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!
What was the main cause of the shift to a postindustrial society?
A. Citizens started to prefer manual labor to jobs that required providing a service.
B. The economic situation allowed people to work from home instead of going to work.
C. The focus became the control of information instead of traditional manufacturing.
D. People started to purchase computers and cell phones in mass quantities.
Answer: i thnk its B
Explanation:
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
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 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 .
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 .
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.
The 1360-kg car and the 2100-kg tow truck are stationary. The muddy surface on which the car s tires rest exerts negligible friction forces on them. What is the tension in the tow cable
Answer:
5.91 kN
Explanation:
The schematic view of the free body diagram of the car is shown in the image attached below.
Let's first calculate the weight of the car.
W = mg
W = 1360 × 9.81
W = 13341.6 N
W = 13.3416 kN
From the image, Using the equilibrium equations, we have:
[tex]\sum F_x = 0\\[/tex]
T cos (18° - 10°) - 13.3416 sin 26° = 0
T cos 8° - 5.84857 = 0
T cos 8° = 5.84857
[tex]T = \dfrac{5.84857}{cos \ 8^0}[/tex]
[tex]T = \dfrac{5.84857}{0.9903}[/tex]
T = 5.905856811 N
T ≅ 5.91 kN
The period of a pendulum is directly proportional to the mass of the bob.
true or false.
Answer:
false
Explanation:
i just took the test <3
The period of a pendulum is not directly proportional to the mass of the bob. The period of pendulum is independent of the mass of the bob. Thus, the given statement is false.
What is the period of pendulum?A simple pendulum is the one which consists of a small metal ball called as the bob or a mass which is suspended from a fixed point by a long piece of thread such that the bob is free to swing back and forth from its mean position under the influence of the gravity.
The time period of a simple pendulum is the time taken by it to complete one oscillation. The formula for the time period or period (T) of a simple pendulum is T = 2π (√L/g), where L is the length of the pendulum thread and g is the acceleration due to gravity.
Learn more about Pendulum here:
https://brainly.com/question/14759840
#SPJ6
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:
A 25 kg box on a horizontal frictionless surface is moving to the right at a speed of 5.0 m s. The box hits and remains attached to one end of a spring of negligible mass whose other end is attached to a wall. As a result, the spring compresses a maximum distance of 0.60 m. (a) i. The spring does work on the box from the moment the box first hits the spring to the moment the spring first reaches its maximum compression. Indicate whether the work done by the spring is positive, negative, or zero.
Answer:
Explanation:
Work done by the spring is negative .
Work done by force F creating displacement d is given by the following expression .
Work = F x d
Both force and displacement are vector quantity .
When direction of force and direction of displacement is same , work is positive . When direction of force and direction of displacement is opposite , work is negative .
When spring is compressed , it exerts a restoring or opposing force in a direction opposite to the direction of displacement of box . Hence here force is opposite to displacement . Restoring force acts opposite to displacement . Hence work done by spring on box is negative .
(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)
A 0.877-kg ball has an elastic, head-on collision with a second ball that is initially at rest. The second ball moves off with half the original speed of the first ball. (a) What is the mass of the second ball
Answer: the mass of the second ball is 2.631 kg
Explanation:
Given that;
m1 = 0.877 kg
Initial velocity = V0
Initial momentum = m1 × V0
final velocity of m1 is u1, final velocity of m2 is u2 = v0/2
now final momentum = m1 × u1 + m2 × u2
using momentum conservation;
m1×V0 = m1×u1 + m2×v0/2
m1×(v0 - u1) = m2×V0/2 ----- let this be equation 1
Now, for elastic collision;
m1×v0²/2 = m1×u1²/2 + m2×(v0/2)²/2
m1×(v0² - u1²) = m2×(v0/2)² --------- let this be equation 2
now; equation 2 / equation 1
: V0 + u1 = v0/2
2V0 + 2u1 = V0
2u1 = V0 - 2V0
u1 = -V0/2
now we insert in equ 1
m1×3V0/2= m2×V0/2
m1 × 3 = m2
m2 = 0.877 × 3
m2 = 2.631 kg
Therefore, the mass of the second ball is 2.631 kg
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
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.
Explain why gravity is the only force acting on a falling ball
Answer:
Gravity is the only force acting on a falling ball because of free fall. Since gravity is unbalanced it accelerates an object and the velocity increases as an object falls due to gravity pushing it. The acceleration due to gravity is 9.8 m/s^2.
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)
A dragster going at 15 m/s increases its velocity to 25 m/s in in 2 seconds. What is its acceleration?
Answer:
5m/s²
Explanation:
Given parameters:
Initial velocity = 15m/s
Final velocity = 25m/s
Time taken = 2s
Unknown:
Acceleration = ?
Solution:
Acceleration is the rate of change of velocity with time taken. It is mathematically expressed as;
Acceleration = [tex]\frac{Final velocity - Initial velocity }{Time taken }[/tex]
Acceleration = [tex]\frac{25 - 15}{2}[/tex] = 5m/s²
Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 480 N/C. If the particles are free to move, what are their speeds (in m/s) after 48.4 ns
Answer:
The speed of electron is 4.08 x 10⁶ m/sThe speed of proton is 2,225.82 m/sExplanation:
Given;
electric field, E = 480 N/C
mass of electron, Me = 9.11 x 10⁻³¹ kg
mass of proton, Mp = 1.67 x 10⁻²⁷ kg
time of motion, t = 48.4 ns = 48.4 x 10⁻⁹ s
initial velocity of the particles, u = 0 (initially at rest)
let the speed of each particle after 48.4 ns = v
the magnitude of charge of the particles, q = 1.6 x 10⁻¹⁹ C
The force experienced by each particle is calculated as;
F = Eq
F = (480 N/C) x (1.6 x 10⁻¹⁹ C)
F = 7.68 x 10⁻¹⁷ N
The speed of each particle after 48.4 ns is calculated as;
[tex]F = ma\\\\F = \frac{m(v-u)}{t} \\\\F = \frac{m(v-0)}{t}\\\\F = \frac{mv}{t} \\\\mv = Ft\\\\v = \frac{Ft}{m} \\\\For \ electron;\\\\v_e = \frac{Ft}{m_e} \\\\v_e = \frac{7.68 \times 10^{-17} \ \times \ 48.4 \times 10^{-9}}{9.11 \times 10^{-31}} \\\\v_e = 4.08 \times 10^6 \ m/s[/tex]
[tex]For \ proton;\\\\v_p = \frac{Ft}{m_p} \\\\v_p = \frac{7.68 \times 10^{-17} \ \times \ 48.4 \times 10^{-9}}{1.67 \times 10^{-27}} \\\\v_p = 2,225 .82 \ m/s[/tex]
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
A He-Ne laser produces 633 nm light.
When it passes through a diffraction
grating with d = 5.55 x 10^-6 m, what is
the angle of its third order (m = 3)
maximum?
Answer:
20.0086
Explanation:sin^-1(3×633×10^-9)÷5.55×10^-6Answer:
20.0086
Explanation:
Acellus
all 4 questions plz ( pls hurry)
Popeye, of mass 75.0 kg, has just downed a can of spinach. He accelerates quickly and stops Bluto, of mass 720 kg (Bluto is very dense), who is charging in at 15.0 m/s. What was Popeye's speed
Answer:
Popeye's speed is 144 m/s.
Explanation:
Given;
mass of Popeye, m₁ = 75 kg
mass of Bluto, m₂ = 720 kg
velocity of Bluto, v₂ = 15 m/a
velocity of Popeye, v₁ = ?
Apply the principle of conservation of linear momentum;
m₁v₁ = m₂v₂
v₁ = (m₂v₂) / m₁
[tex]v_1 = \frac{720\ \times \ 15}{75} \\\\v_1 = 144 \ m/s[/tex]
Therefore, Popeye's speed is 144 m/s.
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.
The average velocity of a tennis ball is measured during serving and found to be 216 km/hr.
Calculate the time elapsed between hitting the ball and meeting the ground if the distance covered by the ball during that interval is 20m
PLEASE ANSWER WITH EXPLANATION :)
Answer:
The time between hitting the ball and meeting the ground is [tex]t=0.33\: s[/tex]
Explanation:
We have a semi parabolic motion here. We know that the initial speed 216 km/h or 60 m/s in the x-direction.
So we can use the following equation:
[tex]v=\frac{\Delta x}{\Delta t}[/tex]
[tex]60=\frac{20}{t}[/tex]
[tex]t=\frac{20}{60}[/tex]
[tex]t=0.33\: s[/tex]
hope it helps you!
what does correlation mean
a mutual relationship or connection between two or more things.
Use the following information about earth and its moon to determine the distance between them
answer is the SECOND one on Edge
3.8 x 10^8m
the chart would’ve been helpful though.
Answer:
B
Explanation:
edge2020
what is phyical quantity
Answer:
physical quantity
Explanation:
All quantity in term of which laws of physics are described and can be measured is called physical quantity
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].
