Positive
In an LED, the longer lead is always connected to the positive terminal
Most modern vehicles have padded dashboards. This reduces collision injuries by...
1: reducing the driver's change in momentum
2:increasing the driver's change in momentum
3:increasing the force of impact
4:increasing the time of impact
Answer:
Option 4. is correct
Explanation:
A dashboard refers to a control panel that lies in front of the driver of the vehicle. Padded dashboards are designed to reduce injuries including face injuries and chest injuries to the driver as well as the front passenger on collisions.
Most modern vehicles have padded dashboards. This reduces collision injuries by increasing the time of impact.
Therefore,
Option 4. is correct
a ford explorer traveled 100 miles the next day for 5 hours. What was the average speed of this vehicle?
Answer:
25 miles per hour
Explanation:
It was 20 miles per hour the next day. We don't have enough information to calculate the average speed for the whole trip.
What is the weight of a 3000g girl who is swimming in the Pacific Ocean?
Answer: 3000g ;-;
she feels lighter in the water but the water doesn't actually take away her weight
Explanation:
A wet towel spread out and hung outside on a day without wind dries faster than an identical wet towel left rolled up in a plastic bag.
Explain why.
Even without wind the sun help evaporate the water so it would dry faster. The other towel is in a bag so the moist and water has nowhere to go, therefor staying in the towel.
The teacher asked the children to turn the scales sideways and adjust the scales to zero. Jason pushes on the top of the scales and Kerry pushes on the base. They hold the scales still. Jason’s push measures 80 N. What is the size of Kerry’s push? *
Answer: J
Explanation:
Jason pushes on the top of the scales and Kerry pushes on the base. They hold the scales still. Jason's push measures 80N. What is the size of Kerry's push?
he pushes zero
An object with mass 100 kg moved in outer space. When it was at location <8, -30, -4> its speed was 5.5 m/s. A single constant force <220, 460, -200> N acted on the object while the object moved from location <8, -30, -4> m to location <14, -21, -7> m. Then a different single constant force <100, 260, 210> N acted on the object while the object moved from location <14, -21, -7> m to location <17, -27, -3> m. What is the speed of the object at this final location
Answer:
v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s
Explanation:
We can solve this problem using the kinematic relations, we have a three-dimensional movement, but we can work as three one-dimensional movements where the only parameter in common is time (a scalar).
X axis.
They indicate the initial position x = 8 m, its initial velocity v₀ = 5.5 m / s, the force Fx₁ = 220 N x₁ = 14 m, now the force changes to Fx₂ = 100 N up to the point xf = 17 m. The final speed is wondered.
As this movement is in three dimensions we must find the projection of the initial velocity in each axis, for this we can use trigonometry
the angle fi is with respect to the in z and the angle theta with respect to the x axis.
sin φ = z / r
Cos φ = r_p / r
z = r sin φ
r_p = r cos φ
the modulus of the vector r can be found with the Pythagorean theorem
r² = (x-x₀) ² + (y-y₀) ² + (z-z₀) ²
r² = (14-8) 2 + (-21 + 30) 2+ (-7 +4) 2
r = √126
r = 11.23 m
Let's find the angle with respect to the z axis (φfi)
φ = sin⁻¹ z / r
φ = sin⁻¹ ( [tex]\frac{-7+4}{11.23}[/tex] )
φ = 15.5º
Let's find the projection of the position vector (r_p)
r_p = r cos φ
r_p = 11.23 cos 15.5
r_p = 10.82 m
This vector is in the xy plane, so we can use trigonometry to find the angle with respect to the x axis.
cos θ = x / r_p
θ = cos⁻¹ x / r_p
θ = cos⁻¹ ( [tex]\frac{14-8}{10.82}[/tex])
θ = 56.3º
taking the angles we can decompose the initial velocity.
sin φ = v_z / v₀
cos φ = v_p / v₀
v_z = v₀ sin φ
v_z = 5.5 sin 15.5 = 1.47 m / z
v_p = vo cos φ
v_p = 5.5 cos 15.5 = 5.30 m / s
cos θ = vₓ / v_p
sin θ = v_y / v_p
vₓ = v_p cos θ
v_y = v_p sin θ
vₓ = 5.30 cos 56.3 = 2.94 m / s
v_y = 5.30 sin 56.3 = 4.41 m / s
we already have the components of the initial velocity
v₀ = (2.94 i ^ + 4.41 j ^ + 1.47 k ^) m / s
let's find the acceleration on this axis (ax1) using Newton's second law
Fₓx = m aₓ₁
aₓ₁ = Fₓ / m
aₓ₁ = 220/100
aₓ₁ = 2.20 m / s²
Let's look for the velocity at the end of this interval (vx1)
Let's be careful if the initial velocity and they relate it has the same sense it must be added, but if the velocity and acceleration have the opposite direction it must be subtracted.
vₓ₁² = v₀ₓ² + 2 aₓ₁ (x₁-x₀)
let's calculate
vₓ₁² = 2.94² + 2 2.20 (14-8)
vₓ₁ = √35.04
vₓ₁ = 5.92 m / s
to the second interval
they relate it to xf
aₓ₂ = Fₓ₂ / m
aₓ₂ = 100/100
aₓ₂ = 1 m / s²
final speed
v_{xf}² = vₓ₁² + 2 aₓ₂ (x_f- x₁)
v_{xf}² = 5.92² + 2 1 (17-14)
v_{xf} =√41.05
v_{xf} = 6.41 m / s
We carry out the same calculation for each of the other axes.
Axis y
acceleration (a_{y1})
a_{y1} = F_y / m
a_{y1} = 460/100
a_[y1} = 4.60 m / s²
the velocity at the end of the interval (v_{y1})
v_{y1}² = v_{oy}² + 2 a_{y1{ (y₁ -y₀)
v_{y1}2 = 4.41² + 2 4.60 (-21 + 30)
v_{y1} = √102.25
v_{y1} = 10.11 m / s
second interval
acceleration (a_{y2})
a_{y2} = F_{y2} / m
a_{y2} = 260/100
a_{y2} = 2.60 m / s2
final speed
v_{yf}² = v_{y1}² + 2 a_{y2} (y₂ -y₁)
v_{yf}² = 10.11² + 2 2.60 (-27 + 21)
v_{yf} = √ 71.01
v_{yf} = 8.43 m / s
here there is an inconsistency in the problem if the body is at y₁ = -27m and passes the position y_f = -21 m with the relationship it must be contrary to the velocity
z axis
first interval, relate (a_{z1})
a_{z1} = F_{z1} / m
a_{z1} = -200/100
a_{z1} = -2 m / s
the negative sign indicates that the acceleration is the negative direction of the z axis
the speed at the end of the interval
v_{z1}² = v_{zo)² + 2 a_{z1} (z₁-z₀)
v_{z1}² = 1.47² + 2 (-2) (-7 + 4)
v_{z1} = √14.16
v_{z1} = -3.76 m / s
second interval, acceleration (a_{z2})
a_{z2} = F_{z2} / m
a_{z2} = 210/100
a_{z2} = 2.10 m / s2
final speed
v_{fz}² = v_{z1}² - 2 a_{z2} | z_f-z₁)
v_{fz}² = 3.14² - 2 2.10 (-3 + 7)
v_{fz} = √6.94
v_{fz} = 2.63 m / s
speed is v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s
You are planning a deep space exploration and want to take your cat with you. If your spacecraft achieves a maximum speed of 87% of the speed of light, how much additional energy is required to bring your cat (during the acceleration only)? Express this quantity in units of Petajoules (1015).Assume you start from zero velocity and your magical cat does not require additional food or litter. The cat weighs 8.5 lbf. Use LaTeX: 3\times10^8 3 × 10 8 m/s for the speed of light.
Answer:
E = 1.7 10² PJ
Explanation:
Let's use the special relativity relations, specifically the energy of a body is
E = γ mc²
γ = [tex]\sqrt{ 1 - \frac{v^2}{c^2} }[/tex]
where m is the rest mass
for that case they tell us that the speed of the body is 87% of the speed of light
= 0.87
let's calculate
γ = [tex]\sqrt{1 - 0.87^2}[/tex]
γ = 0.49305
let's reduce the mass of the jack to SI units
W = 8.5 lb (4.448 N / 1lb) = 37.808 N
W = mg
m = W / g
m = 37.808 / 9.8
m = 3.86 kg
let's look for energy
E = 3.86 (3 10⁸ )² 0.49305
E = 1.7 10¹⁷ J
let's reduce to take PJ
E = 1.7 10¹⁷ J ([tex]\frac{ 1 PJ}{10^{15} J}[/tex] )
E = 1.7 10² PJ
Which molecules are in put in photosynthesis
Answer:
During the process of photosynthesis, cells use carbon dioxide and energy from the Sun to make sugar molecules and oxygen. These sugar molecules are the basis for more complex molecules made by the photosynthetic cell, such as glucose.
Explanation:
yes.
Answer:
Sunlight, Carbon Dioxide, and Water
Explanation:
Technically minerals are in there too but when I learned this it was just Sunlight, Carbon Dioxide, and Water
A single-phase 60-Hz overhead power line is symmetrically supported on a horizontal cross arm. Spacing between the centers of the conductors acing between the centers of the conductors (say, a and b) is 2.5 m. A telephone line is also symmetrically supported on a horizontal cross arm 1.8 m directly below the power line. Spacing between the centers of these conductors (say, c and d) is 1.0 m.
4 x 10^7 ln âDad. Dln/Dnk. Dtet
where, for example, Dud denotes the distance in meters between conductors a and.
a. Hence, compute the mutual inductance per kilometer between the power line and the telephone line.
b. Find the GO-Hz voltage per kilometer induced in the telephone line when the power line carries 150 A.
Complete question is;
A single-phase 60-Hz overhead power line is symmetrically supported on a horizontal cross arm. Spacing between the centers of the conductors acing between the centers of the conductors (say, a and b) is 2.5 m. A telephone line is also symmetrically supported on a horizontal cross arm 1.8 m directly below the power line. Spacing between the centers of these conductors (say, c and d) is 1.0 m.
The mutual inductance per unit length between circuit a-b and circuit c-d is given as 4 x 10^(-7) ln √((D_ad × D_bc)/(D_ac × D_bd)) H/m
where, for example, D_ad denotes the distance in meters between conductors a and d.
a. Hence, compute the mutual inductance per kilometer between the power line and the telephone line.
b. Find the 60-Hz voltage per kilometer induced in the telephone line when the power line carries 150 A
Answer:
A) M = 1.01 × 10^(-4) H/km
B) v_cd = 5.712 V/km
Explanation:
A) From the distances given in the question, we can deduce that;
D_ac = √(((2.5/2) - (1/2))² + 1.8²)
D_ac = 1.95 m
Also;
D_ad = √(((2.5/2) + (1/2))² + 1.8²)
D_ad = 2.51 m
I_a and I_b are put of phase by 180°. Thus, due to a and b, the flux linkages to c and d is given as;
φ_cd = 4 x 10^(-7)I_a( ln (2.51/1.95))
Mutual inductance per km is given as;
M = φ_cd/I_a
Thus;
M = 4 x 10^(-7)( ln (2.51/1.95))
M = 1.01 × 10^(-7) H/m
Per km;
M = 1.01 × 10^(-7) × 1000
M = 1.01 × 10^(-4) H/km
B) voltage per km is gotten by;
v_cd = ωMI
Now, ω = 2πf = 2π × 60 = 377 rad/s
Thus;
v_cd = 377 × 1.01 × 10^(-4) × 150
v_cd = 5.712 V/km
The parallel plates in a capacitor, with a plate area of 9.30 cm2 and an air-filled separation of 4.50 mm, are charged by a 7.80 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 9.60 mm. Neglecting fringing, find (a) the potential difference between the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the work required to separate the plates.
Answer:
a) ΔV ’= 1.66 10¹ V= 16.6 V, b) U = 55.64 10⁻¹² J, c) U_f = 1.18 10⁻¹⁰ J
d) W = 6.236 10⁻¹¹ J
Explanation:
Capacitance can be found for a parallel plate capacitor
C = ε₀ [tex]\frac{A}{d}[/tex]
Let's reduce the magnitudes to the SI system
A = 9.30 cm² (1 m / 10² cm) 2 = 9.30 10⁻⁴ m²
c = 4.50 mm (1 m / 1000 mm) = 4.50 10⁻³ m
Co = 8.85 10⁻¹² 9.30 10⁻⁴ /4.50 10⁻³
Co = 1.829 10⁻¹² F
when the plates separate at d = 9.60 10⁻³ m, the capcitance changes to
C = ε₀ \frac{A}{d_1}
C = 8.85 10⁻¹² 9.30 10⁻⁴/9.60 10⁻³
C = 8.57 10⁻¹³ F
a) the potential difference
C =
since the capacitor is not discharged, let's look for the initial charge
Co = \frac{Q}{ \Delta V}
Q = C₀ ΔV
Q = 1.829 10⁻¹² 7.80
Q = 14.2662 10⁻¹² C
when the condensate plates are separated
C = \frac{Q}{ \Delta V' }
ΔV ’= Q / C
ΔV ’= 14.266 10⁻¹² / 8.57 10⁻¹³
ΔV ’= 1.66 10¹ V= 16.6 V
b) the stored energy is
U = ½ C ΔV²
for initial separation
U = ½ C₀ ΔV²
U = ½ 1.829 10⁻¹² 7.80²
U = 55.64 10⁻¹² J
c) The energy for end separation;
U_f = ½ C DV’2
U_f = ½ 8.57 10⁻¹³ 16,6²2
U_f = 1.18 10⁻¹⁰ J
d) The work
as there are no losses, the work is equal to the variation of the energy
W = ΔU = U_f -U₀
W = 1.18 10⁻¹⁰ - 55.64 10-12
W = 6.236 10⁻¹¹ J
Which two options are forms of kinetic energy?
A. Thermal energy
OO
O B. Nuclear energy
c. Sound energy
O D. Gravitational energy
O E. Chemical energy
Answer:
Themal energy is the best answer from you
Your friend said that the star in this picture with the highest apparent magnitude must definitely have the highest absolute brightness as well.
Answer:
A white dwarf, also called a degenerate dwarf
Explanation:
sorry if im wrong im kind of du-m
Which type of biological molecule would contain fats?
A) Amino Acids.
C) Nucleic Acids.
B) Lipids.
D) Carbohydrates.
B
Explanation:
lipids contains fat
hope it helps
An angry physics student releases a wrecking ball as shown. The wrecking ball is just about to hit the building at the final time. Neglect loss. Identify the energy types present at each time for the highlighted object or system relative to the reference
level shown. Also identify if loss occurs and/or work is done by a non-conservative force between each time
(if necessary).
Answer:
the force between the building and the ball is non-conservative (friction-type force)
Explanation
Explanation:For this exercise the student must create an impulse to move the ball towards the building, in this part he performs positive work since the applied force and the displacement are in the same direction.
When the ball moves it has a kinetic energy and if its height increases or decreases its potential energy also changes, but the sum of being must be equal to the initial work.
When the ball arrives and collides with the building, non-conservative forces, of various kinds; rubbing, breaking, etc. It transforms this energy into a part of heat and another in mechanical energy that the building must absorb, let us destroy its wall
Consequently, the force between the building and the ball is non-conservative (friction-type force
An organ pipe is 1.6 m long and open at both ends. What are the first three harmonic frequencies of this pipe?
Answer:
first harmonic = 56.7 Hz
Third harmonic = 170 Hz
Fifth harmonic = 283.3 Hz
Explanation:
. An object 8.5 cm high is placed 28 cm from a converging lens. The focal length of the lens is 12 cm. Calculate the image distance, di. Calculate the image height, hi.
The converging lens is also called a concave lens. The height of the image formed by the lens is 2.55 cm.
Using the lens formula;
1/f = 1/u + 1/v
f = focal length of the lens
u = object distance
v = image distance
Note that the focal length of a converging lens is positive
Substituting values;
1/12 = 1/28 + 1/v
1/v = 1/12 - 1/28
v = 8.4 cm
Magnification= image height/object height = image distance/object distance
image height = ?
object height = 8.5 cm
image distance = 8.4 cm
object distance = 28 cm
So
image height/8.5 = 8.4/28
image height = 8.5 × 8.4/28
image height = 2.55 cm
Learn more: https://brainly.com/question/11969651
A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 ×102 T, what is the number of turns per meter for this solenoid?
Correct question:
A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?
Answer:
the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.
Explanation:
Given;
length of solenoid, L= 0.35 m
diameter of the solenoid, d = 0.04 m
current through the solenoid, I = 5.0 A
magnetic field in the center of the solenoid, 2.8 x 10⁻² T
The number of turns per meter for the solenoid is calculated as follows;
[tex]B =\mu_o I(\frac{N}{L} )\\\\B = \mu_o I(n)\\\\n = \frac{B}{\mu_o I} \\\\n = \frac{2.8 \times 10^{-2}}{4 \pi \times 10^{-7} \times 5.0} \\\\n = 4.5 \times 10^3 \ turns/m[/tex]
Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.
3. The car's mass is 400 kg. It moves at a velocity of 20 m/s. Calculate the car's momentum. *
(10 Points)
0.05 kg.m/s
8000 kg.m/s
80,000 kg.m/s
20 kg.m/s
Answer:
momentum=mass×velocity
momentum =400kg×20m/s=8000kg.m/s
Which waves can be seen by people?
O radio waves
O microwaves
O visible light waves
O gamma rays
Answer:
Visible light waves
Explanation:
What is the wavelength of microwaves with a frequency of 3x10^10 Hz?
Answer:
0.01 m
Explanation:
Since the speed of light is 3.0×10^8 m/s
Use the equation,
Wavelength = speed ÷ frequency
Wavelength = 3.0×10^8 ÷ 3×10^10
Wavelength = 0.01m
Sound energy travels through matter in waves. Do sound waves travel faster through air or water? Explain your answer.
I uploaded the answer to a file hosting. Here's link:
tinyurl.com/wpazsebu
Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible compared to the gravitational attraction between the ships. Construct a problem in which you place identical excess charges on the space ships to exactly counter their gravitational attraction. Calculate the amount of excess charge needed. Examine whether that charge depends on the distance between the centers of the ships, the masses of the ships, or any other factors. Discuss whether this would be an easy, difficult, or even impossible thing to do in practice.
Answer:
q = 8.61 10⁻¹¹ m
charge does not depend on the distance between the two ships.
it is a very small charge value so it should be easy to create in each one
Explanation:
In this exercise we have two forces in balance: the electric force and the gravitational force
F_e -F_g = 0
F_e = F_g
Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.
Let's write Coulomb's law and gravitational attraction
[tex]k \frac{q_1q_2}{r^2} = G \frac{m_1m_2}{r^2}[/tex]
In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.
k q² = G m²
q = [tex]\sqrt{ \frac{G}{k} }[/tex] m
we substitute
q = [tex]\sqrt{ \frac{ 6.67 \ 10^{-11}}{8.99 \ 10^{9}} }[/tex] m
q = [tex]\sqrt{0.7419 \ 10^{-20}}[/tex] m
q = 0.861 10⁻¹⁰ m
q = 8.61 10⁻¹¹ m
This amount of charge does not depend on the distance between the two ships.
It is also proportional to the mass of the ships with the proportionality factor found.
Suppose the ships have a mass of m = 1000 kg, let's find the cargo
q = 8.61 10⁻¹¹ 10³
q = 8.61 10⁻⁸ C
this is a very small charge value so it should be easy to create in each one
what is the quantum and its types?
Answer:
It is the physics that explains how everything works, the nature of the particles that make up matter and the forces with which they interact.
Its types: Electromagnetism, the strong nuclear force, and the weak nuclear force.
Hope this help :)
Which three terms are needed to describe the energy a BASE Jumper has as
she falls toward the ground?
A. Potential
B. Kinetic
C. Gravitational
D. Electromagnetic
Answer:
A.Potential
B.Kinetic
C.Gravitational
Explanation:
I took the test
This figure shows a sinusoidal wave that is traveling from left to right, in the +x-direction. Assume that it is described by a frequency of 57.1 cycles per second, or hertz (Hz).
7.60 cm4.80 cm
A sinusoidal wave lies on an unlabeled coordinate system. One of the wave's maxima lies on the vertical axis. The horizontal distance from the first maximum to the first minimum is labeled 4.80 cm and the vertical distance between a maximum and a minimum is labeled 7.60 cm.
(a)
What is the wave's amplitude (in cm)?
cm
(b)
What is the wavelength (in cm)?
cm
(c)
Calculate the wave's period (in s).
s
(d)
Compute the speed of this wave (in m/s).
m/s
Answer:
a) A = 3.80 cm, b) λ = 9.60 cm, c) T = 1.75 10⁻² s, d) v = 5.48 m / s
Explanation:
The wave is a way of transporting energy and moment without the need to transport the material. They are described by expressions of the type
x = A sin (kx - wt)
where the amplitude A is the distance from the point of zero intensity to the maximum.
Frequency is the number of times the wave oscillates per unit of time
the wavelength is the distance necessary for the wave to start repeating.
a) In the exercise it tells us that the vertical distance from a machismo to a minimum that is worth 7.60 cm
when checking the definition of amplitude is from zero to a maximum, therefore the value given is twice the amplitude
2A = 7.60
A = 3.80 cm
b) the distance between a minimum and the next maximum is 4.80 cm
Using the definition of wavelength the given value corresponds to half wavelength
λ/ 2 = 4.80
λ = 9.60 cm
c) frequency and period are related
f = 1 / T
T = 1 / f
we calculate
T = 1 / 57.1
T = 0.0175 s
T = 1.75 10⁻² s
d) the speed of the wave is related to the frequency and the wavelength
v = λ f
v = 0.0960 57.1
v = 5.48 m / s
An old mining tunnel disappears into a hillside. You would like to know how long the tunnel is, but it's too dangerous to go inside. Recalling your recent physics class, you decide to try setting up standing-wave resonances inside the tunnel. Using your subsonic amplifier and loudspeaker, you find resonances at 5.0 Hz and 6.4 Hz , and at no frequencies between these. It's rather chilly inside the tunnel, so you estimate the sound speed to be 333 m/s .
Answer:
L = 116.6 m
Explanation:
For this exercise we approximate the tunnel as a tube with one end open and the other closed, at the open end there is a belly and at the closed end a node, therefore the resonances occur at
λ = 4L 1st harmonic
λ = 4L / 3 third harmonic
λ = 4L / 5 fifth harmonic
General term
λ = 4L / n n = 1, 3, 5,... odd
n = (2n + 1) n are all integers
They indicate that two consecutive resonant frequencies were found, the speed of the wave is related to the wavelength and its frequency
v = λ f
λ = v / f
we substitute
[tex]\frac{v}{f} = \frac{4L}{n}[/tex]
L = [tex]n \frac{ v}{4f}[/tex]
for the first resonance n = n
L = (2n + 1) [tex]\frac{v}{4f_1}[/tex]
for the second resonance n = n + 1
L = (2n + 3) [tex]\frac{v}{4f_2}[/tex]
we have two equations with two unknowns, let's solve by equating
(2n + 1) \frac{v}{4f_1}= (2n + 3) \frac{v}{4f_2}
(2n + 1) f₂ = (2n +3) f₁
2n + 1 = (2n + 3) [tex]\frac{f_1}{f_2}[/tex]
2n (1 - \frac{f_1}{f_2}) = 3 \frac{f_1}{f_2} -1
we substitute the values
2n (1- [tex]\frac{5}{6.4}[/tex]) = 3 [tex]\frac{5}{6.4}[/tex] -1
2n 0.21875 = 1.34375
n = 1.34375 / 2 0.21875
n = 3
remember that n must be an integer.
We use one of the equations to find the length of the Tunal
L = (2n + 1) \frac{v}{4f_1}
L = (2 3 + 1) [tex]\frac{333}{4 \ 5.0}[/tex]
L = 116.55 m
22. What force is necessary to accelerate a 2500kg car from rest to 20m/s over 10s?
(6 Points)
2N
250N
5000N
50000N
Answer:
50000N
Explanation:
Force = mass × acceleration
= 2500 × 20
= 50000N
A 21 kg mountain lion carries a 3kg cub in it's mouth as it jumps from rest on the ground to the top of a 2 m talk rock. It takes 1 seconds for the mountain lion to jump and reach the top. How much power did the mountain lion exert? I need help to solve for power
Answer:
The power exerted by the mountain lion is 1,472.35 W.
Explanation:
Given;
mass of mountain, m₁ = 21 kg
mass of the cub, m₂ = 3 kg
height jumped by the mountain lion, h = 2 m
time taken for the mountain lion to jump, t = 1 s
Determine the weight of the lions on the top rock;
W = F = (m₁ + m₂)g
F = (21 + 3) x 9.8
F = (24) x 9.8
F = 235.2 N
Determine the final velocity of the mountain rock as it jumped to the top;
v² = u² + 2gh
where;
u is the initial velocity = 0
h is the height jumped = 2 m
v² = 0 + 2 x 9.8 x 2
v² = 39.2
v = √39.2
v = 6.26 m/s
The power exerted by the mountain lion is calculated as;
P = Fv
P = 235.2 x 6.26
P = 1,472.35 W
Therefore, the power exerted by the mountain lion is 1,472.35 W.
When an object is moving on a surface with a lot of texture, how does this affect the amount of friction created?
A. The friction is stopped.
B. Less friction is created.
C. More friction is created.
D. The amount of friction is the same
Answer:
b) less friction is created
Answer:
B
Explanation:
A sphere of uniform density with mass 15 kg and radius 0.6 m is spinning, making one complete revolution every 0.3 s. The center of mass of the sphere has a speed of 3 m/s. (a) What is the rotational kinetic energy of the sphere
Answer:
[tex]473.56\ \text{J}[/tex]
Explanation:
T = Time taken to complete one revolution = 0.3 s
m = Mass of sphere = 15 kg
r = Radius of sphere = 0.6 m
I = Moment of inertia of sphere = [tex]\dfrac{2}{5}mr^2[/tex]
Angular speed of the sphere is
[tex]\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{0.3}\\\Rightarrow \omega=20.94\ \text{rad/s}[/tex]
Rotational kinetic energy is given by
[tex]K_r=\dfrac{1}{2}I\omega^2=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\omega^2\\\Rightarrow K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 15\times 0.6^2\times 20.94^2\\\Rightarrow K_r=473.56\ \text{J}[/tex]
The rotational kinetic energy of the sphere is [tex]473.56\ \text{J}[/tex].