Answer:
b) the earth did not orbit the sun
Explanation:
The sun appears to move through the background stars due to parallax . Parallax is a phenomenon when the near object appears to move faster than the distant object . If we travel in a train , the near object like electric poles on rail track or trees and fields nearby appear to move faster against the distant background .
Hence when the earth moves around the sun , the sun appears to move against the background the stars which are far away . Had earth remained stationary at a place on its orbit around the sun , the sun would have appeared stationary against the background the stars .
Work done of frictional force from instant
Answer:
[tex]-100\ J[/tex]Step-by-step explanation:
1. Find acceleration:
[tex]m=2\ kg[/tex] [tex]F=-5\ N[/tex] [tex]a=\frac{F}{m}[/tex] (Newton's second law)[tex]a=\frac{-5}{2} =-2.5\ \frac{m}{s^{2}}[/tex]2. Find distance traveled:
[tex]v_0=10\ \frac{m}{s}[/tex] [tex]v=0[/tex][tex]a=-2.5\ \frac{m}{s^{2} }[/tex] [tex]v^2-v_0^2=2ad[/tex] (Kinematic equation)[tex]-100=-5d[/tex] [tex]d=20\ m[/tex]3. Find work done by friction:
[tex]W=Fd[/tex] (Work formula when angle between Force and Displacement vectors are 0°) [tex]W=-5\times20=-100\ J[/tex]A carpenter on the roof of a building accidentally drops herhammer. As the hammer falls it passes two windows of equal height,a) Is the increase in seed of the hammer as it drops past window 1greater than, less than or equal to the increase in speed as itdrops past window 2? b) Choose the best explanation from thefollowing:
I. The greater seed at window 2 results in a greater increase inspeed.
II. Constant acceleration means the hammer speeds up the sameamount for each window.
III. The hammer spends more time dropping past window 1.
Answer:
Explanation:
a) the speed increment of the hammer as it drops past the first window, is greater than that of the speed of the hammer as it drops past the second window. This can also be translated as saying that the hammer spent more time at the second window.
b) III
The best answer would be answer III, The hammer spends more time dropping past window 1, which I had already included in my explanation in (a) above.
Regular exercise is positively related to wellnes t or f
the four strings of a bass guitar are 0.865 m long and are tuned to the notes g (98 hz), d (73.4 hz), a (55 hz), and e (41.2 hz). in one bass guitar, the g and d strings have a linear mass density of 4.8 g/m, and the a and e strings have a linear mass density of 29.8 g/m.what is the total force exerted by the strings on the neck?
Answer: Total force = 636,554.55N
Explanation: To determine tension of strings, wave speed on a string is necessary. Speed is found by:
v = f.λ
f is frequency
λ is wavelength
For the strings, wavelength equals to:
[tex]\lambda = 2L[/tex]
L is the length of the bass guitar string
Then, wave speed:
[tex]v=f.2L[/tex]
Tension on a string is
[tex]v=\sqrt{\frac{F_{T}}{\mu} }[/tex]
[tex]v^{2}=\frac{{F_{T}} }{\mu}[/tex]
[tex]F_{T} = v^{2}\mu[/tex]
[tex]F_{T} = (2f\lambda)^{2}\mu[/tex]
[tex]F_{T} = 4(f\lambda)^{2}\mu[/tex]
μ is linear mass density
For g string:
[tex]F_{T} = 4(98.0.865)^{2}.4.8[/tex]
[tex]F_{T}[/tex] = 137970.3N
For d string:
[tex]F_{T} = 4(73.4.0.865)^{2}.4.8[/tex]
[tex]F_{T}=[/tex] 77397.25N
For a string:
[tex]F_{T} = 4(55.0.865)^{2}.29.8[/tex]
[tex]F_{T}=[/tex] 269795N
For e string:
[tex]F_{T} = 4(41.2.0.865)^{2}.29.8[/tex]
[tex]F_{T}=[/tex] 151392N
Total force = 137,970.3 + 77,397.25 + 269,795 + 151,392
Total force = 636,554.55N
Total force exerted on the neck by the strings is 636,554.55N.
This question involves the concepts of the tension force in strings, linear mass density, and frequency.
The total force exerted by strings on the guitar is "359.4 N".
The tension force exerted by each string is given as:
[tex]F_T=v^2\mu[/tex]
where,
F_T = tension force = ?
v = speed = (frequency)(wavelength) = fλ
μ = linear mass density
Therefore,
[tex]F_T=f^2\lambda^2\mu[/tex]
but for strings in this case:
[tex]\lambda = 2(Length of string) = 2(0.865\ m)=1.73\ m[/tex]
Therefore,
[tex]F_T=f^2(1.3\ m)^2\mu[/tex]
For string g:
[tex]F_{Tg}=(98\ Hz)^2(1.3\ m)^2(4.8\ x\ 10^{-3}\ kg/m)\\F_{Tg}=77.9\ N[/tex]
For string d:
[tex]F_{Td}=(73.4\ Hz)^2(1.3\ m)^2(4.8\ x\ 10^{-3}\ kg/m)\\F_{Td}=43.7\ N[/tex]
For string a:
[tex]F_{Ta}=(55\ Hz)^2(1.3\ m)^2(29.8\ x\ 10^{-3}\ kg/m)\\F_{Ta}=152.3\ N[/tex]
For string e:
[tex]F_{Te}=(41.2\ Hz)^2(1.3\ m)^2(29.8\ x\ 10^{-3}\ kg/m)\\F_{Te}=85.5\ N[/tex]
So, the total force will be the sum of all tension forces:
[tex]F=F_{Tg}+F_{Td}+F_{Ta}+F_{Te}[/tex]
F = 77.9 N + 43.7 N + 152.3 N + 85.5 N
F = 359.4 N
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which is an accurate description of what occurs when a log burns
Answer:
charcoal
Explanation:
Answer:
charcoal
Explanation:
Mark as Brainliest
Which element is most likely to be shiny? sulfur (S) boron (B) calcium (Ca) carbon (C)
Answer:
C.
Explanation:
The element that is most likely to be shiny is calcium. The correct option is 3.
Calcium (Ca) is the most likely to be sparkly of the elements shown. Calcium is a metallic element with a glossy or shining look when newly cut or polished.
It belongs to the alkaline earth metals group, which have metallic characteristics and a distinctive lustre.
Sulphur (S), boron (B), and carbon (C), on the other hand, are non-metallic elements with a dull look.
They come in a variety of forms, including powders, crystals, and amorphous solids, although they are not noted for their metallic lustre.
Thus, the answer is 3. calcium.
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Your question seems incomplete, the probable complete question is:
Which element is most likely to be shiny?
sulfur (S) boron (B) calcium (Ca) carbon (C)A sound wave is observed to travel through a liquid with a speed of 1400 m/s. The specific gravity of the liquid is 1.8.
Determine the bulk modulus for this fluid.
Answer:
21
Explanation:
I think its 21 because its 21
A model rocket accelerates upward from the ground with a constant acceleration, reaching a height of 50 m in 8 s.What is the speed (in m/s) at a height of 50 m?A: 1.07×101 B: 1.25×101 C: 1.46×101 D: 1.71×101 E: 2.00×101
Answer:
The value is [tex]v = 12.5 \ m/s[/tex]
Explanation:
From the question we are told that
The height is [tex]h = 50 \ m[/tex]
The time taken is [tex]t = 8 \ s [/tex]
From the equation of motion we have that
[tex]s = ut + \frac{1}{2} * a * t^2[/tex]
Here u = 0 because the rocket started at rest
[tex]50 = 0 + \frac{1}{2} * a * 8^2[/tex]
=> [tex]a = \frac{100}{64}[/tex]
=> [tex]a = 1.5625 \ m/s^2[/tex]
Also from the kinematic equation we have that
[tex]v = u + at[/tex]
=> [tex]v = 0 + (8 * 1.5625)[/tex]
=> [tex]v = 12.5 \ m/s[/tex]
who is known as father of science?
GalileoGalilei is known as the father of science
We can reasonably model a 75 W incandescent light bulb as a sphere 6.0 cm in diameter. Typically, only about 5 % of the energy goes to visible light: the rest foes largely to nonvisible infrared radiation.(a) What is the visible light intensity (in W/m^2) at the surface of the bulb?(b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with this intensity?
Answer:
a
[tex]I = 6637 \ W/m^2[/tex]
b
[tex]E_{max} = 500 \ N/m[/tex]
And
[tex]B_{max} = 1.67*10^{-6} \ T[/tex]
Explanation:
From the question we are told that
The power is [tex]P = 75 \ W[/tex]
The diameter is [tex]d = 6.0 \ cm = 0.06 \ m [/tex]
Generally the radius is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{ 0.06}{2}[/tex]
=> [tex]r = 0.03 \ m[/tex]
Generally the area of the sphere is mathematically evaluated as
[tex]A = 4 \pi r^2[/tex]
=> [tex]A = 4 * 3.142 * (0.03)^2[/tex]
=> [tex]A = 0.0113 \ m^2[/tex]
Generally the total Intensity of the incandescent light bulb is mathematically represented as
[tex]I= \frac{P}{A}[/tex]
=> [tex]I = \frac{75}{ 0.0113}[/tex]
=> [tex]I = 6637 \ W/m^2[/tex]
Given that 5% of the energy goes to visible light
Then the intensity that goes visible light is
[tex]I_v = 0.05 * 6637[/tex]
[tex]I_v = 332 \ W/m^2[/tex]
The amplitude of the electric field at the surface is mathematically represented as
[tex]E_{max} = \sqrt{\frac{2 * I_v}{\epsilon_o * c } }[/tex]
=> [tex]E_{max} = \sqrt{\frac{2 * 332}{ 8.85*10^{-12} * 3.0*10^8} }[/tex]
=> [tex]E_{max} = 500 \ N/m[/tex]
The amplitude of the magnetic field at the surface is mathematically represented as
[tex]B_{max} = \frac{E_{max}}{c}[/tex]
=> [tex]B_{max} = \frac{ 500}{3.0*10^8}[/tex]
=> [tex]B_{max} = 1.67*10^{-6} \ T[/tex]
Light of wavelength 600 nm illuminates a diffraction grating. The second order maximum is at an angle of 65 degrees. a) List your known variables. b) What is the spacing between slits in this grating? c) How many lines per millimeter does this grating have?
Answer:
A. Known variables include
Wavelength = 600nm
Theta= 65°
m= 2
B.
d= m x wavelength / sin theta
= 2 * 600*10^-9 /sin 65°
= 1.3*10^-6m
C.
N = 1/d
So N = 1/1.3*10^-6m
=76.9 lines per micro meter
An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph for the values of t that are separated by the step Δt = 2s.
Answer:
Explanation:
We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:
[tex]v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt[/tex]
[tex]v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt[/tex]
Where:
[tex]v_{a}[/tex], [tex]v_{b}[/tex] - Initial and final velocities, measured in meters per second.
[tex]t_{a}[/tex], [tex]t_{b}[/tex] - Initial and final times, measured in seconds.
[tex]a(t)[/tex] - Acceleration, measured in meters per square second.
Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:
Region I (t = 0 s to t = 4 s)
[tex]v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt[/tex]
[tex]v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)[/tex]
[tex]v_{4} = -6\,\frac{m}{s}[/tex]
Region II (t = 4 s to t = 6 s)
[tex]v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt[/tex]
[tex]v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)[/tex]
[tex]v_{6} = -4\,\frac{m}{s}[/tex]
Region III (t = 6 s to t = 10 s)
[tex]v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt[/tex]
[tex]v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)[/tex]
[tex]v_{10} = 4\,\frac{m}{s}[/tex]
Finally, we draw the object's velocity graph as follows. Graphic is attached below.
The velocity of a body under constant acceleration increases steadily with time
Please find attached the required velocity graph for values of t that are separated by Δt = 2s
The reasons the attached graph is correct are given as follows:
At t = 0 secondsThe initial velocity of the object at t = 0 is v = 2.0 m/s
The first point on the graph is (0, 2.0)
From t = 0 s, to t = 4 sThe acceleration from t = 0, to t = 4, a₁ = -2 m/s²
The velocity at t = 4 s, v₂ = 2.0 + (-2)×4 = -6
Therefore, the next point on the graph is (4, -6)
From t = 4 s to t = 6 sFrom t = 4 to t = 6, the acceleration, a₂ = 1 m/s²
Therefore, v₃ = -6 + 1 × 2 = -4
The third point on the velocity graph is (6, -4)
From t = 6 s to t = 10 sFrom t = 6 s to t = 10 s, we have, the acceleration, a₃ = 2 m/s²
The velocity, v₄ = -4 + 4 × 2 = 4
Therefore, the fourth point on the velocity graph is (10, 4)
With the above points, the velocity graph can be plotted using MS Excel
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a 20 ft shipping container on a cargo ship has a mass of 24000 kg and a volume of 33.2m3. what is the density of the shipping container
Answer:
722.89
Explanation:
mass=24000kg
volume=33•2
density=?
now,
density=mass/volume
=24000/33•2
=722•89
density=722•89 kg/m^3
A child on a bridge throws a rock straight down to the water below. The point where the child released the rock is 74 m above the water and it took 2.7 s for the rock to reach the water. Determine the rock's velocity (magnitude & direction) at the moment the child released it. Also determine the rock's velocity (magnitude & direction) at the moment it reached the water. Ignore air drag.
The rock's altitude y at time t, thrown with initial velocity v, is given by
[tex]y=74\,\mathrm m+vt-\dfrac12gt^2[/tex]
where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity.
After t = 2.7 s, the rock reaches the water (0 altitude), so
[tex]0=74\,\mathrm m+v(2.7\,\mathrm s)-\dfrac12g(2.7\,\mathrm s)^2[/tex]
[tex]\implies v=-\dfrac{74\,\mathrm m-\frac g2(2.7\,\mathrm s)^2}{2.7\,\mathrm s}\approx-14.177\dfrac{\rm m}{\rm s}[/tex]
so the rock was thrown with a velocity with magnitude 14 m/s and downward direction.
Its velocity at time t is [tex]v-gt[/tex] (with no horizontal component), so that at the moment it hits the water, its velocity is
[tex]v-g(2.7\,\mathrm s)\approx-40.637\dfrac{\rm m}{\rm s}[/tex]
That is, its final velocity has an approximate magnitude of 41 m/s, also directed downward.
Suppose we have Cl-, Na+, and Ca2+ ions in an aqueous solution (with dielectric constant κ = 80.4—we will discuss this later). Consider the situation where a sodium ion is between a chlorine ion and a calcium ion as shown:
(Cl-) (Na+) (Ca2+)
If the sodium ion is 1.50 nm from the chlorine ion and 3.00 nm from the calcium ion, find the electric force on the Na+ ion.
Answer:
a) 1.19 x 10^7 N/C
b) 2 x 10^-12 N
Explanation:
field due to Cl on (9.0xE¡)(1.6x10-19 C) (1.5Å—10-9m) -6.4x10 N/C field due to Ca+3 ion, 2(90x10 (1.6x10-19c) ((4.5-1.5)Å—10-9 m)' magnitude of net field at given point without dielectric E E+E 6.4x108 +3.2x108-9.6x10 N/C magnitude of net field at given point with di electric K 80.4 the force on sodium ion at this point, F-Edq= (1.19å—107 N/C)(1.6å—10-19C)= 2.0Å—10-12 N
Which of the following is the best way to make a conclusion? A. The experiment must be manipulated until the results show what you want B. Estimate results to where they should be C. Choosing the results you like best D. Comparing data from the experiment to
Answer:
D
Explanation:
You should compare your results and data with your original hypothesis. It is okay to have a hypothesis that was shown to be false. That's science. Things do not always work out. You can always discuss what you could have done differently and things that went wrong. There is no right answer in science!
If an initially neutral insulator is charged on its left end, would you get an electrical shock by touching its right end?
Answer:
No there won't be.
Explanation:
An insulator does not conduct electricity readily. An insulator that is charged at one end will have the charges remain at that end, since it does not allow the free flow of charges through it. If you touch the other end of the insulator that is void of electric charges, there would be no charge transfer between you and the insulator, and hence no electric shock.
A system of 1470 particles, each of which is either an electron or a proton has a net charge of -5.456 x 10^-17 C. (a) How many electrons are in this system? (b) What is the mass of this system?
Answer:
a
[tex]N_e = 906 \ electrons[/tex]
b
[tex]M = 9.43 *10^{-25} \ kg[/tex]
Explanation:
From the question we are told that
The total number of particles is n = 1470 particles
The the total amount of charge on the particles is [tex]Q = - 5.456*10^{-17} \ C[/tex]
Generally this total number of particles can be mathematically represented as
[tex]n = N_p + N_e[/tex]
Where [tex]N_p \ and \ N_e[/tex] represent the number of proton and electron respectively
=> [tex]N_p = 1470 - N_e[/tex]
Also the total charge of these particles can be mathematically represented as
[tex]Q = (N_p - N_e ) e[/tex]
Here e is the charge on a single electron or proton with value
The negative sign is due to the fact the electrons are negative signed
[tex]e = 1.60 *10^{-19} \ C[/tex]
[tex]-5.456*10^{-17} = ( 1470 -2N_e )* 1.60 *10^{-19} [/tex]
[tex]N_e = 906[/tex]
Thus
[tex]N_p = 1470 - 906[/tex]
[tex]N_p = 564[/tex]
Generally the mass of the system is mathematically represented as
[tex]M = N_e * M_e + N_p * M_p[/tex]
Here [tex]M_e \ and \ M_p \ are \ mass \ of \ electron \ and \ proton \ with \ values \\ M_e = 9.1 *10^{-31} \ kg \ and \ M_p = 1.67 *10^{-27} \ kg \ respectively[/tex]
So
[tex]M = 906 * 9.1 *10^{-31} + 564 * 1.67 *10^{-27}[/tex]
[tex]M = 9.43 *10^{-25} \ kg[/tex]
What are some superheroes that resemble neurotransmitter functions. Dopamine: Acetylcholine: Endorphins: GABA: Glutamates: Norepinephrine: Serotonin:
Answer:
Serotonin
___________
Doug rubs a piece of fur on a hard rubber rod, giving the rod a negative charge. Which of the following statements best describes what happens in this process?A) Protons are removed from the rod.B) Electrons are added to the rod.C) The fur is also charged negatively.D) The fur is left neutral.
Answer:
B is correct. Electrons are added to the rod.
Explanation:
Because the fur lose electrons to the rode and because positively charged while the rod because negative
What are the components of vector a
Answer:
Option (A)
Explanation:
From the picture attached,
Given : Vector A with magnitude 12 m.
To Find : Components of vector A.
There are two components of any vector.
1). Horizontal component
2). Vertical component
If vector A represents the velocity, horizontal component of a vector decides the horizontal motion and vertical component decides the vertical motion.
To find these components we draw a right triangle OBA as shown in the figure,
From ΔOBA,
Sin(37)° = [tex]\frac{\text{AB}}{\text{OA}}[/tex]
= [tex]\frac{{A_x}}{{A}}[/tex]
[tex]A_x=A\text{Sin}(37)[/tex]
= 12Sin(37)°
= 7.22 m
[tex]A_y=A\text{Cos}(37)[/tex]
= 12Cos(37)°
= 9.58 m
Therefore, Option (1) is the correct option.
Refer to the data. How many males are taller than 175 cm and approximately what percentage of the total is that?
Answer:
Males over 175 cm = 10 Percentage of total = 50%Explanation:
Males measuring;
161 - 165 = 2
166 - 170 = 1
171 - 175 = 7
176 - 180 = 1
181 - 185 = 6
186 - 190 = 3
Males taller than 175 cm = 1 + 6 + 3 = 10
Total number of Males = 2 + 1 + 7 + 1 + 6 + 3 = 20
Percentage of total over 175 cm
= 10 / 20
= 50%
Which is one physical property that all stars have? pick one They are made of gases. They shine very brightly. They have a triangle shape. They contain iron in their cores.
Answer: They are made of gases
The physical property that all stars have will be they are made of gases. option 1 is correct.
What is a star?A star is a heavenly body made up of a brilliant spheroid of plasma held together by gravity. The Sun is the closest star to Earth.
Many additional stars may be seen with the normal eye at night, but due to their great distance from Earth, they appear as stationary points of light in the sky.
The physical property that all stars have will be they are made of gases.
Hence option 1 is correct.
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The acceleration of a particle is given by a(t)= -2.00 m/s^2 + (3 m/s^3)t. Required:a. Find the initial velocity vo such that the particle will have the same x-coordinate at t=4.00 s as it had at t=0. b. What will be the velocity at t=4.00 s ?
Answer:
Explanation:
a(t)= -2.00 m/s^2 + (3 m/s^3)t.
dv / dt = -2.00 m/s^2 + (3 m/s^3)t.
dv = (-2.00 m/s^2 + (3 m/s^3)t.)dt
v = - 2t + 3 t² / 2 + c , where c is a constant
for initial velocity t = 0
v0 = c
v = - 2t + 3 t² / 2 + v0
ds / dt = - 2t + 3 t² / 2 + v0
ds = (- 2t + 3 t² / 2 + v0)dt
s = - 2t²/2 + 3 t³/6 + vot + c₁
At t = 0
s = c₁
At t = 4
s = -16 + 32 + 4v0 + c₁
= 4v0 + c₁ + 16
Given
4v0 + c₁ + 16 = c₁
v0 = - 4 m /s
Putting this value in the equation of velocity
v = - 2t + 3 t² / 2 - 4
At t = 4
v = -8 + 24 - 4
= 12 m / s
What is the car's average velocity (in m/s) in the interval between t = 1.0 s
to t = 1.5 s?
Answer:
1.4 m/s
Explanation:
From the question given above, we obtained the following data:
Initial Displacement (d1) = 0.9 m
Final Displacement (d2) = 1.6 m
Initial time (t1) = 1.5 secs
Final time (t2) = 2 secs
Velocity (v) =..?
The velocity of an object can be defined as the rate of change of the displacement of the object with time. Mathematically, it can be expressed as follow:
Velocity = change of displacement /time
v = Δd / Δt
Thus, with the above formula, we can obtain the velocity of the car as follow:
Initial Displacement (d1) = 0.9 m
Final Displacement (d2) = 1.6 m
Change in displacement (Δd) = d2 – d1 = 1.6 – 0.9
= 0.7 m
Initial time (t1) = 1.5 secs
Final time (t2) = 2 secs
Change in time (Δt) = t2 – t1
= 2 – 1.5
= 0.5 s
Velocity (v) =..?
v = Δd / Δt
v = 0.7/0.5
v = 1.4 m/s
Therefore, the velocity of the car is 1.4 m/s
A student throws a heavy red ball horizontally from a balcony of a tall building with an initial speed v0. At the same time, a second student drops a lighter blue ball from the same balcony. Neglecting air resistance, which statement is true?
Answer:
The correct statement must be: both balls hit the floor at the same time
Explanation:
This is a kinematics exercise. The ball thrown horizontally does not have vertical speed and the ball that is released does not have vertical speed, therefore both take the same time to reach the ground, if we neglect the air resistance
The correct statement must be: both balls hit the floor at the same time
A silver block of silver block of density 10.5 g/cm3 has a volume of 30 cm3. Which of the following is the correct mass of the block
➝ Density of block = 10.5 g/cm³
➝ Volume of block = 30 cm³
We have to find mass of block[tex].[/tex]
➠ Density is defined as mass of substance per unit volume[tex].[/tex]
[tex]\dag\:\boxed{\bf{Density=\dfrac{Mass}{Volume}}}[/tex]
[tex]:\implies\sf\:Mass=Density\times Volume[/tex]
[tex]:\implies\sf\:Mass=10.5\times 30[/tex]
[tex]:\implies\boxed{\boxed{\bf{\red{Mass=315\:g}}}}[/tex]
A boat with a jet engine demonstrates Newton’s law of action-reaction . Why is this true?
Answer:Newton’s law also states that larger bodies with heavier masses exert more gravitational pull, which is why those who walked on the much smaller moon experienced a sense of weightlessness, as it had a smaller gravitational pull. To help explain his theories of gravity and motion, Newton helped create a new, specialized form of mathematics.
Explanation:I'm very stupid so I don't think this is the right answer
A bowling ball of mass 3 kg is dropped from the top of a tall building. It safely lands on the ground 3.5 seconds later. Neglecting air friction, what is the height of the building in meters? (Give the answer without a unit and round it to the nearest whole number)
Answer:
The height of the building is 60 m.
Explanation:
Given;
mass of the mass of the ball, m = 3 kg
time of motion, t = 3.5 s
The velocity of the ball is given by;
v = u + gt
where;
u is the initial velocity of the ball = 0
v = 0 + 9.8 x 3.5
v = 34.3 m/s
When the ball hits the ground, energy is conserved;
mgh = ¹/₂mv²
gh = ¹/₂v²
h = (0.5 v²) / g
h = (0.5 x 34.3²) / (9.8)
h = 60.025 m
h = 60 m
Therefore, the height of the building is 60 m.
The height of the building is 60 m.
calculation of building height:The velocity of the ball should be provided by
v = u + gt
here,
u is the initial velocity of the ball = 0
v = 0 + 9.8 x 3.5
v = 34.3 m/s
Now
When the ball hits the ground, energy is conserved;
mgh = ¹/₂mv²
gh = ¹/₂v²
h = (0.5 v²) / g
h = (0.5 x 34.3²) / (9.8)
h = 60.025 m
h = 60 m
Learn more about friction here: https://brainly.com/question/14455351
which sentences describes an object that has kinetic energy
Answer:
A boat sails across the Ocean
Explanation:
Out of all the options on Edmentum none of them where moving so it would only make sense that this is the only one with kinetic energy.