Answer:
57.0 m
Explanation:
Assume the time it takes for the sound to reach the top of the well is negligible.
Given:
v₀ = 0 m/s
a = 9.8 m/s²
t = 3.41 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (3.41 s) + ½ (9.8 m/s²) (3.41 s)²
Δy = 57.0 m
Answer:
57 m
Explanation:
6. What IMFA holds the molecules of water together?
A. hydrogen bond
C. ionic bond
B. nonpolar covalent bond
D. polar covalent bond
Which of the following best describes weight? The amount of space an object takes up The amount of matter in an object The force on an object due to its mass and gravity The force on an object due to its speed and location
Explanation:
The force on an object due to it's mass and gravity.
A glass rod rubbed with a tissue paper would strip away electrons from the atoms in glass rod. This would acquire positive charge on the glass rod. What charge is generated at the leaves in electroscope (initially neutral) when this charged glass rod is brought towards the metal ball
Answer:
when this charged glass rod is brought towards the metal ball it will acquire a charge opposite to that of charge body brought close to it without touching it but it will acquire the same charge if the charged object touches it
Answer:
B.The glass and the paper have different charges.
Explanation:
Show that the force vector D=(2.0i-4.0j+k)N is orthogonal to the force vector G=(3.0i+4.0j+10k)N.
Answer:
Explanation:
For two vectors to be orthogonal (perpendicular) the product of both vectors must be zero. Given the vectors D=(2.0i-4.0j+k)N and G=(3.0i+4.0j+10k)N, to show that they are orthogonal, we will take their dot product as shown;
D.G = (2.0i-4.0j+k).(3.0i+4.0j+10k)
Note that i.1 = j.j = k.k = 1 and dot product of different component is zero.
D.G = 2.0(3.0) (i.i) + (-4.0)(4.0)j.j + 1(10)k.k
D.G = 6.0(1) -(16)(1)+10(1)
D.G = 6-16+10
D.G = -10+10
D.G = 0
Since the dot product of the two vectors is zero, this shows that force vector D is orthogonal to force vector G.
Which of the following quantities would be acceptable representations of weight?
a. 12.0 lb
b. 0.34 g
c. 120 kg
d. 1600 kN
e. 0.34 m
f. 411 cm
Explanation:
Weight of an object is shows the force of gravity acting on it. It is calculated mass times acceleration due to gravity.
lb is the unit of avoirdupois weight. It means that 12 lb shows weight.
gram, kg is the unit of mass. It means 0.34 g and 120 kg shows mass of an object.
m and cm are the units of length. It means 0.34 m and 411 cm shows the length of the object.
N is the unit of weight.
Hence, 12.0 lb and 1600 kN are acceptable representations of weight.
For each experiment involving nanotechnology, Gerry
and Lena will test a(n)
Gerry and Lena work in a modern physics laboratory
and study nanotechnology, which is a new type of
technology that allows control over individual particles as
small as molecules or atoms. Even though they work
with new technology, they still follow the scientific
method, so their experiments attempt to find
relationships between independent and dependent
variables.
Answer: Hypothesis
Explanation:
For each experiment involving nanotechnology, Gerry and Lena will test a hypothesis.
Gerry and Lana will use the Scientific method for their nanotechnology research and a key stage in the scientific method is to come up with a Hypothesis.
A hypothesis gives a researcher direction because it is a theory that they formulate that is meant to explain the phenomenon that they are researching.
They will then test this theory to either disprove or approve it and in so doing will be able to come up with conclusions to the research. Gerry and Lana will therefore have to test a hypothesis for each experiment in order to continue with their research.
1. Which of the following are quantitative observations? (Select all that apply) a) The sky is blue b) The toy car is about 3 inches long c) It is 250,000 miles from the earth to the moon. d) The wooden cart has a mass of 18.654 g. e) When at rest, the pendulum points toward the center of the earth.
Answer: las respuestas correctas son b, c y d.
indican cantidad
Emily takes a trip, driving with a constant velocity of 85.5 km/h to the north except for a 22 min rest stop. If Emily's average velocity is 68.8 km/h to the north, how long does the trip take? Show your work
Answer:
1½ hours
Explanation:
Let the amount of time she drove at 85.5 km/hr be t, then
The distance that she rode would be 85.5 * t
This is means Emily has traveled the same distance at 68.8 km/hr without any stoppage whatsoever. Thus, we can create an equation as such
85.5 * t = 68.8 (t + 22/60), the 22/60 being, converting 22 minutes into hours. We proceed further as
85.5t = 68.8t + 1513.6/60, multiplying all sides by 60, we have
5130t = 4128t + 1513.6, now we collect like terms
5130t - 4128t = 1513.6
1002t = 1513.6
t = 1513.6/1002
t = 1.51 hrs, or in minutes, 1.51 * 60
t = 90.63 minutes. 1 hour and 36 minutes
So the total time the trip took, 22 minutes in addition, or say, 1 hour 58 minutes.
Part A.)Six boxes held at rest against identical walls.
Rank the boxes on the basis of the magnitude of the normal force acting on them.
Rank from largest to smallest. To rank items as equivalent, overlap them.
1) 130N--->7kg
2) 150N--->1kg
3) 150N--->7kg
4) 120N--->3kg
5) 140N--->5kg
6) 140N--->3kg
(Since the boxes are at rest, Newton's 2nd law dictates that the horizontal forces on each box must add up to zero. You can use this information to determine the normal forces. If two boxes are both pushed against the wall by the same force, then they should experience the same normal force.)
Part B.) Rank the boxes on the basis of the frictional force acting on them.
Rank from largest to smallest. To rank items as equivalent, overlap them.
1) 130N--->7kg
2) 150N--->1kg
3) 150N--->7kg
4) 120N--->3kg
5) 140N--->5kg
6) 140N--->3kg
Answer:
Explanation:
When a body is held against a vertical wall , to keep them in balanced position , normal force is applied on their surface . this force creates normal reaction which acts against the normal force and it is equal to the normal force as per newton's third law . Ultimately friction force is created which is proportional to normal force and it acts in vertically upward direction . It prevents the body from falling down .
Hence normal force = reaction force .
From second law also net force is zero , so if normal force is N and reaction force is R
R - N = mass x acceleration = mass x 0 = 0
R = N .
Ranking normal force from highest to smallest
150 N , 130 N , 120 N
B )
Frictional force is equal to the weight of the body because the body is held at rest .
Ranking of frictional force form largest to smallest
7 kg , 5 kg , 3 kg , 1 kg .
Here frictional force is irrespective of the normal force acting on the body because frictional force adjusts itself so that it becomes equal to weight in all cases here because it always balances the weight of the body .
Based on the magnitude of normal force acting on them, the ranking from largest to smallest will be:
150N--->7kg = 150N--->1kg.140N--->5kg = 140N--->3kg.130N--->7kg. 120N--->3kg.Based on the masses, the frictional force acting on the boxes will be such that the ranking from largest to smallest will be:
150N--->7kg = 130N--->7kg. 140N--->5kg 140N--->3kg = 120N--->3kg 150N--->1kgNormal force magnitudeThe normal forces acting on the box will depend on the force acting on the box. This means that the larger the force acting on the box, the larger the normal force.
This is why the box with a force of 150 N will be have the highest normal force acting on it.
Frictional force magnitudeThe frictional force acting on the boxes will depend on the mass. Boxes will lager masses will have more frictional force acting on them.
The boxes of 7kg will therefore have the largest frictional forces acting on them.
Find out more on frictional forces at https://brainly.com/question/24386803.
calculate the rate of change (slope) of the graph shown.
Answer:5
Explanation:
You do change in y over change in X
A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° above horizontal. Assume the ball encounters no air resistance, and use a Cartesian coordinate system with the origin located at the ball's initial position.
(a) Create an expression for the football’s horizontal velocity, vfx, when caught by a receiver in terms of v0, θ, g, and h.
(b) The receiver catches the football at the same height as released by the quarterback. Create an expression for the time, t f, the football is in the air in terms of v0, θ, g, and h.
(c) The receiver catches the ball at the same vertical height above the ground it was released. Calculate the horizontal distance, d in meters, between the receiver and the quarterback.
Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ = [tex]v_{oy}[/tex] / vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m
(a) The horizontal velocity in the projectile motion is always constant.it is the horizontal component of velocity by which the object is thrown. The expression for the football horizontal velocity will be [tex]H = \frac{u^{2} sin^{2}\theta}{2g}[/tex].
(b) The amount of time it takes for the body to project and land is the time of flight.
(c) The horizontal distance traveled by the ball is defined by the ball is called the range of the ball. The horizontal distance traveled will be 16.7 m.
What is a range of projectile?The horizontal distance is covered by the body when the body is thrown at some angle is known as the range of the projectile. It is given by the formula
(a) Let the velocity at which the ball passes will be v. Resolve the velocity components into two components one is the horizontal component
[tex]\rm{u_x = ucos\theta}[/tex]
[tex]\rm{u_y = usin\theta}[/tex]
let the x distance is traveled in the horizontal direction so,
[tex]y = u_y \times t[/tex]
[tex]\rm{H = u_y t+\frac{1}{2} gt^2}[/tex]
[tex]\rm{H = usin\theta (\frac{usin\theta}{g}) +\frac{1}{2} g(\frac{usin\theta}{g}) ^2}[/tex]
[tex]H = \frac{u^{2} sin^{2}\theta}{g} -\frac{u^{2}sin^{2}\theta }{2g}[/tex]
[tex]H =\frac{u^{2}sin^{2}\theta }{2g}[/tex]
This is the required relation for the horizontal velocity.
(b)
Newton's equation of motion
[tex]\rm{v = u +gt}[/tex]
[tex]\rm{v_y = u_y +gt}[/tex]
[tex]\rm{v_y = 0}[/tex]
[tex]{u_y = usin\theta}[/tex]
[tex]\rm{usin\theta = gt}[/tex]
[tex]t= \rm{ \frac{usin\theta}{g} }[/tex]
These are the required relation for time t.
(c)
The range of the projectile is given as
[tex]R =\frac{u^{2}sin{2}\theta }{g}[/tex]
[tex]R =\frac{(13.5)^{2}sin64^0 }{9.81}[/tex]
R = 16.7 m
So the horizontal distance covered will be 16.7 m.
To learn more about the range of projectile refer to the link ;
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1. At the starting gun a runner accelerates from rest at 2.0 m/s2 for 2.2 s. What is the runners speed 1.2 s after she starts running?
2. A skier starts from rest and accelerates down a slope at 2.2 m/s2 . How much time is required for the skier to reach a speed of 9.0 m/s ?
Explanation:
1. Given:
v₀ = 0 m/s
a = 2.0 m/s²
t = 1.2 s
Find: v
v = at + v₀
v = (2.0 m/s²) (1.2 s) + 0 m/s
v = 2.4 m/s
2. Given:
v₀ = 0 m/s
v = 9.0 m/s
a = 2.2 m/s²
Find: t
v = at + v₀
9.0 m/s = (2.2 m/s²) t + 0 m/s
t ≈ 4.1 s
A man on a road trip drives a car at different constant speeds over several legs of the trip. He drives for 55.0 min at 60.0 km/h, 18.0 min at 80.0 km/h, and 60.0 min at 60.0 km/h and spends
25.0 min eating lunch and buying gas.
Complete Question
A man on a road trip drives a car at different constant speeds over several legs of the trip. He drives for 55.0 min at 60.0 km/h, 18.0 min at 80.0 km/h, and 60.0 min at 60.0 km/h and spends
25.0 min eating lunch and buying gas
What is the total distance traveled over the entire trip (in km)
Answer:
The value is [tex]D = 139.02 \ km[/tex]
Explanation:
From the question we are told that
For [tex] t_1 = 55 min = \frac{55}{60} = 0.917 \ h[/tex] the speed is [tex]v_1 = 60 \ km/h[/tex]
For [tex] t_2 = 18 min = \frac{18}{60} = 0.3 \ h[/tex] the speed is [tex]v_2 = 80 \ km/h[/tex]
For [tex] t_3 = 60 min = \frac{60}{60} = 1 \ h[/tex] the speed is [tex]v_3 = 60\ km/h[/tex]
The time taken to have lunch is [tex]t _l = 25 \ min = \frac{25}{60} = 0.42 \ h[/tex]
Generally the total distance traveled over the entire trip (in km) is mathematically represented as
[tex]D = t_1 * v_1 + t_2 * v_2 + t_3 * v_3[/tex]
=> [tex]D = 0.917 * 60 + 0.3 * 80 + 1 * 60[/tex]
=> [tex]D = 55.02 + 24 + 60[/tex]
=> [tex]D = 139.02 \ km[/tex]
Three tiny charged metal balls are arranged on a straight line. The middle ball is positively charged and the two outside balls are negatively charged. The two outside balls are separated by 20 cm and the middle ball is exactly halfway in between. (HINT: Draw a picture; in your picture, the distance between the two outermost balls should be 20 cm.) The absolute value of the charge on each ball is the same, 1.45 μCoulombs (the meaning of μ, which is read as "micro", is 10-6). Give your answers in newtons.
a) What is the magnitude of the attractive force on either outside ball due ONLY to the positively-charged middle ball?
(b) What is the magnitude of the repulsive force on either outside ball due ONLY to the other outside ball?
(c) What is the magnitude of the net force on the outside of the ball
Answer:
[tex](a) 189.23 N[/tex], [tex](b) 47.31 N[/tex] and [tex](c) 141.92 N[/tex].
Explanation:
Three balls are shown in figure having charge [tex]q=1.45 \mu C[/tex]. The middle ball, [tex]B[/tex], is positively charged having charge [tex]+q[/tex], and the remaining two outside balls, [tex]A[/tex] and [tex]C[/tex], are negatively charged having charged [tex]-q[/tex] as shown.
[tex]AC=20 cm[/tex] and [tex]AB=BC=10[/tex] cm as B is the mid-point of AC.
Let [tex]d_1=AC=20\times 10^{-3}m[/tex] and [tex]d_2=AB=BC=10\times 10^{-3}m[/tex]
From Coulomb's law, the magnitude of the force, [tex]F[/tex], between two point charges having magnitudes [tex]q_1 \& q_2[/tex], separated by distance, [tex]d[/tex], is
[tex]F=\frac {1}{4\pi\epsilon_0}\frac {q_1q_2}{d^2}\;\cdots (i)[/tex]
where, [tex]\epsilon_0[/tex] is the permittivity of free space and
[tex]\frac {1}{4\pi\epsilon_0}=9\times 10^9[/tex] in SI units.
This force is repulsive for the same nature of charges and attractive for the different nature of charges.
Now, Using equations(i),
(a) The magnitude of attraction force between balls A and B is
[tex]F_{AB}=F_{BC}= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_2)^2}[/tex]
[tex]\Rightarrow F_{AB}= 9\times 10^9}\frac {1.45\times 10^{-6}\times1.45\times 10^{-6}}{\left(10\times 10^{-3}\right)^2}[/tex]
[tex]\Rightarrow F_{AB}=189.23 N[/tex]
(a) The magnitude of the repulsive force between balls A and C is
[tex]F_{AC}= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_1)^2}[/tex]
[tex]\Rightarrow F_{AC}= 9\times 10^9}\frac {1.45\times 10^{-6}\times1.45\times 10^{-6}}{\left(20\times 10^{-3}\right)^2}[/tex]
[tex]\Rightarrow F_{AC}=47.31 N[/tex]
(c) The magnitude of the net force, [tex]F_{net}[/tex], on the outside of the ball is,
[tex]F_{net}=189.23-47.31 N[/tex]
[tex]\Rightarrow F_{net}=141.92 N[/tex]
Hello How to obtain the voltage and current for the resistors? I don't care about what the result is, I care about HOW Thanks
Answer:
its right the way it is
Explanation:
if there is a multiple choice then pick 20 v
A car is moving at 35 mph and comes to a stop in 5 seconds.
Find the acceleration of the car.
Answer:
-10.267
Explanation:
Initial speed > 35 miles per hour
Final speed > 0 miles per hour
Time > 5 seconds
you have to divide the velocity by the time which should be 35 / 5 = 7
so the answer should be 7 but I'm not sure
define rolling friction
Answer:
Definition - The friction that occurs when an object rolls across a rolling friction is easier to overcome than sliding friction.
Ex - Anything with wheels (cars, bicycle,etc) or ball rolling.
Explanation:
I think this answer will help .... if so please follow me.
A long straight conducting rod carries a current I with a non-uniform current density J = ar2, and has a radius R. The value of the constant is 28.5 A/mm4 and the radius of the rod is 5.20 mm. Determine the magnitude of the magnetic field at the following points.(a) r1 = R/2(b) r2 = 2R
The magnetic field is just a unit vector that explains the electromagnetic influence on electric currents, current flow, and magnetic fluids.
Magnetic field:Current density [tex]J = ar^2[/tex]
value of the constant is= [tex]28.5 \ \frac{A}{mm^4}[/tex]
radius = 5.20 mm
magnetic permeability [tex]\mu = 4\pi \times 10^7 \ \frac{N}{A^2}[/tex]
calculating the area element for the straight circular conduction rod
:
[tex]d_A=2\pi r dr[/tex]
Calculating the current, which is carried in the rod
[tex]dI = \int dA \vec{J}[/tex]
Calculating the above equation with the limit value that is 0 to r.
[tex]I^{1}=\int_{0}^{r} ar^2 \times 2 \pi \cdot r d_r[/tex]
[tex]=2\pi a\int_{0}^{r} r^3 d_r \\\\=\frac{\pi ar^4}{2}[/tex]
The calculated current value which is carried by the rod is [tex]\boxed{\frac{\pi ar^4}{2}}[/tex]
In option (a)
calculating the magnitude of the magnetic field at the point [tex]r_1= \frac{R}{2}[/tex]
[tex]\to B=\frac{\mu_{0} I^{1}}{2 \pi r}\\\\\to B \times 2\pi r= \mu_{0} (\frac{\pi a}{2}r^4)\\\\\to B \times 2\pi = \mu_{0} (\frac{\pi a}{2}r^3)\\\\\to B= \frac{\mu_{0}}{4}r^3 a\\\\[/tex]
Substituting the above value:[tex]\frac{R}{2} \ for \ r[/tex]
[tex]B= \frac{\mu_{0}}{4}(\frac{R}{2})^3 a[/tex]
[tex]B= \frac{4 \pi \times 10^{-7}}{4}(\frac{5.2 \times 10^{-3}}{2})^3 \frac{28.5}{10^{-12}}[/tex]
[tex]= \frac{4 \pi \times 10^{-7}}{4} \times \frac{140.608 \times 10^{-9}}{8} \times \frac{28.5}{10^{-12}}\\\\= \frac{ 3.14 \times 10^{-7}}{1} \times \frac{140.608 \times 10^{-9}}{8} \times \frac{28.5}{10^{-12}}\\\\=\frac{1572.87624 \times 10^{-16}}{ 10^{-12}}\\\\=0.157 \ \ T[/tex]
Thus, the magnitude of the magnetic field at the point [tex]r_1 =\frac{R}{2}[/tex] is [tex]\boxed{0.157 \ T}[/tex]
In option (b)
In this, we calculate the magnitude of the magnetic field at the point [tex]r_2= 2R[/tex]
[tex]\to B=\frac{\mu_{0} I^{1}}{2 \pi r} \\\\\to B \times 2\pi r= \mu_{0} (\frac{\pi a}{2}R^4)\\\\\to B \times 2\pi 2R = \mu_{0} (\frac{\pi a}{2}R^3)\\\\\to B= \frac{\mu_{0}}{8}R^3 a\\\\[/tex]
Substituting the values
[tex]B= \frac{4 \pi \times 10^{-7}}{8}(5.2 \times 10^{-3})^3(\frac{28.5}{10^{-12}})[/tex]
[tex]= \frac{4 \times 3.14 \times 10^{-7}}{8} \times (5.2)^3 \times (10^{-3})^3 \times\frac{28.5}{10^{-12}}\\\\= \frac{ 3.14 \times 10^{-7}}{2} \times 140.608 \times 10^{-9} \times\frac{28.5}{10^{-12}}\\\\= 6291.50495 \times 10^{-4}\\\\= 0.629 \ \ or \ \ 0.63\\[/tex]
Thus, the magnitude of the magnetic field at the point [tex]r_2 = 2R[/tex] is [tex]\boxed{0.63 \ \ T}[/tex]
Find out more about Magnetic fields here:
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A gnat takes off from one end of a pencil and flies around erratically for 26.1 s before landing on the other end of the same pencil. If the gnat flew a total distance of 2.15 m, and the pencil is 0.0463 m long, find the gnat's average speed and the magnitude of the gnat's average velocity.
Answer:
2.15m/26.1s is = average speed
0.08 m/s
What elements on the periodic table are metalloids?
Answer:
The term 'mettaloid' is normally applied to a group of between six and nine elements ;
BoronSilicon GermaniumArsenicAntimonyTellurium, and possibly BismuthPolonium Astatineare found near the center of the P-block or main block of the periodic table.
Explanation:
The density of water is about 1 gram per milliliter. A milliliter is a cubic centimeter (i.e., cm3 ). A red blood cell has a density similar to water and is shaped like a one micrometer thick disk with a diameter of about 10 micrometers. About what is the mass in grams of a red blood
Answer:
The mass in grams of a red blood cell is about 7.85 × 10⁻¹¹ grams
Explanation:
To find the mass in grams of a red blood cell,
From,
[tex]Density = \frac{Mass}{Volume}[/tex]
Then,
[tex]Mass = Density \times Volume[/tex]
From the question,
Density of a red blood cell is similar to that of water
Density of water = 1 g/mL = 1 g/ cm³
Then, Density of a red blood cell = 1 g/cm³
Now, we will find the volume a red blood cell.
From the question,
A red blood cell is shaped like a one micrometer thick disk with a diameter of about 10 micrometersSince the shape is like that of a thick disc, we can determine the volume by using the formula for volume of a cylinder.
Hence,
Volume of a red blood cell = [tex]\pi r^{2}h[/tex]
Where [tex]\pi[/tex] Is a constant (Take [tex]\pi[/tex] = 3.14)
[tex]r[/tex] is the radius
and [tex]h[/tex] is the thickness
Diameter of a red blood cell = 10 micrometers
Then, radius of a red blood cell = 10/2 micrometers = 5 micrometers
[tex]r[/tex] = 5 micrometers = 5 × 10⁻⁶ meters
and [tex]h[/tex] = 1 micrometer = 1 × 10⁻⁶ meters
Hence,
Volume of a red blood cell = 3.14 × (5 × 10⁻⁶)² × 1 × 10⁻⁶
∴ Volume of a red blood cell = 7.85 × 10⁻¹⁷ cubic meter (m³)
Convert this to cubic centimeter
(NOTE: 1 cubic meter = 1000000 cubic centimeter)
Hence,
Volume of a red blood cell = 7.85 × 10⁻¹¹ cubic centimeter (cm³)
Now, for the mass
[tex]Mass = Density \times Volume[/tex]
Density of a red blood cell = 1 g/cm³
Volume of a red blood cell = 7.85 × 10⁻¹¹ cubic centimeter (cm³)
Then,
Mass = 1 g/cm³ × 7.85 × 10⁻¹¹ cm³
Mass = 7.85 × 10⁻¹¹ g
Hence, the mass in grams of a red blood cell is about 7.85 × 10⁻¹¹ grams
what describes a feature of the physical properties of all substances
Physical properties of matter tells us what a substance is or what a substance can do when it is not undergoing a chemical change.
These properties are observable with our senses or instruments or some pieces of apparatus.
Some of these properties are :
Color
Odor
hardness
texture
boiling point
Melting point
density
viscosity
They differ from chemical properties which are shown by the reaction of a substance with another.
A point charge q is placed at the center of a spherical Gaussian surface. The electric fix through the surface will change if:
a. the shape or the surface to a changed to a cube with the same volume as the original sphere.
b. the point charge is moved off center (but still inside the sphere) the point charge is moved to allocation just outside the sphere.
c. if a second point charge is placed just outside the sphere.
d. the radius of the surface is doubled.
Answer:
. the point charge is moved off center (but still inside the sphere) the point charge is moved to allocation just outside the sphere.
Explanation:
Because
Gaussian surface is given as
ϕ= qclose/E0
So Q enclosed charge will be same if w the charge inside the Gaussian surface is moved, so flux will not change.
What is the difference between an observation and an inference?
Explanation:
observation is something what you could observe from your organs like eyes ears etc and also it is what you observed during an event for an experiment but inference is what you decide to do after observation or an event.
the act of inferring (to derive by reasoning). Observation = an act or instance of noticing or perceiving.
Thank u!
4. I wonder if the amount of homework I complete has an effect on my science test scores. a. Question: b. Hypothesis + Prediction: c. Independent Variable (IV): d. Dependent Variable (DV): e. Controlling Variable (CV): f. Experiment:
Answer:
a. Question
Explanation:
The statement above is more of a question rather than a hypothesis or the other options. Consider, the statement has no direct initial assumption or hint of personally researching for answers.
In other words, the speaker could be asking another individual his opinion whether the amount of homework completed has an effect on science test scores.
Please help!?!?!?!?!?!?!
Answer:
D
Explanation:i think but dont get mad if im wrong
How does the lighter material of the bamboo affect the force created by the rider on the bike?
Newton's second law allows us to find the result for the effect of building a lower mass bike is:
An increase in the acceleration of the bicycle.
Newton's second law states that force is equal to the product of mass and acceleration of the body.
∑ F = m a
Where the bold letters indicate vectors, m is the mass and a the acceleration.
They indicate that the bicycle is made of bamboo, so it has less mass, therefore the acceleration is:
[tex]a = \frac{\sum F}{m}[/tex]
if the mass decreases, more acceleration is recorded, therefore from the kinematic reactions the bicycle reaches higher speeds.
v = v₀ + a t
The bicycles parts of the rest.
v = a t
In conclusion, using Newton's second law we can find the result for the effect of building a bicycle with alower mass is:
An increase in the acceleration of the bicycle.
Learn more here: brainly.com/question/8522449
A busy chipmunk runs back and forth along a straight line of acorns that has been set out between its burrow and a nearby tree. At some instant, it moves with a velocity of −1.09 m/s−1.09 m/s . Then, 2.99 s2.99 s later, it moves with a velocity of 1.75 m/s1.75 m/s . What is the chipmunk's average acceleration during the 2.99 s2.99 s time interval
Answer:
0.59 seconds
Explanation:
If a car can go from 20m/s to 40m/s in 4.0 secs, what would be it’s acceleration?
A transverse sinusoidal wave is moving along a string in the positive direction of an x axis with a speed of 89 m/s. At t = 0, the string particle at x = 0 has a transverse displacement of 4.2 cm from its equilibrium position and is not moving. The maximum transverse speed of the string particle at x = 0 is 16 m/s. (a) What is the frequency of the wave? (b) What is the wavelength of the wave? If the wave equation is of the form y(x, t) = ym sin(kx ± ωt + φ), what are (c) ym, (d) k, (e) ω, (f) φ, and (g) the correct choice of sign in front of ω?
Answer:
The answer is below
Explanation:
Given that:
y = transverse displacement = 4.2 cm = 0.042 m at x = 0 and t = 0.
Speed = v = 89 m/s, maximum transverse speed of the string particle = [tex]u_m[/tex] = 16 m/s.
ω = [tex]u_m[/tex] / [tex]y_m[/tex] = 16 / 0.42 = 380.95 rad/s
a) Frequency = ω/2π = 380.95 / 2π = 60.63 Hz
b) Wavelength (λ) = speed / frequency
λ = v / f = 89/63.66= 1.468 m
c) Using the wave equation:
[tex]y=y_msin(kx \pm wt \pm \phi)\\y=0.042,t=0,x=0\\\\Hence\\y_m=0.042\ m[/tex]
d) Wave number k is given by:
k = 2π / λ = 2π / 1.468 = 4.28 rad/s
e) The angular velocity is given by:
ω = [tex]u_m[/tex] / [tex]y_m[/tex] = 16 / 0.42 = 380.95 rad/s
f) Using the wave equation:
[tex]y=y_msin(kx \pm wt \pm \phi)\\\\y=0.042,t=0,x=0,y_m=0.042\\\\Hence\\0.042=0.042sin(4.28(0)\pm 380.95(0)\pm \phi)\\\\sin\phi=1\\\\\phi=\frac{\pi}{2} \\\\y=0.042sin(4.28x\pm 380.95t\pm \frac{\pi}{2})[/tex]
g) Since the wave is in the positive x direction, hence ω is negative
