Speed:
Measured in distance / timeIt tells the speed of a bodyIt is a scalar quantitySpeed is always positiveIt's unit is meter per secondAcceleration:
Measured in velocity / timeIt tells the change in the velocityImplied on change in a velocityIt is a vector quantityIt's unit is meter per second square
A conductor is placed in a steady external electric field. Which of the following is FALSE?
a) All excess charge is distributed on the surface of the conductor
b) The electric field inside the conductor is the same as the external electric field
c) The electric field is zero inside the conductor
d) the electric field just outside the surface of the conductor is perpendicular to the surface
Answer:
a
Explanation:
because the electric field doesn't effect the conductor and its goes into storage for later
A smooth circular hoop with a radius of 0.400 m is placed flat on the floor. A 0.325-kg particle slides around the inside edge of the hoop. The particle is given an initial speed of 8.50 m/s. After one revolution, its speed has dropped to 5.50 m/s because of friction with the floor.
(a) Find the energy transformed from mechanical to internal in the particle "hoop" floor system as a result of friction in one revolution.
(b) What is the total number of revolutions the particle makes before stopping? Assume the friction force remains constant during the entire motion.
Answer:
a) W = - 6.825 J, b) θ = 1.72 revolution
Explanation:
a) In this exercise the work of the friction force is negative and is equal to the variation of the kinetic energy of the particle
W = ΔK
W = K_f - K₀
W = ½ m v_f² - ½ m v₀²
W = ½ 0.325 (5.5² - 8.5²)
W = - 6.825 J
b) find us the coefficient of friction
Let's use Newton's second law
fr = μ N
y-axis (vertical) N-W = 0
fr = μ W
work is defined by
W = F d
the distance traveled in a revolution is
d₀ = 2π r
W = μ mg d₀ = -6.825
μ = [tex]\frac{ -6.825}{d_o \ mg}[/tex]
The total work as the object stops the final velocity is zero v_f = 0
W = 0 - ½ m v₀²
W = - ½ 0.325 8.5²
W = - 11.74 J
μ mg d = -11.74
we subtitle the friction coefficient value
( [tex]\frac{-6.8525 }{d_o mg}[/tex]) m g d = -11.74
6.825 [tex]\frac{d}{d_o}[/tex] = 11.74
d = 11.74/6.825 d₀
d = 1.7201 2π 0.400
d = 4.32 m
this is the total distance traveled, the distance and the angle are related
θ = d / r
θ = 4.32 / 0.40
θ = 10.808 rad
we reduce to revolutions
θ = 10.808 rad (1rev / 2π rad)
θ = 1.72 revolution
1. Una pelota rueda hacia la derecha siguiendo una trayectoria en línea recta de modo que recorre una distancia de 10m en 5 s , después cambia su trayectoria cuando es lanzada hacia arriba 25m durante 7 s. Calcular la velocidad y la rapidez al punto final (altura maxima) al que llegó la pelota.
2. Una mariposa vuela en línea recta hacia el sur recorriendo una distancia de 15 m durante 28 s, después cambia de dirección hacia el Oeste recorriendo una distancia de 50 m en un tempo de 80 s ¿cuál es la velocidad y rapidez de la mariposa?
3.- Una persona camina durante 21 minutos hacia el este de su casa una distancia de 1500 m y después cambia su dirección hacia el Norte recorriendo una distancia de 3350 m en un tiempo 32 minutos llegando al supermercado. ¿Calcula la velocidad y rapidez de la persona?
4.- Un automóvil se mueve al Oeste recorriendo una distancia de 80 km en 1.2 horas, posteriormente cambia su trayectoria hacia el Sur, recorriendo una distancia de 120 km en un tiempo 1.6 hora. ¿Calcula la velocidad y rapidez del automóvil?
Answer:
https://youtu.be/ymHHdoCGJOU
The velocity of a body is given by the equation v=a+bx, where 'x' is displacement. The unit of b is
Answer:
2bsnsnsnns181991oiwiw
During normal beating, the heart creates a maximum 4.10-mV potential across 0.350 m of a person's chest, creating a 1.00-Hz electromagnetic wave. (a) What is the maximum electric field strength created? V/m (b) What is the corresponding maximum magnetic field strength in the electromagnetic wave? T (c) What is the wavelength of the electromagnetic wave?
Explanation:
Given that,
Maximum potential, V = 4. mV
Distance, d = 0.350 m
Frequency of the wave, f = 100 Hz
(a) The maximum electric field strength created is given by:
[tex]E=\dfrac{V}{d}\\\\E=\dfrac{4.1\times 10^{-3}}{0.350 }\\\\E=0.0117\ V/m[/tex]
(b) The corresponding maximum magnetic field strength in the electromagnetic wave is given by :
[tex]B=\dfrac{E}{c}\\\\B=\dfrac{0.0117}{3\times 10^8}\\\\B=3.9\times 10^{-11}\ T[/tex]
(c) The wavelength of the electromagnetic wave can be calculated as :
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{100}\\\\=3\times 10^6\ m[/tex]
So, the wavelength of the electromagnetic wave is [tex]3\times 10^6\ m[/tex].
A car is traveling at 118 km/h when the driver sees an accident 85 m ahead and slams on the brakes. What minimum constant deceleration is required to stop the car in time to avoid a pileup
Answer:
The constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²
Explanation:
From the question, the car is traveling at 118 km/h, that is the initial velocity, u = 118km/h
The distance between the car and the accident at the moment when the driver sees the accident is 85 m, that is s = 85 ,
Since the driver slams on the brakes and the car will come to a stop, then the final velocity, v = 0 km/h = 0 m/s
First, convert 118 km/h to m/s
118 km/h = (118 × 1000) /3600 = 32.7778 m/s
∴ u = 32.7778 m/s
Now, to determine the deceleration, a, required to stop,
From one of the equations of motion for linear motion,
v² = u² + 2as
Then
0² = (32.7778)² + 2×a×85
0 = 1074.3841 + 170a
∴ 170a = - 1074.3841
a = - 1074.3841 / 170
a = - 6.3199
a ≅ - 6.32 m/s²
Hence, the constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²
Suppose a 60-turn coil lies in the plane of the page in a uniform magnetic field that is directed out of the page. The coil originally has an area of 0.325 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.60 T
Answer:
emf = 312 V
Explanation:
In this exercise the electromotive force is asked, for which we must use Faraday's law
emf = [tex]- N \frac{d \Phi }{dt}[/tex]- N dfi / dt
Ф = B. A = B A cos θ
bold type indicates vectors.
They indicate that the magnetic field is constant, the angle between the normal to the area and the magnetic field is parallel by local cosine values 1
It also indicates that the area is reduced from a₀ = 0.325 me² to a_f = 0 in a time interval of ΔT = 0.100 s, suppose that this reduction is linear
emf = -N B [tex]\frac{dA}{dT}[/tex]
emf = - N B (A_f - A₀) / Dt
we calculate
emf = - 60 1.60 (0 - 0.325) /0.100
emf = 312 V
The direction of this voltage is exiting the page
To get up on the roof, a person (mass 70.0kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 meters from the bottom. What are the magnitudes of the forces on the ladder at the top and bottom
The magnitude of the forces acting at the top are;
[tex]\mathbf{F_{Top, \ x}}[/tex] = 132.95 N
[tex]\mathbf{F_{Top, \ y}}[/tex] = 0
The magnitude of the forces acting at the bottom are;
[tex]\mathbf{F_{Bottom, \ x}}[/tex] = [tex]\mathbf{ F_f}[/tex] = -132.95 N
[tex]\mathbf{F_{Bottom, \ y}}[/tex] = 784.8 N
The known parameters in the question are;
The mass of the person, m₁ = 70.0 kg
The length of the ladder, l = 6.00 m
The mass of the ladder, m₂ = 10.0 kg
The distance of the base of the ladder from the house, d = 2.00 m
The point on the roof the ladder rests = A frictionless plastic rain gutter
The location of the center of mass of the ladder, C.M. = 2 m from the bottom of the ladder
The location of the point the person is standing = 3 meters from the bottom
g = The acceleration due to gravity ≈ 9.81 m/s²
The required parameters are;
The magnitudes of the forces on the ladder at the top and bottom
The strategy to be used;
Find the angle of inclination of the ladder, θ
At equilibrium, the sum of the moments about a point is zero
The angle of inclination of the ladder, θ = arccos(2/6) ≈ 70.53 °C
Taking moment about the point of contact of the ladder with the ground, B gives;
[tex]\sum M_B[/tex] = 0
Therefore;
[tex]\sum M_{BCW}[/tex] = [tex]\sum M_{BCCW}[/tex]
Where;
[tex]\sum M_{BCW}[/tex] = The sum of clockwise moments about B
[tex]\sum M_{BCCW}[/tex] = The sum of counterclockwise moments about B
Therefore, we have;
[tex]\sum M_{BCW}[/tex] = 2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81
[tex]\sum M_{BCCW}[/tex] = [tex]F_R[/tex] × √(6² - 2²)
Therefore, we get;
2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81 = [tex]F_R[/tex] × √(6² - 2²)
[tex]F_R[/tex] = (2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81)/(√(6² - 2²)) ≈ 132.95
The reaction force on the wall, [tex]F_R[/tex] ≈ 132.95 N
We note that the magnitude of the reaction force at the roof, [tex]F_R[/tex] = The magnitude of the frictional force of bottom of the ladder on the floor, [tex]F_f[/tex] but opposite in direction
Therefore;
[tex]F_R[/tex] = [tex]-F_f[/tex]
[tex]F_f[/tex] = - [tex]F_R[/tex] ≈ -132.95 N
Similarly, at equilibrium, we have;
∑Fₓ = [tex]\sum F_y[/tex] = 0
The vertical component of the forces acting on the ladder are, (taking forces acting upward as positive;
[tex]\sum F_y[/tex] = -70.0 × 9.81 - 10 × 9.81 + [tex]F_{By}[/tex]
∴ The upward force acting at the bottom, [tex]F_{By}[/tex] = 784.8 N
Therefore;
The magnitudes of the forces at the ladder top and bottom are;
At the top;
[tex]\mathbf{F_{Top, \ x}}[/tex] = [tex]F_R[/tex] ≈ 132.95 N←
[tex]\mathbf{F_{Top, \ y}}[/tex] = 0 (The surface upon which the ladder rest at the top is frictionless)
At the bottom;
[tex]\mathbf{F_{Bottom, \ x}}[/tex] = [tex]F_f[/tex] ≈ -132.95 N →
[tex]\mathbf{F_{Bottom, \ y}}[/tex] = [tex]F_{By}[/tex] = 784.8 N ↑
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You are working on a project to make a more efficient engine. Your team is investigating the possibility of making electrically controlled valves that open and close the input and exhaust openings for an internal combustion engine. Determine the stability of the valve by calculating the force on each of its sides and the net force on the valve.
The valve is made of a thin but strong rectangular piece of non-magnetic material that has a current-carrying wire along its edges. The rectangle is 0.35 cm x 1.83 cm. The valve is placed in a uniform magnetic field of 0.15 T such that the field lies in the plane of the valve and is parallel to the short sides of the rectangle. The region with the magnetic field is slightly larger than the valve. When a switch is closed, a 1.7 A current enters the short side of the rectangle on one side and leaves on the opposite short side of the rectangle. At the suggestion of a colleague, who is hoping to ensure different currents along the sides of the valve, resistors have been included along the wire on each of the short sides of the valve. The value of the resistor on one side is twice that on the other side.
Answer:
The answer is "0.00466 N".
Explanation:
[tex]F=(B \times i) L\\\\[/tex]
therefore the smaller side is parallel to magnetic field
[tex]\therefore \\\\F= B i L\ \sin\ 'o'=0 \ N[/tex]
calculating the force on the layer side:
[tex]\to F=0.15 \times 1.7 \times 0.0183 \times \sin 90^{\circ}=0.00466\ N\\\\[/tex]
Therefore [tex]F_o[/tex] the net force on the rectangular loop [tex]= 0.00466 \ N[/tex]
In many cartoon shows, a character runs of a cliff, realizes his predicament and lets out a scream. He continues to scream as he falls. If the physical situation is portrayed correctly, from the vantage point of an observer at the foot of the cliff, the pitch of the scream should be Group of answer choices
Answer:
Increasing until terminal velocity is reached
Explanation:
Provided the scream is a constant pitch at the source, Doppler effect will make the pitch increase as the velocity of the source towards the listener increases.
A car is traveling at 104 km/h when the driver sees an accident 50 m ahead and slams on the brakes. What minimum constant deceleration is required to stop the car in time to avoid a pileup
a = - 8.34 m/sec² ( deceleration or negative)
Equations for UAM ( uniformly accelerated motion) are:
vf = v₀ ± a*t and s = s₀ + v₀*t + (1/2)*a*t²
In our case, the motion is with deceleration, then
vf = v₀ - a*t and s = s₀ + v₀*t - (1/2)*a*t²
working on these equatios we get:
vf = v₀ - a*t (1) s - s₀ = v₀*t - (1/2)*a*t² (2)
v₀ - vf = a*t
t = (v₀ - vf)/a
By substitution of (1) in equation (2)
s - s₀ = v₀ * (v₀ - vf)/a - (1/2) * a* [(v₀ - vf)/a]²
s - s₀ = (v₀² - v₀*vf)/a - (1/2) * a* (1/a²)* (v₀ - vf)²
s - s₀ = 1/a * ( v₀² - v₀*vf ) - 1/a* (1/2) * (v₀ - vf)²
s - s₀ = 1/a* [ ( v₀² - v₀*vf ) - (1/2) * (v₀ - vf)²]
a * (s - s₀ ) = v₀² - v₀*vf - v₀²/2 - vf²/2 + v₀*vf
a * (s - s₀ ) = (1/2) * v₀² - (1/2)*vf²
a * (s - s₀ ) = (1/2) * ( v₀² - vf²)
We find an expression to calculate the minimum deceleration to stop the car in time to avoid crashing
s₀ = 50 meters s = 0 v₀ = 104 Km/h vf = 0
1 Km = 1000 m and 1 h = 3600 sec
v₀ = 104 Km/h = 28.88 m/sec
a = (1/2) [ (28.88)² - 0 ] / 0 - 50
a = - 8.34 m/sec² ( deceleration or negative)
which planet composed entirely of hydrogen and helium?
Answer:
The composition of Jupiter is similar to that of the Sun—mostly hydrogen and helium. Deep in the atmosphere, pressure and temperature increase, compressing the hydrogen gas into a liquid. This gives Jupiter the largest ocean in the solar system—an ocean made of hydrogen instead of water.
A chair of weight 85.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 40.0 N directed at an angle of 35.0deg below the horizontal and the chair slides along the floor.
Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair.
Answer:
N = 107.94 N
Explanation:
For this exercise we must use Newton's second law.
Let's set a reference system with the x-axis parallel to the ground and the y-axis vertical
X axis
Fₓ = ma
ej and
N -F_y - W = 0
let's use trigonometry to decompose the applied force
cos -35 = Fₓ / F
sin -35 = F_y / F
Fₓ = F cos -35
F_y = F sin -35
Fₓ = 40.0 cos -35 = 32.766 N
F_y = 40.0 sin -35 = -22.94 N
we substitute
N = Fy + W
N = 22.94 + 85
N = 107.94 N
when do things move faster? Day or Night?
When the lightbulbs were used as the resistors, you observed only a flash of light, as opposed to a continuous glow. Explain why that behavior is expected. After all, the light bulb is directly connected to the power supply.
Solution :
Whenever the lightbulbs are used as resistors, we throw the switch to the left. This allows the current to flow through the circuit which causes the bulb to glow and also the capacitor gets charged. When the capacitor gets fully charged, the electric field becomes constant between its two plates. Now there is no displacement current induced in the plates of the capacitor. The capacitor works as an open switch and the bulb gets switched off.
And thus the bulb flashes for the moment as opposed to continuous glow.
A ball is thrown upward from the edge of a cliff with an initial velocity of 6 m/s. (a) How fast is it moving 0.5 s later? In what direction? (b) How fast is it moving 2 s later? In what direction?
Answer:
Explanation:
Kinematic equation
v = u + at
If UP is assumed to be the positive direction and we let gravity be 10 m/s² which will be in the downward direction so will be negative.
a) v = 6 + (-10)(0.5) = 1 m/s the result is positive, so upward
b) v = 6 + (-10)(2) = -14 m/s the result is negative, so downward
Andrea's near point is 20.0 cm and her far point is 2.0 m. Her contact lenses are designed so that she can see objects that are infinitely far away. What is the closest distance that she can see an object clearly when she wears her contacts?
Answer:
the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm
Explanation:
Given the data in the question,
near point = 20 cm
far point = 2 m = 200 cm
Now, for an object that is infinitely far away, the image is at is its far point.
so using the following expression, we can determine the focal length
1/f = 1/i + 1/o
where f is the focal length, i is the image distance and o is the object distance.
here, far point i = 2 m = 200 cm and v is ∞
so we substitute
1/f = 1/(-200 cm) + 1/∞
f = -200 cm
Also, for object at its closest point, the image appear at near point,
so
1/f = 1/i + 1/o
we make o the subject of formula
o = ( i × f ) / ( i - f )
given that near point i = 20 cm
we substitute
o = ( -20 × -200 ) / ( -20 - (-200) )
o = 4000 / 180
o = 22.2 cm
Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm
Which of the following quantities is measured by the area under the velocity time graph? (a) Magnitude of velocity (b) Magnitude of acceleration (c) Magnitude of displacement (d) Average Speed
Answer:
c.
magnitude of displacement
An ink-jet printer steers charged ink drops vertically. Each drop of ink has a mass of 10-11 kg, and a charge due to 500,000 extra electrons. It goes through two electrodes that gives a vertical acceleration of 104 m/s2. The deflecting electric field is _____ MV/m.
Answer:
E = 1.25 MV / m
Explanation:
For this exercise let's use Newton's second law
F = m a
where the force is electric
F = q E
we substitute
q E = m a
E = m a / q
indicate there are 500,000 excess electrons
q = 500000 e
q = 500000 1.6 10⁻¹⁹
q = 8 10⁻¹⁴ C
the mass is m = 10⁻¹¹ kg and the acceleration a = 10⁴ m / s²
let's calculate
E = 10⁻¹¹ 10⁴ / 8 10⁻¹⁴
E = 0.125 10⁷ V / m = 1.25 10⁶ V / m
E = 1.25 MV / m
Light from two lasers is incident on an opaque barrier with a single slit of width 4.0 x 10^-4 m. One laser emits light of wavelength 480 nm and the other is 640 nm. A screen to view the light intensity pattern is 2.0 m behind the barrier. The distance from the center of the pattern to the nearest completely dark spot (dark for both colors) is ____ cm. (include 2 digits after the decimal point)
Answer:
a) y = 2.4 x 10⁻³ m = 0.24 cm
b) y = 3.2 x 10⁻³ m = 0.32 cm
Explanation:
The formula of Young's Double Slit experiment will be used here:
[tex]y = \frac{\lambda L}{d}\\\\[/tex]
where,
y = distance between dark spots = ?
λ = wavelength
L = distance of screen = 2 m
d = slit width = 4 x 10⁻⁴ m
a) FOR λ = 480 nm = 4.8 x 10⁻⁷ m:
[tex]y = \frac{(4.8\ x\ 10^{-7}\ m)(2\ m)}{4\ x\ 10^{-4}\ m}[/tex]
y = 2.4 x 10⁻³ m = 0.24 cm
a) FOR λ = 640 nm = 6.4 x 10⁻⁷ m:
[tex]y = \frac{(6.4\ x\ 10^{-7}\ m)(2\ m)}{4\ x\ 10^{-4}\ m}[/tex]
y = 3.2 x 10⁻³ m = 0.32 cm
Which term defines seeking some way of achieving a goal by overcoming obstacles or finding an answer to a question?
Answer:
Problem solving
hope this helps :)
The temperature of a body falls from 30°C to 20°C in 5 minutes. The air
temperature is 13°C. Find the temperature after a further 5 minutes.
Answer:
15.88°C I am not 100% sure this is right but I am 98% sure this IS right
Light falling on a metal surface causes electrons to be emitted from the metal by the photoelectric effect.
As we decrease the frequency of this light, but do not vary anything else (there may be more than one correct answer),
A: the number of electrons emitted from the metal increases.
B: the maximum speed of the emitted electrons decreases.
C: the maximum speed of the emitted electrons does not change.
D: the work function of the metal increases.
(B)
Explanation:
The speed of the ejected electrons depends on the frequency of the incident radiation. The closer the energy of the incident photons to the work function of the metal, the slower is the speed of the ejected electrons. Intensity of the incident radiation has no effect on the speed of the ejected electrons, only its frequency.
Light falling on a metal surface causes electrons to be emitted from the metal by the photoelectric effect. As we decrease the frequency of this light, but do not vary anything else B: the maximum speed of the emitted electrons decreases.
What is photoelectric effect?
Photoelectric effect is the phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation. The effect is often defined as the ejection of electrons from a metal plate when light falls on it.
According to Photoelectric effect the kinetic energy of the photo electrons emitted depend on the frequency of incident light , the more is the frequency the more is the kinetic energy of emitted electron and hence high will be the velocity of the emitted electrons and vise versa
since , in question the frequency has been decreased hence , the kinetic energy must be decreased therefore velocity will also get decreased
hence , correct option will be B: the maximum speed of the emitted electrons decreases.
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The lines in the emission spectrum of hydrogen result from __________.
a. energy given off in the form of visible light when an electron moves from a higher energy state to a lower energy state
b. protons given off when hydrogen burns
c. electrons given off by hydrogen as it cools
d. electrons given off by hydrogen when it burns
e. decomposing hydrogen atoms.
Answer:
Option (a) is correct.
Explanation:
The lines in the emission spectrum of hydrogen is due to the transfer of electrons form higher energy levels to the lower energy levels.
When the electrons transfer from one level of energy that is higher level of energy to the other means to the lower level of energy then they emit some photons which having the frequency or the wavelength in the visible region.
An energy efficient light bulb uses 15 W of power for an equivalent light output of a 60 W incandescent light bulb. How much energy is saved each month by using the energy efficient light bulb instead of the incandescent light bulb for 4 hours a day? Assume that there are 30 days in one month
A. 7.2 kW⋅hr
B. 21.6 kW⋅hr
C. 1.8 kW⋅hr
D. 5.4 kW⋅hr
E. 1.35 kW⋅hr
Answer: (d)
Explanation:
Given
15 W is equivalent to 60 W light that is, it save 45 W
So, for 4 hours it is, [tex]4\times 45=180\ W.hr[/tex]
For 30 days, it becomes
[tex]\Rightarrow 180\times 30=5400\ W.hr\\\Rightarrow 5.4\ kWh[/tex]
Thus, [tex]5.4\ kWh[/tex] is saved in 30 days
option (d) is correct.
A toy car of mass 600g moves through 6m in 2 seconds.The average kinetic energy og the toy car is?
Explanation:
kE =1/2mv²
1/2(0.6×(3m/s)²
1/2(0.6×9m/s)
2.7J I think this is the answer
The average kinetic energy of the toy car is 2.7 J.
What is kinetic energy?
The energy of the body by the virtue of its motion is known as the kinetic energy of the body. It is defined as the product of half of mass and square of the velocity.
Given data;
Mass of car is,m= 600 g = 0.6 kg
d is the distance travelled = 6 m
T is the time travelled = 2 sec
The velocity of the car is found as;
v = d /t
V = 6m / 2 sec
V = 3 m/sec
KE =1/2mv²
KE = 1/2 × 0.6 kg ×( 3 m/sec )²
KE = 2.7 J
Hence, the average kinetic energy of the toy car is 2.7 J.
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If a body travels 6km in 30 minutes in a fixed direction, calculate it's velocity.
Plz show me the process too.
We know
[tex]\boxed{\large{\sf Velocity=\dfrac{Distance}{Time}}}[/tex]
[tex]\\ \Large\sf\longmapsto Velocity=\dfrac{6}{\dfrac{1}{2}}[/tex]
[tex]\\ \Large\sf\longmapsto Velocity=6\times 2[/tex]
[tex]\\ \Large\sf\longmapsto Velocity=12km/h[/tex]
Is this the right answer??
We should keep km and min in smallest SI unit
15- A racehorse coming out of the gate accelerates from rest to a velocity f 15.0 m/s due west in 1.80 s. What is its average acceleration?
Answer: (15 - 0)/1.8 = 8. 33m/s^2
Explanation:
The acceleration of the racehorse is 8.33 m/s²
The given parameters;
initial velocity of the racehorse, u = 0
final velocity of the racehorse, v = 15 m/s
time of motion of the horse, t = 1.8 s
The acceleration of the racehorse is calculated from change in velocity per change in time of motion as shown below;
[tex]a = \frac{\Delta v}{\Delta t} = \frac{v-u}{t} \\\\a = \frac{15 - 0}{1.8} \\\\a = 8.33 \ m/s^2[/tex]
Thus, the acceleration of the racehorse is 8.33 m/s²
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A car's bumper is designed to withstand a 6.12 km/h (1.7-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.210 m while bringing a 810 kg car to rest from an initial speed of 1.7 m/s.
Answer:
5572.8 N
Explanation:
Applying,
F = ma.............. Equation 1
Where F = Force, m = mass of the car, a = acceleration.
We can find a by applying,
v² = u²+2as............. Equation 2
Where v = final velocity, u = initial velocity, a = acceleration, = distance.
From the question,
Given: v = 0 m/s (come to rest), u = 1.7 m/s, s = 0.210 m
Substitute these value into equation 2
0² = 1.7²+2×0.21×a
a = -1.7²/(2×0.21)
a = -2.89/0.42
a = -6.88 m/s²
Also given: m = 810 kg
Substitute these value into equation 1
F = 810(-6.88)
F = -5572.8 N
Hence the force on the bumber is 5572.8 N
A(n) _______________ absorbs energy and then emits electromagnetic radiation based on its _______________. Classical physics predicted that at a high enough temperature, _______________ light would be emitted. Instead, white light was emitted, resulting in the ultraviolet _______________. The photoelectric effect occurs when light shining on a metal creates a(n) _______________. However, only light of a certain minimum _______________ causes electrons to flow. Gas atoms excited by an electric current emit bands of colors of light in a(n) _______________. Each narrow band of light is associated with _______________ of a specific energy.
Answer:
Blackbody radiator, temperature, ultraviolet, catastrophe, electric current, frequency, spectrum, photons
Explanation:
# a p e x
1 and 2 ) A blackbody radiator is an object that absorbs energy, then emits electromagnetic radiation based on the temperature of the object. This comes directly from the definition in the passages.
3 and 4 ) Ultraviolet catastrophe describes when old physicists assumed as frequency increased the waves would go from visible to ultraviolet because that is what comes next on the spectrum. Instead of this happening, the light became white light and it was an apparent 'catastrophe'
Appropriate words for blank position shown below,
Blackbody radiatortemperature ultravioletcatastropheelectric currentfrequency spectrum photonsA blackbody radiator is defined as an object that absorbs all electromagnetic radiation that falls on it at all frequencies over all angles of incidence.
Ultraviolet light is a type of electromagnetic radiation that makes black-light posters glow
A movement of positive or negative electric particles produce current.
Frequency is defined as the number of cycles or vibrations undergone during one unit of time by a body in periodic motion.
Photons are particles which transmit light.
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