Explanation:
Q1. Given:
v = 0 m/s
a = -5.5 m/s²
t = 3.5 s
Find: Δx
Δx = vt − ½ at²
Δx = (0 m/s) (3.5 s) − ½ (-5.5 m/s²) (3.5 s)²
Δx ≈ 33.7 m
Q2. Given:
Δx = 400 m
v₀ = 7.0 m/s
v = 35 m/s
Find: a
v² = v₀² + 2aΔx
(35 m/s)² = (7.0 m/s)² + 2a (400 m)
a = 1.47 m/s²
A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, what is the mass of the air in the classroom?
Answer:
the mass of the air in the classroom = 2322 kg
Explanation:
given:
A classroom is about 3 meters high, 20 meters wide and 30 meters long.
If the density of air is 1.29 kg/m3
find:
what is the mass of the air in the classroom?
density = mass / volume
where mass (m) = 1.29 kg/m³
volume = 3m x 20m x 30m = 1800 m³
plugin values into the formula
1.29 kg/m³ = mass
1800 m³
mass = 1.29 kg/m³ ( 1800 m³ )
mass = 2322 kg
therefore,
the mass of the air in the classroom = 2322 kg
If A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, then the mass of the air in the classroom is 2322Kg.
What is density??Density is the ratio of mass to volume. it tells how much mass a body is having for its unit volume. for example egg yolk has 1027kg/m³ of density, means if we collect numbers of egg yolk and keep it in a container having volume 1 m³ then total amount of mass it is having will be 1027kg. Density is a scalar quantity. when we add egg yolk into the water, egg yolk has greater density than water( 997 kg/m³), because of higher density of egg yolk it contains higher mass in same volume as water. hence due to higher mass higher gravitational force is acting on the egg yolk therefore it goes down on the inside the water. water will float upon the egg yolk. same situation we have seen when we spread oil in the water. ( in that case water has higher density than oil. thats why oil floats on the water).
Given,
Height = 3 m
Width = 20 m
length= 30 m
Density of air = 1.29kg/m³
The volume of the room = 3×20×30 m³
Volume V = 1800m³
By formula,
Density = Mass/Volume
1.29kg/m³ = Mass/1800m³
Mass of the air = 1.29×1800 = 2322 Kg
The mass of the air is classroom is 2322Kg.
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Write the following numbers in proper scientific notation:
1.) 36513
2.) 0.00050
3.) 8004
4.) 0.00000000406
Answer:
0.00050, 0.000000000406, 36513
An electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C. (a) What is the speed of the electron 1.3 ns after entering this region? (b) How far does the electron travel during the 1.3 ns interval?
Answer:
1.) 11 km/s
2.) 9.03 × 10^-5 metres
Explanation:
Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.
Electron q = 1.6×10^-19 C
Electron mass = 9.11×10^-31 Kg
(a) What is the speed of the electron 1.3 ns after entering this region?
E = F/q
F = Eq
Ma = Eq
M × V/t = Eq
Substitute all the parameters into the formula
9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19
V = 7.68×10^-18 /7.0×10^-22
V = 10971.43 m/s
V = 11 Km/s approximately
(b) How far does the electron travel during the 1.3 ns interval?
The initial velocity U = 64 km/s
S = ut + 1/2at^2
S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2
S =8.32×10^-5 + 7.13×10^-6
S = 9.03 × 10^-5 metres
what is machinery
............
You walk 20 m north then 30 m west for a total timer four minutes what is the magnitude of your average velocity in (m/s)
Answer:
The average velocity is 0.15 m/s
Explanation:
Use the definition of average velocity as the distance traveled divided the time it took.
Since the movement was on the plane from the origin (0, 0) to the point (-30, 20) corresponding to 30 m west and 20 m north, we calculate the distance using the distance between two points on the plane:
[tex]distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} = \sqrt{(-30)^2+20^2} =\sqrt{1300} \approx 36.06\,\,m[/tex]
Then the magnitude of the average velocity can be estimated via the quotient between distance divided time, but since the units required are meters per second, we first convert the four minute time into seconds: 4 * 60 = 240 seconds.
Then the average velocity becomes:
[tex]v_{ave}=\frac{distance}{time} =\frac{36.06}{240} =0.15\,\,m/s[/tex]
seagull is flying at a rate of 20 miles per hour south, it encounters wind blowing 20 miles per hour north. What is the resultants
Answer:
the winds will make the bird stop
Explanation:
is basically 20 - 20
Which statement supports Newton’s first law of motion?
Answer:
Newton's first law of motion is that an object at rest will stay at rest, unless acted upon by an external force.
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The "external force" can be anything, from a gust of wind, to a person moving the object. His first law also corresponds with the definition of inertia.
Hope this helps!
Which atom is most likely to accept electrons to form an ionic bond? a mercury ion with a negative 2 charge a potassium ion with a negative 1 charge radon, a noble gas with 8 valence electrons sulfur, a nonmetal with 6 valence electrons
Sulfur, a nonmetal with 6 valence electrons atom is most likely to accept electrons to form an ionic bond option (D) correct.
What is an ionic bond?Ionic bonds, also known as electrovalent bonds, are a type of connection created in a chemical molecule by the electrostatic attraction of ions with opposing charges.
As we know,
It has been shown that the top of the electronegativity scale is defined by the fundamental atomic particles F>O>N.
Ionic bonds need an electron, often a nonmetal, and an electron, typically a metal. Metals display ionic bonding because there aren't many-electron in outer orbitals.
Thus, sulfur, a nonmetal with 6 valence electrons atom is most likely to accept electrons to form an ionic bond option (D) correct.
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Genes are section of DNA that code for a particular trait. Genes are
Answer:
a gene is the basic physical and functional unit of heredity.
Genes are section of DNA that code for a particular trait. Genes are basic unit of inheritance in all living things specifying the physical and biological traits.
What are genes ?The precise meaning of the term "gene" has long been a topic of discussion in science. Here's an easy way to think about it: Our cells and tissues are built with proteins as the foundation. And genes are the area of our genome where the instructions for constructing those proteins are stored.
For instance, the human genome has over 20,000 genes that code for proteins. It's interesting to note that just 1.5% of the complete human genome contains all of the information for those 20,000 protein-coding genes.
A broader definition of a gene includes DNA sequences, sometimes called RNA genes, that include instructions for building an RNA molecule that performs a function other than directly encoding a protein.
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A Van de Graaff generator is charged so that a proton at its surface accelerates radially outward at 1.35 ✕ 1012 m/s2. Find the following. (a) the magnitude of the electric force on the proton at that instant magnitude N direction ---Select--- (b) the magnitude and direction of the electric field at the surface of the generator magnitude N/C direction ---Select---
Answer:
(a).
We know that force is
F = m a
So
F = (1.67 x 10^(-27) x (1.38 x10^12)
F = 2.3 x 10^-15 N facing the radially outward direction
(b).
Similarly Force for charge is
F = q E = m a
So relating
E = F/q = 2.3x 10^-15 /(1.6 10^ -19
E = 144.75 N/C facing the radially outward direction
The answer to the question is
(a) The force on the proton is [tex]2.254\times10^{-15}\,N[/tex] directed radially outwards.
(b) The electric field at the surface of the generator is [tex]14031.25\, N/C[/tex] directed radially outwards.
The answer can be explained as shown below.
Given that a proton accelerates radially outward at [tex]1.35\times 10^{12}\,m/s^2[/tex].
ie; [tex]a=1.35\times 10^{12}\,m/s^2[/tex]We know the mass of a proton, [tex]m_p = 1.67\times10^{-27}\,kg[/tex].From Newton's second law we have,
[tex]F=mg[/tex]But here, [tex]m=m_p[/tex]
So, the electric force on the proton is;
[tex]F = m_p\, a= (1.67\times10^{-27}\,kg)\, \times(1.35\times 10^{12}\, m/s^2)=2.254\times10^{-15}\,N[/tex]The force on the proton is [tex]2.254\times10^{-15}\,N[/tex] directed radially outwards.Also, we know that, in electrostatics,
[tex]F=Eq\\ \\\implies E=\frac{F}{q}[/tex]The charge of a proton is, [tex]q=1.6\times 10^{-19}\,C[/tex]Therefore, the electric field is given by,
[tex]E=\frac{2.254\times 10^{-15}N}{1.6\times 10^{-19}\,C} = 14031.25\, N/C[/tex]The electric field at the surface of the generator is [tex]14031.25\, N/C[/tex] directed radially outwards.Learn more about the electric force here:
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why Newtown made the laws of motion
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:
Answer:
Newton was convinced the planets must obey the same physical laws that are observed on Earth. This meant there must be an unseen force acting on them. He knew from experiment that, in the absence of an applied force, a moving body will continue in a straight line forever.
Explanation:
At a time of 30 seconds a runner passes a distance marker labeled "125 meters." If the velocity of the runner is +5.0 m/s, when did the runner pass the distance marker for 75 meters?
Answer:
Explanation
He runs at 5m/s, so in 30 s he should be at 150m. So you have to do 125m - 150m and you'll get -25m, this is his initial position. They want to know the time when he hits 75m, so you would do 75 + 25, and get 100. Then do 100m / 5m/s, and you will get 25 seconds.
A ball rolls off a 1.0 m high table and lands on the floor 3.0 m away from the table. a. How long is the ball in the air?
Answer:
The ball is in the air for approximately 0.45 seconds
Explanation:
We only need to use the information on the height of the table to find the time, since the vertical movement is a movement under the acceleration of gravity, and with no initial velocity in the y-direction (recall that the ball rolls off a 1.0 m high table)
therefore the equation of motion for the vertical component is:
[tex]y_f-y_i=\frac{1}{2}\, g\,t^2[/tex]
which for our information becomes:
[tex]y_f-y_i=\frac{1}{2}\, g\,t^2\\1 = 4.9\,t^2\\t^2=\frac{1}{4.9} \\t=\sqrt{\frac{1}{4.9}} \\t\approx 0.45\,s[/tex]
A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A that splits at point B and attaches to the ship at points C and D. The two rope segments BC and BD angle away from the center of the ship at angles of ϕ = 26.0 ∘ and θ = 21.0 ∘, respectively. The tugboat pulls with a force of 1200 lb . What are the tensions TBC and TBD in the rope segments BC and BD?
Answer:
The tensions in [tex]T_{BC}[/tex] is approximately 4,934.2 lb and the tension in [tex]T_{BD}[/tex] is approximately 6,035.7 lb
Explanation:
The given information are;
The angle formed by the two rope segments are;
The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°
The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°
Therefore, we have;
The tension in rope segment BC = [tex]T_{BC}[/tex]
The tension in rope segment BD = [tex]T_{BD}[/tex]
The tension in rope segment AB = [tex]T_{AB}[/tex] = Pulling force of tugboat = 1200 lb
By resolution of forces acting along the line A_F gives;
[tex]T_{BC}[/tex] × cos(26.0°) + [tex]T_{BD}[/tex] × cos(21.0°) = [tex]T_{AB}[/tex] = 1200 lb
[tex]T_{BC}[/tex] × cos(26.0°) + [tex]T_{BD}[/tex] × cos(21.0°) = 1200 lb............(1)
Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;
[tex]T_{BC}[/tex] × sin(26.0°) + [tex]T_{BD}[/tex] × sin(21.0°) = 0...........................(2)
Which gives;
[tex]T_{BC}[/tex] × sin(26.0°) = - [tex]T_{BD}[/tex] × sin(21.0°)
[tex]T_{BC}[/tex] = - [tex]T_{BD}[/tex] × sin(21.0°)/(sin(26.0°)) ≈ - [tex]T_{BD}[/tex] × 0.8175
Substituting the value of, [tex]T_{BC}[/tex], in equation (1), gives;
- [tex]T_{BD}[/tex] × 0.8175 × cos(26.0°) + [tex]T_{BD}[/tex] × cos(21.0°) = 1200 lb
- [tex]T_{BD}[/tex] × 0.7348 + [tex]T_{BD}[/tex] ×0.9336 = 1200 lb
[tex]T_{BD}[/tex] ×0.1988 = 1200 lb
[tex]T_{BD}[/tex] ≈ 1200 lb/0.1988 = 6,035.6938 lb
[tex]T_{BD}[/tex] ≈ 6,035.6938 lb
[tex]T_{BC}[/tex] ≈ - [tex]T_{BD}[/tex] × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb
[tex]T_{BC}[/tex] ≈ -4934.1733 lb
From which we have;
The tensions in [tex]T_{BC}[/tex] ≈ -4934.2 lb and [tex]T_{BD}[/tex] ≈ 6,035.7 lb.