Answer:
dependent: the outcome of the experience
independent variable: everything literaly.
Independent is where you change some variables and see the result
Dependent is literaly the result, or the outcome dependent on the exprience.
Explanation:
I got u.
A spring attached to a mass is at rest in the initial position (not shown). The spring is compressed in position A and is then released, as shown in position B. Which equation describes conservation of energy in position A?
Answer:
Explanation:
When the spring is compressed, it is compressed to its amplitude (whereas equilibrium is the spring's natural length with no mass attached to it and displacement is the spring's reaction to a mass hung on the end of it without any "extra" pushing or pulling on the mass). It is at the amplitude where the spring experineces max potential energy, which is choice 2, E = mph
Answer:
its c
Explanation:
...
Example Problem
The potential energy of an object is given by U(x) = 8x2 - x4, where U is in joules and x is in
(a) Determine the force acting on this object.
(b) At what positions is this object in equilibrium?
(c) Which of these equilibrium positions are stable and which are unstable?
metres.
111 Unit 2 Concepts and Definitions Prof Mark Lester
Exam Part B Example
A neutron of mass m moving with velocity v collides head-on and elastically with a stationary nucleus of mass M.
(a) Show that the velocity of the nucleus after the collision, U, is given by
U= 2m v (m+M)
(b) Hence show that the neutron loses a fraction f of its energy where
f= 4mM (m+M)
10marks 5 marks
(c) A fast neutron enters a target of carbon nuclei which may be assumed to have masses 12 times that of the neutron. How many head-on collisions will it take
before the neutron loses 95% of its energy?
4 marks
(d) Suggest one reason why in a real reactor a neutron is likely to make more
collisions with the moderator nuclei before losing this much energy
2
1 mark
Answer:
Part A
a) F = -16x + 4, b) x = 0.25 m, c) STABLE
Explanation:
Part A
a) Potential energy and force are related
F = [tex]- \frac{dU}{dx}[/tex]- dU / dx
F = - (8 2x -4)
F = -16x + 4
b) The object is in equilibrium when the forces are zero
0 = -16x + 4
x = 4/16
x = 0.25 m
c) An equilibrium position is called stable if with a small change in position, the forces make it return to the initial position, in case the forces make it move away it is called unstable.
In this case there is only one equilibrium point
by changing the position a bit
x ’= x + Δx
we substitute
F ’= - 16 x’ + 4
F ’= - 16 (x + Δx) + 4
F ’= (-16x +4) - 16 Δx
at equilibrium position F = 0
F ’= 0 - 16 Δx
we can see that the body returns to the equilibrium position, therefore it is STABLE
PART B
This is an exercise in body collisions, let's define the system formed by the two bodies in such a way that the forces during the collisions are internal and the moment is conserved
initial instant. Before the shock
p₀ = m v
final instant. After the crash
p_f = (m + M) v_f
We have two possibilities: an elastic collision in which the bodies separate, each one maintaining its plus, and an INELASTIC collision where the neutron is absorbed by the nucleus and the final mass is M '= m + M, in this case they indicate that the collision is elastic
p₀ = pf
mv = mv ’+ M v_f
in the case of the elastic collision, the kinetic energy is conserved
K₀ = K_f
½ m v² = ½ m v’² + ½ M v_f²
we write the system of equations
mv = mv ’+ M v_f (1)
m (v² -v'²) = M v_f ²
m (v - v ’) = M v_f
m (v-v ’) (v + v’) = M v_f
v + v ’= v_f
we substitute in equation 1 and solve
v ’=[tex]\frac{m -M }{m+M } \ vo[/tex]
v_f = [tex]\frac{2m}{m+M} \ v_o[/tex]
the mechanical energy of the neutron is
initial
Em₀ = K = ½ m v²
final moment
Em_f = K + U = ½ m v_f ² + U
U is the energy lost in the collision
total energy is conserved
Em₀ = Em_f
½ m v² = ½ m v_f ² + U
U = ½ m (v² -v_f ²)
U = ½ m [v² - ( [tex]\frac{m-M}{m+M}[/tex] v)² ]
U = ½ m v² [1- ( [tex]\frac{m-M}{m+M}[/tex] )² ]
U = ½ m v2 [ [tex]\frac{2M}{m+M}[/tex]]
U = [tex]\frac{2 mM}{m +M } \ v^2[/tex]
Let's do the same calculations for the nucleus
initial Em₀ = 0
final Em_f = K + U = ½ M v_f ² + U
Em₀ = Em_f
0 = K + U
U = -K
U = - ½ M v_f ²
U = - ½ M [ [tex]\frac{2m}{m+M} \ v[/tex] ]²
U = [tex]\frac{2 m M }{m+M} \ v^2[/tex]
We can see that we obtain the same result, that is, the potential energy lost by the neutron is equal to the potential energy gained by the nucleus.
b) the fraction of energy lost
f = U / Em₀
f = 4 m M / m + M
c) let's calculate the fraction of energy lost in a collision
m = 1.67 10⁻²⁷ kg
M = 12 1.67 10⁻²⁷= 20 10⁻²⁷ kg
f = 4 1.6 20 / (1.6+ 20) 10⁻²⁷
f = 5.92 10⁻²⁷ J
the energy of a fast neutron is greater than 1 eV
Eo = 1 eV (1.67 10⁻¹⁹ J / 1eV) = 1.67 10⁻¹⁹ J
Let's use a direct portion rule if in a collision f loses in how many collisions it loses 0.95Eo
#_collisions = 0.95 Eo / f
#_collisions = 0.95 1.67 10⁻¹⁹ / 5.92 10⁻²⁷
#_collisions = 2.7 10⁷ collisions
Carlos is playing darts. He throws a dart at the bullseye, but it hits the outer ring of the dartboard instead. He aims for the bullseye again and hits it with a second dart. Carlos threw three more darts which landed close to his first shot. What does the second throw indicate about his accuracy and precision? The second throw shows poor accuracy but high precision. The second throw shows accuracy but poor precision. The second throw shows both accuracy and precision. The second throw shows poor accuracy and poor precision.
Answer:
The second throw shows both accuracy and precision.
Explanation:
The second throw shows both accuracy and precision because he knew in which degree and the amount of force that can be applied to the dart so that it hits the center of the bullseye or close to it. Due to knowing of these two things the Carlos is able to hit the dart close to the target at the second time so in this way he got both accuracy and precision at the second turn of throwing darts.
Answer:
B-The second throw shows accuracy but poor precision.
Explanation:
how to solve for resistors
9514 1404 393
Answer:
A1 = 3A, A2 = 1.5A
Effective resistance = 2Ω
Explanation:
When the switch is closed, the voltage across each resistor is 6V, so the current through it (A2) is ...
A2 = 6V/(4Ω) = 1.5A
There are two parallel paths, each with that current, so the current from the battery is ...
A1 = A2 +A2 = 1.5A +1.5A = 3.0A
Then the effective resistance is ...
Reff = 6V/(3.0A) = 2.0Ω
The solution to the circuit is ...
A1 = 3A, A2 = 1.5A
Effective resistance = 2Ω
Tony walks at an average speed of 70 m/min from his home to
school. If the distance between his home and the school is
2100 m, how much time does it take for Tony to walk to
school?
min
Answer:
The answer is 30 min
Explanation:
t = s/v
v = 70 m/min, s = 2100 m
t = 2100/70 = 30 min.
Hope it helps you! ＼(^ᴥ^)／
Define equilibrium and explain first condition of equilibrium. Pease make it fast...
Answer:
First Condition of Equilibrium
For an object to be in equilibrium, it must be experiencing no acceleration. This means that both the net force and the net torque on the object must be zero. Here we will discuss the first condition, that of zero net force. ... Fnet=0 F net = 0.
10. If the mass of the Earth is... increased by a factor of 2, then the Fgrav is ______________ by a factor of _______. ... increased by a factor of 3, then the Fgrav is ______________ by a factor of _______. ... decreased by a factor of 4, then the Fgrav is ______________ by a factor of _______.
Answer:
If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.
If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.
If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4
Explanation:
In order to solve this question, we must take into account that the force of gravity is given by the following formula:
[tex]F_{g0}=G \frac{mM_{E0}}{r^{2}}[/tex]
So if the mass of the earth is increased by a factor of 2, this means that:
[tex]M_{Ef}=2M_{E0}[/tex]
so:
[tex]F_{gf}=G \frac{2mM_{E0}}{r^{2}}[/tex]
Therefore:
[tex]\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}[/tex]
When simplifying we end up with:
[tex]\frac{F_{gf}}{F_{g0}}=2[/tex]
so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.
If the mass of the earth is increased by a factor of 3
So if the mass of the earth is increased by a factor of 2, this means that:
[tex]M_{Ef}=3M_{E0}[/tex]
so:
[tex]F_{gf}=G \frac{3mM_{E0}}{r^{2}}[/tex]
Therefore:
[tex]\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}[/tex]
When simplifying we end up with:
[tex]\frac{F_{gf}}{F_{g0}}=3[/tex]
so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.
If the mass of the earth is decreased by a factor of 4
So if the mass of the earth is decreased by a factor of 4, this means that:
[tex]M_{Ef}=\frac{M_{E0}}{4}[/tex]
so:
[tex]F_{gf}=G \frac{mM_{E0}}{4r^{2}}[/tex]
Therefore:
[tex]\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}[/tex]
When simplifying we end up with:
[tex]\frac{F_{gf}}{F_{g0}}=\frac{1}{4}[/tex]
so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.
Define Circular motion and its types !
[tex] \green{\huge{\red{\boxed{\green{\mathfrak{QUESTION}}}}}} [/tex]
Define Circular motion and its types
[tex] \huge\green{\boxed{\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}}}[/tex]
[tex]{ \red{ \bold {\mathbb {\textit{Circular \: motion}}}}}[/tex]
MOTION OF THE BODY CIRCULAR PATH IS CALLED CIRCULAR MOTION
A tengential is always responsible for change in speedIf the Direction of acc and velocity is same then body's speed incIf they r in oppsite direction then Body's speed dec .If speed becomes constant then tangential acc become zero .By the basis of tangial acc the circular motion is divided into two types uniform circular motion and non uniform circular motion[tex]{ \blue {\bold{UNIFORM \: CIRCULAR}}} \\ \green{ \bold{ MOTION}}[/tex]
when object in a circular path in a constant speed or constant circular speed is called uniform circular motion. In case of uniform circular motion acc tendial will be zero In case of uniform circular motion acc centripetal will be equal to net acc.[tex] { \green{ \bold{ NON UNIFORM}}} \\ {\blue{ \bold{CIRCULAR \: MOTION}}}[/tex]
Centripetal acc is not equal net acc Circular acc and tendial acc is not equal to zero .A battery of emf 5V and internal resistance 2ohm is joined to a resistor of 8ohm.Calculate the terminal potential difference.
Answer:
4V
Explanation:
First, we calculate the total resistance to the given battery cell of emf 5V. The total resistance is the sum of all the resistance in the cell i.e.
Total resistance = 2Ω + 8Ω = 10Ω
Using ohms law equation to calculate the current passing through the battery cell:
V = IR
Where; V = voltage, I = current, R = resistance
5V = I × 10Ω
I = 5/10
I = 0.5A
Terminal voltage is calculated by the us of the following equation:
V=emf−IR
Where; R is internal resistance
V = 5 - (0.5 × 2)
V = 5 - 1
V = 4V
Therefore, the potential difference across the terminals of the battery cell is 4V
In Boolean Algebra zero represent
1) Zero potential
2) Ground potential
3) low potential
4) Both 1 &2
Answer:
Option 1 & 2
Explanation:
The area of mathematics known as Boolean algebra concerns with operations on logical quantities with binary variables. To express truths, Boolean features are transformed as binary numbers: 1 = true and 0 = false. Boolean algebra concerns with logistical processes, whereas fundamental algebra deals with machine based.
what is gamma rays brust
Explanation:
A gamma - Ray Brust (GRB) is often the brightest source in the sky, often brighter than everything else combined.Answer:
A solenoid is a type of electromagnet, the purpose of which is to generate a controlled magnetic field through a coil wound into a tightly packed helix. The coil can be arranged to produce a uniform magnetic field in a volume of space when an electric current is passed through it.
4. An object is thrown from from the ground upward with an initial speed of 3.75 m/s. How long will the object be in the air before it lands on the ground?
Answer:
Explanation:
There's an easy way to answer this and then an easier way. I'll do both since I'm not sure what you're doing this for: physics or calculus. Calculus is the easier way, btw.
Going with the physics version first, here's what we know:
a = -9.8 m/s/s
v₀ = 3.75 m/s
t = ??
That's not a whole lot...at least not enough to directly solve the problem. What we have to remember here is that at the max height of a parabolic path, the final velocity is 0. So we can add that to our info:
v = 0 m/s. Use the one-dimensional equation that utilizes all that info and allows us to solve for time:
v = v₀ +at and filling in:
0 = 3.75 + (-9.8)t and
-3.75 = -9.8t so
t = .38 seconds. This is how long it takes to get to its max height. Another thing we need to remember (which is why calculus is so much easier!) is that at the halfway point of a parabolic path (the max height), the object has traveled half the time it takes to make the whole trip. In other words, if .38 is how long it takes to go halfway, then 2(.38) is how long the whole trip takes:
2(.38) = .76 seconds. Now onto the calculus way:
The position function is
[tex]s(t)=-4.9t^2+3.75t[/tex] The first derivative of this is the velocity function and, knowing that when the velocity is 0, the time is halfway gone, we will find the velocity function and then set it equal to 0 and solve for t:
v(t) = -9.8t + 3.75 and
0 = -9.8t + 3.75 and
-3.75 = -9.8t so
t = ,38 and multiply that by 2 to find the time the whole trip took:
2(.38) = .76 seconds.
what is physical change ?
Answer:
Physical Changes :- The substance in which no new substance is formed are called physical changes.
The molecular composition of the substance are totally same.
For example :- Crushing a mineral into powder.
How many more neutrons are in a I SOTOPE of copper-14 than in standard carbon atom
Answer:
2 more neutrons
Explanation:
To obtain the answer to the question, let us calculate the number of neutrons in carbon–14 and standard carbon (i.e carbon–12). This can be obtained as follow:
For carbon–14:
Mass number = 14
Proton number = 6
Neutron number =?
Mass number = Proton + Neutron
14 = 6 + Neutron
Collect like terms
14 – 6 = Neutron
8 = Neutron
Neutron number = 8
For carbon–12:
Mass number = 12
Proton number = 6
Neutron number =?
Mass number = Proton + Neutron
12 = 6 + Neutron
Collect like terms
12 – 6 = Neutron
6 = Neutron
Neutron number = 6
SUMMARY:
Neutron number of carbon–14 = 8
Neutron number of carbon–12 = 6
Finally, we shall determine the difference in the neutron number. This can be obtained as follow:
Neutron number of carbon–14 = 8
Neutron number of carbon–12 = 6
Difference =?
Difference = (Neutron number of carbon–14) – (Neutron number of carbon–12)
Difference = 8 – 6
Difference = 2
Therefore, carbon–14 has 2 more neutrons than standard carbon (i.e carbon–12)
5- Clasifica los siguientes cambios de la materia, anotando delante de cada uno cambio físico (F) o cambio químico (Q): • Disolver azúcar en agua • Freir una chuleta • Arrugar un papel • El proceso de la digestión • Secar la ropa al sol • Congelar una paleta de agua • Hacer un avión de papel • Oxidación del cobre • Romper un lápiz • Prender fuegos artificiales • Excavar un hoyo • Quemar basura
Answer:
1) Disolver azúcar en agua - Cambio químico - Es un caso de una solución en donde el solvente es el azúcar y el soluto es el agua.
2) Freir una chuleta - Cambio físico - Es un proceso de cocinado por un transferencia de calor y transferencia de masa.
3) Arrugar un papel - Cambio físico - Se aplica fuerzas externas para deformar el papel.
4) El proceso de la digestión - Cambio químico - Reducción de los alimentos a desechos y la absorción de nutrientes por el contacto con jugos gástricos o ambientes intestinales.
5) Secar la ropa al sol - Cambio físico - El secado es un fenómeno de transferencia de masa.
6) Congelar una paleta de agua - Cambio físico - Cambio del estado líquido al estado sólido por transferencia de calor.
7) Hacer un avión de papel - Cambio físico - Aplicación de fuerzas externas para plegar y doblar la hoja de papel.
8) Oxidación del cobre - Cambio químico - Proceso de corrosión por contacto con iones que se transportan en el ambiente.
9) Romper un lápiz - Cambio físico - Proceso de ruptura por esfuerzo normal a causa de un momento como consecuencia de una fuerza externa aplicada sobre el lápiz.
10) Prender fuegos artificiales - Cambio químico - Reacción química de reducción-oxidación.
11) Excavar un hoyo - Cambio físico - Remoción de tierra por trabajo físico.
12) Quemar basura - Cambio químico - Reacción de combustión.
Explanation:
A continuación, veremos que representa cada caso:
1) Disolver azúcar en agua - Cambio químico - Es un caso de una solución en donde el solvente es el azúcar y el soluto es el agua.
2) Freir una chuleta - Cambio físico - Es un proceso de cocinado por un transferencia de calor y transferencia de masa.
3) Arrugar un papel - Cambio físico - Se aplica fuerzas externas para deformar el papel.
4) El proceso de la digestión - Cambio químico - Reducción de los alimentos a desechos y la absorción de nutrientes por el contacto con jugos gástricos o ambientes intestinales.
5) Secar la ropa al sol - Cambio físico - El secado es un fenómeno de transferencia de masa.
6) Congelar una paleta de agua - Cambio físico - Cambio del estado líquido al estado sólido por transferencia de calor.
7) Hacer un avión de papel - Cambio físico - Aplicación de fuerzas externas para plegar y doblar la hoja de papel.
8) Oxidación del cobre - Cambio químico - Proceso de corrosión por contacto con iones que se transportan en el ambiente.
9) Romper un lápiz - Cambio físico - Proceso de ruptura por esfuerzo normal a causa de un momento como consecuencia de una fuerza externa aplicada sobre el lápiz.
10) Prender fuegos artificiales - Cambio químico - Reacción química de reducción-oxidación.
11) Excavar un hoyo - Cambio físico - Remoción de tierra por trabajo físico.
12) Quemar basura - Cambio químico - Reacción de combustión.
1. 20kg of water is ejected horizontally in 10s; the speed of the water leaving the nozzle is 30m/s. Calculate the force experienced by a fire-fighter holding the hose.
Answer 60 NEWTON
Explanation:
FORCE = MASS * acceleration
acceleration= VELOCITY / TIME
acceleration= 30 / 10 = 3 M/S2
FORCE = MASS * acceleration
FORCE = 20 *3 = 60 NEWTON
How do the rocks in the mantle move?
A. They do not move.
B. They move in convection currents.
C. They flow like water.
D. They move on top of the crust.
Answer:
b. they move in convection currents.
Explanation:
i learned this in 4th grade
Your parallel capacitors are 15 μf and 20 μf. The series capacitors are 10 μf and 12 μf. This circuit is connected to a 14 v battery, also determine the potential energy and the voltage across each capacitor
Answer:
a. i. 6.608 V ii. 5.507 V iii. 1.89 V iv. 1.89 V
b. i. 0.22 mJ ii. 0.182 mJ iii. 0.027 mJ iv. 0.036 mJ
Explanation:
a. The voltage across each capacitor
Since the 15 μf and 20 μf capacitors are in parallel, their total capacitance is C = 15 μf + 20 μf = 35 μf.
Also, since C is in series with the 10 μf and 12 μf which are in series, their total capacitance, C' is gotten from 1/C' = 1/10 μf + 1/12 μf + 1/35 μf
1/C' = (12 + 42 + 35)/420 /μf
1/C' = 89/420 /μf
C' = 420/89 μf
C' = 4.72 μf
The total charge in the circuit' is thus Q = C'V where V = voltage = 14 V
So, Q = C'V = 4.72 μf × 14 V = 66.08 μC
Since the 10 μf and 12 μf are in series, Q is the charge flowing through them.
Since Q = CV and V = Q/C
i. The voltage across the 10 capacitor is
V = 66.08 μC/10 μF = 6.608 V
ii. The voltage across the 12 capacitor is
V = 66.08 μC/12 μF = 5.507 V
The voltage across the 15 μF and 20 μF capacitors.
Since the capacitors are in parallel, the voltage across them is the voltage across their combined capacitance, C
So, V = Q/C = 66.08 μC/35 μF = 1.89 V
iii. The voltage across the 15 μF capacitor is 1.89 V
iv. The voltage across the 20 μF capacitor is 1.89 V
b. The potential energy of each capacitor
i. The potential energy of the 10 μF capacitor
E = 1/2CV² where C = Capacitance = 10 μF = 10 × 10⁻⁶ F and V = voltage across capacitor = 6.608 V
E = 1/2CV²
E = 1/2 × 10 × 10⁻⁶ F(6.608 V)²
E = 5 × 10⁻⁶ F(43.666) V²
E = 218.33 × 10⁻⁶ J
E = 0.21833 × 10⁻³ J
E = 0.21833 mJ
E ≅ 0.22 mJ
ii. The potential energy of the 12 μF capacitor
E = 1/2CV² where C = Capacitance = 12 μF = 12 × 10⁻⁶ F and V = voltage across capacitor = 5.507 V
E = 1/2CV²
E = 1/2 × 12 × 10⁻⁶ F(5.507 V)²
E = 6 × 10⁻⁶ F(30.327) V²
E = 181.96 × 10⁻⁶ J
E = 0.18196 × 10⁻³ J
E = 0.18196 mJ
E ≅ 0.182 mJ
iii. The potential energy of the 15 μF capacitor
E = 1/2CV² where C = Capacitance = 15 μF = 15 × 10⁻⁶ F and V = voltage across capacitor = 1.89 V
E = 1/2CV²
E = 1/2 × 15 × 10⁻⁶ F(1.89 V)²
E = 7.5 × 10⁻⁶ F(3.5721) V²
E = 26.79 × 10⁻⁶ J
E = 0.02679 × 10⁻³ J
E = 0.02679 mJ
E ≅ 0.027 mJ
iv. The potential energy of the 15 μF capacitor
E = 1/2CV² where C = Capacitance = 20 μF = 15 × 10⁻⁶ F and V = voltage across capacitor = 1.89 V
E = 1/2CV²
E = 1/2 × 20 × 10⁻⁶ F(1.89 V)²
E = 10 × 10⁻⁶ F(3.5721) V²
E = 35.721 × 10⁻⁶ J
E = 0.035721 × 10⁻³ J
E = 0.035721 mJ
E ≅ 0.036 mJ
How long will it take a car to acceleration from 15.2ms to 23.Ms if the car has an average acceleration of 3.2m\s
Answer: 2.43 s
Explanation:
Initial velocity is [tex]u=15.2\ m/s[/tex]
Final velocity [tex]v=23\ m/s[/tex]
Average acceleration is [tex]a_{avg}=3.2\ m/s[/tex]
Average acceleration is change in velocity in the given amount of time
[tex]\therefore a_{avg}=\dfrac{v-u}{t}\\\\\Rightarrow 3.2=\dfrac{23-15.2}{t}\\\\\Rightarrow t=\dfrac{7.8}{3.2}\\\\\Rightarrow t=2.43\ s[/tex]
Thus, 2.43 s is required to acquire that average acceleration with 23 m/s velocity .
Question 23 of 23
Suppose a current flows through a copper wire. Which two things occur?
O A. The field is parallel to the direction of flow of the current.
B. An electric field forms around the wire.
OC. A magnetic field forms around the wire.
U
D. The field is perpendicular to the direction of flow of the current.
SUBM
Answer:
The field is parallel to the direction of flow of the current.
what force to be required to accelerate a car of mass 120 kg from 5 m/s to 25m/s in 2s
Answer:1200
Explanation:
F=ma =m(Vf-Vi)/t
F=120(25-5)/2 =1200N
Answer:
[tex]f = m \frac{v1 - v2}{t} \\ = 120 \times \frac{25 - 5}{2} \\ = 120 \times 10 \\ = 1200N \\ thank \: you[/tex]
help me daddyz Alice did an experiment to find the relationship between the angle at which a ray of light strikes a mirror and the angle at which the mirror reflects the light. She placed a ray box in front of a mirror. She changed the angle at which the light from the ray box struck the mirror and noted the corresponding angle at which the mirror reflected the light. Which of the following is the control variable in this experiment? The ray box used as the source of light The direction along which the light moves Angle at which the light from the ray box strikes the mirror Angle at which the mirror reflects the light from the ray box
Answer:
The ray box used as the source of light
Answer: its A
Explanation: i just did the test and got it right also known. The ray used as the source of light
A car moving east at a velocity of 16.0 m/s collides with a stationary truck with exactly twice the mass. If the two vehicles lock together, calculate the velocity of their combined mass immediately after collision
Answer:
5.33ms-¹
Explanation:
that is the procedure above
A car moving along a racetrack has a centripetal acceleration of 11.3 m/ s2. If the speed of the car is 30.0 m/s, what is the distance between the car and the center of the track?
Answer:
Explanation:
The formula for centripetal acceleration is
[tex]a_c=\frac{v^2}{r}[/tex] and filling in:
[tex]11.3=\frac{(30.0)^2}{r}[/tex] and solving for r:
[tex]r=\frac{(30.0)^2}{11.3}[/tex] gives us that
r = 79.6 m
what happens to gravitational force when distance is quarter from the original
please answer it faster help me
Answer:
The force of gravitational attraction between them also decreas
Which of the following best distinguishes the relationship between applied research and basic research?
A Applied research commercializes the discoveries of basic research and brings them to the general public.
B Basic research takes place in academia, and complex research takes place in industry.
C Basic research provides fundamental knowledge that can be used to conduct applied research.
D Applied research is not published for peer review, but basic research must be.
Answer:
C. Basic research provides fundamental knowledge that can be used to conduct applied research.
Explanation:
Applied research is used to solve practical and specific problems. Basic research is used to advance theories and scientific knowledge; it might not result in an invention or a solution, but can be used to gain knowledge about a specific subject.
A car initially traveling at 60 km/h accelerates at a constant rate of 2.0 m/s2. How much time is required for the car to reach a speed of 90 km/h
Answer:
Time, t = 4.165 seconds.
Explanation:
Given the following data;
Initial velocity = 60 km/h
Final velocity = 90 km/h
Acceleration = 2 m/s²
Conversion:
60 km/hr to meters per seconds = 60*1000/3600 = 60000/3600 = 16.67 m/s
90 km/hr to meters per seconds = 90*1000/3600 = 90000/3600 = 25 m/s
To find the time, we would use the first equation of motion;
[tex] V = U + at[/tex]
Where;
V is the final velocity.
U is the initial velocity.
a is the acceleration.
t is the time measured in seconds.
Making time, t the subject of formula, we have;
[tex] t = \frac{V - U}{a}[/tex]
Substituting into the equation, we have;
[tex] t = \frac{25 - 16.67}{2}[/tex]
[tex] t = \frac{8.33}{2}[/tex]
Time, t = 4.165 seconds.
During the rainy season, we can observe lighting in the sky. Due to lighting, the atmospheric
nitrogen combines with atmospheric oxygen to form nitric oxide. Which among the following is a
correct statement concerning this process?
O A physical change has taken place in the atmosphere during lighting.
Ques
O No change occurred, as lighting is natural event.
A chemical change has taken place to form nitric oxide.
O All of the choices
Answer:
A chemical change has taken place to form nitric oxide.
When the spacecraft is at the halfway point, how does the strength of the gravitional force on the spaceprobe by Earth compre with the strength
Solution :
When the spacecraft is at halfway point, the distance from the Earth as well as Mars are same. We have to account the masses of the planets. The gravitational force that is exerted by the Earth is greater because of its combined mass with the space probe.
The mass of Earth is greater than the mass of Mars. Therefore, the force of Earth is more than Mars.
in parallel combination of electrical appliances Total Electric Power a. increase b. decrease c. remain same
Answer:
In a parallel combination of electrical appliances total electric power will increase
Answer is A it will increase