Answer:
The wavelength will be "1.182 μm".
Explanation:
The given values are:
Wavelength
[tex]\lambda=591 \ nm[/tex]
or,
[tex]=591\times 10^-9 \ m[/tex]
Distance,
[tex]d = 3.00 m[/tex]
[tex]n = 1[/tex]
Distance of fringe from center,
[tex]y = 4.84 \ mm[/tex]
We have to find the wavelength of first order dark fringe,
[tex]\lambda = ?[/tex]
As we know,
⇒ [tex]y_{bright} =\frac{1\times \lambda\times L}{d}[/tex]
On putting the given values in the formula, we get
[tex]0.00484=\frac{1\times (591\times 10^{-9})\times 3}{d}[/tex]
On applying the cross multiplication, we get
[tex]\lambda = \frac{0.00484\times 000036632}{0.5\times 3}[/tex]
[tex]=1182\times 10^{-9}[/tex]
or,
[tex]=1.182 \ \mu m[/tex]
Your _________ is simply your weight-to-height ratio which can be calculated using certain body measurements.
O health
O size
O weight loss
O BMI
Answer:
SiZe
Explanation:
bad bunny is the best!!!
The Chernobyl reactor accident in what is now Ukraine was the worst nuclear disaster of all time. Fission products from the reactor core spread over a wide area. The primary radiation exposure to people in western Europe was due to the short-lived (half-life 8.0 days) isotope 131I
131 I, which fell across the landscape and was ingested by grazing cows that concentrated the isotope in their milk. Farmers couldn't sell the contaminated milk, so many opted to use the milk to make cheese, aging it until the radiactivity decayed to acceptable levels. How much time must elapse for the activity of a block of cheese containing 131I 131 I to drop to 1.0% of its initial value?
Answer:
The correct answer is "53.15 days".
Explanation:
Given that:
Half life of [tex]131_{I}[/tex],
[tex]T_{\frac{1}{2} }= 8 \ days[/tex]
Let the initial activity be "[tex]R_o[/tex]".and, activity to time t be "R".To find t when R will be "1%" of [tex]R_o[/tex], then
⇒ [tex]R=\frac{1}{100}R_o[/tex]
As we know,
⇒ [tex]R=R_o e^{-\lambda t}[/tex]
or,
∴ [tex]e^{\lambda t}=\frac{R_o}{R}[/tex]
By putting the values, we get
[tex]=\frac{R_o}{\frac{R}{100} }[/tex]
[tex]=100[/tex]
We know that,
Decay constant, [tex]\lambda = \frac{ln2}{T_{\frac{1}{2} }}[/tex]
hence,
⇒ [tex]\lambda t=ln100[/tex]
[tex]t=\frac{ln100}{\lambda}[/tex]
[tex]=\frac{ln100}{\frac{ln2}{8} }[/tex]
[tex]=53.15 \ days[/tex]
In addition to producing images, ultrasound can be used to heat tissues of the body for therapeutic purposes. An emitter is placed against the surface of the skin; the amplitude of the ultrasound wave at this point is quite large. When a sound wave hits the boundary between soft tissue and bone, most of the energy is reflected. The boundary acts like the closed end of a tube which can lead to standing waves. Suppose 0.7 MHz ultrasound is directed through a layer of tissue at a bone 0.55 cm below the surface. Remember, sound waves in the body travel at 1540 m/s. Explain.
Answer:
ΔT = 4.058 10²⁰ [tex]\frac{S_o^2}{r^2 \ c_e}[/tex]
Explanation:
In this experiment the system can be approximated as a tube with one end open and the other closed.
The open end is where the ultrasound emitter is and the closed end where the limit between the tissue and the bone is, the length of the tube is L = 0.55 cm = 5.5 10⁻³ m
a node is formed at the closed end and a belly at the open end, so the resonance has the form
λ = 4L 1st harmonic
λ = 4/3 L third harmonic
λ = 4/5 L fifth harmonic
λ = 4L / (2n + 1) n = 0, 1, 2, (2n + 1)
This wave is a standing wave therefore energy density remains in place
P = 1/2 ρ v (w S₀)²
angular velocity is related to frequency
w = 2π f
we substitute
E = P = 2π² ρ v f² S₀²
if this energy per unit area is transformed into heat
E = m c_e DT
let's use the concept of density
ρ = m / V
m = ρ V
if there are no losses in the system
½ π² ρ v f² S₀² = ρ V c_e ΔT
ΔT = [tex]\frac{\pi ^2 \ v \f^2 S_o^2}{2V \ c_e}[/tex]
When analyzing this expression the temperature increase is
* quadratic at the frequency and maximum amplitude of the wave
* proportional to the speed of the wave in the tissue
* inversely proportional to tissue volume
we can approximate the volume of the tissue to the volume of a cylinder tube
V = π r² L
ΔT = [tex]\frac{\pi \ v \ f^2 S_o^2 }{r^2 \ L \ c_e}[/tex]
we calculate
ΔT = π 1450 (0.7 10⁶)² S₀² /( r² 5.5 10-3 c_e)
ΔT = 4.058 10²⁰ [tex]\frac{S_o^2}{r^2 \ c_e}[/tex]
use a trigonometric equation to determine the leg of this triangle
C=90°
A=30°
c=10m
What is a?
Answer: 5
Explanation: B is for sure 60°, c* cosB = 10*1/2 =5
the density of brick is 1,600 kg/m3. what is the mass of a brick with a volume of 0.0006 m3? WILL MARK BRAINLIEST
Answer:
.0000004
Explanation:
The mass of a brick with a volume of 0.0006 m³ and a density of 1600 kg/m³ is 0.96kg.
HOW TO CALCULATE MASS?The mass of a substance can be calculated by multiplying the density of the substance by its volume. That is;
Mass = density × volume
According to this question, the density of brick is 1,600 kg/m3 and it has a volume of 0.0006m³. The mass is calculated as follows:
Mass = 0.0006 × 1600
Mass = 0.96kg
Therefore, the mass of a brick with a volume of 0.0006m³ and a density of 1600 kg/m³ is 0.96kg.
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two 0.5 kg carts, one red and one green, sit about half a meter apart on a low friction track, you push on the red one with the constant force of 4N for 0.17m and then remove your hand. the cart moves 0.33 m on the track and then strikes the green cart. what is the work done by you on the two cart system?
Answer:
The work done by you on the two cart system is 2 N-m
Explanation:
Work done is the product of force and displacement.
W = F * D
Substituting the given values we get -
W =
[tex]4 * (0.17+0.33)\\= 2[/tex]
The work done by you on the two cart system is 2 N-m
A flat circular coil of wire having 400 turns and diameter 6.0 cm carries a current of 7.0 A. It is placed in a magnetic field of with the plane of the coil making an angle of 30° with the magnetic field. What is the magnitude of the magnetic torque on the coil?
Answer:
6.8 N.m
Explanation:
The computation of the magnitude of the magnetic torque on the coil is given below:
Given that
n = 400
d = 6.0 cm
Current is I = 7.0 A
Angle is [tex]\theta[/tex] = 30 degree
Now
We know that
the magnitude of the magnetic torque is
= nIABsin[tex]\theta[/tex]
= (400) (7.0) π ÷ 4 (0.06m)^2 sin(90° - 30°)
As
[tex]\theta[/tex] = (90° - Ф)
= (400) (7.0) π ÷ 4 (0.06m)^2 sin 60°
= 6.8 N.m
A bat emits a 40 kHz chirp to locate flying insects. If the speed of sound is 340 m/s and a bat hears the echo from the moth after 0.6 seconds, then how far away is the moth?
102 m
Explanation:
The time 0.6 sec is the time it took for the sound to travel from the bat to the moth and back. So it took 0.3 sec for the sound to reach the moth. From the definition of speed, the distance of the moth d to the bat is given by
v = d/t ---> d = vt = (340 m/s)(0.3 sec) = 102 m
A self-driving car traveling along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 25.0 m/s. Then the vehicle travels for 39.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s.
(a) How long is the self-driving car in motion (in s)?
(b) What is the average velocity of the self-driving car for the motion described? (Enter the magnitude in m/s.) m/s
Answer:
[tex]56.5\ \text{s}[/tex]
[tex]21.13\ \text{m/s}[/tex]
Explanation:
v = Final velocity
u = Initial velocity
a = Acceleration
t = Time
s = Displacement
Here the kinematic equations of motion are used
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{25-0}{2}\\\Rightarrow t=12.5\ \text{s}[/tex]
Time the car is at constant velocity is 39 s
Time the car is decelerating is 5 s
Total time the car is in motion is [tex]12.5+39+5=56.5\ \text{s}[/tex]
Distance traveled
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{25^2-0}{2\times 2}\\\Rightarrow s=156.25\ \text{m}[/tex]
[tex]s=vt\\\Rightarrow s=25\times 39\\\Rightarrow s=975\ \text{m}[/tex]
[tex]v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-25}{5}\\\Rightarrow a=-5\ \text{m/s}^2[/tex]
[tex]s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-25^2}{2\times -5}\\\Rightarrow s=62.5\ \text{m}[/tex]
The total displacement of the car is [tex]156.25+975+62.5=1193.75\ \text{m}[/tex]
Average velocity is given by
[tex]\dfrac{\text{Total displacement}}{\text{Total time}}=\dfrac{1193.75}{56.5}=21.13\ \text{m/s}[/tex]
The average velocity of the car is [tex]21.13\ \text{m/s}[/tex].
What is the centripetal force for a roller coaster if the mass is 10 kg and the normal force is 25 N?
Answer:
Fc = 123 Newton
Explanation:
Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.
Mathematically, net force is given by the formula;
[tex] Fnet = Fapp + Fg[/tex]
Where;
Fnet is the net force.
Fapp is the applied force.
Fg is the force due to gravitation.
Given the following data;
Normal force = 25N
Mass = 10kg
To find the centripetal force;
From the net force, we have the following formula;
Fc = N + mg
Where;
Fc is the centripetal force.
N is the normal force.
mg is the the weight of the object.
Substituting into the formula, we have;
Fc = 25 + 10(9.8)
Fc = 25 + 98
Fc = 123 Newton
Plz help w answer 1:/ confused ash
Answer:
I would say d I had the same question yesterday and I got it correct so hope that helps
A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotation axis of 18kg⋅m2. She then tucks into a small ball, decreasing this moment of inertia to 3.6kg⋅m2. While tucked, she makes two complete revolutions in 1.2s.
Required:
If she hadn't tucked at all, how many revolutions would she have made in the 1.5 s from board to water?
Answer:
θ₁ = 0.5 revolution
Explanation:
We will use the conservation of angular momentum as follows:
[tex]L_1=L_2\\I_1\omega_1=I_2\omega_2[/tex]
where,
I₁ = initial moment of inertia = 18 kg.m²
I₂ = Final moment of inertia = 3.6 kg.m²
ω₁ = initial angular velocity = ?
ω₂ = Final Angular velocity = [tex]\frac{\theta_2}{t_2} = \frac{2\ rev}{1.2\ s}[/tex] = 1.67 rev/s
Therefore,
[tex](18\ kg.m^2)\omega_1 = (3.6\ kg.m^2)(1.67\ rev/s)\\\\\omega_1 = \frac{(3.6\ kg.m^2)(1.67\ rev/s)}{(18\ kg.m^2)}\\\\\omega_1 = \frac{\theta_1}{t_1} = 0.333\ rev/s\\\\\theta_1 = (0.333\ rev/s)t_1[/tex]
where,
θ₁ = revolutions if she had not tucked at all = ?
t₁ = time = 1.5 s
Therefore,
[tex]\theta_1 = (0.333\ rev/s)(1.5\ s)\\[/tex]
θ₁ = 0.5 revolution
What is the answer to this problem
Answer:
Material that allow the electrons to move freely in order to produce a current
Please mark as brainliest if answer is right
Have a great day, be safe and healthy
Thank u
XD
These steps are followed when using the half-life of carbon-14 to determine
the age of an object that contains carbon. What is the correct order of these
steps?
A. Use the half-life of carbon-14 to determine the number of half-lives
that have passed.
B. Measure the ratio of parent nuclei to daughter nuclei.
C. Use the number of half-lives that have passed to determine the age
of the object.
A. A,B,C
B. A,C,B
0 0
C. B, A,C
D. C, A, B
Answer: a different one is a.b.c
Explanation: still for ape.x
The correct order to determine the age of the an object using carbon-14 is C, A, B. Thus, option D is correct.
What is half life?
The half-life time is defined as the time taken by the radioactive element to reduce one half of its initial value. It is denoted by t(1/2).
To measure the age of an object, a radioactive isotope called carbon-14 is used. The half-life of carbon-14 is 5,730 years. All the objects in the universe consumes carbon in their lifetime and hence, carbon-14 is used to measure the age of the objects.
The process of determining the age of objects using carbon-14 is called Radiocarbon dating. All living organisms consume carbon in means of food and from atmosphere and when the plant and animals dies, the radioactive carbon atoms start decaying.
When it starts decaying, by using Carbon-14 the age of an object is calculated. The age is estimated by measuring the amount of carbon-14 present in the sample and comparing this carbon with the reference Carbon-14 isotope.
The amount of carbon in preserved plants is identified by:
f(t) = 10e {₋ct}
t = time in years when the plant dies( t= 0)
c = the amount of carbon-14 remaining in preserved plants.
The steps include to find the age of an object is :
1. Use the number of half-lives that have passed to determine the age of the object.
2. Use the half-life of carbon-14 to determine the number of half-lives that have passed.
3.Measure the ratio of parent nuclei to daughter nuclei.
Hence, from these steps the age of an object is determined. Therefore the correct solution is D) C, A, B.
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8. Consider a capacitor that is made of two large conducting plates that are rectangular in shape (1 cm by 6 cm), aligned parallel to each other, and separated by an air-filled gap of 0.001 cm. This capacitor is included in a circuit where a battery provides 15,000 V of potential difference. When the capacitor is fully charged in this circuit, what is the c
Answer: [tex]7.96\ \mu C[/tex]
Explanation:
Given
The dimension of the plate is [tex]1\ cm\times 6\ cm[/tex]
The gap between the plate is [tex]0.001\ cm[/tex]
Voltage applied [tex]V=15,000\ V[/tex]
The capacitance of the capacitor is
[tex]C=\dfrac{\epsilon_o A}{d}\\\\C=\dfrac{8.85\times 10^{-12}\times 1\times 6\times 10^{-4}}{10^{-5}}\\\\C=53.1\times 10^{-11}\ F[/tex]
Charge acquired by the capacitor
[tex]\Rightarrow Q=CV\\\Rightarrow Q=53.1\times 10^{-11}\times 15,000\\\Rightarrow Q=796.5\times 10^{-8}\\\Rightarrow Q=7.96\times 10^{-6}\ C[/tex]
A ray of light traveling in water hits a glass surface. The index of refraction of the water is 1.33, and that of the glass is 1.50. At what angle with the plane of the surface must the incident ray strike the glass in order that the polarization of the reflected ray is the greatest
Answer:
[tex]\mu=41.5\textdegree[/tex]
Explanation:
From the question we are told that:
Water index of refraction [tex]i_w=1.33[/tex]
Glass index of refraction [tex]i_g=1.50[/tex]
Generally the equation for Brewster's law is mathematically given by
[tex]\theta=tan^{-1}(\frac{i_g}{i_w})[/tex]
[tex]\theta=tan^{-1}(\frac{1.50}{1.33})[/tex]
[tex]\theta=48.44 \textdegree[/tex]
Therefore Angle of incident to plane \mu (normal at 90 degree to the surface)
[tex]\mu=90\textdegree-\theta[/tex]
[tex]\mu=90\textdegree-48.44\textdegree[/tex]
[tex]\mu=41.5\textdegree[/tex]
Pedro is planning to model how changes in weather affect evaporation from lakes for his first experiment he wants to test how humidity affects the evaporation rate. he places one beaker with 300 mL of water in a dry area. and places another beaker with 300 mL of water near a humidifier which of the following variables does Pedro need to control during his experiment
A. humidity only
B. humidity and evaporation rate
C. volume of water and tempature
D. volume of water only
Pedro needs to control the variables such as volume of water and temperature during his experiment. So, option C.
What is meant by humidity ?The amount of water vapor in the air is known as humidity. The humidity will be high if there is a lot of water vapour in the atmosphere.
Water can evaporate even at very low temperatures, but as the temperature rises, the rate of evaporation increases.
More surface molecules per unit of volume may be able to escape from a substance with a larger surface area, so it will evaporate more quickly.
The control variables in an experiment are the variables that the experimenter intends to keep constant always so as to limit their effect on the measurements of the relationship between the dependent and the independent variable.
Therefore, in order to have a proper measurement of the effect of humidity on evaporation rate, other variables such as temperature, and the volume of the water in the experiment investigations which affect evaporation rate by the provision of heat, (temperature) and their heat capacity, the volume, etc. should be controlled.
Hence,
Pedro needs to control the variables such as volume of water and temperature during his experiment. So, option C.
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What’s the answer to 19 please please help me
Answer:
A.
Explanation:
A po sa tingin ko
sana po makatulong
when a light beam emerges from water into air, the average light speed ____.
a. increases
b. decreases
c. remains the same
d. not provided
Answer:
Increases
Explanation:
The light is emerging from a denser medium to rarer medium so the average light speed increases.
When a light beam emerges from water into air, the average light speed increases.
What is refraction?The bending of a ray passes at an angle from one medium to another in which the speed will be different, as when light passes from air into water and also from water into air can be known as refraction.
When light travels from water to air, the angle from water to air the light gets speeds up and then its changes direction. The light turns away from the normal line. Because the light moves from a denser medium to the rarer medium. But from air to water the speed will be decreased because in that case the light travels from the rarer medium to the denser medium.
Hence, the light beam when passes, the average speed gets increased.
So, Option A is the correct answer.
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18. Un avión de rescate de animales que vuela hacia el este a 36.0 m/s deja caer una paca de
heno desde una altitud de 60.0 m. Si la paca de heno pesa 175 N, ¿cuál es el momentum
de la paca antes de que golpee el suelo?
Answer:
Definimos momento como el producto entre la masa y la velocidad
P = m*v
(tener en cuenta que la velocidad es un vector, por lo que el momento también será un vector)
Sabemos que el peso de la paca de heno es 175N, y el peso es masa por aceleración gravitatoria, entonces.
Peso = m*9.8m/s^2 = 175N
m = (175N)/(9.8m/s^2) = 17.9 kg
Ahora debemos calcular la velocidad de la paca justo antes de tocar el suelo.
Sabemos que la velocidad horizontal será la misma que tenía el avión, que es:
Vx = 36m/s
Mientras que para la velocidad vertical, usamos la conservación de la energía:
E = U + K
Apenas se suelta la caja, esta tiene velocidad cero, entonces su energía cinética será cero y la caja solo tendrá energía potencial (Si bien la caja tiene velocidad horizontal en este punto, por la superposición lineal podemos separar el problema en un caso horizontal y en un caso vertical, y en el caso vertical no hay velocidad inicial)
Entonces al principio solo hay energía potencial:
U = m*g*h
donde:
m = masa
g = aceleración gravitatoria
h = altura
Sabemos que la altura inicial es 60m, entonces la energía potencial es:
U = 175N*60m = 10,500 N
Cuando la paca esta próxima a golpear el suelo, la altura h tiende a cero, por lo que la energía potencial se hace cero, y en este punto solo tendremos energía cinética, entonces:
10,500N = (m/2)*v^2
De acá podemos despejar la velocidad vertical justo antes de golpear el suelo.
√(10,500N*(2/ 17.9 kg)) = 34.25 m/s
La velocidad vertical es 34.25 m/s
Entonces el vector velocidad se podrá escribir como:
V = (36 m/s, -34.25 m/s)
Donde el signo menos en la velocidad vertical es porque la velocidad vertical es hacia abajo.
Reemplazando esto en la ecuación del momento obtenemos:
P = 17.9kg*(36 m/s, -34.25 m/s)
P = (644.4 N, -613.075 N)
Please help me ASAP!!
Answer:
the ball lose kentic energy and gains potential energy rolling upward
Answer:
c
Explanation:
when a ball is rolling down it loses potential and gains kinetic, but in this case since the ball is going upwards it is losing kinetic energy and gaining potentail like a roller coaser
Help pls... 100points
1 Answer:
The pitch of a sound depends on
a. frequency
b. intensity
c. amplitude
d. quality
2 Answer:
The loudness of a sound depends on
a. frequency
b. intensity
c. overtones
d. timbre
3 Answer:
The number of sound waves per unit time is called
a. amplitude
b. wavelength
c. speed
d. frequency
4 Answer:
The blending of fundamental tone and overtones produces a sound's
a. intensity
b. pitch
c. timbre
d. loudness
5 Answer:
Sonar is used
a. to locate objects in water
b. to determine underwater distances
c. for both a and b
d. for neither a nor b
6 Answer:
Sounds of frequencies below 20 Hz are called
a. fundamentals
b. overtones
c. ultrasonic sounds
d. infrasonic sounds
7 Answer:
A sound is music if it has
a. a definite pitch
b. a pleasing quality
c. rhythm
d. all of these characteristics
8 Answer:
The stirrup is a part of the
a. outer ear
b. middle ear
c. inner ear
d. cochlea
9 Answer:
An example of a percussion instrument is a (an)
a. trumpet
b. harp
c. drum
d. flute
10 Answer:
The speed of sound depends on
a. temperature
b. density
c. elasticity
d. all of these factors
11 Answer:
Sounds with frequencies above 20,000 Hz are called _____.
12 Answer:
The science of sound is called _____.
13 Answer:
High-frequency waves are used in a sound navigation and ranging
system called _____.
14 Answer:
Sounds produced when sections of a string vibrate separately are
called _____.
15 Answer:
Sound without pleasing quality, identifiable pitch, and repeating
patterns is called _____.
16 Answer:
The speed of sound decreases with increasing temperature.
a. true
b. false
17 Answer:
Sound waves are transverse.
a. true
b. false
18 Answer:
The decibel is a unit of pitch.
a. true
b. false
19 Answer:
A change in the pitch of a sound due to motion of source or
receiver (observer) is called the Doppler effect.
a. true
b. false
20 Answer:
Guitars and cellos are examples of percussion instruments.
a. true
b. false
Answer:
Explanation:
1. a as pitch depends on frequency
2. b as loudness depends on intensity
3. d as that is definition of frequency
4. c as that is definition of timbre
5. c as sonar can do both
6. d as that is definition of infrasound
7. d
8. b as that is one of three bones in the middle ear
9. c as you hit the drum to make sound
10. d
11. ultrasound
12. acoustics
13. ultrasound
14. harmonics? or resonance?
15. noise
16. b
17. b
18. b
19. a
20. b
g 2. In a laboratory experiment on standing waves a string 3.0 ft long is attached to the prong of an electrically driven tuning fork which vibrates perpendicular to the length of the string at a frequency of 60 Hz. The weight (not mass) of the string is 0.096 lb. a) [5 pts] What tension must the string be under (weights are attached to the other end) if it is to vibrate in four loops
Answer:
The tension in string will be "3.62 N".
Explanation:
The given values are:
Length of string:
l = 3 ft
or,
= 0.9144 m
frequency,
f = 60 Hz
Weight,
= 0.096 lb
or,
= 0.0435 kgm/s²
Now,
The mass will be:
= [tex]\frac{0.0435}{9.8}[/tex]
= [tex]0.0044 \ kg[/tex]
As we know,
⇒ [tex]\lambda=\frac{2L}{n}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{2\times 0.9144}{4}[/tex]
⇒ [tex]=0.4572 \ m[/tex]
or,
⇒ [tex]v=f \lambda[/tex]
⇒ [tex]=0.4572\times 60[/tex]
⇒ [tex]=27.432 \ m/s[/tex]
Now,
⇒ [tex]v=\sqrt{\frac{T}{\mu} }[/tex]
or,
⇒ [tex]T=\frac{m}{l}\times v^2[/tex]
On putting the above given values, we get
⇒ [tex]=\frac{0.0044}{0.9144}\times (27.432)^2[/tex]
⇒ [tex]=\frac{752.51\times 0.0044}{0.9144}[/tex]
⇒ [tex]=3.62 \ N[/tex]
A boy is pulling a sled with a net force of 10 N. If the mass of the sled is 20 kg, what is the acceleration of the sled?
Answer:
0.5 m/s
Explanation:
acceleration= force times mass
Given the amount of force applied on the sled as well as its mass, the acceleration of the sled is 0.5m/s².
What is force?A force is simply referred to as either a push or pull of an object resulting from the object's interaction with another object.
From Newton's Second Law, force is expressed as;
F = m × a
Where is mass of object and a is the acceleration
Given the data in the question;
Force applied F = 10N = 10kgm/s²Mass of the sled = 20kgAcceleration a = ?F = m × a
10kgm/s² = 20kg × a
a = 10kgm/s² ÷ 20kg
a = 0.5m/s²
Given the amount of force applied on the sled as well as its mass, the acceleration of the sled is 0.5m/s².
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While diving in cancun Mexico where the seawater has a density of 1,015 kg/m3 Nana observed that her pressure meter device reading was 3.75 atm. The reading at sea level is standard 1.0 atm. At what depth is she diving when the meter read 3.75 atm g
Answer:
The depth of the diver is 28.01 m
Explanation:
Given;
density of the seawater, ρ = 1,015 kg/m³
standard sea level pressure, P₀ = 1.0 atm = 101,325 Pa
the final reading of her pressure, P₁ = 3.75 atm = 379968.75 Pa
acceleration due to gravity, g = 9.8 m/s²
Let the depth she was diving at the final pressure = h
This depth is calculated as;
P₁ = P₀ + ρgh
P₁ - P₀ = ρgh
[tex]h = \frac{ P_1 \ - \ P_o}{\rho g} = \frac{379968.75 \ - \ 101325}{1015 \ \times \ 9.8} = 28.01 \ m[/tex]
Therefore, the depth of the diver is 28.01 m
A balloon with a charge of 4.0 µC is held a distance of 0.70 m from a second balloon having the same charge. Calculate the magnitude of the repulsive force. (Hint: µ is a metric prefix that means micro or 10 -6 so 4.0 µC means 4.0 microcoulombs
Answer:
Im sorry I do not know how to do this, I hope you will be able to figure it out
Explanation:
Please help I’ll mark you brainliest
Answer:
Percentage:
Rr = 50% because it's 2/4 (for both or 25% each since you have them separate)
rr = also 50%, because it's also 2/4.
Phenotype:
Rr = heterozygous
rr = "hozygous" recessive
In addition, RR is "hozygous" dominant
Explanation:
They said the hozygous is a swearword LOL.
b
Jasmine plays a note of wavelength 22 cm on her clarinet. The speed of
sound in air is 340 m/s. What is the frequency of this note? Give your
answer to 3 significant figures.
Answer:
Frequency = 1,550Hz
Explanation:
To solve this we can use the equation: [tex]f=\frac{v}{\lambda}[/tex]
(frequency = velocity/wavelength).
We are given the information that the wavelength is 22cm and the speed is 340m/s. The first step is to make sure everything is in the correct units (SI units), and to convert them if needed. The SI Units for velocity and wavelength are m/s and m respectively. This means we need to convert 22cm into meters, which we can do by dividing by 100, (as there are 100cm in a meter). 22/100 = 0.22m
Now we can substitute these values into the formula and calculate to solve:
[tex]f=\frac{340}{0.22} \\\\f=1545.454...[/tex]
Simplify to 3 significant figures:
f = 1,550Hz
(Which I believe is just below a G6 if you were interested)
Hope this helped!
Which diagram shows magnets that will attract each other? 2 bar magnets side by side with their long axes vertical, both red S on top and blue N on bottom. 2 bar magnets top to bottom with their long axes vertical, the top one with red S on top and blue N on bottom and the bottom magnet with blue N on top and red S on bottom. 2 bar magnets top to bottom with their long axes vertical, the top one with blue N on top and red S on bottom and the bottom magnet with red S on top and blue N on bottom. 2 bar magnets top to bottom with their long axes vertical, the top one with red S on top and blue N on bottom and the bottom magnet with red S on top and blue N on bottom.
2 bar magnets top to bottom with their long axes vertical, the top one with red S on top and blue N on bottom and the bottom magnet with red S on top and blue N on bottom. this diagram shows magnets that will attract each other. Hence option D is correct.
What is Magnet ?A permanent magnet is an item constructed of magnetised material that generates its own persistent magnetic field. A refrigerator magnet, for example, is commonly used to hold notes on a refrigerator door. Ferromagnetic (or ferrimagnetic) materials are those that can be magnetised and are strongly attracted to a magnet. These include the elements iron, nickel, and cobalt, as well as their alloys, some rare-earth metal alloys, and naturally occurring minerals such as lodestone. Although ferromagnetic (and ferrimagnetic) materials are the only ones that are strongly attracted to a magnet and are widely thought to be magnetic, all other substances respond weakly to a magnetic field via one of many different forms of magnetism.
Hence option D is correct.
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On a distance-time graph, the faster the object, the _______ the slope of the line. *
A.curvier
B.flatter
C.lower
D.steeper
The following words are used to describe the direction of a moving object EXCEPT... *
A. East
B. North
C.South
D. Today
Answer:
..In a distance-time graph, the slope or gradient of the line is equal to the speed of the object. The steeper the line (and the greater the gradient) the faster the object is moving. Calculate the speed of the object represented by the green line in the graph, from 0 to 3 s.
d) steeper
D)Today