Answer:
0.252 m/s
Explanation:
Applying the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
Note: From the question, The collision between the car and the train is an inelastic collision and as such, both move with a common velocity after collision.
mu+m'u' = V(m+m')................... Equation 1
Where m = mass of the train, u' = initial velocity of the train, m' = mass of the car, u' = initial velocity of the car, V = common velocity after collision.
make V the subject of the equation
V = (mu+m'u')/(m+m')............... Equation 2
Given: m = 270000 kg, u = 0.325 m/s, m' = 52500 kg, u' = -0.12 m/s
Substitute these values into equation 2
V = [(270000×0.325)+{52500(-0.12)}]/(270000+52500)
V = 81450/322500
V = 0.252 m/s
Show that energy dissipated due to motion of a conductor in the magnetic field is due to mechanical energy.
Explanation:
let us use the explanation below to get the intuition so desired;
According to Faraday's law of electro magnetic induction, when ever a coil/conductor is made to rotate in a magnetic field, voltage or emf is created and current is produced, in the long run energy has be produced or converted.
The conversion of this energy is made possible by the motion of the coil/conductor is the magnetic field, just by the motion of the conductor cutting through the magnetic field, thus creating electro motive force(E.M.F) hence producing current, and ultimately energy is created
Integrated Concepts:_______.
(a) Calculate the ratio of the highest to lowest frequencies of electromagnetic waves the eye can see, given the wavelength range of visible light is from 380 to 760 nm.
(b) Compare this with the ratio of highest (20,000 Hz) to lowest (20 Hz) frequencies the ear can hear.
a). frequency = (speed) / (wavelength)
The speed of light is around 3 x 10⁸ m/s.
For 380 nm (violet light), frequency = (3 x 10⁸ m/s) / (380 x 10⁻⁹ m)
Frequency = 7.89 x 10¹⁴ Hz
For 760 nm (red light), frequency = (3 x 10⁸ m/s) / (760 x 10⁻⁹/s)
Frequency = 3.94 x 10¹⁴ Hz
The ratio is 2 .
That's 1 octave, or 0.3 of a decade.
b). The ratio of highest/lowest sounds is (20,000 Hz/20 Hz) = 1,000
That's 3 decades, or about 10 octaves.
===> Speaking logarithmically ( ! ), ears are sensitive to a range of sound frequencies that's 10 times as wide as the range of light frequencies that eyes can detect.
Una persona lanza una pelota hacia arriba con una velocidad de 15 metros por segundo. - Calcule: o Altura máxima que alcanza la pelota o Tiempo en el aire.
Answer:
Ok, sabemos que la velocidad inicial de la pelota es 15m/s.
Desconocemos la posición inicial a la que es lanzada la pelota, pero vamos a suponer que es a una altura igual a cero, es decir, la pelota es lanzada al ras del suelo.
Una vez lanzada, la única fuerza actuando en la pelota es la gravitatoria, entonces la aceleración de la pelota es:
a = -g = -9.8m/s^2
El signo negativo es por que esta aceleración apunta hacia abajo.
Ahora, para la velocidad, necesitamos integrar sobre el tiempo.
v(t) = (-9.8m/s^2)*t + v0
donde v0 = 15m/s
v(t) = (-9.8m/s^2)*t + 15m/s.
De aca podemos obtener el tiempo en el que la pelota llega a la altura máxima, que es el punto donde la velocidad es igual a cero.
0 = (-9.8m/s^2)*t + 15m/s.
t = (15/9.8)s = 1.53 s
Ahora, para la ecuación de la posición integramos la ecuación de la velocidad sobre el tiempo:
p(t) = (1/2)(-9.8m/s^2)*t^2 + 15m/s*t + p0
donde p0 es la pocision inicial, pero arriba dijimos que era igual a cero, entonces la ecuación queda:
p(t) = (-4.5m/s^2)*t^2 + 15m/s*t
ahora reemplazamos t por el tiempo que encontramos antes, y descubrimos que:
p(1.53s) = (-4.5m/s^2)*(1.53s)^2 + 15m/s*1.53s = 12.41m
La máxima altura que alcanza la pelota es 12.41 metros arriba del punto desde el que se la lanzo.
Ahora, el tiempo total que esta en el aire puede ser calculado de tal forma que la posición vuelva a ser cero, es decir, la pelota llega a la misma altura desde la que fue lanzada inicialmente (y es agarrada por la persona, podemos suponer)
Entonces:
p(t) = 0 = (-4.5m/s^2)*t^2 + 15m/s*t
Ahora resolvemos la eq cuadrática, usando la eq. de Bhaskara:
[tex]t = \frac{-15 +- \sqrt{15^2 - 4*(-4.5)*0} }{-2*4.5} = \frac{-15 +-15}{-9.8}[/tex]
Entonces las soluciones son:
t = (-15 + 15)/-9.8 = 0s
t = (-15 - 15)/-9.8 = 3.06s
Tomamos la segunda solución, ya que la primera corresponde al tiempo inicial.
Entonces concluimos con que la pelota estuvo 3.06 segundos en el aire.
Radioisotopes often emit alpha particles, beta particles, or gamma rays. The distance they travel through matter increases in order from alpha to gamma. Each radioisotope has a characteristic half-life, which is the time needed for half of a sample of radioisotope to undergo nuclear decay. Which quality is desirable for a radioisotope that is used for medical imaging of a specific organ
Answer:
It emits alpha and beta particles.
Explanation:
EDG 20
Answer:
the answer is b, calcium-45
Explanation:
What rule should be used to transform a table of data to represent the
reflection of f(x) over the yaxis?
Answer:
Multiply the x values with -1.
Explanation:
By multiplying the numbers by one, you are changing them to be the opposite of their original state.
You multiply the numbers that are in the x-value column because you are reflecting the image over the y-axis.
Hope this helped and good luck!
Answer:
Multiply each x-value in the table by -1
Explanation:
The maximum wavelength For photoelectric emissions in tungsten is 230 nm. What wavelength of light must be use in order for electron with maximum energy of 1.5ev to be ejection
Answer:
λ = 1.8 x 10⁻⁷ m = 180 nm
Explanation:
First we find the work function of tungsten by using the following formula:
∅ = hc/λmax
where,
∅ = work function = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λmax = maximum wavelength for photoelectric emission = 230 nm
λmax = 2.3 x 10⁻⁷ m
Therefore,
∅ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.3 x 10⁻⁷ m)
∅ = 8.64 x 10⁻¹⁹ J
Now we convert Kinetic Energy of electron into Joules:
K.E = (1.5 eV)(1.6 x 10⁻¹⁹ J/1 eV)
K.E = 2.4 x 10⁻¹⁹ J
Now, we use Einstein's Photoelectric Equation:
Energy of Photon = ∅ + K.E
Therefore,
Energy of Photon = 8.64 x 10⁻¹⁹ J + 2.4 x 10⁻¹⁹ J
Energy of Photon = 11.04 x 10⁻¹⁹ J
but,
Energy of Photon = hc/λ
where,
λ = wavelength of light = ?
Therefore,
11.04 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ
λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(11.04 x 10⁻¹⁹ J)
λ = 1.8 x 10⁻⁷ m = 180 nm
Find the net force on q3. Include the direction ( +or-).
q1= -53.0 uC; q2=105 uC; q3= -88.0 uC; q1 to q2= 0.50m; q2 to q3= 0.95m
I will give brainliest to whoever gets the correct answer!
Answer:
72.16 N
Explanation:
Given:
q₁ = -53.0 μC
q₂ = 105 μC
q₃ = -88.0 μC
q₁ to q₂ = 0.50 m
q₂ to q₃ = 0.95 m
To find:
Net force on q₃
Solution:
First compute net electric field on q₃
E = F/q = k.Q/d²
The formula of electric field at q₃:
E = k.Q / r²
Where
r is distance
Q is magnitude of charge
k is a constant with a value of 8.99 x 10⁹ N m²/C²
When
q₂ to q₃ = 0.95m and
q₂ = 105 μC then
Find electric field due to charge q₂
E = ( (8.99 x 10⁹)x( 105 x 10⁻⁶) ) / 0.95²
= (8990000000 x 0.000105) / 0.9025
= 943950 / 0.9025
= 1045927.977839
= 1.046 x 10⁶ N/C
This interprets that it will act or point away from q₂
q₁ to q₂= 0.50m
q₂ to q₃ = 0.95m and
q₁ = -53 μC then
Find electric field due to charge q₁
E = (8.99 x 10⁹) x (53 x 10⁻⁶) / (0 .50 + 0.95)²
= (8990000000 x 0.000053) / (1.45)²
= 476470 /2.1025
= 226620.689655
= 0.227 x 10⁶ N/C
This interprets that it will act or point towards q₁
Since these fields are opposite in direction.
Compute Net Field
Net Field = 1.046 x 10⁶ - 0.227 x 10⁶ N/C
= 1046000 - 227000
= 819000
= 0.819 x 10⁶
≈ 0.82 x 10⁶
This interprets that it will act or point away from q₂
Compute force on q3
q₃ E = 88 x 10⁻⁶ x 0.82 x 10⁶
= 88000000 x 820000
= 72160000000000
= 72.16 N
Force on -ive charge in a field is always in a direction opposite to direction of field
So this interprets that direction of this field will be towards q₂.
Un autobús viaja en una carretera a una velocidad de 70 km/h y acelera durante 30 segundos hasta llegar a su límite de velocidad, que son 95 km/h. ¿Cuál fue su aceleración?
Answer:
a = 30 km / h²
Explanation:
Dado que
Velocidad inicial, u = 70 km / h
Velocidad final, v = 95 km / h
Tiempo, t = 30 s = 0.1 h
Lo sabemos
v = u + a t
a = aceleración
Ahora poniendo los valores en la ecuación anterior
[tex]95 = 70 + a \ times 0.1 [/tex]
[tex]a = \ dfrac {95-70} {0.1} = 30 \ km / h ^ 2 [/tex]
Por lo tanto, la aceleración será
a = 30 km / h²
what is a primary source?
Answer:
A primary source is an original document such as diaries, speeches, manuscripts, letters, interviews, records, eyewitness accounts, autobiographies. Empirical scholarly works such as research articles, clinical reports, case studies, dissertation
Explanation:
Answer:
Scientific paper
Explanation:
A p e x
Your car's 32.5 W headlight and 2.00 kW starter are ordinarily connected in parallel in a 12.0 V system. What power (in W) would one headlight and the starter consume if connected in series to a 12.0 V battery
Answer:
Explanation:
the resistance of a electrical device
R = V² / P where V is volt and P is power .
The devices are in parallel so same volt will apply on them
So R₁ = 12² / 32.5 = 4.431 ohm
R₂ = 12² / 2 x 10³ = .072 ohm
when they are in series
Common Current in them = 12 / 4.431 + .072
= 2.6649 A
power consumed by first device when they are in series
= current² x resistance
= 2.6649² x 4.431 = 31.46 W
power consumed by other
= 2.6649² x .072 = .511 W
Calculate the acceleration of a mobile that at 4s is 32m from the origin, knowing that its initial speed is 10m / s.
Answer:
5.5 m/s^2
Explanation:
I believe this is the answer > using the formula a= v-v0/t
Hope this helps!
Answer:
-1 m/s²
Explanation:
Given:
Δx = 32 m
v₀ = 10 m/s
t = 4 s
Find: a
Δx = v₀ t + ½ at²
32 m = (10 m/s) (4 s) + ½ a (4 s)²
a = -1 m/s²
The definition of parallel lines requires the terms line and plane while the definition of perpendicular lines requires the undefined terms of line and point.
Answer with explanation:
Complete question is provided in the attachment below.
There are 3 undefined term in geometry :1) A Point 2) A line 3) A Plane.
When two lines are parallel they never meet , so the requirement to define them is lines and a plane on which they lie.
While when two lines are perpendicular , they intersect each other at a point by making a right angle between them.
So it required lines and a point to define it.
Three books are at rest, in equilibrium, on a horizontal table as
shown. The weight of each book, which is equal to the force
gravity exerts in the downward direction, is given. What is the
net force on the middle book?
Answer:
Net force = 0
Explanation:
Short answer: if the middle book is not acceleration in any direction, the net force is zero.
Long answer: refer to the attached free-body diagram (FBD) to see why the net force is zero. Skills to draw FBD are essential in solving problems in statics.
Top book exerts -mg (downwards) on the middle book.
Middle book exerts -mg (downwards) on the bottom book.
Total downward force is -mg-mg = -2mg
By Newton's third law, when an object is not in motion, reaction equals action of -2mg in the opposite direction, therefore the reaction (upwards) is +2mg.
This makes the net force of -2mg (downwards) + 2mg (upwards) =0
On a horizontal table, three books are resting and balanced. The net force on the middle book will be equal to zero.
What is equilibrium?Equilibrium is the condition of an object when two or more opposing forces, whether internal, external or a combination of both, act on the body and cancel one another out to maintain the object in the same state as it was. The Latin term for weight or balance, libra, serves as the origin of the word equilibrium.
According to the question :
The net force is 0 if the middle book does not accelerate in any direction.
The Middle book is impacted negatively (downwards) by the top book.
The middle book pulls down on the lower book by -mg.
The total downward force is -2mg = (-mg-mg)
According to Newton's third rule, a reaction when an item is not moving equals an action of -2mg in the opposite direction, hence the reaction (upwards) is a response of +2mg.
Consequently, the net force of -2mg (downwards) + 2mg (upwards) = 0 is created.
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The total mass of eight identical
building blocks is 31.52 kg. Find the
mass of 1 block.
Answer:
3.94
Explanation:
divide total mass by the number of blocks since they are identical
Answer:
3.94
Explanation:
You want to find the mass of one block. Since we know there is 8 blocks with the same mass, you can divide the total mass by 8 since the mass is equally distributed within the 8 blocks
A square loop 24.0 cm on a side has a resistance of 6.10Ω. It is initially in a 0.665-T magnetic field, with its plane perpendicular to magnetic field B but is removed from the field in 40.0ms. Calculate the electric energy dissipated in this process.
Answer:
[tex]E=6.01\times 10^{-3}\ J[/tex]
Explanation:
It is given that,
Side of a square loop, l = 24 cm = 0.24 m
Resistance of loop, R = 6.1 ohms
Initial magnetic field is 0.665 T and final magnetic field is 0 as the field is removed in 40 ms
We need to find the electrical energy dissipated in this process.
Due to change in magnetic field, the loop will induce a voltage. The induced voltage is given by :
[tex]V=-\dfrac{dB}{dt}\\\\V=\dfrac{BA}{t}[/tex]
If I is induced current then,
[tex]V=IR[/tex]
[tex]I=\dfrac{V}{R}\\\\I=\dfrac{BA}{tR}[/tex]
Power is given by voltage times current. So,
[tex]P=\dfrac{(BA)^2}{(t^2R)}[/tex]
Now, energy is given by the product of power and time. So,
[tex]E=\dfrac{(BA)^2}{(t^2R)}\times t\\\\E=\dfrac{(BA)^2}{(tR)}[/tex]
Now putting all the values in above formula. So,
[tex]E=\dfrac{(0.665\times (0.24)^2)^2}{(40\times 10^{-3}\times 6.1)}\\\\E=6.01\times 10^{-3}\ J[/tex]
So, the electrical energy of [tex]6.01\times 10^{-3}\ J[/tex] is dissipated in this process.
The electrical energy dissipated throughout this process will be "6.01 × 10⁻³ J".
Magnetic fieldAccording to the question,
Square loop's side, l = 24 cm or,
= 0.24 m
Loop's resistance, R = 6.1 ohms
Initial magnetic field = 0.665 T
Final magnetic field = 0
We know the relation,
→ V = - [tex]\frac{dB}{dt}[/tex]
= [tex]\frac{BA}{t}[/tex]
Also we know,
Current, V = IR
I = [tex]\frac{V}{R}[/tex]
= [tex]\frac{BA}{tR}[/tex]
Now, Energy, E = [tex]\frac{(BA)^2}{t^2R}[/tex] × t or,
= [tex]\frac{(BA)^2}{tR}[/tex]
By substituting the values,
= [tex]\frac{(0.665\times (0.24)^2)^2}{40\times 10^{-3}\times 6.1}[/tex]
= 6.01 × 10⁻³ J
Thus the response above is correct.
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A negative test charge experiences a force to the right as a result of an electric field. Which is the best conclusion to
draw based on this description?
O The electric field points to the left because the force on a negative charge is opposite to the direction of the field.
O The electric field points to the right because the force on a negative charge is in the same direction as the field.
O No conclusion can be drawn because the sign of the charge creating the field is unknown.
O No conclusion can be drawn because the amount of charge on the test charge is unknown.
[tex]\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}[/tex]
Actually Welcome to the Concept of the Electrostatics.
Basically here we can see that, the Negative test charge usually a Electron is facing a force from the right side, this we can say that a Electric field as like charge is creating a repulsive force from the right to left side.
thus,
1) The electric field points to the left because the force on a negative charge is opposite to the direction of the field.
The magnetic field perpendicular to a single 13.2-cm diameter circular loop of copper wire decreases uniformly from 0.670 T to zero. If the wire is 2.25 mm in diameter, how much charge moves past a point in the coil during this operation? (rhoCu = 1.68 x 10-8 Ω.m)
Answer:
5.23 C
Explanation:
The current in the wire is given by I = ε/R where ε = induced emf in the wire and R = resistance of wire.
Now, ε = -ΔΦ/Δt where ΔΦ = change in magnetic flux = AΔB and A = area of loop and ΔB = change in magnetic field intensity = B₂ - B₁
B₁ = 0.670 T and B₂ = 0 T
ΔB = B₂ - B₁ = 0 - 0.670 T = - 0.670 T
A = πD²/4 where D = diameter of circular loop = 13.2 cm = 0.132 m
A = π(0.132 m)²/4 = 0.01368 m² = `1.368 × 10⁻² m²
ε = -ΔΦ/Δt = -AΔB/Δt = -1.368 × 10⁻² m² × (-0.670 T)/Δt= 0.9166 × 10⁻² Tm²/Δt
Now, the resistance R of the circular wire R = ρl/A' where ρ = resistivity of copper wire = 1.68 x 10⁻⁸ Ω.m, l = length of wire = πD and A' = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m
R = ρl/A' = 1.68 x 10⁻⁸ Ω.m × π × 0.132 m÷π(2.25 × 10⁻³ m)²/4 = 0.88704/5.0625 = 0.1752 × 10⁻² Ω = 1.752 × 10⁻³ Ω
So, I = ε/R = 0.9166 × 10⁻² Tm²/Δt1.752 × 10⁻³ Ω
IΔt = 0.9166 × 10⁻² Tm²/1.752 × 10⁻³ Ω = 0.5232 × 10 C
Since ΔQ = It = 5.232 C ≅ 5.23 C
So the charge is 5.23 C
The induced emf through wire depends on the current flow (indirectly with charge flow as well).
The value of charge moving past a point in the coil during its operations is 5.23 C.
What is the magnetic field?The region in a space where a particle experiences some magnetic force, is known as the magnetic field.
Given data -
The diameter of the circular loop is, d = 13.2 cm = 0.132 m.
The change in magnetic field strength is, ΔB = 0.670 T.
The new diameter of the wire is, d' = 2.25 mm = 2.25 × 10³ m.
The resistivity of the wire is, [tex]\rho = 1.68 \times 10^{-8} \;\rm \Omega.m[/tex].
The current in the wire is given by the following expression,
I = ε/R
Here,
ε is the induced emf in the wire.
R is the resistance of the wire.
And the expression for the induced emf of the wire is given as,
ε = -ΔΦ/Δt
Here,
ΔΦ is the change in magnetic flux. And its expression is,
ε = A × ΔB
Here,
A is the area of the loop. And its value is, A = πd²/4.
Solving as,
A = π(0.132 m)²/4
A = 0.01368 m²
A = `1.368 × 10⁻² m²
Now, calculating the induced emf as,
ε = ΔΦ/Δt
ε = A × ΔB/Δt
ε = 1.368 × 10⁻² m² × (0.670 T)/Δt
ε = 0.9166 × 10⁻² Tm²/Δt
Now, the resistance R of the circular wire is,
R = ρ × L/A'
Here,
L is the length of the wire and its value is. L = πd .
And A' is the new cross-sectional area of wire,
A' = πd'²/4
So, the resistance is,
R = ρ × L/A'
R = 1.68 x 10⁻⁸ ×( π × 0.132 m) ÷ π(2.25 × 10⁻³ m)²/4 =
R = 0.88704/5.0625
R = 0.1752 × 10⁻² Ω
R = 1.752 × 10⁻³ Ω
Now, the current flow (I) in the wire is given as,
I = ε/R
I = 0.9166 × 10⁻² Tm²/Δt1.752 × 10⁻³ Ω
And obtaining the value of charge from the expression of current as,
Q = IΔt
Q = 0.9166 × 10⁻² Tm²/1.752 × 10⁻³ Ω
Q = 0.5232 × 10 C
Q = 5.23 C
Thus, we can conclude that the value of charge moving past a point in the coil during its operations is 5.23 C.
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70kg man runs up a flight of staurs in 4 sec . The vertical height of the stairs is 4.5 m . Calculate his power
Answer:
771.75 wattExplanation:
Given,
Mass ( m ) = 70 kg
Distance ( d ) = 4.5 m
Time taken ( t ) = 4 seconds
Power = ?
Now, applying the formula to find power:
[tex]power = \frac{work \: done}{time \: taken} [/tex]
[tex] = \: \frac{f \: \times \: d}{t} [/tex]
[tex] = \frac{m \: \times \: g \: \times \: d}{t} [/tex]
Plugging the values:
[tex] = \frac{70 \times 9.8 \times 4.5}{4} [/tex]
Calculate the products
[tex] = \frac{3087}{4} [/tex]
Divide:
[tex] = 771.75 \: watt[/tex]
Hope this helps...
Best regards!!
Suppose the maximum safe intensity of microwaves for human exposure is taken to be 1.00 W/m2. (a) If a radar unit leaks 50.0 W of microwaves (other than those sent by its antenna) uniformly in all directions, how far away (in cm) must you be to be exposed to an intensity considered to be safe
Answer:
The safe distance is 199 cm approximately 200 cm
Explanation:
Safe intensity = 1.00 W/m^2
wattage of radar leaked radar = 50.0 W
safe distance from the microwave will be = ?
We know that the intensity of a wave radiated uniformly in all direction is given as
[tex]I[/tex] = [tex]\frac{W}{A}[/tex]
where
W is the wattage of the leaked radar
A is the radial area, which is the area of a sphere that encapsulates the region through which this wave spreads uniformly.
From the equation above,
[tex]A[/tex] = [tex]\frac{W}{I}[/tex] = 50/1 = 50 m^2
But the area of this sphere [tex]A[/tex] = [tex]4\pi r^{2}[/tex]
where
r is the safe distance from the radar source
substituting for the area, we have
50 = 4 x 3.142 x [tex]r^{2}[/tex]
50 = 12.568 [tex]r^{2}[/tex]
[tex]r^{2}[/tex] = 50/12.568 = 3.978
r = [tex]\sqrt{3.978}[/tex] = 1.99 m = 199 cm ≅ 200 cm
Which is the best example of muscular endurance
Answer:
personally I'd say C by do not know if that is the exact answer
Find the force. 10 points. Will give brainliest!
Answer:
8996kg*m/s/s
Explanation:
Given:
a=26m/s/s
m=346kg
Required:
f=?
Formula:
f=m*a
Solution:
f=346kg*26m/s/s
f=8996kg*m/s/s
Hope this helps ;) ❤❤❤
Answer:
[tex]\boxed{F = 8996 \ Newton}[/tex]
Explanation:
Given:
Mass = m = 346 kg
Acceleration = a = 26 m/s²
Required:
Force = F = ?
Formula:
F = ma
Solution:
F = 346 * 26
F = 8996 Newton
A 1000 kg car travels on a highway with a speed of 30 m/s. The driver sees a roadblock and applies the brakes, which provide a
constant braking force of 4 kN. What is the acceleration of the car?
A. 4 m/s
Answer:
-4m/s
Explanation:
use the formula
[tex]f = ma[/tex]
where f-force
m-mass
a-accleration
so
1kN=1000N
so apply
4000=1000×a
a=4m/s
(the negative is because the car was braking)
Answer:
Your answer is -4 m/s^2
Explanation:
Set Up: Let +x be the direction the car is traveling.
List the known & unknown quantities:
m = mass of the car = 1000 kg
υ = 30 m/s
Fx = –4 kN = –4000 N (negative since it is a braking force)
ax = acceleration =?
Solve: Use Newton’s second law of motion.
Fx=max
ax=Fx/m = −4000 N /1000 kg = −4000 kg·m/s^2 / 1000 kg =−4m/s^2
List the submultiples and multiple units of length, mass, and time with respect to real-life situations. How are these units are related to S I unit of the above mentioned physical quantity?
Answer:
Explanation:
In physics, there are two types of physical quantities namely the fundamental and the derived quantities. Fundamental quantities are independent quantities on which derived quantities depends on. Length, mass and time are examples of fundamental quantities.
The SI unit of length is meters. A meter is a multiple unit. Its submultiple units are centimetres (10⁻²metres), kilometres (10³metres), decimetres (10⁻¹metres) etc
The SI unit of mass is kilogram (kg). The only sub multiple unit used in real-life situation is grams.
1 kg = 100 grams
The SI unit of time is seconds. The multiple units are the minutes, hours, weeks, days and years.
1 minute = 60 seconds
1 hour = 3600 seconds
1 day = 24 * 3600 = 86,400 secs
Consider a solenoid of length L, N windings, and radius b (L is much longer than b). A current I is flowing through the wire. If the radius of the solenoid were doubled (becoming 2b), and all other quantities remained the same, the magnetic field inside the solenoid would
Answer:
The magnetic field remains the same.
Explanation:
If a solenoid has length L, N windings, and radius b, then the magnetic field inside the solenoid is given by :
[tex]B=\mu_o NI[/tex]
[tex]\mu_o[/tex] is magnetic permeability
If the radius of the solenoid were doubled and all other quantities remained the same, the magnetic field inside the solenoid would remains the same as it is independent of the radius of the solenoid.
Ask Your Teacher An electric utility company supplies a customer's house from the main power lines (120 V) with two copper wires, each of which is 34.0 m long and has a resistance of 0.109 Ω per 300 m. (a) Find the potential difference at the customer's house for a load current of 116 A.
Answer:
The potential difference at the customer's house is 117.1 V.
Explanation:
a) The potential difference at the customer's house can be calculated as follows:
[tex] \Delta V_{h} = \Delta V_{p} - \Delta V_{l} [/tex]
Where:
[tex]V_{h}[/tex]: is the potential difference at the customer's house
[tex]V_{p}[/tex]: is the potential difference from the main power lines = 120 V
[tex]V_{l}[/tex]: is the potential difference from the lines
[tex] \Delta V_{h} = \Delta V_{p} - IR [/tex]
The resistance, R, is:
[tex]\frac{0.109 \Omega}{300 m}*2*34.0 m = 0.025 \Omega[/tex]
Now, the potential difference at the customer's house is:
[tex]\Delta V_{h} = 120 V - 116A*0.025 \Omega = 117.1 V[/tex]
Therefore, the potential difference at the customer's house is 117.1 V.
I hope it helps you!
Power is _________________the force required to push something the work done by a system the speed of an object the rate that the energy of a system is transformed the energy of a system
Answer:
[defined as]
Explanation:
it is the missing word
A long, rigid conductor, lying along an x axis, carries a current of 5.0 A in the negative x direction. A magnetic field B: is present, given by B: ???? 3.0iˆ ???? 8.0x2jˆ, with x in meters and B: in milliteslas. Find, in unit-vector notation, the force on the
Answer:
Explanation:
magnetic field B = (3 i + 8 x 2 j )x 10⁻³ T
= (3 i + 16 j )x 10⁻³ T
L = - i ( unit length of conductor )
Force F = I ( L x B ) , I is current
= 5 [ - i x ( 3i + 16 j ) 10⁻³]
= 5 ( - 16 k ) x 10⁻³
F = - 80 x 10⁻³ k
A 24 cm radius aluminum ball is immersed in water. Calculate the thrust you suffer and the force. Knowing that the density of aluminum is 2698.4 kg / m3
Answer:
W =1562.53 N
Explanation:
It is given that,
Radius of the aluminium ball, r = 24 cm = 0.24 m
The density of Aluminium, [tex]d=2698.4\ kg/m^3[/tex]
We need to find the thrust and the force. The mass of the liquid displaced is given by :
[tex]m=dV[/tex]
V is volume
Weight of the displaced liquid
W = mg
[tex]W=dVg[/tex]
So,
[tex]W=dg\times \dfrac{4}{3}\pi r^3\\\\W=2698.4\times 10\times \dfrac{4}{3}\times \pi \times (0.24)^3\\\\W=1562.53\ N[/tex]
So, the thrust and the force is 1562.53 N.
PLEASE HELP ASAP. IT'S URGENT
Answer:
Q1 acceleration = 16m/s²
Q2 Net force = 9N North
Explanation:
Q1 Using the formula F=ma
Q2 R = F1 + F2
Which statement best explains the relationship between current, voltage, and resistance?
a.If we increase the amount of voltage applied, and do not change the resistance, this will result in a decrease in current.
b.If we decrease the current applied, and do not change the resistance, we increase the voltage.
c.If we increase the amount of voltage applied, and do not change the resistance, we will also increase the current.
d.If we decrease the amount of current, this will not affect the amount of voltage, only the amount of resistance.
Explanation:
Ohm's law gives the relationship between current, voltage, and resistance. Its mathematical form is given by :
V = IR
I is current and R is resistance
Resistance resists the flow of electric current in a circuit. When the amount of applied voltage is increased, it will not change the resistance. It increases the current. Hence, the correct option is (C).