ₂ = 37.98º
the difference in degree traveled is
Δθ = θ₁- θ₂
Δθ = 62.29 - 37.98
Δθ = 24.31º
as in the exercise they indicate that every 15º there is an hour
t = 24.31º (1h / 15º)
t = 1.62 h
A 120-gram toy airplane flies in a straight line at a speed of 1.3 m/s. How much kinetic energy does the airplane have?
Answer:
0.1014 J
Explanation:
This is the answer if the toy airplane was 120-GRAMS
if you meant KG then it's 101.4 J
In xray machines, electrons are subjected to electric fields as great as 6.0 x 10^5 N/C. Find
an electron's acceleration in this field.
Answer:
a = 1.055 x 10¹⁷ m/s²
Explanation:
First, we will find the force on electron:
[tex]E = \frac{F}{q}\\\\F = Eq\\[/tex]
where,
F = Force = ?
E = Electric Field = 6 x 10⁵ N/C
q = charge on electron = 1.6 x 10⁻¹⁹ C
Therefore,
[tex]F = (6\ x\ 10^5\ N/C)(1.6\ x\ 10^{-19}\ C)\\[/tex]
F = 9.6 x 10⁻¹⁴ N
Now, we will calculate the acceleration using Newton's Second Law:
[tex]F = ma\\a = \frac{F}{m}\\[/tex]
where,
a = acceleration = ?
m = mass of electron = 9.1 x 10⁻³¹ kg
therefore,
[tex]a = \frac{9.6\ x\ 10^{-14}\ N}{9.1\ x\ 10^{-31}\ kg}\\\\[/tex]
a = 1.055 x 10¹⁷ m/s²
A vertical straight wire 35.0 cmcm in length carries a current. You do not know either the magnitude of the current or whether the current is moving upward or downward. If there is a uniform horizontal magnetic field of 0.0300 TT that points due north, the wire experiences a horizontal magnetic force to the west of 0.0180 NN. Find the magnitude of the current.
Answer:
[tex]1.714\ \text{A}[/tex]
Explanation:
F = Magnetic force = 0.018 N
B = Magnetic field = 0.03 T
L = Length of wire = 35 cm
[tex]\theta[/tex] = Angle between current and magnetic field = [tex]90^{\circ}[/tex]
Magnetic force is given by
[tex]F=IBL\sin\theta\\\Rightarrow I=\dfrac{F}{BL\sin\theta}\\\Rightarrow I=\dfrac{0.018}{0.03\times 35\times 10^{-2}\times \sin90^{\circ}}\\\Rightarrow I=1.714\ \text{A}[/tex]
The magnitude of the current is [tex]1.714\ \text{A}[/tex].
ocean currents are always cold true or false
True or False: An electric circuit has to be complete for the load to
receive electricity.
To understand the meaning and possible applications of the work-energy theorem. In this problem, you will use your prior knowledge to derive one of the most important relationships in mechanics: the work-energy theorem. We will start with a special case: a particle of mass m moving in the x direction at constant acceleration a. During a certain interval of time, the particle accelerates from vi to vf, undergoing displacement s given by s=xf−xi.
A. Find the acceleration a of the particle.
B. Evaluate the integral W = integarvf,vi mudu.
Answer:
a) the acceleration of the particle is ( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) / 2as
b) the integral W = [tex]\frac{1}{2}[/tex]m( [tex]_f[/tex]² - v[tex]_i[/tex]² )
Explanation:
Given the data in the question;
force on particle F = ma
displacement s = x[tex]_f[/tex] - x[tex]_i[/tex]
work done on the particle W = Fs = mas
we know that; change in energy = work done { work energy theorem }
[tex]\frac{1}{2}[/tex]m( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) = mas
[tex]\frac{1}{2}[/tex]( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) = as
( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) = 2as
a = ( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) / 2as
Therefore, the acceleration of the particle is ( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) / 2as
b) Evaluate the integral W = [tex]\int\limits^{v_{f} }_{v_{i} } mvdv[/tex]
[tex]W = \int\limits^{v_{f} }_{v_{i} } mvdv[/tex]
[tex]W =m[\frac{v^{2} }{2} ]^{vf}_{vi}[/tex]
W = [tex]\frac{1}{2}[/tex]m( [tex]_f[/tex]² - v[tex]_i[/tex]² )
Therefore, the integral W = [tex]\frac{1}{2}[/tex]m( [tex]_f[/tex]² - v[tex]_i[/tex]² )
True or false? A system must contain more than one object.
Answer:
true
Explanation:
normally -No system has ever performed well with one object.
A system must contain more than one object is a true statement.
What is system?A system is a group of interacting or interrelated objects that act according to a set of rules to form a unified whole.
Normally, no system has ever performed well with one object.
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ASAP 20 POINTS!!
The air also contained a small amount of argon
As the temperature of the air decreased from 20C to -190 C the argon changed
Explain the changes in arrangement and movement of the particles of the argon as the temperature of the air decreased
Answer:
See explanation
Explanation:
Let us recall that temperature is a measure of the average kinetic energy of the molecules of a body.The higher the temperature, the higher the kinetic energy of the molecules of the body.
As temperature decreases, the kinetic energy of the molecules of a substance also decreases rapidly and the magnitude of intermolecular interaction between molecules of the substance increases.
Hence, as argon gas is cooled from 20°C to -190°C the kinetic energy of the gas molecules decreases an the magnitude of intermolecular interaction increases hence the gas changes into liquid and subsequently changes into a solid at -190°C.
pls helpone phyics question lots of points!
Answer:
PE = 0.73J
Explanation:
Remember that in conservative spring systems,
Total energy = potential + kinetic energy.
On the y-axis lies the kinetic energy and the question asks for the potential energy.
PE + KE must always equal the same result.
In this case, KE + PE = 1
So rearranging the equation,
PE = 1 - KE
KE = 0.27 (as we can see from the graph)
Therefore,
PE = 1 - 0.27 = 0.73J
Bonus tip: The graphs of potential and kinetic energy will look the exact opposite in this case. When the KE graph is at 0J, the PE graph is at 1J and vice versa. And they always cross over at 0.5J
A 0.86kg grenade is tossed on the ground with a velocity of 6 m/s West. If the grenade explodes into 2 pieces,
one that has a mass of 32kg and travels East at 10 m/s.
(A) What is the mass of the second piece
(B) What is the velocity of the second piece?
Answer:
(A) 0.54 kg
(B) 15.5 m/s west
Explanation:
Mass is conserved.
M = m₁ + m₂
0.86 kg = 0.32 kg + m₂
m₂ = 0.54 kg
Momentum is conserved (take east to be positive).
Mv = m₁v₁ + m₂v₂
(0.86 kg)(-6 m/s) = (0.32 kg)(10 m/s) + (0.54 kg) v₂
v₂ = -15.5 m/s
. If block A has a velocity of 0.6 m/s to the right, determine the velocity of cylinder
Answer:
As we can see, a string is attached with block A, and three string is folded with ply which is attached with B
x
B
=3x
A
Now differentiate with respect to x
V
B
=3V
A
Given,
V
A
=0.6m/s(totheright)
So,
V
B
=0.6×3
=1.8m/s(downward)
Explanation:
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A 0.100 kg limestone cube is released from rest, and proceeds to slide down a frictionless ramp. At the bottom of the ramp, the limestone cube makes an elastic collision with a steel cube whose mass is 0.200 kg, which is initially at rest. At what vertical height should the limestone cube be placed such that the steel cube has a velocity of 1.50 m/s after the collision
Answer:
the height at which the limestone cube must be placed is 0.23 m.
Explanation:
Given;
mass of limestone cube, m₁ = 0.1 kg
initial velocity of the limestone cube, u₁ = 0
mass of steel cube, m₂ = 0.2 kg
initial velocity of the steel cube, u₂ = 0
final velocity of the steel cube, v₂ = 1.5 m/s
Apply the principle of conservation of energy to determine the height of the limestone cube.
Potential energy of the limestone cube at top = Kinetic energy of steel cube at base
m₁gh = ¹/₂m₂v₂²
where;
h is the height at which the limestone cube is placed
[tex]h = \frac{m_2v_2^2}{2m_1g} \\\\h = \frac{0.2\times 1.5 ^2}{2 \times 0.1 \times 9.8} \\\\h =0.23 \ m[/tex]
Therefore, the height at which the limestone cube must be placed is 0.23 m.
Consider a long, thin rod with a length of 3 m rotating about it's end. This rod has a moment of inertia of 12 kg·m2 about this pivot.
What is the mass of the rod? Give your answer in kilograms to two decimal places.
Answer:
The mass of the rod is 16 kg.
Explanation:
Given that,
The length of a rod, L = 3 m
The moment of inertia of the rod, I = 12 kg-m²
We need to find the mass of the rod. The moment of inertia of the rod of length L is given by :
[tex]I=\dfrac{ML^2}{12}[/tex]
Where
M is mass of the rod
[tex]M=\dfrac{12I}{L^2}\\\\M=\dfrac{12\times 12}{(3)^2}\\\\M=16\ kg[/tex]
So, the mass of the rod is 16 kg.
When the disks collide and stick together, their temperature rises. Calculate the increase in internal energy of the disks, assuming that the process is so fast that there is insufficient time for there to be much transfer of energy to the ice due to a temperature difference. (Also ignore the small amount of energy radiated away as sound produced in the collisions between the disks.)
Answer:
ΔT = [tex]\frac{\Delta K}{(m_1+m_2) c_e }[/tex]
Explanation:
This is an interesting problem, no data is given, so the result is a general expression.
Suppose that the disks are initially rotating with angular velocity w₁ and w₂, as well as that they have radii r₁ and r₂ and masses m₁ and m₂
we start the problem finding odl final angular velocity of the discs together, for this we define a system formed by the two discs, in this case the torques during the collision are internal and the angular momentum is conserved
initial instant. Just before the crash
L₀ = L₁ + L₂
with
L₁ = I₁ w₁
the moment of inertia of a disc with an axis passing through its center is
I₁ = ½ m₁ r₁²
we substitute
I₀ = ½ m₁ r₁² w₁ + ½ m₂ r₂² w₂
final instant. Right after the crash
L_f = I w
in angular momentum it is a scalar quantity, so it is additive
I = I₁ + I₂
angular momentum is conserved
L₀ = L_f
I₁ w₁ + I₂ w₂ = I w
w = [tex]\frac{ I_1 w_1 + I_2 w_2 }{I}[/tex] (1)
We already have the angular velocities of the system, let's find the kinetic energy of it
initial
K₀ = K₁ + K₂ = ½ I₁ w₁² + ½ I₂ w₂²
final
K_f = K = ½ I w²
the variation of the kinetic energy is the loss in the increase of the temperature of the system, they indicate us that we neglect the other possible losses
ΔK = K_f -K₀
ΔK = ½ I w² - (½ I₁ w₁² + ½ I₂ w₂²) (2)
In this chaos we know all the values for which the numerical value of ΔK can be calculated, the symbolic substitution gives expressions with complicated
Now if all this variation of energy turns into heat
Q = ΔK
m_{total} c_e ΔT = ΔK
where the specific heat of the bear discs must be known, suppose they are of the same material
ΔT = [tex]\frac{\Delta K}{(m_1+m_2) c_e }[/tex] (3)
to make a special case, we suppose some data
the discs have the same mass and radius, disc 2 is initially at rest and the discs are made of bronze that has c_e = 380 J / kg ºC
we look for the angular velocity
I₁ = I₂ = I₀
I = 2 I₀
we substitute in 1
w = [tex]\frac{I_o w_1 + I_o 0 }{2I_o}[/tex] I₀ w₁ + I₀ 0 / 2Io
w = w₁ /2
we look for the variation of the kinetic energy with 2
ΔK = ½ (2I₀) (w₁ /2)² - (½ I₀ w₁² + ½ I₀ 0)
ΔK = ¼ I₀ w₁² -½ I₀ w₁²
ΔK = - ¼ I₀ w₁²
the negative sign indicates that the kinetic energy decreases
We look for the change in Temperature with the expression 3
ΔT = [tex]\frac{ \Delta K}{(m_1 +m_2) c_e}[/tex]ΔK / (m1 + m2) ce
ΔT = [tex]\frac{ \frac{1}{4} I_o w_1^2 }{ 2m c_e}[/tex]
ΔT = [tex]\frac{1}{8} \frac{ (\frac{1}{2} m r_1^2 ) w_1^2 }{ m c_e}[/tex]
ΔT = [tex]\frac{1}{16} r_1^2 w_1^2 / c_e[/tex]
in this expression all the terms are contained
The increase in internal energy of the disks will be [tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex].
What is internal energy?The energy contained within a thermodynamic system is known as its internal energy. It's the amount of energy required to build or prepare a system in any given internal state.
The given data in the problem is;
[tex]\rm \omega_1[/tex] is the angular velocity of disk 1
[tex]\rm \omega_2[/tex] is the angular velocity of disk 2
r₁ is the radius of disk 1
r₂ is the radius of disk 2
m₁ is the mass of disk 1
m₂ is the mass of disk 2
Momentum before the collision;
[tex]\rm L_1 = I_1 \omega_1[/tex]
The moment of inertia of disc 1
[tex]\rm i_1 = \frac{1}{2} m_1r_1^2[/tex]
The momentum gets conserved;
[tex]\rm L_0 = L_f \\\\ I_1 \omega_1 + I_2\omega_2 = I \omega \\\\ \rm \omega= \frac{I_1 \omega_1 + I_2\omega_2}{I}[/tex]
The change in the kinetic energy is;
[tex]\traingle KE= K_f - K_0 \\\\ \traingle KE= \frac{1}{2} I \omega^2-(\frac{1}{2} I_1\omega_1^2 + (\frac{1}{2} I_2\omega_2^2 )[/tex]
The change in the energy gets converted into heat;
[tex]\rm Q= \triangle k \\\\\ m_{total } c_e dt = \triangle k[/tex]
The change in the temperature is
[tex]\triangle T= \frac{\triangle k }{(m_1+m_2)c_e}[/tex]
The internal energy change is found by;
[tex]\rm \triangle E = mc_v dt[/tex]
[tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex]
Hence the increase in internal energy of the disks will be [tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex].
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Define the average acceleration of a particle
between two given instants.
PLEASE HELP ASAP!!!!
What is the kinetic energy of a 160 kg object that is moving at a speed of 20 m/s?
A. 3,200 J
B. 32,000 J
С. 180 J
D. 64,000 J
A piston-cylinder device initially contains 0.6 kg of water with a volume of 0.1 m3 . The mass of the piston is such that it maintains a constant pressure of 1000 kPa. The cylinder is connected through a valve to a supply line that carries steam at 5 MPa and 500 o C. Now the valve is opened and steam is allowed to flow slowly into the cylinder until the volume of the cylinder doubles and the temperature in the cylinder reaches 280 o C, at which point the valve is closed. The pressure remains constant during the process. Determine:
Answer: Hello the missing piece of your question is attached
question : Determine mass of steam that has entered ( in kg )
answer : 0.206 kg
Explanation:
V1 = 0.1 m^3 ,
v' = V1 / m1 = 0.1 / 0.6 = 0.167 m^3/kg
V2 = 0.2 m^3
using the steam tables
at ; P = 1000 kPa, v' = 0.167 m^3/kg
U1 = 2321 KJ/kg
at ; P = 1000 kPa , T2 = 280°C
v'2= 0.2481 m^3kg
U2 = 2760.6
at ; P = 5MPa , T = 500°C
h1 = 3434.7 KJ/Kg
calculate final mass ( m2 )
M2 = V2 / v'2
= 0.2 / 0.2481 = 0.806 kg
therefore the mass added = m2 - m1
= 0.806 - 0.6 = 0.206 kg
A twin-sized air mattress used for camping has dimensions of 100 cm by 194 cm by 14 cm when blown up. The weight of the mattress is 4 kg. How heavy a person (in N) could the air mattress hold if it is placed in freshwater
Answer:
[tex]2625.156\ \text{N}[/tex]
Explanation:
Dimensions of mattress 100 cm by 194 cm by 14 cm
[tex]m_m[/tex] = Mass of mattress = 4 kg
[tex]\rho[/tex] = Density of water = [tex]1000\ \text{kg/m}^3[/tex]
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Volume of mattress
[tex]V=100\times 194\times 14=271600\ \text{cm}^3=0.2716\ \text{m}^3[/tex]
Weight of water displaced is equal to the buoyant force
Mass of water
[tex]m=\rho V\\\Rightarrow m=1000\times 0.2716\\\Rightarrow m=271.6\ \text{kg}[/tex]
Mass of person would be
[tex]m_p=m-m_m=271.6-4\\\Rightarrow m_p=267.6\ \text{kg}[/tex]
Weight of the person would be
[tex]w=m_pg\\\Rightarrow w=267.6\times 9.81\\\Rightarrow w=2625.156\ \text{N}[/tex]
The air mattress could hold a person that weighs up to [tex]2625.156\ \text{N}[/tex].
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Calculate the amount of torque of an object being pushed by 6 N force along a circular path of a radius of 1x10^-2 mat 30 degree angle
Answer:
[tex]\tau=0.03\ N-m[/tex]
Explanation:
Given that,
Force acting, F = 6N
The radius of the path, [tex]r=10^{-2}\ m[/tex]
Angle, [tex]\theta=30^{\circ}[/tex]
We need to find the amount of torque acting on the object. The formula for torque is given by :
[tex]\tau=Fr\sin\theta\\\\\tau=6\times 10^{-2}\times \sin(30)\\\\\tau=0.03\ N-m[/tex]
So, the required torque is equal to 0.03 N-m.
Energy is transferred between the ocean and the air to make sure that the temperature in the air is higher than the temperature on the surface False True
What are the two main processes carried out by the excretory system?
At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown straight upward from Earth's surface with an initial velocity of 19.6 m/s. They move along nearby lines and pass each other without colliding. When the second ball is at its highest point, what is the velocity of the center of mass of the two-ball system
Answer:
The velocity of the center of mass of the two-ball system is 13.1 m/s.
Explanation:
Given;
mass of the first ball, m₁ = 0.5 kg
mass of the second ball, m₂ = 0.25 kg
initial velocity of the second ball, u₂ = 19.6 m/s
At the highest point the velocity of the second ball, v₂ = 0
The highest point reached by the second ball is calculated as;
v₂² = u₂² - 2gh
0 = u₂² - 2gh
2gh = u₂²
h = u₂² / 2g
h = (19.6²) / (2 x 9.8)
h = 19.6 m
The final velocity of the first ball when it had traveled 19.6 m down;
v₁² = u₁² + 2gh
v₁² = 0 + 2gh
v₁ = √2gh
v₁ = √(2 x 9.8 x 19.6)
v₁ = 19.6 m/s
The velocity of the center of mass of the two-ball system is calculated as;
[tex]v = \frac{m_1v_1 \ + \ m_2v_2}{m_1 \ + \ m_2} \\\\v = \frac{0.5\times 19.6 \ + \ 0.25\times 0}{0.5 \ + \ 0.25} \\\\v = 13.1 \ m/s[/tex]
When a wave enters a new medium from an angle, both the speed and the ________ change
a
The frequency
b
The amplitude
c
The energy
d
The angle
Answer:
B: Amplitude
Explanation:
When a wave travels from one medium to the other from an angle, the things that change are amplitude, wavelength, intensity and velocity.
The frequency doesn't change because the frequency depends upon the source of the wave and not the medium by which the wave is propagated.
Answer:
The angle
Explanation:
A 4768-kg roller coaster train full of riders approaches
Answer: ?
Explanation:
What type of forces is acting on an object that prevents the object from moving?
use the Group of answer choices
a Balanced forces
b Friction forces
c Gravity forces
d Unbalanced forces
Answer: Friction Forces
Explanation: (I took the same test and got the answer)
_________________________________________
I hope this helps!
Suppose that you changed the area of the bottom surface of the friction cart without changing its mass, by replacing the Teflon slab with one that was smaller but thicker. The contact area would shrink, but the normal force would be the same as before. Would this change the friction force on the sliding cart
Answer:
in this case the weight of the vehicle does not change , consequently the friction force should not change
Explanation:
The friction force is a macroscopic manifestation of the interactions of the molecules between the two surfaces, this force in the case of solid is expressed by the relation
fr = μ N
W-N= 0
N = W
as in this case the weight of the vehicle does not change nor does the Normal one, consequently the friction force should not change
Object A is negatively charged. Object A and Object B
attract. Object B and Object C repel. Object C and Object
D repel. What type of charge does Object B, Object C, and
Object D possess?
Answer:
Malrpr00qpq9owoowopwiaahaulaqkkkala9asoLHahababajjajalls
Explanation:
hhoootyiñlf7ogffyiklmhf
Consider the following four objects: a hoop, a flat disk, a solid sphere, and a hollow sphere. Each of the objects has mass M and radius R. The axis of rotation passes through the center of each object, and is perpendicular to the plane of the hoop and the plane of the flat disk. Which of these objects requires the largest torque to give it the same angular acceleration?
a. the solid sphere.
b. the hollow sphere.
c. the hoop.
d. the flat disk.
e. both the solid and the hollow spheres.
Answer:
b) The hollow sphere.
Explanation:
The moment of Inertia of a solid rotating object is related with the torque applied and the angular acceleration caused by the torque, by the following relationship, that resembles Newton's 2nd Law for point masses:[tex]\tau = I * \alpha (1)[/tex]
As it can be seen, for a given angular acceleration α, the larger the moment of inertia, the larger the torque needed to give the object the same angular acceleration.From the proposed solids, the one that has the largest moment of Inertia, is the hollow sphere, which moment of Inertia can be written as follows:[tex]I_{hsph} = \frac{2}{3} *M*R^{2} (2)[/tex]
So, the hollow sphere requires the largest torque to give the object the same angular acceleration.