Write down the condition required for the thermonuclear fusion.
Answer:
The main condition necessary for a controlled thermonuclear fusion:
The ions must be held together in close proximity at high temperature with a confinement time long enough to avoid cooling.
A 2.98-kg object oscillates on a spring with an amplitude of 8.05 cm. Its maximum acceleration is 3.55 m/s2. Calculate the total energy.
Answer:
a = ω^2 A formula for max acceleration (ignoring sign)
V = ω A formula for max velocity
V^2 = ω^2 A^2 = a A from first equation
E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J
(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule
What is the unit of frequency
Answer:
hertz (Hz)
The number of periods or cycles per second is called frequency. The SI unit for frequency is the hertz (Hz).
Answer: the answer is hertz! if you’re on plato it’s option C. have a nice day! :)
Explanation:
What are the characteristics of high energy wave?
A. Low frequencies and short wavelengths.
B. High frequencies and long wavelengths.
C. Low frequencies and long wavelengths.
D. High frequencies and short wavelengths
Answer:
D. High frequency and short wavelengths.
Explanation:
If a wave is high in energy it will have a higher frequency.
High frequency = short wavelengths
A crane is lifting a 500 lb piano with a velocity of 2fts''. Express the power of the crane in hp. (1 hp = 550ft-lbs)
Answer:
goalindia
goalindiaGiven:
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kg
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 W
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg⇒ v = 1357.72/(226.8 × 9.8)
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg⇒ v = 1357.72/(226.8 × 9.8)⇒ v = 1357.72/2222.64
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg⇒ v = 1357.72/(226.8 × 9.8)⇒ v = 1357.72/2222.64⇒ v = 0.61 m/s
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg⇒ v = 1357.72/(226.8 × 9.8)⇒ v = 1357.72/2222.64⇒ v = 0.61 m/s- So, the velocity of the car will be 0.61❤
The spring constant of Spring A is twice as great as the spring constant of Spring B. Both springs are stretched the same amount. How does the
force the Spring A applies compare to the force Spring B applies?
Answer:
FA = 2FB
Force on spring A is twice the Force on spring B
Explanation:
F = kx
FB = (kB)x
FA = (kA)x
FA= (2kB)x
FA = 2(kB)x
FA = 2FB
The force [tex]F_A[/tex] needed to stretch spring A is going to be twice as much as the force [tex]F_B[/tex] needed to stretch spring B.
Explanation:
We know that the spring constants are related as
[tex]k_A = 2k_B[/tex]
The force [tex]F_A[/tex] needed to stretch spring A is given by
[tex]F_A = -k_Ax[/tex]
Also, the force [tex]F_B[/tex] needed to stretch spring is
[tex]F_B = -k_Bx[/tex]
Taking the ratio of the forces, we get
[tex]\dfrac{F_A}{F_B} = \dfrac{-k_Ax}{-k_Bx} = \dfrac{k_A}{k_B}[/tex]
Since [tex]k_A = 2k_B,[/tex] the equation above becomes
[tex]\dfrac{F_A}{F_B} = \dfrac{2k_B}{k_B} = 2[/tex]
or
[tex]F_A = 2F_B[/tex]
This shows that since the spring constant of spring A is twice as large as that of spring B, the force needed is going to be twice as large.
A guitar string 63.6 cm long vibrates with a standing wave that has five antinodes. Which harmonic is this
Answer:
fifth harmonic
Explanation:
A device is rated at 1.3kW when connected to a 120 V source. The equivalent resistance of this device in ohm is:
a- 18.3
b- 12.0
c- 11.1
d- 14.4
Answer:
D.
correct me if im wrong
brainlest plsss<333
How much energy is needed to change the temperature of 50g of water 15°c
Explanation:
This question is not feasible. There is no way to calculate the energy needed because the question is missing the final temperature
25.Figure 22.22 shows a plot
of potential versus position
along the x-axis. Make a plot
of the x-component of the
electric field for this situation.
The relationship between the potential and the electric field allows to find the results for the value of the electric field as a function of the distance is:
In the attachment we see the graph of the electric field as a function of distance.
Electric potential is defined by the change in potential energy of a test charge between two points, between the value of the test charge.
dV = - E . ds
E = [tex]- \frac{dV}{ds} \ \hat s[/tex]
Where the bold letters indicate vectors, V is the potential difference, E the electric field and s the path.
Let's apply this expression for each section of the given graph:
1) section from x₀ = 0 to x_f = 2 m, the potential is V₀ = 2 V is constant.
The derivative of a constant is zero.
E = 0
2) Section between x₀ = 2 and x_f = 4 m, the potential varies linearly from V₀ = 2 v to V_f = -2 V.
We look for the equation of the line.
V-V₀ = m (x- x₀)
We carry out the derivative.
E = - m i ^
The slope (m) is:
[tex]m= \frac{V_f - V_o}{x_f- x_o}[/tex]
Let's calculate.
[tex]m= \frac{-2 -2}{4-2} = \ -2 \ V/m[/tex]
Let's substitute.
E = [tex]2 \hat i \ V/m[/tex]
3) From x₀ = 4 to x_f = 4.5 m, the potential varies from V₀ = -2 to V_f = 0.
We look for the equation of the line and we derive.
E = - m i ^
Let's substitute.
[tex]m = \frac{0-(-2)}{4.5-4} = \ 4 V/m[/tex]
E = - 4 [tex]\hat i[/tex] V / m
4) From x₀ = 4.5 m to x_f = 6m. The potential is constant and the derivative of a constant is zero.
E = 0
5) From x₀ = 6m to x_f = 8 m, the potential changes linearly from v₀ = 0 to V_f = 1 V
We look for the equation of the line and we derive.
E = - m i ^
[tex]m = \frac{1-0}{8-6} = \ 0.5 \ V/m[/tex]
E = - 0.5 [tex]\hat i[/tex] V/m
6) From x₀ = 8m to x_f = 9m, the potential changes linearly from V₀ = 1 V to V_f = -1.
We look for the equation of the line and we derive.
E = - m i ^
[tex]m = \frac{-1-1}{9-8} = \ -2 \ V/m[/tex]
Let's substitute.
E = 2 [tex]\hat i[/tex] V/m
7) From x₀ = 9m to x_f = 10 m, the potential changes linearly from V₀ = -1 V to V_f = -2 V
We look for the equation of the line and we derive.
E = - m i ^
[tex]m = \frac{-2+1}{10-9} = \ -1 \ V/m[/tex]
Let's substitute.
E = 1 [tex]\hat i[/tex] V/m
In the attachment we can see these Electric fields as a function of distance.
In conclusion, the relationship between the potential and the electric field we can find the results for the value of the electric field as a function of the distance is:
In the attachment we see the graph of the electric field as a function of distance.
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A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 m in 5.00 s. (a) what is the mass of the block office? (b) If the worker stops pushing at the end of 5.00 s, how far does the block move in the next 5.00 s?
Answer:
(a) 91 kg (2 s.f.) (b) 22 m
Explanation:
Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.
(a)
[tex]s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}[/tex]
Subsequently,
[tex]F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})[/tex]
*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.
(b) To find the final velocity of the ice block at the end of the first 5 seconds,
[tex]v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}[/tex]
According to Newton's First Law which states objects will remain at rest
or in uniform motion (moving at constant velocity) unless acted upon by
an external force. Hence, the block of ice by the end of the first 5
seconds, experiences no acceleration (a = 0) but travels with a constant
velocity of 4.4 [tex]m \ s^{-1}[/tex].
[tex]s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}[/tex]
Therefore, the ice block traveled 22 m in the next 5 seconds after the
worker stops pushing it.
Explain how the removal of heat energy affects the speed of the particles in a substance
Answer:
The removal of heat energy slows the speed of particles
Explanation:
When you add heat to a substance, the heat energy gets transferred to kinetic energy, and the molecules began to move a greater distance at a greater speed. When you remove heat, the opposite happens.
The 0.01 kg marble is dropped from rest at A through the smooth glass tube and accumulate in the basket at C as shown in Figure Q2(b). Determine: i) the velocity of the marble at B ii) the horizontal distance R of the basket from the end of the tube, and iii) the speed at which the marble falls into the basket.
Crazy Wally Ok Ok ok hhahahaha
8.
When minerals seep into and replace the organic cells of a dead organism, creating a
mineral replica of the organism is called ...
Petrification
b. Tar Seeps
C. Amber
d. Mummification
Answer:
a. petrification
Explanation:
tar seeps = natural trade that, because of its close proximity to the ground surface, seeps from the cracks in the Earth or between rocks forming pits or pools (tar pits)
amber = fossilized resin produced by extinct coniferous trees, typically yellow in color
mummification = a process in which the skin and flesh of a corpse can be preserved by embalming and drying
5. Which statement about the acceleration of an object is corr
a. The acceleration of an object is directly proportional to t
external force acting on the object and inversely propo
mass of the object.
b. The acceleration of an object is directly proportional to
external force acting on the object and directly propor
mass of the object.
c. The acceleration of an object is inversely proportional
external force acting on the object and inversely prop
mass of the object.
d. The acceleration of an object is inversely proportiona
external force acting on the object and directly propo
mass of the object.
Answer:
a
Explanation:
you will find out that an object that has more weight requires more energy to move than the one with less weight
A change
1. Ricardo has purchased a forklift for his business. The forklift can put out 4950 W of power. If the forklift is operating at full capacity, how much work can it do in 2.40 seconds?
O 2060)
O 4950)
O 6490)
O 11,900
The work done if the forklift is operating at full capacity is 11,900 J.
We have to recall that power is defined as the rate of doing work. The rate of doing work is defined as;
Power = Work done/time taken
When;
Power = 4950 W
Time taken = 2.40 s
Work done = Power × time taken
Work done = 4950 W × 2.40 s
Work done = 11,900 J
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The block in the figure below has a mass of 5.1 kg and it rests on an incline of angle . You pull on the rope with a force F = 34 N. Assume the incline is smooth and determine the angle of the incline if the block moves with constant speed.
42.9°
Explanation:
Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:
[tex]x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;[/tex]
[tex]\Rightarrow mg\sin{\theta} = F[/tex]
Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at [tex]\theta.[/tex] Solving for the angle, we get
[tex]\sin{\theta} = \dfrac{F}{mg}[/tex]
or
[tex]\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)[/tex]
[tex]\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right][/tex]
[tex]\;\;\;=42.9°[/tex]
Andy Petite pitches a 0.8 kg baseball with a velocity of 67 m/s. Josh Hamilton
hits the ball back at a velocity of -44 m/s. Determine the impulse Josh
Hamilton's bat delivered to the baseball and enter to the nearest kg-m/s with
the correct sign.
The Impulse delivered to the baseball is 89 kgm/s.
To solve the problem above, we use the formula of impulse.
⇒ Formula:
I = m(v-u)................. Equation 1Where:
I = Impulse delivered to the baseballm = mass of the baseballv = Final velocity of the baseballu = initial speed of the baseballFrom the question,
⇒ Given:
m = 0.8 kgu = 67 m/sv = -44 m/s⇒ Substitute these values into equation 1
I = 0.8(-44-67)I = 0.8(-111)I = -88.8I ≈ -89 kgm/sNote: The negative tells that the impulse is in the same direction as the final velocity and therefore can be ignored.
Hence, The Impulse delivered to the baseball is 89 kgm/s.
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A 2.00 kg rock is dropped from the top of a 30.0 m high building. Calculate the ball’s momentum at the time that it strikes the ground.
Explanation:
We use the Theorem of conservation of mechanical energy for finding the velocity when it strikes the ground:
Ei = Ef
Ki + Ui = Kf + Uf
Ui = Kf
m g h = 1/2 m v^2
v = sqrt(2gh)
So the momentum will be:
p = mv = m * sqrt(2gh)
You are given two vectors A⃗ =−3.00i^+5.00j^ and B⃗ =7.00i^+2.00j^. Let counterclockwise angles be positive.
a)What angle θA, where 0∘≤θA<360∘, does A⃗ make with the +x-axis?
Express your answer in degrees.
b)What angle θB, where 0∘≤θB<360∘, does B⃗ make with the +x-axis?
Express your answer in degrees.
c)Vector C⃗ is the sum of A⃗ and B⃗ , so C⃗ =A⃗ +B⃗ . What angle θC, where 0∘≤θC<360∘, does C⃗ make with the +x-axis?
Express your answer in degrees.
Answer:
Explanation:
We can subtract directly the corresponding components and check using the parallelogram rule.
Explanation:
Have a look:
enter image source here
Where, graphically, I used the fact that:
→
A
−
→
B
=
→
A
+
(
−
→
B
)
For the magnitude we use Pythagoras (with the components) to get:
∣
∣
∣
→
A
−
→
B
∣
∣
∣
=
√
(
−
1
)
2
+
(
5
)
2
=
√
1
+
25
=
√
26
≈
5.1
For the direction I can see that will be
90
∘
from the
x
axis up to the
y
axis, plus the little bit passed the
y
axis given as:
θ
=
arctan
(
1
5
)
=
11.3
∘
giving in total: angle
=
90
∘
+
11.3
∘
=
101.3
∘
What is the significance of Isaac's name?
Answer:
ghgivjgifigo ra together aigig disgust u hoodie
1. What is the frequency of light waves with wavelength of 5 x 10-⁷ m?
Answer:
Speed of light =m/s
wavelength = m
frequency = ?
we have
Speed = frequency × wavelength
[tex]3* 10^8[/tex] = frequency × [tex]5 * 10^{-7}[/tex]
Frequency = [tex]\frac{3*10^8}{5*10^{-7}}=6*10^{14}[/tex]hz
Taking into account the definition of wavelength, frecuency and propagation speed, the frequency of light waves with wavelength of 5×10⁻⁷ m is 6×10¹⁴ Hz.
Definition of wavelengthFirst of all, wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).
Definition of frequencyOn the other side, frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).
Definition of propagation speedFinally, the propagation speed is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave disturbance propagates along its displacement.
The propagation speed relate the wavelength (λ) and the frequency (f) inversely proportional using the following equation:
v = f× λ
All electromagnetic waves propagate in a vacuum at a constant speed of 3×10⁸ m/s, the speed of light.
Frequency of light waves with wavelength of 5×10⁻⁷ mIn this case, you know:
v= 3×10⁸ m/sf= ?λ= 5×10⁻⁷ mReplacing in the definition of propagation speed:
3×10⁸ m/s = f× 5×10⁻⁷ m
Solving:
3×10⁸ m/s ÷ 5×10⁻⁷ m= f
f= 6×10¹⁴ Hz
In summary, the frequency of light waves with wavelength of 5×10⁻⁷ m is 6×10¹⁴ Hz.
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5. Layer of Earth consisting of crust & upper layer of mantle ________
Answer:
lithosphere
Explanation:
hope this helps you!!
i need help with the problem below
Answer:
Explanation:
a) F = ma
a = F/m
a = 9(800) / 1 x 10⁹ = 7.2 x 10⁻⁶ m/s
b) t = v/a
t = 200 / 7.2 x 10⁻⁶
t = 2.8 x 10⁷ s about 10½ months
c) v² = u² + 2as
s = (v² - u²) / 2a
s = (200² - 0²) / (2( 7.2 x 10⁻⁶))
s = 2.8 x 10⁹ m nearly 7 times around the earth
And all this assumes NO FRICTION.
Someone help me please !!!! Will mark Brianliest !!!!!!!!!!!!!!!!
Answer:
Decant it.
Explanation:
Pour the water/sugar solution off the sand. When the sand wants to start coming out as well, Stop and add fresh water to the beaker, stir to rinse the remaining solution into a less concentrated solution and decant again.
Repeat the dilution process until the mix is essentially sand and water, then drive the remaining water from the sand by drying.
In what way does Isaac represent us?
Answer:
I believe The Knowledge that we apply everyday or based on these methods and discoveries may represent us
Explanation:
a
A person throws a ball up into the air, and the ball falls back towar
would the kinetic energy be the lowest? (1 point)
at a point before the ball hits the ground
when the ball leaves the person's hand
o when the ball is at its highest point
at a point when the ball is still rising
Answer:
when the ball is at its highest point
Explanation:
Provided the ball returns to where it was thrown. The velocity, and therefore kinetic energy, will be momentarily zero at the highest point of the throw.
A crate is released on a frictionless plank inclined at angle q with
respect to the horizontal. Which of the following relationships is true?
(Assume that the x-axis is parallel to the surface of the incline.)
a. Fy = Fg
c. Fy = Fx
b. Fx = 0
d. none of the above
None of the given options is correct based on the relationship for the components of the crate's weight.
The given parameter:
Angle of inclination, = qThe vertical component of the force on the crate is calculated as follows;
[tex]F_y = W \times cos(q)\\\\F_y = F_gcos(q)[/tex]
The horizontal component of the crate is calculated as follows;
[tex]F_x = F_g \times sin(q)\\\\F_x = F_g sin(q)[/tex]
Thus, we can conclude that none of the given options is correct based on the relationship for the components of the crate's weight.
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A small ball with a mass of 0.6 kg and a velocity of 12 m/s hits another ball with the same mass. The firs forward and hits a third ball with a mass of 0.2 kg. If the system is closed, what is the velocity of the third ball?
4 m/s
36 m/s
30 m/s
1.44 m/s
It takes about 4.4 Newtons to lift 1 pound.
How many Newtons would it take to lift 2 pounds?
If it takes 4.4 Newton to lift 1 pound, then to lift 2 pounds the force required will be equal to 8.8 Newton.
What is Force?A force in physics is an input that has the power to change an object's motion. A mass-containing object's velocity can vary, or accelerate, as a result of a force. Intuitively, a push or a pull can also be used to describe forces.
Being such a vector quantity, a force does have magnitude and direction. The SI unit metric newton is used to measure it (N). The letter F stands for force.
According to Newton's second law's original formulation, an object's net force is equal to the speed that its momentum is changing over time.
As per the given data in the question,
It takes 4.4 N to lift 1 pound,
Let the total force required to lift 2 pounds is x.
x = (4.4 × 2)/1
x = 8.8 N
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