Answer:
D. Mars is a long distance away
On a horizontal frictionless surface a mass M is attached to two light elastic strings both having length l and both made of the same material. The mass is displaced by a small displacement Δy such that equal tensions T exist in the two strings, as shown in the figure. The mass is released and begins to oscillate back and forth. Assume that the displacement is small enough so that the tensions do not change appreciably. (a) Show that the restoring force on the mass can be given by F = -(2T∆y)/l (for small angles) (b) Derive an expression for the frequency of oscillation.
Answer:
(a) By small angle approximation, we have;
F = -2×T×Δy/l
(b) [tex]The \ frequency \ of \ oscillation, \ f = \dfrac{1}{2\cdot \pi }\cdot\sqrt{\dfrac{2 \cdot T}{l \cdot M} }[/tex]
Explanation:
(a) The diagram shows the mass, M, being restored by two equal tension, T acting on the elastic strings l, such the restoring force, F acts along the path of motion of the mass, with distance Δy
Therefore, the component of the tension T that form part of the restoring force is given as follows;
Let the angle between the line representing the extension of the elastic strings T and the initial position of the string = ∅
Then we have;
String force, [tex]F_{string}[/tex] = T×sin∅ + T×cos∅ + T×sin∅ - T×cos∅ = 2×T×sin∅
Whereby the angle is small, we have;
sin∅ ≈ tan∅ = Δy/l
Which gives;
[tex]F_{string}[/tex] = 2×T×sin∅ = 2×T×Δy/l (for small angles)
Restoring force F = [tex]-F_{string}[/tex] = -2×T×Δy/l
F = -2×T×Δy/l
(b) Given that the the tensions do not change appreciably as the mass, M, oscillates from Δy we have;
By Hooke's law, F = -k×x
Whereby Δy corresponds to the maximum displacement of the mass, M from the rest position, which gives;
Which gives;
F = M×a = -k×Δy
a = -k×Δy/M
d²(Δy)/dt² = -k×Δy/M
When we put angular frequency as follows;
ω² = k/M
We get;
d²(Δy)/dt² = -ω²×Δy
Which gives;
Δy(t) = A×cos(ωt + Ф)
The angular frequency is thus, ω = √(k/M)
Period of oscillation = 2·π/ω = 2·π/√(k/M)
The frequency of oscillation, f = 1/T = √(k/M)/(2·π)
Where:
k = 2·T/l, we have;
f = √(k/M)/(2·π) = √(2·T/l)/m)/(2·π)
The frequency of oscillation is given as follows;
[tex]f = \dfrac{1}{2\cdot \pi }\cdot\sqrt{\dfrac{2 \cdot T}{l \cdot M} }[/tex]
A polonium isotope with an atomic mass of 211.988868 u undergoes alpha decay, resulting in a daughter isotope with an atomic mass of 207.976652 u. Ignoring any recoil of the daughter, find the kinetic energy of the emitted alpha particle in megaelectronvolts (MeV).
Answer:
K = 9.53 MeV
Explanation:
The kinetic energy that the alpha particle has emitted, is the energy in excess after removing the resting energy of the atoms and the helium nucleus that forms the alpha particle
Since energy and masses are related and cannot be
m₀ c² = [tex]m_{f}[/tex] c² + m_He c²+ K
K = c² (m₀ - m_{f} - m_He)
the mass of the Helium atom is 4 u
K = (3 10⁸)² (211,988868 -207.976652 - 4,002) 1,661 10⁻²⁷
K = 14,949 10⁻¹¹ (0.0102)
K = 1,527 10⁻¹² J
let's reduce 1 J = 6,242 10¹² MeV
K = 9.53 MeV
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
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.
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
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.
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
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
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!
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).
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.
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
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
A very thin film of soap, of thickness 170 nm, in between air seems dark. On the other hand, when placed on top of glass some visible light is seen to shine from the film. How can this happen and what is the smallest visible light that creates constructive interference when we place the film on top of glass
Answer:
λ₀ = 2 d n
Explanation:
A soap film is a layer where the lus is reflected on the surface and on the inside of the film, these two reflected rays can interfere with each other either constructively or destructively.
Let's analyze the general conditions of this interference,
* When the ray of light reaches the surface of the film it is reflected, as the index of refraction of the air is less than the index of the film, the reflected ray has a phase change of 180º
* When the ray penetrates the film, its wavelength changes due to the refractive index of the film.
λ = λ₀ / n
where lick is the wavelength in the vacuum or air and n index of refraction of the film, in general this interference is observed perpendicular to the film, so the sine veils 1. the expression for constructive interference taking in what previous remains
2d = (m + ½) λ
the expression for destructive interference remains
2d = m λ
2d = m λ₀ / n
When the film is placed on a glass plate whose index of refraction is greater than the index of refraction of the film, in the reflection in the lower part of the film another phase difference of 180º is created, for which we have a difference of total phase of 180 +180 = 360º, which is equivalent to no phase difference, therefore the two previous equations are interchanged.
Therefore where we had destructive interference now a cosntructive interference happens we can see the reflected light.
Find us the wavelength that this constructive interference creates
2d n = m λ₀
λ₀ = 2 d n / m
To find the minimum wavelength, suppose we observe the first interference pattern m = 1
λ₀ = 2 d n
where d is the thickness of the film and n the index of refraction of the same
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
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.
Which of the following best represents a double replacement reaction?
A. AB → A + B
B. A + B → AB
C. A + BC → AC + B
D. AC + BD → AD + BC
Answer:
D. AC+BD -> AD +BC
Explanation:
I got it right
A double displacement reaction is also called the double replacement reaction or double decomposition reaction. Here AC + BD → AD + BC is a double displacement reaction. The correct option is D.
What is double displacement reaction?A double displacement reaction is defined as a chemical reaction in which the positive and negative ions of two ionic compounds exchange their places to form two new compounds. The general form of a double displacement reaction is given as:
AC + BD → AD + BC
Here 'A' and 'B' are positively charged cations, whereas 'C' and 'D' are negatively charged anions. The double displacement reactions generally occur between the substances in an aqueous solution. In order for a reaction to occur, one of the products is usually a solid precipitate, a gas or a molecular compound like water.
Thus the correct option is D.
To know more about double replacement reaction, visit;
https://brainly.com/question/13870042
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The gas in a balloon has T=280K and V=0.0279m^3. If the temperature increases to 320K at constant pressure, what is the new volume of the balloon? (Hint: n and P are constant) (Unit= m^3)
Answer:
[tex]\boxed{ V_{2}= 0.03189 m^3}[/tex]
Explanation:
According to Charles Law
=> [tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}[/tex]
Where [tex]V_{1}[/tex] = 0.0279 m³, [tex]T_{1}[/tex] = 280 K and [tex]T_{2}[/tex] = 320 K
=> [tex]\frac{0.0279}{280} = \frac{V_{2}}{320}[/tex]
=> [tex]V_{2}[/tex] = 0.03189 m³
Which property describes the amount of energy that flows past a given area
per unit of time?
A. Wavelength
B. Speed
c. Intensity
D. Pitch
Answer:
c. Intensity
Explanation:
Wavelength is a distance (meters).
Speed is distance per time (meters / second).
Intensity is power per area (Watts / square meter).
Pitch is frequency (cycles / second).
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
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:
7.Why are we able to drink cold drink with straw from a bottle?
Answer:
BETTER EXPLANATION
Explanation:
When you drink from a straw, you create a little space of low pressure inside your mouth and in the top of the straw. Then the air outside the straw pushes down on the surface of the drink and forces the liquid up through the straw and into your mouth.
A stone is dropped from a height of 49m and simultaneously another ball is thrown upward from the ground with a speed of 40m/s. when and where do the two stone meet
Answer:
S1 = 1/2 g t^2 distance stone falls in time t
S2 = Vy t - 1/2 g t^2 distance thrown stone rises in time t
H = 49 = S1 + S2 = Vy t
t = 49 / 40 sec time when stones meet
Check:
Stone 1 falls: 1/2 g t^2 = 1/2 * 9.8 * (49 / 40)^2 = 7.35 m
Stone 2 rises : 40 * (49 / 40) - 1/2 * 9.8 (49 / 40)^2 = 41.65 m
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:
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
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.
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!!
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
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
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.
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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.