Answer:
d. 268 s
Explanation:
Constant Speed Motion
An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.
Expressed in a simple equation, we have:
[tex]\displaystyle v=\frac{d}{t}[/tex]
Where
v = Speed of the object
d = Distance traveled
t = Time taken to travel d.
From the equation above, we can solve for d:
d = v . t
And we can also solve it for t:
[tex]\displaystyle t=\frac{d}{v}[/tex]
Two cars are initially separated by 5 km are approaching each other at relative speeds of 55 km/h and 12 km/h respectively. The total speed at which they are approaching is 55+12 = 67 km/h.
The time it will take for them to meet is:
[tex]\displaystyle t=\frac{5}{67}[/tex]
t = 0.0746 hours
Converting to seconds: 0.0746*3600 = 268.56
The closest answer is d. 268 s
Objects 1 and 2 attract each other with a gravitational force
of 18.0 units. If the mass of Object 1 is halved AND the
mass of object 2 is tripled, then the new gravitational force
will be units.
Answer:
the answer is 72.0 units:) thank me later!!!!!!!
A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has force constant 815 N/m . The coefficient of kinetic friction between the floor and the block is μk=0.40. The block and spring are released from rest and the block slides along the floor.
Required:
What is the speed of the block when it has moved a distance of 0.0200 m from its initial position? (At this point the spring is compressed 0.0100 m.)
Answer:
v = 0.41 m/s
Explanation:
In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.So, we can write the following general equation, taking the initial and final values of the energies:[tex]\Delta K + \Delta U = W_{ffr} (1)[/tex]
Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.⇒ Kf = 1/2*m*vf² (2)The change in the potential energy, can be written as follows:[tex]\Delta U = U_{f} - U_{o} = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)[/tex]
where k = force constant = 815 N/m
xf = final displacement of the block = 0.01 m (taking as x=0 the position
for the spring at equilibrium)
x₀ = initial displacement of the block = 0.03 m
Regarding the work done by the force of friction, it can be written as follows:[tex]W_{ffr} = - \mu_{k}* F_{n} * \Delta x (4)[/tex]
where μk = coefficient of kinettic friction, Fn = normal force, and Δx =
horizontal displacement.
Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:Fn = Fg= m*g (5)Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:[tex]\frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o}) (6)[/tex]
[tex]\frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x) -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)[/tex]
Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:[tex]\frac{1}{2} * 2.00 kg* v^{2} = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)[/tex]
[tex]v =\sqrt{(0.326-0.1568} = 0.41 m/s (9)[/tex]all 4 questions plz ( pls hurry)
A weight lifter lifts a 360-N set of weights from ground level to a position over his head, a vertical distance of 1.95 m. How much work does the weight lifter do, assuming he moves the weights at constant speed
Answer:
W= 702 J
Explanation:
Given that,
A weight lifter lifts a 360-N set of weights from ground level to a position over his head, a vertical distance of 1.95 m.
We need to find the work the weight lifter do. Work done by an object is given by the formula as follows :
W = Fd
Putting all the values,
W = 360 N × 1.95 m
= 702 J
So, the required work done is 702 J.
a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the fraction of the legth tof the rod above water
Answer:
[tex]\frac{h_{liquid} }{ h_{body} }[/tex] = 5/9
Explanation:
This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.
B = ρ_liquid g V_liquid
let's write the translational equilibrium condition
B - W = 0
let's use the definition of density
ρ_body = m / V_body
m = ρ_body V_body
W = ρ_body V_body g
we substitute
ρ_liquid g V_liquid = ρ_body g V_body
[tex]\frac{\rho_{body} }{\rho_{liquid} } } = \frac{V_{liquid} }{V_{body} } }[/tex]
In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar
V = A h_bogy
Thus
[tex]\frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }[/tex]
we substitute
5/9 = [tex]\frac{h_{liquid} }{ h_{body} }[/tex]
A 845kg dragster (very fast car) accelerates from 2m/s to 30m/s in 0.9s. Determine the average force exerted on the dragster.
Plis I need help
Answer:
Explanation:
Just use the Force formula.
F = M . A
Acceleration Formula
A = V - Vo / Time
So...
F = 845 . (30 - 2 / 0.9)
F = 845 . 20
F = 16900 N
Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus the bullet rises to a height of 10 cm along a circular arc with a 18 cm radius.
Assume: The entire track is frictionless. A bullet with a m1 = 30 g mass is fired horizontally into a block of wood with m2 =
4.8 kg mass.
The acceleration of gravity is 9.8 m/s^2.
Calculate the total energy of the composite system at any time after the collision. Answer in units of J.
Taking the same parameter values as those in Part 1, determine the initial velocity of the bullet. Answer in units of m/s.
Answer:
1)4.7334J
2)225.4m/s
Explanation:
v= the Velocity of both the bullet and the block after collision=?
H= Height of the bullet along circular arc= 10cm=0.1m
g= acceleration due to gravity= 9.81m/s^2
R= Radius of the circular arc= 18cm= 0.18m
m= Mass of the bullet= 30g= 0.03kg
M= Mass of the block = 4.8 kg
Using the law of conservation of energy
Potential energy of the system= Kinectic energy of the system
1/2 mv^2= mgh..............eqn(1)
But we have two mass m and M
We can write eqn(1) as
0.5(m+M)v^2= (m+M)gh ...........eqn(2)
If we make "v" subject of the formula we have
v = √2gh
Then substitute the values we have
= √2 x 9.81 x 0.1 = 1.40m/s
1) We can now calculate the total energy of the system after collision as
KE = 1/2(m+M)v^2
= 1/2 x (0.03+4.8) x (1.40)^2
KE = 4.7334J
Hence, the total energy of the composite system at any time after the collision is 4.7334J
2)to determine the initial velocity of the bullet.
From law of momentum conservation, which can be expressed as
m1u1+m2u2=(m1+m2)v
Where the initial Velocity of the bullet u1= ?
Final velocity of the bullet = 0
the Velocity of both the bullet and the block after collision=v= 1.40m/s
(0.03×u1) +(u×0)= (4.8+0.03)1.4
0.03u1=6.762
U1=225.4m/s
Hence, the initial velocity of the bullet is 225.4m/s
The total energy of the composite system at any time after the collision is 4.7334 J and initial velocity of the bullet is 225.4 m/s.1`
What is conservation of momentum?
Momentum of a object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity.
When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the law of conservation of momentum.
Total energy of the composite system at any time after the collision-
The mass of the bullet is 30g and the mass of the wood block is 4.8 kg. Thus the total mass of these two is,
[tex]m=0.03+4.8\\m=4.83 \rm kg[/tex]
As the compound system of the block plus the bullet rises to a height of 10 cm. Thus, the speed of it is,
[tex]v=\sqrt{2gh}\\v=\sqrt{2\times9.81\times0.1}\\v=1.4\rm m/s[/tex]
Now, the total energy of the composite system at any time after the collision is equal to the kinetic energy. Therefore,
[tex]E=\dfrac{1}{2}mv^2\\E=\dfrac{1}{2}4.83(1.4)^2\\E=4.7334\rm J[/tex]
Total energy of the composite system at any time after the collision is 4.7334 J.
The initial velocity of the bullet-The block was at rest initial, thus it has initial velocity of it is zero. Thus, the initial momentum of the bullet is equal to the final momentum of bullet and block. Therefore,
[tex]0.03\times u_o=4.83\times1.4\\u_o=225.4\rm m/s[/tex]
Thus, the total energy of the composite system at any time after the collision is 4.7334 J and initial velocity of the bullet is 225.4 m/s.1`
Learn more about the conservation of momentum here;
https://brainly.com/question/7538238
Which property of a solid measures how resistant the material is to deformation?
A. Elasticity
B. Hardness
C. Plasticity
D. resilience
Answer: the answer is a
Explanation:
Which of the following is NOT true about essential body fat? A. The human body would not function normally without essential body fat. B. Essential body fat accounts for about 3% of men's total weight. C. The percentage of essential body fat is the same for both males and females. D. Essential body fat is found in one's organs, bones, and muscles. Please select the best answer from the choices provided. A B C D Mark this and return
Answer:
I believe it is A. The human body would not function normally without essential body fat.
Explanation:
Answer:
Cccccccccccc
Explanation:
A person weighing 785 Newtons on the surface of the Earth would weigh 47 Newtons on the surface of Pluto. What is the magnitude of the gravitational acceleration on the surface of Pluto?
Answer:
3.72 N / kg .
Explanation:
The weight of the object can be calculated using the expression below
Weight = mg
Weight= 785 Newtons
M = mass
g= acceleration due to gravity on Earth is 9.8 m/s² .
So, substitute the values we have,
So on Earth . . .
785 N = m x 9.8
mass= 785/9.8
=80.1kg
We can calculate when on On Mars, as
Weight= mg
298N= 80.1 kg x g( on Mars)
Acceleration of gravity of that Mars =
298 N/ 80.1
= 3.72 N / kg .
Hence, the magnitude of the gravitational acceleration on the surface of Pluto is 3.72 N / kg .
Once again, move the balloon to the right and let it go. Note how fast the balloon moves. Next, brush the balloon against the entire sweater. Allow all the electrons to transfer. Again, move the balloon all the way to the right, let it go, and note how fast it moves. Is there a difference in how fast the balloon moves when the balloon has more electrons and the sweater has fewer electrons?
Answer:
Yes. The balloon moves faster when it has more electrons and the sweater has fewer electrons
Explanation:
From Plato. Hope this helps!
Answer:
Yes. The balloon moves faster when it has more electrons and the sweater has fewer electrons
Explanation:
Edmentum
A soccer has been kicked as far as it can get with an initial momentum of 153 kg*m/s and the
ball weights 1.8 kg. What is the velocity of the ball?
Answer:
85 m/s
Explanation:
The formula for momentum is product of weight and velocity
p=mv where m is mass and v is velocity
Given that;
Momentum = 153 kg*m/s
Weight = 1.8 kg
Velocity = ?
p=mv
153 = 1.8 * v
153/1.8 = v
85 m/s
Two resistors, A and B, are connected in parallel across of a 6V battery. The current through B is found to be 2.0 A. When the two resistors are connected in series to the 6V battery, a voltmeter connected across the resistor A measures a voltage of 4V. Find the resistances of A and B.
Answer:
Resistance of A is [tex]6\ \Omega[/tex] and B is [tex]3\ \Omega[/tex]
Explanation:
The voltage across both the resistances will be the same as they are connected in parallel.
V = Voltage = 6 V
[tex]I_B=2\ \text{A}[/tex]
Resistance is given by
[tex]R_B=\dfrac{V}{I_B}\\\Rightarrow R_B=\dfrac{6}{2}\\\Rightarrow R_B=3\ \Omega[/tex]
[tex]V_B=V_b-V_A\\\Rightarrow V_B=6-4\\\Rightarrow V_B=2\ \text{V}[/tex]
Series connection
[tex]V_A=4\ \text{V}[/tex]
The current is constant in series connection
[tex]I=\dfrac{V_B}{R_B}\\\Rightarrow I=\dfrac{2}{3}\ \text{A}[/tex]
[tex]R_A=\dfrac{V_A}{I}\\\Rightarrow R_A=\dfrac{4}{\dfrac{2}{3}}\\\Rightarrow R_A=6\ \Omega[/tex]
The resistance of A is [tex]6\ \Omega[/tex] and B is [tex]3\ \Omega[/tex].
a plane flies from Addis Ababa to Gamble is 400km. in 2 hrs, then straight back to Gamma is 200km. in 1hrs what is the average speed, average velocity, total distance and total displacement
Answer:
a) Average speed = 200 Km/hr
b) Average velocity = 0 m/s
c) Total distance = 800 Km
d) Total displacement = 0 Km
Explanation:
a) Let the speed of the plane from Addis Ababa to Gamble be represented by [tex]S_{1}[/tex], and from Gamble to Addis Ababa by [tex]S_{2}[/tex].
Average speed = [tex]\frac{S_{1} + S_{2} }{2}[/tex]
[tex]S_{1}[/tex] = [tex]\frac{distance}{time}[/tex] = [tex]\frac{400}{2}[/tex]
= 200 Km/hr
[tex]S_{2}[/tex] = [tex]\frac{distance}{time}[/tex] = [tex]\frac{200}{1}[/tex]
= 200 Km/hr
Average speed = [tex]\frac{200 + 200}{2}[/tex]
= 200 Km/hr
b) Velocity = [tex]\frac{displacement}{time}[/tex]
Average velocity = [tex]\frac{total displacement}{total time taken}[/tex]
Since the plane flies from Addis Ababa to Gamble, then back to Addis Ababa, its total displacement is zero.
So that,
Average velocity = 0 m/s
c) Total distance = 400 Km + 400 Km
= 800 Km
d) Total displacement = 400 Km + (-400 Km)
= 0 Km
I need help this question
Answer:
[tex]93\:\mathrm{kg}[/tex]
Explanation:
We can use Newton's Universal Law of Gravitation to solve this:
[tex]F=G\frac{m_1m_2}{r^2}[/tex], where [tex]F[/tex] is force, [tex]G[/tex] is gravitational constant [tex]6.67\cdot 10^{-11}[/tex], [tex]m_1[/tex]and [tex]m_2[/tex] represent the masses of both objects, and [tex]r[/tex] represents the distance between their center of masses.
Plugging in our given values, we have:
[tex]5.2\cdot 10^{-8}=6.67\cdot 10^{-11}\cdot \frac{97\cdot m_2}{3.4^2},\\m_2=92.9=\fbox{$93\:\mathrm{kg}$}[/tex](two significant figures).
Answer:
the answer is 93kg
Explanation:
the answer is 93kg
An ocean wave traveling in one direction has a wavelength of 1.0 m and a frequency of 1.25 Hz. Take the direction of wave propagation lo be the positive x direction.
(a) What is the speed (in m/s) of this ocean wave?
(b) Assuming that this wave is harmonic, and its amplitude is 2.0 m, what equation would describe its motion? Let the displacement at t = 0 s and x = 0 m be a maximum.
(c) What will be the height of the wave 3.0 m from the origin at t = 10 s?
Answer:
a) [tex]v=1*1.25=1.25\: m/s[/tex]
b) [tex]y(x,t)=2sin(2\pi( x-1.25 t)[/tex]
c) [tex]y(3,10)=1.73 m[/tex]
Explanation:
a) The speed of a wave is given by the following equation:
[tex]v=\lambda f[/tex]
Where:
λ is the wavelength
f is the frequency
[tex]v=1*1.25=1.25\: m/s[/tex]
b) The harmonic wave has the following equation:
[tex]y=Asin(kx-\omega t)[/tex]
A is the amplitude (2 m)
k is the wavenumber (2π/λ)
ω is the angular frequency (2πf)
[tex]y(x,t)=2sin(2\pi x-2\pi*1.25 t)[/tex]
[tex]y(x,t)=2sin(2\pi( x-1.25 t)[/tex]
c) Here we need to find the heigth at x=3 m and t =10 s, so we need to find y(3,10).
[tex]y(3,10)=2sin(2\pi(3-1.25*10)[/tex]
[tex]y(3,10)=2sin(2\pi(3-1.25*10)[/tex]
[tex]y(3,10)=1.73 m[/tex]
I hope it helps you!
1. The Age of the Dinosaurs
Dinosaurs existed about 250 million years ago to 65 million years ago. This era is broken up into three periods known as the Triassic, Jurassic and Cretaceous periods. The Triassic Period lasted for 35 million years from 250-205 million years ago. Planet Earth was a very different place back then. All the continents were united to form one huge land mass known as Pangaea. The Jurassic Period was the second phase. The continents began shifting apart. The time scale for this famous period is from 205 to 138 million years ago. The Cretaceous Period was the last period of the dinosaurs. It spanned a time from 138 million to about 65 million years ago. In this period the continents fully separated. However, Australia and Antarctica were still united.
What type of text structure is this?
Please explain why your answer is correct:
If a student wishes to conduct an experiment to prove the conservation of momentum between two colliding objects, what are the minimum quanitities the student must record in order to complete this experiment and explain why you think they must record those.
A small steel ball falls from rest through a distance of 3m. When calculating the time of fall, air resistance can be ignored because
Answer:
First, let's write the movement equations for this ball.
The only force acting on the ball will be the gravitational acceleration (because we ignore the air resistance) then the acceleration equation is:
a(t) = -9.8m/s^2
Where the minus sign is because the ball is falling down.
To get the velocity of the ball, we need to integrate over time to get:
v(t) = -(9.8m/s^2)*t + v0
Where v0 is the initial velocity of the ball. Because it falls from rest, we can conclude that the initial velocity is 0 m/s, then the velocity equation is:
v(t) = -(9.8m/s^2)*t
For the position equation we need to integrate again, here we get:
p(t) = -(1/2)*(9.8m/s^2)*t^2 + p0
Where p0 is the initial position. In this cse we know that the ball falls from a height of 3m, then po = 3m
The position equation is:
p(t) = -(1/2)*(9.8m/s^2)*t^2 + 3m
The ball will hit the ground when p(t) = 0m, then we need to solve the equation:
p(t) = -(1/2)*(9.8m/s^2)*t^2 + 3m = 0m
for t.
-(1/2)*(9.8m/s^2)*t^2 + 3m = 0m
3m = (1/2)*(9.8m/s^2)*t^2
3m*2 = (9.8m/s^2)*t^2
6m/(9.8m/s^2) = t^2
√(6m/(9.8m/s^2)) = t = 0.78s
The ball needs 0.78 seconds to hit the ground.
The correct answer is (d) the weight of steel ball is much larger than air resistance.
since the density of steel ball is quite higher than that of air. The weight of even a small steel ball will be much larger than the air resistance acting opposite to the motion of the steel ball. Hence in the case of a freely falling steel ball the air resistance can be neglected.
Learn more about air resistance:
https://brainly.com/question/2575108
A 0.5 kg rock is dropped from a height of 1.0 m above the ground. Approximately how much kinetic energy will be stored in the rock after it has fallen halfway to the ground.
Answer:
2.45 J
Explanation:
The following data were obtained from the question:
Mass (m) = 0.5 kg
Height (h) = 1 m
Kinetic energy (KE) =?
Next, we shall determine the velocity of the rock after it has fallen half way. This can be obtained as follow:
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 1/2 = 0.5 m
Final velocity (v) =?
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 0.5)
v² = 9.8
Take the square root of both side
v = √9.8
v = 3.13 m/s
Finally, we shall determine the kinetic energy of the rock after it has fallen half way. This can be obtained as follow:
Mass (m) = 0.5 kg
Velocity (v) = 3.13 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 0.5 × 3.13²
KE = 0.25 × 9.8
KE = 2.45 J
Therefore, the kinetic energy of the rock after it has fallen half way is 2.45 J
Estimate the weight of a 1000kg car that is accelerating at 3 m/s/s.
Answer:
W = 9800 N
Explanation:
Given that,
Mass of a car, m = 1000 kg
Acceleration of the car, a = 3 m/s²
We need to find the weight of the car. Weight of an object is given by the product of mass and acceleration due to gravity on the Earth.
W = mg
Put all the values,
W = 1000 kg × 9.8 m/s²
= 9800 N
So, the weight of the car is 9800 N.
(1 point) Unknown resistor in voltage divider Suppose that a power supply is connected across two resistors R1 and R2 that are connected in series. The power supply voltage E, the voltage across the second resistor V2, and the first resistance R1 are all known, but R2 is not known. Find an expression for R2 in terms of E,R1,andV2
Answer:
R₂ = V₂R₁/(E + V₂)
Explanation:
By voltage divider, V₂ = ER₂/(R₁ + R₂)
So, cross-multiplying, we have
V₂(R₁ + R₂) = ER₂
expanding the bracket, we have
V₂R₁ + V₂R₂ = ER₂
collecting like terms, we have
V₂R₁ = ER₂ + V₂R₂
factorizing, we have
V₂R₁ = (E + V₂)R₂
dividing through by (E + V₂), we have
R₂ = V₂R₁/(E + V₂)
13. The
are beautiful lights in the sky caused from solar energy entering the
atmosphere after moving along the magnetic field of the Earth.
Answer:
Are you talking about aurora borealis?
Explanation:
An 80kg astronaut traveled to the moon, where gravity is one-sixth (116) as
strong as Earth's gravity. What is the astronaut's weight on the moon?
F=mg.g=(1/6)(9.8m/s?)
Answer:
Wmoon = 131 [N]
Explanation:
We know that the weight of a body is equal to the product of mass by gravitational acceleration.
Since we are told that the gravitational acceleration of the moon is equal to one-sixth of the acceleration of Earth's gravitation. Then we must multiply the value of Earth's gravitation by one-sixth.
[tex]w_{moon}=\frac{1}{6} *m*g\\w_{moon}=\frac{1}{6} *80*9.81\\w_{moon}=130.8 [N] = 131 [N][/tex]
A fireworks mortar is launched straight upward from a pool deck platform 4 m off the ground at an initial velocity of 61 m/sec. The height of the mortar can be modeled by where h(t) is the height in meters and t is the time in seconds after launch. What is the maximum height
Answer:
22.48m
Explanation:
Given that the pool deck platform is 4 m above the ground.
The initial velocity of mortar in the upward direction, u=21 m/s
At the maximum height, the velocity of the mortar will become zero.
So, the final velocity, v=0
Acceleration due to gravity, g = 9.81 m/[tex]s^2[/tex] in the downward direction.
By using the equation of motion [tex]v^2=u^2+2as\\[/tex]
On putting all the values, we have
[tex]0^2=21^2+2(-9.81)s\\\\s=21^2/(2\times 9.81)[/tex]
s= 22.48 m
Hence, the mortar will reach a maximum height of 22.48m.
Which layer of the atmosphere is most likely to protect life on Earth from ultraviolet radiation
Answer:
ozone layer
Explanation:
The ozone layer forms a thin shield in the upper atmosphere protecting life on earth from UV rays. Ozone is a naturally occurring gas that is found in two layers of the atmosphere
2 objects have a total momentum of 400kg m/s, they collide. Object A’s mass is5kg & object B’s mass is 11kg. After the collision Object B is moving at 15m/s.What is the velocity of Object A AFTER the collision?
Answer:
Explanation:
We shall apply law of conservation of momentum .
Momentum before collision = momentum after collision .
Momentum before collision = 400 kg m/s
Momentum after collision = 5 x v + 11 x 15
where v is velocity of A after the collision .
5 x v + 11 x 15 = 400
5 v = 400 - 165
5v = 235
v = 47 m /s .
Need help to get this question right!!
Answer:
16 times as strong
Explanation:
From the question given above, the following assumptions were made:
Initial Force (F₁) = F
Initial distance apart (r₁) = r
Final distance apart (r₂) = ¼r
Final force (F₂) =?
Next, we shall obtain a relationship between the force and the distance apart. This can be obtained as follow:
F = GM₁M₂ / r²
Cross multiply
Fr² = GM₁M₂
If G, M₁ and M₂ are kept constant, then,
F₁r₁² = F₂r₂²
Finally, we determine the new force as follow:
Initial Force (F₁) = F
Initial distance apart (r₁) = r
Final distance apart (r₂) = ¼r
Final force (F₂) =?
Fr² = F₂ × (¼r)²
Fr² = F₂ × r²/16
Fr² = F₂r² / 16
Cross multiply
16Fr² = F₂r²
Divide both side by r²
F₂ = 16Fr² / r²
F₂ = 16F
From the calculations made above, we can see that the new force is 16 times the original force.
Thus, the new force is 16 times stronger.
In which situation is maximum work considered to be done by a force?
A.
The angle between the force and displacement is 180°.
B.
The angle between the force and displacement is 90°.
C.
The angle between the force and displacement is 60°.
D.
The angle between the force and displacement is 45°.
E.
The angle between the force and displacement is 0°.
Answer:
A) When the angle between the Force (F) and Displacement (x) is 0°, because, Work done (W) is directly proportional to the Cosine of the Angle between the Force applied and the resultant displacement of the subject.
W = F•x cos ∅
If ∅ = 0°,
W = F•x ===> Maximum Work Done.
If ∅ = 45°,
W = F•x/√2
If ∅ = 90°,
W = 0
If ∅ = 180°,
W = –F•x ===> Minimum Work Done.
a student attaches a 0.5 kg object to a 0.7 m string and rotates the object around her head and parallel to the ground. how much tension force is required to make the object rotate with a speed of 12 m/s?
The object would have a centripetal acceleration a of
a = (12 m/s)² / (0.7 m) ≈ 205.714 m/s²
so that the required tension in the string would be
T = (0.5 kg) a ≈ 102.857 N ≈ 100 N
(rounding to 1 significant digit)
