Answer:
Just add all the resistors together
1+2+3=6
6Ω is the total resistance
Explanation:
Which orbit has the highest energy?
n = 1
n = 2
n = 3
Answer:
3
Explanation:
The closer an orbit is to the nucleus the fewer energy
Mark puts an object that has a mass of 58.2kg in a cart that has a mass of 73.00 kg.
Answer:
131.2 kg, to have the same precision as 58.2 kg
Explanation:
A car with mass mc = 1225 kg is traveling west through an intersection at a magnitude of velocity of vc = 9.5 m/s when a truck of mass mt = 1654 kg traveling south at vt = 8.6 m/s fails to yield and collides with the car. The vehicles become stuck together and slide on the asphalt, which has a coefficient of friction of μk = 0.5.
A) Write an expression for the velocity of the system after the collision, in terms of the variables given in the problem statement and the unit vectors i and j.
B) How far, in meters, will the vehicles slide after the collision?
Answer:
a) v(f) = -4i - 5j
b) 4.18 m
Explanation:
The equation to be used for this question is
v(c)m(c) + v(t)m(t) = [m(c) + m(t)] v(f)
if we rearrange and make v(f) subject of formula, then
v(f) = v(c)m(c) + v(t)m(t) / [m(c) + m(t)]
One vehicle is headed towards south and the other vehicle, west when they collide they will travel together in a southwestern direction. This means that both vehicles are traveling in the negative direction taking a standard frame of reference. Thus, we can write the equation in component form by substituting the values as
v(f) = 1225(-9.5i) + 1654(-8.6j) / 1225 + 1654
v(f) = -11637.5i - 14224.4j / 2879
v(f) = -4i - 5j m/s
From the answer,
v(f) = √(4² + 5²)
v(f) = √41
v(f) = 6.4 m/s
And we know that
KE = ½mv²
Fd = umgd
And, KE = Fd, so
½mv² = umgd
½v² = ugd
Making d the subject of formula,
d = v²/2ug
d = 6.4² / 2 * 0.5 * 9.8
d = 41 / 9.8
d = 4.18 m
(a) The velocity of the system after collision is 4.04 i + 4.9 j.
(b)The distance traveled by the vehicles after collision is 1.73 m.
The given parameters;
mass of the car, Mc = 1225 kgvelocity of the car, Vc = 9.5 m/smass of the truck, Mt = 1654 kgvelocity of the truck, Vt = 8.6 m/sApply the principle of conservation of linear momentum to determine the velocity of the system after collision;
[tex]m_1u_x_1 + m_2 u_y_2 = V(m_1 + m_2)\\\\V = \frac{(1225\times 9.5)_x \ + \ (1654 \times 8.6)_y}{m_1 + m_2} \\\\V = \frac{(1225\times 9.5)_x \ + \ (1654 \times 8.6)_y}{1225+ 1654} \\\\V= \frac{(11,637.5)_x \ + \ (14,224.4)_y}{2879} \\\\V = 4.04x \ + 4.94y\\\\V = 4.04i \ + 4.9 j[/tex]
The magnitude of the final velocity of the system is calculated as;
[tex]V = \sqrt{v_x^2 + v_y^2} \\\\V = \sqrt{(4.04)^2 + (4.9)^2} \\\\V = 6.35 \ m/s[/tex]
The change in the mechanical energy of the system;
[tex]\Delta K.E = K.E_f - K.E_i\\\\[/tex]
The initial kinetic energy of the cars before collision is calculated as;
[tex]K.E_i = \frac{1}{2} m_1u_1_x^2 \ + \frac{1}{2} m_1u_2_y^2 \\\\K.E_i = \frac{1}{2} (1225)(9.5)^2\ + \frac{1}{2} (1654)(8.6)^2\\\\K.E_i = 55,278.13_x \ + \ 61,164.92_y\\\\K.E_i = \sqrt{55,278.13^2 \ + \ 61,164.92^2} \\\\K.E_i = \sqrt{6,796,819,094.9} \\\\K.E_i = 82,442.82 \ J[/tex]
The final kinetic energy of the system;
[tex]K.E_f = \frac{1}{2} (m_1 + m_2)V^2\\\\K.E_f = \frac{1}{2} (1225 + 1654)(6.35)^2\\\\K.E_f = 58,044.24 \ J[/tex]
The change in kinetic energy is calculated as;
[tex]\Delta K.E = K.E_f -K.E_i\\\\\Delta K.E= (58,044.24) - (82,442.82)\\\\\Delta K.E = -24,398.58 \ J[/tex]
Apply the principle of work-energy theorem, to determine the distance traveled by the vehicles after collision;
[tex]W = \Delta K.E\\\\- \mu Fd = - 24,398.58\\\\\mu mgd= 24,398.58\\\\d = \frac{24,398.58}{\mu mg} \\\\d = \frac{24,398.58}{0.5 \times 9.8(1225 + 1654)} \\\\d = 1.73 \ m[/tex]
Thus, the distance traveled by the vehicles after collision is 1.73 m.
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A charge of +80 µC is placed on the x axis at x = 0. A second charge of –50 µC is placed on the x axis at x = 50 cm. What is the magnitude of the electrostatic force on a third charge of 4.0 µC placed on the x axis at x = 30 cm?
Answer:
Explanation:
77
The magnitude of the electrostatic force on a third charge of 4.0 µC placed on the x-axis at x = 30 cm is equal to 77 N.
What is coulomb's law?According to Coulomb’s law, the force of repulsion or attraction between two charged bodies is the multiplication of their charges and is inversely proportional to the square of the distance between them.
The magnitude of the electric force can be written as follows:
[tex]F =k \frac{q_1q_2}{r^2}[/tex]
where k has a value of 9 × 10⁹ N.m²/C².
Given, the first charge, q₁ = + 80 ×10⁻⁶ C
The second charge , q₂ = - 50 × 10⁻⁶ C
The third charge, q₃ = + 4 ×10⁻⁶ C
The distance between these two charges (q₁ and q₃) , r = 0.30 m
The distance between these two charges (q₂ and q₃) , r = 0.20m
The magnitude of the electrostatic force on the third charges will be:
[tex]F =9\times 10^9 (\frac{80\times 10^{-6}C\times 4 \times 10^{-6}C}{(0.30)^2} ) +9\times 10^9 (\frac{50\times 10^{-6}C\times 4 \times 10^{-6}C}{(0.20)^2} )[/tex]
F = 77 N
Therefore, the magnitude of the electrostatic force is equal to 77 N.
Learn more about Coulomb's law, here:
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A block of mass m = 2.5 kg is attached to a spring with spring constant k = 730 N/m. It is initially at rest on an inclined plane that is at an angle of θ = 21° with respect to the horizontal, and the coefficient of kinetic friction between the block and the plane is μk = 0.19. In the initial position, where the spring is compressed by a distance of d = 0.11 m, the mass is at its lowest position and the spring is compressed the maximum amount. Take the initial gravitational energy of the block as zero.
A) What is the block's initial mechanical energy?
B) If the spring pushes the block up the incline, what distance, L, in meters will the block travel before coming to rest? The spring remains attached to both the block and the fixed wall throughout its motion.
Answer:
A) Em = 4.41 J
B) L = 0.33m
Explanation:
A) The total mechanical energy of the block is the elastic potential energy due to the compressed spring. The gravitational energy is zero. Then you have:
[tex]E_m=\frac{1}{2}k(\Delta x)^2[/tex]
k: constant's spring = 730 N/m
Δx: distance of the compression = 0.11m
You replace the values of k and Δx:
[tex]E_m=\frac{1}{2}(730N/m)(0.11m)^2=4.41\ J[/tex]
B) To find the distance L traveled by the block you take into account that the total mechanical energy of the block is countered by the work done by the friction force, and also by the work done by the gravitational energy.
Then, you have:
[tex]E_m-W_f-W_g=0\\\\W_f=(\mu Mg cos\theta)L\\\\W_g=(Mgsin\theta)L[/tex]
μ: coefficient of kinetic friction = 0.19
g: gravitational acceleration = 9.8m/s^2
M: mass of the block = 2.5kg
θ: angle of the inclined plane = 21°
You replace the values of all parameters:
[tex]E_m-W_f-W_g=0\\\\4.41-(0.19)(2.5kg)(9.8m/s^2)(cos21\°)L-(2.5kg)(9.8m/s^2)(sin21\°)L=0\\\\4.41-4.34L-8.78L=0\\\\4.41-13.12L=0\\\\L=0.33m[/tex]
hence, the distance L in which the block stops is 0.33m
(a) The block's initial mechanical energy on the given position is 4.42 J.
(b) The distance traveled by the block when it is pushed by the spring before coming to rest is 1.02 m.
The given parameters;
mass of the block, m = 2.5 kgspring constant, k = 730 N/mangle of inclination, θ = 21°coefficient of friction, μ = 0.19compression of the spring, x = 0.11 mThe block's initial mechanical energy is calculated as follows;
[tex]E = K.E _i + P.E_i\\\\E = \frac{1}{2} mv^2 \ + \ \frac{1}{2} kx^2\\\\E = \frac{1}{2} m (0)^2 \ + \ \frac{1}{2} \times 730 \times (0.11)^2\\\\E = 4.42 \ J[/tex]
The block will travel up if the energy applied by the spring is greater than the work-done by frictional force on the block.
The work-done on the block by the frictional force is calculated as follows;
[tex]W_f = F_k \times d\\\\W_f= \mu_k F_n \times d\\\\W_f = \mu_k mgcos(\theta) \times d\\\\W_f = (0.19)(2.5)(9.8)cos(21) \times d \\\\W_f = 4.346 d[/tex]
Apply work-energy theorem;
[tex]4.346d = 4.42\\\\d = \frac{4.42}{4.346} = 1.02 \ m[/tex]
Thus, the distance traveled by the block when it is pushed by the spring before coming to rest is 1.02 m.
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Do charges exit a circuit with less energy than they had when they entered the circuit
Ursula is a student of Kantian philosophy who argues that we should never use people as a means to an end unless
Answer: we embrace and are simultaneously respecting their rational autonomy
Explanation:
Ursula is a student of Kantian philosophy who argues that we should never use people as a means to an end unless we embrace and are simultaneously respecting their rational autonomy
Which is not a difference between chemical reactions and nuclear reactions?
A.The subatomic particles involved in the reaction is different.
B.The amount of energy absorbed during the reaction is different.
C.The masses of the reactants and products differ.
D.The amount of energy released during the reaction is different.
Answer:
B.The amount of energy absorbed during the reaction is different
Explanation:
I just took the test