Answer:
a camera employs camera lens to firm some images.
Explanation:
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kinetic energy of an object is quadrupled, momentum will change by what factor?
Answer:
So as KE becomes 4times , Momentum will increase by 2 times.
The spring in the figure has a spring constant of 1400 N/m . It is compressed 17.0 cm , then launches a 200 g block. The horizontal surface is frictionless, but the block's coefficient of kinetic friction on the incline is 0.210.
What distance d does the block sail through the air?
Use the work-energy theorem to find the velocity of the block when it's released by the spring. The work done by the spring on the block as it's restored to equilibrium is
W = 1/2 kx ²
where k is the spring constant and x is the compression of the spring. So
W = 1/2 (1400 N/m) (0.170 m)² = 20.23 J
This is equal to the block's change in kinetic energy ∆K,
W = ∆K
and since it starts from rest, the initial K is zero, leaving us with
W = 1/2 mv ²
where m is the mass of the block and v is its speed, so that
20.23 J = 1/2 (0.200 kg) v ²
==> v ≈ 14.2 m/s
The block slides at this speed across the frictionless surface until it hits the incline which introduces friction.
First, you need to find the length of the incline. It forms a 45° angle, and the underlying 45°-45°-90° triangle has a hypotenuse of length √2 (2.0 m) ≈ 2.83 m.
Next, you need to find the total work done on the block as it slides up the incline. Use Newton's second law to examine the forces acting on the block during this phase:
• the net force acting on the block in the direction perpendicular to the incline is
∑ F = n - mg cos(45°) = 0
where n = mg cos(45°) ≈ 1.39 N is the magnitude of the normal force and mg cos(45°) ≈ 1.39 N is the perpendicular component of the block's weight;
• the net force acting on the block parallel to the surface is
∑ F = -f - mg sin(45°) = ma
where f = µn = 0.210n ≈ 0.291 N is the magnitude of kinetic friction, mg sin(45°) ≈ 1.39 N is the parallel component of the weight, and a is the acceleration of the block.
Only the parallel forces do work on the block, and this work is negative because friction and weight oppose the block's sliding up the incline. The total work done on the block is then
W = (-0.291 N - 1.39 N) (2.83 m) ≈ -4.74 J
Use the work-energy theorem again to find the block's new speed v at the top of the incline:
W = ∆K
==> -4.74 J = 1/2 (0.200 kg) v ² - 1/2 (0.200 kg) (14.2 m/s)²
==> v ≈ 12.4 m/s
And now this becomes a projectile problem. The block travels a horizontal distance x after being launched at an angle of 45° with initial speed 12.4 m/s after time t according to
x = (12.4 m/s) cos(45°) t
Its height y from the 2.0 m-high surface at time t is given by
y = (12.4 m/s) sin(45°) t - 1/2 gt ²
The block lands on the surface when y = 0, which occurs after t ≈ 1.79 s, at which point the block has covered a distance d ≈ 15.7 m.
The block sail through the air at the distance of "15.8 m"
Given:
Spring constant,
1400 N/mMass,
200 gBlock's coefficient,
0.210By using Work energy theorem, we get
→ [tex]W_{spring}+W_g+W_f = KE_f-KE_i[/tex]
By substituting the values, we get
→ [tex]\frac{1}{2}\times 1400\times (0.17)^2- (0.2\times 9.8\times 2)-(0.21\times 0.2\times \frac{9.8}{\sqrt{2} }\times \sqrt{2}\times 2 )= \frac{1}{2}\times 0.2\times V_f^2[/tex]
here,
[tex]V_f = 12.44 \ m/s[/tex]
→ [tex]d = \frac{V_f^2 Sin 2 \Theta}{g}[/tex]
[tex]= \frac{(12.44)^2 Sin 90^{\circ}}{9.8}[/tex]
[tex]= 15.8 \ m[/tex]
Thus the answer above is right.
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Which best describes velocity?
A. 15 miles an hour
B. 15 miles an hour going west.
C. Speeding up from 10 to 15 miles an hour.
D. Going 15 miles an hour for 2 hours.
Answer:
B. 15 miles an hour going west
CR Physical Science B (GP) 20-21 / 1 Motion
1. A distance-time graph indicates that an object travels 2 m in 2 s and then travels another 80 m during the next 40 s.
What is the average speed of the object?
O 4 m/s
O
2 m/s
O 8 m/s
10 m/s
Answer:
Average speed = total distance traveled / total time traveled
V = (2 + 80) / (2 + 40) = 82 / 42 = 2 m / s
What is air
A. A Buchner substance
B. A compound
C. An element
D. A mixture
Air is classified as a mixture. Option D is the correct answer.
Air is a combination of different gases, primarily nitrogen (about 78%), oxygen (about 21%), and small amounts of other gases such as carbon dioxide, argon, and trace elements. These gases are not chemically bonded to each other, but rather exist together in the same space. Option D is the correct answer.
In a mixture, the substances involved retain their individual properties and can be separated by physical means. This is true for air as well. The gases in air can be separated through processes like fractional distillation or filtration. It's important to note that air also contains other components such as water vapor, dust particles, and pollutants, which can vary in concentration depending on the location and environmental conditions. These components further contribute to the complex nature of air as a mixture.
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How high can water be theoretically lifted by a vacuum pump at sea level?
A) less than 10.3 m
B) 10.3 m
C) more than 10.3 m
Explanation:
everything can be found in the picture
How bullet train and a circuit breaker work on the magnet effect of current
An object produces a sound wave with a wavelength 75.0cm. if the speed of sound is 350.0m/s, the frequency of the sound is
Answer:
wavelength=75.0
speed of sound(v)=350 .0m/s
frequency(f)=?
we know,
v=f*wavelengh
350.0 =f*750
f. =350/75
=4.667
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planet smaller than earth but larger than mercury
Answer:
venus......................
2. When you increase your speed from 35 mph to 70 mph, the force of impact you experience will be increased: A. 2 times B. 4 times C. 5 times D. 7 times
When you increase your speed from 35 mph to 70 mph, the force of impact you experience will be increased 4 times.
Hence, the correct option is B.
The force of impact in a collision is directly proportional to the square of the velocity. When the speed is doubled (increased from 35 mph to 70 mph), the force of impact will increase by a factor of four ([tex]2^2[/tex]). This is a fundamental concept in physics known as the kinetic energy formula, which is given by:
Kinetic Energy = (1/2) * mass * [tex]velocity^2[/tex]
Since kinetic energy is directly proportional to the square of velocity, doubling the speed results in four times the kinetic energy and, consequently, four times the force of impact.
Therefore, When you increase your speed from 35 mph to 70 mph, the force of impact you experience will be increased 4 times.
Hence, the correct option is B.
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63 km/h is equal to...
O a) 27.5 m/s
b) 18 m/s
c) 1.75 m/s
d) 1.8 m/s
please hurry
Answer:
17.5 m/s
Explanation:
Refer to the attachment!~
an object falls freely from rest the total distance covered by it in 2s will be
Answer:
Distance, S = 19.6 meters
Explanation:
Given the following data;
Time = 2 seconds
We know that acceleration due to gravity is equal to 9.8 m/s².
Also, the initial velocity of the object is equal to zero because it's starting from rest.
To find the total distance covered by the object, we would use the second equation of motion;
[tex] S = ut + \frac {1}{2}at^{2}[/tex]
Where;
S represents the displacement or height measured in meters. u represents the initial velocity measured in meters per seconds. t represents the time measured in seconds. a represents acceleration measured in meters per seconds square.Substituting into the equation, we have;
[tex] S = 0*2 + \frac {1}{2}*(9.8)*2^{2}[/tex]
[tex] S = 0 + 4.9*4[/tex]
[tex] S = 4.9*4 [/tex]
Distance, S = 19.6 meters
Therefore, the total distance covered by the object is 19.6 meters.
What is the change in internal energy if 70 J of heat is added to a system and
the system does 30 J of work on the surroundings. Uze al-Q-W.
O A. 40 J
O B. -40.3
O C. 100.
D. -1003
Answer:
A. 40 J
Explanation:
Given;
heat added to the system, Q = 70 J
work done by the system, W = 30 J
The change in the internal energy of the system is calculated using the first law of thermodynamic as shown below;
ΔU = Q - W
ΔU = 70 J - 30 J
ΔU = 40 J
Therefore, the change in the internal energy of the system is 40J
Which of these hazmat products are allowed in your FC?
Please choose all that apply.
A GPS unit (lithium batteries)
A subwoofer (magnetized materials)
A can of hairspray (flammable/aerosols)
Fireworks (explosives)
Answer: Hazmat products are allowed in your FC are:
A GPS unit (lithium batteries) A subwoofer (magnetized materials)Explanation:
Hazmat products consist of flammable, corrosive and harmful substances which are actually very hazardous to human health and environment.
Hazardous material allowed in FC are as follows.
Magnetized material products like as speakers.Non-spillable battery products like toy cars.Lithium-ion battery containing products like laptops, mobile phones etc.Non-flammable aerosol.So, hazmat allowed products are GPS unit (lithium batteries) and subwoofer (magnetized materials).
Thus, we can conclude that hazmat products are allowed in your FC are:
A GPS unit (lithium batteries) A subwoofer (magnetized materials)The first and second coils have the same length, and the third and fourth coils have the same length. They differ only in the cross-sectional area. According to theory, what should be the ratio of the resistance of the second coil to the first coil and the fourth coil to the third
Answer:
The ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of second coil.
And
The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of fourth coil.
Explanation:
The resistance of the coil is directly proportional to the length of the coil and inversely proportional to the area of coil and hence inversely proportional to the square of radius of the coil.
So, the ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of second coil.
And
The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of fourth coil.
ADvantage of friction
Answer:
1. Friction enables us to walk freely.
2. It helps to support ladder against wall.
3. It becomes possible to transfer one form of energy to another.
4. Objects can be piled up without slipping.
The brachialis attaches to the forearm .035 m from the elbow at an angle of 32 deg. If the brachialis produces 750 N of force, what is the magnitude of the stabilizing component of the brachialis force
Answer:
[tex]XY=636N[/tex]
Explanation:
From the question we are told that:
Distance [tex]d=0.35m[/tex]
Angle [tex]\theta=32\textdegree[/tex]
Force [tex]F=750N[/tex]
Generally the equation for magnitude of the stabilizing component of the brachialis force is mathematically given by
[tex]XY=Fcos\theta[/tex]
[tex]XY=750cos 32\textdegree[/tex]
[tex]XY=636N[/tex]
The extension produced on the wire by the application of 300 N force is 2 mm. The elastic energy stored on the wire is
Answer:
150000 is the answer to the question
Guys please help me
Answer:
1)[tex]t=2.26\: s[/tex]
2)[tex]S=33.9\: m[/tex]
3)[tex]v=26.77\: m/s[/tex]
4)[tex]\alpha=55.92[/tex]
Explanation:
1)
We can use the following equation:
[tex]y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}[/tex]
Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.
[tex]0=25-0.5*9.81*t^{2}[/tex]
[tex]t=2.26\: s[/tex]
2)
The equation of the motion in the x-direction is:
[tex]v_{ix}=\frac{S}{t}[/tex]
[tex]15=\frac{S}{2.26}[/tex]
[tex]S=33.9\: m[/tex]
3)
The velocity in the y-direction of the stone will be:
[tex]v_{fy}=v_{iy}-gt[/tex]
[tex]v_{fy}=0-(9.81*2.26)[/tex]
[tex]v_{fy}=-22.17\: m/s[/tex]
Now, the velocity in the x-direction is 15 m/s then the velocity will be:
[tex]v=\sqrt{v_{x}^{2}+v_{fy}^{2}}=\sqrt{15^{2}+(-22.17)^{2}}[/tex]
[tex]v=26.77\: m/s[/tex]
4)
The angle of this velocity is:
[tex]tan(\alpha)=\frac{22.17}{15}[/tex]
[tex]\alpha=tan^{-1}(\frac{22.17}{15})[/tex]
[tex]\alpha=55.92[/tex]
Then α=55.92° negative from the x-direction.
I hope it helps you!
44. Water from a fire hose knocks over a wooden shed. Compared with the pressure within the water, the pressure exerted against the shed is
a) less.
b) the same.
c) more.
d) nonexistent.
12) A negatively-charged balloon touching a wooden wall
A) pulls positive charge on the wall surface toward it.
B) pushes negative charge in the wall away from it.
C) polarizes molecules in the wall.
D) all of the above.
Answer:
D) all of the above.
Explanation:
First polarises it, cahrging and discharging occurs at once.
A negatively charged balloon touching a wooden wall then from the given options option D is correct which is all of the above.
What is a charge?Charged matter experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. Positive or negative charges can exist in an electric field (commonly carried by protons and electrons, respectively).
Contrary charges attract one another, while like charges repel one another. A neutral object is one that carries no net charge. Classical electrodynamics, the name given to an early understanding of how charged particles interact, is still accurate for issues that do not call for taking into account quantum phenomena.
In the first step it polarizes molecules in the wall, then charging and discharging in the wall will take place at once.
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concave mirror daily application
Answer:
concave mirror use in headlights and torches
Explanation:
Concave mirrors are used in headlights and torches. The shaving mirrors are also concave in nature since these mirrors can produce enlarged clear images. Doctors use concave mirrors as head mirrors to have a clearer view of eyes, noses, and ears. The dental mirrors used by dentists are also concave.
Một mol khí đang ở điều kiện tiêu chuẩn thì bị nén váo một bình 5 lít. Nhiệt độ khí trong bình là 770 C. Tính áp suất khí?
Đáp án:
p=5,74atm
Giải thích các bước giải:
áp suất khí là:
pV=nRT⇒p=nRTV=1.0,082.(77+273)5=5,74atmpV=nRT⇒p=nRTV=1.0,082.(77+273)5=5,74atm
p=5,74atm
Giải thích các bước giải:
áp suất khí là:
pV=nRT⇒p=nRTV=1.0,082.(77+273)5=5,74atmpV=nRT⇒p=nRTV=1.0,082.(77+273)5=5,74atm
A jet is circling an airport control tower at a distance of 15.8 km. An observer in the tower watches the jet cross in front of the moon. As seen from the tower, the moon subtends an angle of 9.21x10-3 radians. Find the distance traveled (in meters) by the jet as the observer watches the nose of the jet cross from one side of the moon to the other.
Answer:
The Distance traveled (in meters) by the jet as the observer watches the nose of the jet cross from one side of the moon to the other.
=145.5m
Explanation:
The radius of the circling jet r = 15.8km = 15.8*10^3 m
the angle subtended by moon
θ = 9.21*10^-3 rad
Therefore, the distance traveled by jet is
s = rθ
= (15.8*10^3 m)(9.21*10^-3 )
= (15,800)(0.00921)
=145.5m
If you wanted to know how much the temperature of a particular piece of material would rise when a known amount of heat was added to it, which of the following quantities would be most helpful to know?
a. coefficient of linear expansion
b. specific heat
c. initial temperature
d. thermal conductivity
e. density
Answer:
Option (b) is correct.
Explanation:
The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree C, is called specific heat of the substance.
The formula of the specific heat is
H = m c (T' - T)
where, m is the mass, c is the specific heat and T' - T is the change in temperature.
So, to know the rise in temperature, by adding the known amount of heat, the specific heat is required.
So, option (b) is correct.
hoW DO YOU SPELL THIS?
Answer:
Spell what?
Can u please mention that too?
Answer:
what
Explanation: how do you spell what please include this in your next post
A temperature of 200 degrees Fahrenheit is equivalent to approximately A.93.3 degrees Celsius B. 232 degrees Celsius C. 37.8 degrees Celsius D. 840 degrees Celsius
Answer:
you can use G.oogle for this question.
Answer:
93.3 degrees Celsius.
Explanation:
A uniform electric field is oriented in the −x direction. The magnitude of the electric field is 6500 N/C. How will the equipotential surfaces associated with this electric field be oriented?
A 2.5-kg rock is dropped off a 32-m cliff and hits a spring, compressing it 57 cm. What is the spring constant? Round your answer to two significant figures.
The spring constant, k, is
StartFraction N over m EndFraction.
Answer: 4800 N/m
Explanation:
Given
mass of rock [tex]m=2.5\ kg[/tex]
Height of cliff [tex]h=32\ m[/tex]
compression in the spring [tex]x=57\ cm[/tex]
Here, potential energy is converted into kinetic energy which in turn converts to elastic potential energy of the spring
[tex]\Rightarrow mgh=\dfrac{1}{2}kx^2\\\\\Rightarrow k=\dfrac{2mgh}{x^2}\\\\\Rightarrow k=\dfrac{1568}{0.3249}\\\\\Rightarrow k=4826.100\approx 4800\ N/m[/tex]
Answer:
480
Explanation:
2021