Answer:
Q. 1. Newton's Law of gravitation states that all bodies in the universe exerts a force of attraction on all other bodies in the universe with a proportional force to both the product of the masses of the bodies and inversely proportional to the square of the distance between their centers
Mathematically, we have;
[tex]F = G \times \dfrac{m_1 \times m_2}{R^2}[/tex]
Where;
m₁, and m₂ are the masses of the bodies
R = The distance between their centers
G = The gravitational constant = 6.6743 × 10⁻¹¹ N·m²/kg²
The gravitational constant, G, is the Newton's law of gravitation's constant of proportionality between the force of attraction that exist two bodies and the product of their masses divided by the square of the distance between their centers
Q. 2. Newton's law of gravitation in vector form is presented as follows;
[tex]\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{12}[/tex]
The above equation gives the gravitational force of attraction of body 1 on body 2, with the negative sign and unit vector indicating that the force of of gravity is towards body 1
The force of gravity of body 2 on 1 is presented as follows;
[tex]\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{12}^2} \cdot \hat R_{21}[/tex]
The gravitational force of attraction of body 2 on body 1 is therefore, equal in magnitude and opposite in direction of the gravitational force of body 1 on body 2 (towards body 2)
[tex]-\underset{F_{12}}{\rightarrow} = G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{12} = G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot -(\hat R_{21}) = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{21}[/tex]
[tex]-\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{21} = \underset{F_{21}}{\rightarrow}[/tex]
[tex]-\underset{F_{12}}{\rightarrow} = \underset{F_{21}}{\rightarrow}[/tex]
Explanation:
Angie walked a distance of 90 meters east in 70 seconds. What was her
velocity?
A. 0.78 m/s east
B. 1.3 m/s east
O C. 7 m/s east
D. 9 m/s east
4. When setting goals, you should do everything EXCEPT which of the following?
Take into account your current level of activity.
Ask all of your friends what they think your goals should be..
Set a time limit.
Be realistic.
Có một số điện trở giống nhau R0 = 3
. Cần ít nhất bao nhiêu điện trở để có một
đoạn mạch có điện trở Rtđ = 8
Answer:
hlo
Explanation:
hlo olz mark me as brainlest
A force of 5N accelerates a mass by 2 m/s². What will be the acceleration if the force and mass were increased to twice their original value?
Answer:
4m/s4
Explanation:
difference between effort distance and load distance
Answer:
Lever systems are simple machines that change or increase the input force that we apply to the load. The lever provides us with some mechanical...
Answer:
● Effort arm or Effort distance (ED): The perpendicular distance from the fulcrum to the point of effort is called effort arm.
● Load arm or Load distance (LD): The perpendicular distance from the fulcrum to the point of load is called load arm.
sl unit of upthrust and SI unit of pressure
Answer:
The SI unit of upthrust is Newton(N).
The SI unit of preesure is Pascal(P).
Thank You
una caja en reposo se traslada 93 cm con un peso de 67N en un tiempo de 9,89h.¿cual es la aceleración la masa y la fuerza de dicho objeto
Answer:
a. Acceleration, a = 1.47 * 10^{-9} m/s²
b. Mass = 4.57 * 10^{10} kilograms
c. Force = 67.12 Newton
Explanation:
Given the following data;
Distance = 93 cm to meters = 93/100 = 0.93 meters
Weight = 67 N
Time = 9.89 hours to seconds = 35604 seconds
Initial velocity = 0 m/s (since it's starting from rest)
Acceleration due to gravity, g = 9.8 m/s
a. To find the acceleration, we would use the second equation of motion;
[tex] S = ut + \frac{1}{2} at^{2} [/tex]
Where;
S is the distance covered or displacement of an object.
u is the initial velocity.
a is the acceleration.
t is the time.
Substituting the values into the equation, we have;
[tex] 0.93 = 0*35604 + \frac{1}{2} * a*35604^{2} [/tex]
[tex] 0.93 = 0 + \frac{1}{2} * 1267644816a [/tex]
[tex] 0.93 = 633822408a [/tex]
[tex] Acceleration, a = \frac{0.93}{633822408} [/tex]
Acceleration, a = 1.47 * 10^{-9} m/s²
b. To find the mass
Weight = mass * acceleration due to gravity
67 = mass * 1.47 * 10^{-9}
[tex] Acceleration, a = \frac{67}{1.47 * 10^{-9}} [/tex]
Mass = 4.57 * 10^{10} kilograms
c. To find the force;
Force = mass * acceleration
Force = 4.57 * 10^{10} * 1.47 * 10^{-9}
Force = 67.12 Newton
Can someone do this for the football
Estimated volume
(cm3)
Estimated density
(g/cm3)
edge.
Answer:
0.10 g/cm3
TRUST ME GUYS
can uh help in in this question step by step
Convert to m/s
[tex]\\ \sf \longmapsto 72\times \dfrac{5}{18}=5(4)=20m/s[/tex]
Final velocity=v=0m/sTime=2s=t[tex]\\ \sf \longmapsto Acceleration=\dfrac{v-u}{t}[/tex]
[tex]\\ \sf \longmapsto Acceleration=\dfrac{0-20}{2}[/tex]
[tex]\\ \sf \longmapsto Acceleration=\dfrac{-20}{2}[/tex]
[tex]\\ \sf \longmapsto Acceleration=a=-10m/s^2[/tex]
Distance be sUsing second equation of kinematics
[tex]\\ \sf \longmapsto s=ut+\dfrac{1}{2}at^2[/tex]
[tex]\\ \sf \longmapsto s=20(2)+\dfrac{1}{2}(-10)(2)^2[/tex]
[tex]\\ \sf \longmapsto s=40+(-20)[/tex]
[tex]\\ \sf \longmapsto s=40-20[/tex]
[tex]\\ \sf \longmapsto s=20m[/tex]
Now
Mass=m=5000kgUsing newtons second law
[tex]\\ \sf \longmapsto Force=ma[/tex]
[tex]\\ \sf \longmapsto Force=5000(-10)[/tex]
[tex]\\ \sf \longmapsto Force=-50000N[/tex]
Force is in opposite direction so its negative[tex]\\ \sf \longmapsto Force=50kN[/tex]
Answer . The acceleration of the truck is 10m/[tex]s^{2}[/tex], and the distance covered is 40 m. Have attached the picture for solution.
Hope that helps.
Lúc 7g bạn an đi từ nhà đến trường với tóc độ trung bình là 20km/h . Bạn đến trường lúc 7g20. Tính khoảng cách từ nhà tới trường?
Answer:
Distance = 6.667 kilometres
Explanation:
Given the following data;
Speed = 20 km/h
Departure time = 7:00
Arrival time = 7:20
Time taken = 20 minutes
To calculate the distance travelled from home to school;
First of all, we would have to convert the value of time in minutes to hours.
Conversion:
60 minutes = 1 hour
20 minutes = X hours
Cross-multiplying, we have;
X = 20/60 = 1/3 hours
Mathematically, the distance travelled by an object is calculated by using the formula;
Distance = speed * time
Distance = 20 * 1/3
Distance = 20/3 =
Distance = 6.667 kilometres
A 250–g piece of gold is at 19 °C. 5.192 kJ of energy is added to it by heat. The specific heat of gold is 129 J/(kg·°C). Calculate its final temperature.
We heat a 25–g sample of metal from 10 °C to 100 °C. 1.082 kJ of energy is added to it by heat. Calculate
the specific heat of the metal.
Answer:
A. DT is given by Q= MCs DT
m = mass of the substances
Cs= is it's specific heat capacity
Ck= Q
Mk ×DTk
=250 × 9 × 5
129
=Dt = 180.1085271
answer is 180degree C.
Explanation:
B. = 25×10 ×100
1.082
=2500
1.082
= 23105.360 g/kj.
The final temperature is 180 degree. and the specific heat of the metal is 23105.360 g/kj.
How to calculate the specific heat?Q = m . C . ΔT
Q = heat; m = mass; C is the specific heat and
ΔT = Final T° - Initial T°
Q = C lat . m
Q = Heat
m = mass
C lar = Latent heat of fusion
A) DT is given by Q= M Cs DT
where, m = mass of the substances
Cs= is it's specific heat capacity
Ck= Q
Mk × DTk
=250 × 9 × 5
129 =Dt = 180.1085271
Thus, the final temperature is 180 degree.
B) We heat a 25–g sample of metal from 10 °C to 100 °C. 1.082 kJ of energy is added to it by heat = 25×10 ×100
=2500
1.082
Q = 23105.360 g/kj
Hence, the specific heat of the metal is 23105.360 g/kj.
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Sort the processes based on the type of energy transfer they involve. condensation freezing deposition sublimation evaporation melting thermal energy added thermal energy removed
Answer:
condensation - thermal energy removed
freezing -thermal energy removed
deposition - thermal energy removed
sublimation - thermal energy added
evaporation - thermal energy added
melting - thermal energy added
Explanation:
Thermal energy is heat energy. Processes in which heat is added involve the addition of thermal energy while processes in which heat energy is removed involves removal of thermal energy.
Condensation involves a change from gas to liquid, freezing involves a change from liquid to solid while deposition involves the settling of mobile particles at a place. All these processes involve a decrease in energy of particles.
On the other hand, sublimation is a direct change from solid to gas, melting involves a change from solid to liquid while evaporation involves a change from liquid to gas. All these processes occur when energy is added to the particles in a system.
Answer:
condensation - thermal energy removed
freezing -thermal energy removed
deposition - thermal energy removed
sublimation - thermal energy added
evaporation - thermal energy added
melting - thermal energy added
write down any 5 example of conservation of momentum?
Answer:
1) Motion of air mass moving from equator northward (closer to earth axis)
2) Motion of object in orbit
3) Collision of 2 objects
4) Skater changing rotation by extension of arms
5) Motion of rocket due to velocity of expelled gas
1. A bicycle initially moving with a velocity
5.0 m s-1 accelerates for 5 s at a rate of 2 m s? Wh
will be its final velocity ?
Answer:
[tex]\boxed {\boxed {\sf 15 \ m/s \ or \ 15 \ m*s^{-1}}}[/tex]
Explanation:
We are asked to find the final velocity. We are given the acceleration, time, and initial velocity, so we can use the following kinematics formula.
[tex]v_f= v_i+ at[/tex]
In this formula, [tex]v_f[/tex] is the final velocity, [tex]v_i[/tex] is the initial velocity, [tex]a[/tex] is the acceleration, and [tex]t[/tex] is the time.
The bicycle has an initial velocity of 5.0 m *s⁻¹ or m/s, acceleration of 2 m/s², and a time of 5 seconds.
[tex]\bullet \ v_i = 5.0 \ m/s \\\bullet \ a= 2\ m/s^2\\\bullet \ t= 5 \ s[/tex]
Substitute the values into the formula.
[tex]v_f=5.0 \ m/s + ( 2\ m/s^2 * 5 \ s)[/tex]
Solve inside the parentheses.
[tex]\frac {2 \ m}{s^2}* 5 \ s = \frac{ 2 \ m}{s} * 5 = \frac{ 10 \ m}{s} = 10 \ m/s[/tex][tex]v_f= 5.0 \ m/s + (10 \ m/s)[/tex]
Add.
[tex]v_f= 15 \ m/s[/tex]
The units can also be written as:
[tex]v_f= 15 \ m*s^{-1}[/tex]
The bicycle's final velocity is 15 meters per second.
The mass of objects is 4kg and it has a density of 5gcm^-3. what is the volume
Answer:
4kg×5gm^3=60
Explanation:
the object if heavy
this is physics practical
Answer:
well done buddy
Explanation:
I let go of a piece of bread from a balcony. A bird flying 5.0 m overhead sees me drop it, and starts to dive straight down towards the bread the instant that I release it. She catches it after it falls 3.0 m. Assuming she accelerates constantly from rest (v0 = 0) at the time I let go of the bread, what is her acceleration? Show your work
This question can be solved using the equations of motion. There are two scenarios where the equations of motion can be used. The first scenario is the free-fall motion of the piece of bread. The second scenario is the uniformly accelerated motion of the bird.
The acceleration of the bird is "a = 26.13 m/s²".
First, we will calculate the time taken by the bread to fall 3 m. Using the second equation of motion for this free-fall motion:
[tex]h = v_it + \frac{1}{2}gt^2[/tex]
where,
h = height fall = 3 m
vi = initial velocity = 0 m/s
g = acceleration due to gravity = 9.8 m/s²
t = time taken = ?
Therefore,
[tex]3\ m = (0\ m/s)t+\frac{1}{2}(9.8\ m/s^2)t^2\\t = \sqrt{\frac{(3\ m)(2)}{9.8\ m/s^2}}\\\\t = 0.78\ s[/tex]
The bird took the same time to catch the bread. Now applying the second equation of motion to the bird's motion:
[tex]s = v_it + \frac{1}{2}at^2[/tex]
where,
s = distance covered by the bird = 5 m + 3 m = 8 m
vi = initial velocity of the bird = 0 m/s
a = acceleration of the bird = ?
t = time taken = 0.78 s
Therefore, using these values we get:
[tex]8\ m = (0\ m/s)(0.78\ s)+\frac{1}{2}a(0.78\ s)^2\\\\a = \frac{16\ m}{(0.78\ s)^2}[/tex]
a = 26.13 m/s²
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10. Match the following varibles to their relationship in Newton's 2nd Law. Questions 1. Force and Acceleration 2. Mass and Acceleration 3. Speed and Distance Answer Choices A. Direct Relationship B. Inverse Relationship C. Not in Newton's 2nd Law
Explanation:
based on the above information
1.A
2.B
3. C
Many people believe that if the human race continues to use energy as we are now, without change, we'll witness a significant worldwide environmental impact in this century. Research this topic and discuss this possibility. Include concrete examples of specific environmental consequences of global warming.
Answer:
It is correct to say that if the human race continues to use energy as it is now, without change, we will witness negative environmental impacts around the world in this century.
As a concrete example, we can cite the means of transport that use fossil fuels, such as cars and buses, which release polluting gases into the atmospheric layer and cause the greenhouse effect, contributing to global warming.
To solve these problems, it is necessary to raise the awareness of individuals, so that there is more and more interest and search for environmentally responsible solutions, such as the large-scale production of electric cars, which do not pollute the environment.
State the relative position for the earth and sun in a lunar eclipse (in a partial and total eclipse)
Answer:
A lunar eclipse is when the Earth passes between the moon and the sun, casting a shadow on the moon. This can only occur when the sun, Earth and moon are aligned exactly, or very closely so, with the Earth in the middle.
Using your Periodic Table, which element below has the smallest atomic radius? A.) Sodium, B.) Chlorine, C.) Phosphorus, D.) Iron
Chlorine has the smallest atomic radius since the atomic radius decreases as you travel to the right and up
5. a. Answer the following questions. What is density? Write a formula by showing the relation among density mass and volume.
Answer:
Density is how compact something is. The relationship is M/V=D (Mass divided by Volume equals Density).
Explanation:
WHAT IS DENSITY:
Density is the degree of compactness of a substance.
EXAMPLE:
"a reduction in bone density"
FORMULA OF DENSITY:
The formula for density is d = M/V, where d is density, M is mass, and V is volume.
12 x cos 50 = ?
Does anyone have the answer ? I forgot my my calculator.
12 x cos 50 = 7.713451316...
Using your Periodic Table, which of the elements below is most likely to be a solid at room temperature?
A.) potassium, B.) Hydrogen, C.) Neon, D.) Chlorine
The answer is definitely Potassium
A lady walks 10 m to the north, then she turns and continues walking 30 m due east.
Determine her(a) distance covered
(b) displacement.
Answer:
The distance covered is 40 m and the displacement is 31,6m.
Explanation:
The distance covered is the sum of the two distances (10+30). The displacement is equal to the distance of the hipotenusa of the triangle that the two distances (10 m to north and 30m to east) create. Using the Pythagoras theorem the displacent is equal to the Square root of (30^2 +10^2) .
round off 20.96 to 3 significant figures. a.20.9 b.20 c.21.0 d.21
Answer:
option c. 21.0
Explanation:
It was given that to find 3 significant figures. So the answer is 21.0
determjne the density of liquid whose relative density is 1.25 given that the density is 1000kgm-3
Answer:
divide the density of solution by density of water
EXPLANATION:
LIKE:
1.25÷1000kgm-3
Draw a wave that has a wavelength of 3 cm and an amplitude of 1 cm. Label the wavelength, the amplitude, the rest position, and the crest and trough of your wave.
Answer:
Please find attached, the required wave drawn with MS Excel
Explanation:
Functions that represent waves is given as follows
A general form of the wave equation is A·sin(B·x) + D
Where;
B = 2·π/T
T = The period of the wave = 1/f
D = The vertical shift of the wave = 0
A = The amplitude of the wave = 1 for sine wave
v = The wave velocity
λ = The wavelength of the wave
f = The frequency of the wave
v = f·λ
At constant v, λ ∝ 1/f
∴ λ ∝ T
Where T = 3, we have;
B = 2·π/T
∴ B = 2·π/3
Therefore, we have the wave with an amplitude of 1 cm, and wavelength, 3 cm, given as follows
y = sin((2·π/3)·x)
Plotting the above wave with MS Excel, we can get the attached wave
An element is highly conductive, highly reactive, soft, and lustrous. The element most likely belongs to which group?(1 point)
transition metals
noble gases
metalloids
alkali metals
Answer:
Alkali metals
Explanation:
Elements in this group are highly reactive, soft, lustrous and highly conductive.
An element which is highly conductive, highly reactive, soft, and lustrous is most likely an alkali metal.
Alkali metals are in group 1 of the Periodic table which means that they have only a single valence electron.
This causes them to be soft and highly reactive because:
The single valance electron leads to weak bonds amongst the element's atoms which makes them softThe elements want to lose the single valance electron so as to become stable so they will react with other elements to give away the electron.Examples of alkali electrons include:
Lithium Sodium Potassium etcIn conclusion therefore, alkali metals are highly reactive and soft and so the element described above is most likely an alkali metal.
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Một vật không mang điện sẽ bị nhiễm điện dương khí
Answer:
không có điện
Explanation: