Answer:
+1.46×10¯⁶ C
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C
Force (F) = 0.615 N
Distance apart (r) = 0.750 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Charge 2 (q₂) =?
The value of the second charge can be obtained as follow:
F = Kq₁q₂ / r²
0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²
0.615 = 236700 × q₂ / 0.5625
Cross multiply
236700 × q₂ = 0.615 × 0.5625
Divide both side by 236700
q₂ = (0.615 × 0.5625) / 236700
q₂ = +1.46×10¯⁶ C
NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C
Answer:
+1.46
Explanation:
acellus
What is acceleration?
[tex]\huge\boxed{✍ ACCELERATION\hookrightarrow}[/tex]
⎆ Acceleration is defined as the ⇻ rate of change of velocity with respect to time / rate of change of velocity in unit time.
⎆ It's a vector quantity (it has both magnitude & direction).
ʰᵒᵖᵉ ⁱᵗ ʰᵉˡᵖˢ
# ꧁❣ RainbowSalt2²2² ࿐
A ball of mass 8kg falls from rest from a height of 100m. Neglecting air resistance, calculate its total energy after falling a distance of 40m.
Answer:
Total energy, T= 4704 Joules
Explanation:
Given the following data;
Mass = 8kg
Initial height, h1 = 100m
Final height, h2 = 40 m
We know that acceleration due to gravity is equal to 9.8 m/s².
To find the total energy, T;
T = mg(h1 - h2)
T = 8 * 9.8 * (100 - 40)
T = 78.4 * 60
Total energy, T= 4704 Joules
A 1800-kg Jeep travels along a straight 500-m portion of highway (from A to B) at a constant speed of 10 m/s. At B, the Jeep encounters an unbanked curve of radius 50 m. The Jeep follows the road from B to C traveling at a constant speed of 10 m/s while the direction of the Jeep changes from east to south. What is the magnitude of the acceleration of the Jeep as it travels from A to B
Answer:
a = 0 m/s²
Explanation:
From A to B the jeep travels in a straight line with a constant speed of 10 m/s. Acceleration is defined as the rate of change in velocity. In the current scenario, the velocity of the jeep remains constant. The direction of the jeep does not change as it travels in the straight path. Also, the magnitude of the velocity is constant at 10 m/s.
Therefore, there is no change in the velocity of the jeep from A to B. Hence, the will be no acceleration from A to B.
a = 0 m/s²
Describe the processes that take place inside the nucleus for the following decays; Beta- decay and Beta+ decay.
Answer:
Explanation:
(a) Electron Emission (Beta- decay):
When an unstable nucleus decays by the emission of Beta- particle, its charge number ‘Z’ increases by 1 but, its mass number ‘A’ remains unchanged. The transformation is represented by the equation:
[tex]_zX^A\ -------->\ _{Z+1}Y^A\ +\ _{-1}e^0[/tex]
It is called ‘Negative Beta Decay’. It is more common than alpha decay.
Example:
[tex]_6C^{14}\ -------->\ _{7}N^{14}\ +\ _{-1}e^0[/tex]
Note:
There are no electrons in a nucleus so, with the emission of a particle, one of the neutrons is converted to a proton and an electron.
[tex]_0n^1\ --------->\ _1P^1\ +\ _{-1}e^0[/tex]
(b) Positron Emission (Beta+ decay):
When an unstable nucleus decays by the emission of the positron, its charge number ‘Z’ decreases by 1 but, its mass number ‘A’ remains unchanged. The transformation is represented by the equation:
[tex]_zX^A\ -------->\ _{Z-1}Y^A\ +\ _{+1}e^0[/tex]
Examples:
[tex]_{15}P^{30}\ -------->\ _{14}Si^{30}\ +\ _{+1}e^0[/tex]
Note:
Inside the nucleus, only a proton can be transformed into a neutron with the emission of a positron (anti-particle of electron)
[tex]_1P^1\ -------->\ _0n^1\ +\ _{+1}e^0[/tex]
A person is 88 J of work lifting a grocery bag straight off the ground to a height of 1.7 m what is a grocery bags mass
Answer:
5.277 kg
Explanation:
Since the formula for work is W = F * d and we are given distance and work, the force on the grocery bag is 88 = F * 1.7 F = 88 / 1.7 = 51.765 N.
We also know that force follows the equation F = m * a. Since the constant gravitational acceleration on earth is 9.81 m / s^2, we can find the mass through 51.765 = m * 9.81 m = 51.765/9.81 = 5.277 kg
Convert 13.7g to mg
A 1500 kg tractor pulls a 750 kg trailer north and applies a 2250 N force on it. What is the force on the tractor?
Answer:
law of Action and Reaction F = 2250 N
Explanation:
The tractor and the trailer are two bodies that interact, therefore, by the law of Action and Reaction, the force that one applies on the other is equal to the force that the second body (trailer) applies on the first (tractor), but with opposite direction
F = 2250 N
directed from trailer to tractor
23. Lunar Gravity Compare the force holding a
10.0-kg rock on Earth and on the Moon. The
gravitational field on the Moon is 1.6 N/kg.
Explanation:
Given
mass of the rock is 10 kg
Force requires to hold the rock is equal to its weight
Weight is given by the product of mass and acceleration due to gravity
Weight on the earth surface
[tex]\Rightarrow W_e=10\times 9.8\\\Rightarrow W_e=98\ N[/tex]
Weight on the moon surface
[tex]\Rightarrow W_m=1.6\times 10\\\Rightarrow W_m=16\ N[/tex]
So, the force holding the rock on earth is approximately 6 times the force on the moon.
if something is frictionless does it have thermal energy
Answer:
No
Explanation:
because there is no pressure
A 1.5 V battery is connected to a 1,000 μF capacitor in series with a 150 Ω resistor. a. What is the maximum current that flows through the resistor during charging? b. What is the maximum charge on the capacitor? c. How long does the capacitor take to reach a potential of 1.0V?
Answer:
[tex]0.01\ \text{A}[/tex]
[tex]0.0015\ \text{C}[/tex]
[tex]0.0608\ \text{s}[/tex]
Explanation:
[tex]V_0[/tex] = Voltage = 1.5 V
[tex]C[/tex] = Capacitance = [tex]1000\ \mu\text{F}[/tex]
[tex]R[/tex] = Resistance = [tex]150\ \Omega[/tex]
Current is given by
[tex]I=\dfrac{V_0}{R}\\\Rightarrow I=\dfrac{1.5}{150}\\\Rightarrow I=0.01\ \text{A}[/tex]
Current flowing in the resistor is [tex]0.01\ \text{A}[/tex].
Charge is given by
[tex]Q=CV\\\Rightarrow Q=1000\times 10^{-6}\times 1.5\\\Rightarrow Q=0.0015\ \text{C}[/tex]
The charge on the capacitor is [tex]0.0015\ \text{C}[/tex].
Voltage is given by
[tex]V=V_0e^{-\dfrac{t}{RC}}\\\Rightarrow t=-RC\ln\dfrac{V}{V_0}\\\Rightarrow t=-150\times 1000\times 10^{-6}\times\ln\dfrac{1}{1.5}\\\Rightarrow t=0.0608\ \text{s}[/tex]
Time taken to reach 1 V is [tex]0.0608\ \text{s}[/tex].
Please help with all three questions
Answer:
The correct answer is D C A
A metal spherical shell with inner radius 14 cm and outer radius 24 cm has a net charge of 1= -3 nC. At the center of the shell is a small particle with charge Q2= -7 nC. What is the charge density in (C/m2) on the outer surface of the spherical shell?
Answer:
I don't no the answer sorry
Pls help! What is the average speed of an airplane that travels from New York to Los Angeles, a total distance of 4800 km, in 6.0 hours?
A.
800 km/hr
B.
80 km/hr
C.
0.00125 km/hr
D.
28,800 km/hr
Answer:
800 meters per hour
Explanation:
800 meters per hour is the average speed of an airplane that travels from New York to Los Angeles, a total distance of 4800 km, in 6.0 hours.
Any is 160 cm tall and stands 50 cm in front of a plane mirror. The image of Amy is ?
Answer:
Option A. 160 cm tall and 100 cm in front of Amy
Explanation:
To know the the correct answer to the question, we must recognise that the image formed by plane mirror is:
I. Same size as the object.
II. Same distance as the object from the mirror.
III. Laterally inverted.
IV. Virtual
With the above information, we obtained the answer to the question as follow:
Object height = 160 m
Thus,
Image height = 160 m
Distance of object from mirror = 50 cm
Therefore,
Distance of image from the mirror = 50 cm
Distance between the object and image = 50 + 50 = 100 cm
Thus, option A gives the correct answer to the question because the image of Amy is 160 cm tall and 100 cm in front of her.
write the formula of mechanical advantage
Answer:
the formula of mechanical advantage is
MA = load / effort
VR = effort distance / load distance
hope it is helpful to you
Water in an electric teakettle is boiling. The power absorbed by the water is 0.90 kW. Assuming that the pressure of vapor in the
kettle equals atmospheric pressure, determine the speed of effusion of vapor from the kettle's spout if the spout has a cross-
sectional area of 1.60 cm2. Model the steam as an ideal gas.
Answer:
v = 4.233 m/s
Explanation:
By applying the rate of boiling from [tex]Q= mL_v[/tex];
the rate of the boiling can be described as:
[tex]\mathcal{P} = \dfrac{Q}{\Delta t} \\ \\ \mathcal{P} = \dfrac{mL_v}{\Delta t}[/tex]
The mode of the steam (water vapor) as an ideal gas can be illustrated by formula:
[tex]P_oV_o = nRT[/tex] --- (1)
where;
n = number of moles;
[tex]n = \dfrac{mass (m)}{Molar mass (M)}[/tex]
Then; equation (1) can be rewritten as:
[tex]P_oV_o = (\dfrac{m}{M}) RT \\ \\ \dfrac{P_oV}{\Delta T} = \dfrac{m}{\Delta t} ( \dfrac{RT}{M})[/tex]
∴
[tex]\dfrac{m}{\Delta t} = \dfrac{\mathcal{P}}{L_v}[/tex]
Then:
[tex]P_o \times A \times v= \dfrac{\mathcal{P}}{L_v}\Big ( \dfrac{RT}{M }\Big)[/tex]
making (v) the subject of the formula:
[tex]v= \Big ( \dfrac{\mathcal{P} RT}{M\times L_v \times P_o \times A }\Big)[/tex]
Given that:
[tex]\mathcal{P}[/tex] = 0.90 kW = 900 W
R(rate constant) = 8.314 J/mol.K
Temperature at 100° C = 373K
For water vapor:molar mass= 18.015 g/mol ≅ 0.0180 kg/mol
Latent heat of vaporisation [tex]L_v[/tex] = 2.26 × 10⁶ J/kg
Atmospheric pressure [tex]P_o = 1.013 \times 10^6 \ N/m^2[/tex]
Cross sectional area A =1.60 cm² = 1.60 × 10⁻⁴ m²
[tex]v= \Big ( \dfrac{900 W (8.314 \ J/mol.K)(373)}{0.0180 \ kg/mol) (2.26 \times 10^6 \ J/kg) (1.013 \times 10^5 \ N/m^2)(1.60 \times 10^{-4} \ m^2)}\Big)[/tex]
v = 4.233 m/s
Two people are carrying a uniform 704.0 N log through the forest. Bubba is 2.2 m from one end of the log (x), and his partner is 0.9 m from the other end (y). The log is 6.2 m long (z). What weight is Bubba supporting
Answer:
F₁ = 499.61 N , this is the force that Bubba support
Explanation:
The trunk is in equilibrium with the two forces applied by man, let's use the equilibrium relation
let's set a reference frame at the extreme left and assume that the counterclockwise rotations are positive
Let's write the expression for the translational equilibrium
subscript 1 is for Bubba's mass and subscript 2 for his partner
F₁ + F₂ -W = 0
F₁ + F₂ = W
the expression for rotational equilibrium
∑ τ = 0
F₁ 2.2 + F₂ (6.2-0.9) - W 6.2/2 = 0
2.2 F1 + 5.3 F2 = 3.1 W
let's write our system of equations
F₁ + F₂ = W
2.2 F₁ + 5.3 F₂ = 3.1 W
we solve for F₁ in the first equation and substitute in the second
F₁ = W-F₂
2.2 (W- F₂) + 5.3 F₂ = 3.1 W
F₂ ( -2.2 +5.3) = W (3.1 - 2.2)
F₂ = 704 0.9 / 3.1
F₂ = 204.39 N
This is the force that the partner supports
we look for F1
F₁ = W-F₂
F₁ = 704 - 204.39
F₁ = 499.61 N
This is the force that Bubba support
what is acceleration?
Answer:
the rate of change of velocity per unit of time.
Answer:
accelerate is rate of chang of velocity
Two events are observed in a frame of reference S to occur at the same space point, the second occurring 1.80 s after the first. In a frame S′ moving relative to S, the second event is observed to occur 2.05 s after the first. What is the difference between the positions of the two events as measured in S^?
Answer:
The difference between the positions of the two events as measured in = 3.53 *10^8 m/s
Explanation:
As we know -
[tex]\Delta x = -\gamma \mu\Delta t[/tex]
Here,
[tex]\Delta x[/tex] is the difference between the positions of the two events as measured in S^
[tex]\gamma[/tex] [tex]= \frac{1}{\sqrt{1-\frac{\mu^2}{c^2} } }[/tex]
And
[tex]\mu[/tex] = 0.547 c
Substituting the given values in above equation, we get -
[tex]\Delta x = (0.547 c)*\frac{1}{\sqrt{1-\frac{\mu^2}{c^2} } }*2.15\\\Delta x = (0.547 c)*\frac{1}{\sqrt{1-\frac{(0.547 c)^2}{c^2} } }*2.15\\\Delta x = (0.547 *3*10^8)*\frac{1}{\sqrt{(1-\(0.547 )^2 } }*2.15\\\Delta x = 3.53 *10^8[/tex]meter per second
URGENT!!!!!!!!!: If an electromagnetic wave has a frequency of 6 x 10^5 Hz, what is its
wavelength? Use v/f. The speed of light is 3 x 10^8 m/s.
Answer:
A
Explanation:
because the speed divide by the frequency is equal to the wavelength(in meters)
5×10² m
Phenotype is the actual genes that make a person True or False
Answer:
True?
Explanation:
Genotype and phenotype are two fundamental terms in the science of genetics. ... An organism's genotype is the set of genes in its DNA responsible for a particular trait. An organism's phenotype is the physical expression of those genes.
The term "phenotype" refers to the observable physical properties of an organism; these include the organism's appearance, development, and behavior. An organism's phenotype is determined by its genotype, which is the set of genes the organism carries, as well as by environmental influences upon these genes.
This is why I think it's true. You don't have to read the rest,but that's why I know.
Answer:
The answer is true
Explanation:
I dont know but trust is probably true.
Convert 0.00553s to cs
Answer: 0.553
Explanation:
Question 6 of 10
The mass number of a nucleus decreases during which nuclear reactions?
A. Nuclear fusion and alpha decay
B. Nuclear fission and beta decay (electron emission)
C. Nuclear fusion and beta decay (positron emission)
D. Nuclear fission and alpha decay
Answer:
D. Nuclear Fission and alpha decay
A region of vacuum contains both a uniform electric field with magnitude E and a uniform magnetic field with magnitude B.
Part A What is the ratio E/B if the energy density for the magnetic field equals the energy density for the electric field? Express your answer in volts per meter-tesla to three significant figures. IVO ADV A O O ?
Part B If E = 750 V/m, what is B, in teslas, if the magnetic field and electric-field energy densities are equal? Express your answer in teslas. O ALQ * R O 2 ?
Answer:
A) E / B = 2.99 10⁸ V/ mT, B) B = 2.50 10⁻⁶ T
Explanation:
A) the expressions for the energy densities are:
u_E = ½ ε₀ E²
u_B = ½ B² /μ₀
indicate that the two densities are equal
½ ε₀ E² = ½ B² /μ₀
E / B = 1 /[tex]\sqrt{\epsilon_o \ \mu_o }[/tex]
we calculate
E / B = 1 / √( 8.85 10⁻¹² 4π 10⁻⁷
E / B = 1 /√( 11.1212 10⁻¹⁸)
E / B = 0.29986 10⁹9
E / B = 2.99 10⁸ V/ mT
B) for this case E = 750 V / m, ask the magnetic field
E / B = 1 /[tex]\sqrt{\epsilon_o \ \mu_o}[/tex]
B = E [tex]\sqrt{\epsilon_o \ \mu_o}[/tex]
we calculate
B = 750 √(8.85 10⁻¹² 4π 10⁻⁷
B = 750 3.3348 10⁻⁹
B = 2.50 10⁻⁶ T
Energy densities can be expressed as
u_E = ½ ε₀ E²
u_B = ½ B² /μ₀
When the two densities are equal we use the formula
½ ε₀ E² = ½ B² /μ₀
E / B = 1 / μ₀
E / B = 1 / √( 8.85 10⁻¹² 4π 10⁻⁷)
E / B = 1 /√( 11.1212 10⁻¹⁸)
E / B = 0.29986 10⁹
E / B = 2.99 10⁸ V/ mT
If the magnetic field and electric-field energy densities are equal and E = 750 V / m, we can deduce thatE / B = 1 / √ ξ₀ μ₀
B = E √ ξ₀ μ₀
We substitute the values into the equation
B = 750 √(8.85 10⁻¹² 4π 10⁻⁷)
B = 750 3.3348 10⁻⁹
B = 2.50 10⁻⁶ T
Read more on https://brainly.com/question/23360987
A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A constant tangential force of 290 N applied to its edge causes the wheel to have an angular acceleration of 0.854 rad/s2.
(a) What is the moment of inertia of the wheel?
kg · m2
(b) What is the mass of the wheel?
kg
(c) If the wheel starts from rest, what is its angular velocity after 5.80 s have elapsed, assuming the force is acting during that time?
rad/s
Answer:
c
Explanation:
units c mark me brainless k
A man has 887.5 J of kinetic energy while running with a velocity of 5 m/s. What is his mass?
Answer:
The mass of the man is 71 kg
Explanation:
Given;
kinetic energy of the man, K.E = 887.5 J
velocity of the man, v = 5 m/s
The mass of the man is calculated as follows;
K.E = ¹/₂mv²
where;
m is the mass of the man
2K.E = mv²
m = 2K.E / v²
m = (2 x 887.5) / (5)²
m = 71 kg
Therefore, the mass of the man is 71 kg
PLEASE HELP ME WITH THOSE TWO QUESTIONS
A 122 Ohm, 232 Ohm, and 500 Ohm resistors are all connected in series with a 12 V battery. What is the current flowing through them?
A 100 Ohm, 200 Ohm, and 300 Ohm resistor are all connected in series to a 12 V battery. What is the potential difference drop across the 100 Ohm resistor?
Answer:
(a) 0.014 A
(b) 2 V
Explanation:
(a) Applying
V = IR'...................... Equation 1
Where V = Voltge, I = current, R = total resistance.
make I the subject of the equation
I = V/R'................... Equation 2
From the question,
Given: V = 12 V, R' = (122+232+500) ohms (The resistance are connected in series) = 854 ohms
Substitute these values into equation 2
I = 12/854
I = 0.014 A
(b) Applying
V' = V(R₁)/(R₁+R₂+R₃)...................... Equation 3
Where V' = Voltage across the 100 ohms resistor.
From the question,
V = 12V, R₁ = 100 ohm, R₂ = 200 ohm, R₃ = 300 ohm.
Substitute these values into equation 3
V' = (12×100)/(100+200+300)
V' = 1200/600
V' = 2 V
PLS HELPP MEEE
Fiber-optic cables rely on total internal reflection.
Please select the best answer from the choices provided
T
F
Typical novae occur when (a) a red-giant star ejects a planetary nebula; (b) two neutron stars merge, forming a more massive neutron star; (c) an extremely massive star collapses, and also ejects its outer atmosphere; (d) matter accreted from a companion star unstably ignites on the surface of a white dwarf; or (e) a neutron star’s magnetic field becomes strong enough to produce two oppositely directed jets of rapidly moving particles.
Answer:
I think its d matter accreted from a companion star unstable ignites on the surface of a white dwarf.
A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 5.00 m at constant speed. If the coefficient of kinetic friction between the desk and the floor is 0.400, how much work did she do
Answer:
1568 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 80 Kg
Distance (d) = 5 m
Coefficient of kinetic friction (μ) = 0.4
Workdone (Wd) =?
Next, we shall determine the normal reaction. This can be obtained as follow:
Mass (m) = 80 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Normal reaction (N) =?
N = mg
N = 80 × 9.8
N = 784 N
Next, we shall determine force of friction. This can be obtained as follow:
Coefficient of kinetic friction (μ) = 0.4
Normal reaction (N) = 784 N
Force of friction (F) =?
F = μN
F = 0.4 × 784
F = 313.6 N
Finally, we shall determine the work done. This can be obtained as follow:
Distance (d) = 5 m
Force of friction (F) = 313.6 N
Workdone (Wd) =?
Wd = F × d
Wd = 313.6 × 5
Wd = 1568 J
Thus, the workdone is 1568 J