Ben (55kg) is standing on very slippery ice when Junior (25kg) bumps into him. Junior was moving at a speed of 8m/s before the collision. Find the speed of ben and junior as they move across the ice after the collision. Give the answer in m/s. Describe the work you did to get the answer
Answer:
2.5m/s
Explanation:
m1=55kg
m2=25kg
Ben's velocity before collision=0m/s
Junior's velocity before collision= 8m/s
P1=m1v1+m2v2
P1=55kg(0m/s)+25kg(8m/s)
P1=200kg(m/s)
Know that the momentum before the collision = momentum after the collision
P2=(55kg+25kg)V
200kg(m/s)=(75kg)V
200kg(m/s)/75kg=V
V=2.5m/s.
It makes sense bc after they collide, their speed would slow down, and it wouldn't make sense if the momentum would be greater or less after they collide.
a rectangular block has the density of 350g/cm3 the dimensions are 3.5 6.cm 2.5
cm calculate the mass of the bock
Answer:
18375g
Explanation:
[tex]\boxed{density = \frac{mass}{volume} }[/tex]
[tex]∴ mass = density \times volume[/tex]
Let's find the volume of the rectangular block.
Volume
= length ×breadth ×height
= 3.5 ×6 ×2.5
= 52.5cm³
Mass of the block
= 350(52.5)
= 18375g
20 PTS!
If an object is moving at a constant velocity, which must be true?
Its acceleration is zero.
Its acceleration in decreasing.
Its acceleration is increasing,
Its acceleration is a non-zero constant.
[tex]\huge \bf༆ Answer ༄[/tex]
The Correct choice is ~ A
[tex] \textsf{its \: acceleration \: is \: zero}[/tex]If an object moves at a constant velocity, then the change in velocity over the time is 0, Acceleration is defined as rate of change in velocity but since there is no change in velocity, the value of Acceleration is equal to Zero.
A 2.40 kg ball is attached to an unknown spring and allowed to oscillate. The figure below shows a graph of the ball’s position �� as a function of time . What are the oscillation’s: a. period; b. frequency; c. angular frequency; d. amplitude; and e. What is the force constant of the spring?
(a) The period of the oscillation is 0.8 s.
(b) The frequency of the oscillation is 1.25 Hz.
(c) The angular frequency of the oscillation is 7.885 rad/s.
(d) The amplitude of the oscillation is 3 cm.
(e) The force constant of the spring is 148.1 N/m.
The given parameters:
Mass of the ball, m = 2.4 kgFrom the given graph, we can determine the missing parameters.
The amplitude of the wave is the maximum displacement, A = 3 cm
The period of the oscillation is the time taken to make one complete cycle.
T = 0.8 s
The frequency of the oscillation is calculated as follows;
[tex]f = \frac{1}{T} \\\\f = \frac{1 }{0.8} \\\\f = 1.25 \ Hz[/tex]
The angular frequency of the oscillation is calculated as follows;
[tex]\omega = 2\pi f\\\\\omega = 2\pi \times 1.25\\\\\omega = 7.855 \ rad/s[/tex]
The force constant of the spring is calculated as follows;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\ k = \omega ^2 m\\\\k = (7.855)^2 \times 2.4\\\\k = 148.1 \ N/m[/tex]
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If a force of 20kn acts on a circular rod of a diameter 10mm, calculate the stress of the rod
σ = F/A = F/ (πD²/4) = 20000 / (π0.10²/4) = 2.55 MN/m²
the conduction of heat from hot body to cold body is an example of what thermodynamics process?
Answer:
Heat flow
Explanation:
A rollercoaster accelerates from 10 m/s to 100 m/s2 for 25 seconds. What is the acceleration?
Answer:
A roller coasters accelerates from an initial velocity of of 6.0 m/s to a final velocity of 70 m/s over 4 seconds. What's the acceleration? Q. Acceleration only takes place when things speed up. Q. A drag racer accelerated from 0 m/s to 200 m/s in 5 s.
Explanation:
In a collision, a 30 kg ball moving at 3 m/s transfers all of its momentum to a 5kg ball. what is the velocity of the 3kg ball after the collision?
Answer:
Explanation:
Is it a 30 kg ball or a 3 kg ball?
Does not matter as we don't have to do actual calculations
If all of the momentum is transferred out of a mass, the velocity remaining must be ZERO
Need help asap
A scientist has invented a robot to work on the seabed. According to his calculation, the armour of the robot can withstand a maximum pressure of 10⁵ Pa exerted by the sea water. If the density of the sea water is 1025 kg/m3, what is the maximum depth of the seabed that this robot can work? [Given g = 9.81 m/s2 and rho water = 1000 kg/m3]
Answer:
Explanation:
Well which is it ? ρ = 1000 kg/m³ or ρ = 1025 kg/m³?
Obviously the sea is salt water so we can ignore ρ = 1000 kg/m³
1025 kg/m³(d m)(9.81 N/kg) = 1 x 10⁵ N/m² = Pa
d = 9.9450535...
d = 10 meters
That's if we only account for the pressure due to the water. On top of that pressure would be atmospheric pressure which is about 101000 Pa
so the robot would be a hair above its pressure limit before it even got in the water.
A baseball is hit with a speed of 27.0 m/s at an angle of 47.0 ∘ . It lands on the flat roof of a 10.0 m -tall nearby building. Part A If the ball was hit when it was 1.3 m above the ground, what horizontal distance does it travel before it lands on the building?
Answer:
H = Vy t - 1,2 g t^2 formula for height of ball after t sec
H = 10 - 1.3 = 8.7 m
Vy = 27 sin 47 = 19.7 m/s vertical speed of ball
8.7 = 19.7 t - 9.8/2 t^2 height of ball after t sec
4.9 t^2 - 19.7 + 8.7 = 0 rearranging
[19.7 ± (388 - 170)^1/2] / 2 *4.9 = [19.7 ± 14.7] / 9.8 = .51 ,3.5 sec
.51 sec would be on the way up and 3.5 sec on the way down
Sx = 27 * cos 47 * 3.5 = 64.4 m around 200 ft seems reasonable
A 3.1 kg ball is dropped from the top of a 38 m tall building. What is the speed of the ball when it is halfway from the building to the ground? Round your answer to 2 decimal places.
Answer:
19.3m/s
Explanation:
Use third equation of motion
[tex]v^2-u^2=2gh[/tex]
where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity
insert values to get answer
[tex]v^2-0^2=2(9.81m/s^2)(38/2)\\v^2=9.81m/s^2 *38\\v^2=372.78\\v=\sqrt[]{372.78} \\v=19.3m/s[/tex]
A student connects a 21.0 V battery to a capacitor of unknown capacitance. The result is that 52.8 µC of charge is stored on the capacitor. How much energy (in J) is stored in the capacitor?
Answer:
1.108 × [tex]10^{-3}[/tex]J
Explanation:
v=21.0v
Q=52.8× [tex]10^{-6}[/tex]
E=?
V=E/Q
E=v ×Q
=21 ×52.8 ×[tex]10^{-6}[/tex]
=1108.8 ×[tex]10^{-6}[/tex]
E= 1.108 × [tex]10^{-3}[/tex]J
Please help me answer the following image!
How is Compression Force Measured?
don't just copy theanswer please
Compression force can be measured with a force gage or load cell.
Which type of muscle cell can have multiple nuclei
Answer:
Skeletal Muscle cells
Skeletal muscle cells are long, cylindrical, and striated. They are multi-nucleated meaning that they have more than one nucleus. This is because they are formed from the fusion of embryonic myoblasts.
Answer:
skeletal muscle cells can have multiple nuclei
Explanation:
This is because they are formed from the fusion of embryonic myoblasts.
A jet is flying at a speed of 700 kilometers per hour. The pilot encounters turbulence due to a 50-kilometer- per-hour wind blowing at an angle of 47°. Find the resultant force on the plane.
Answer:
F = 0 N
Explanation:
The problem does not talk about any acceleration. We can just assume it always moves at a constant speed so the resultant force will be 0 newtons. Of course you could say that it accelerates for a certain amount of time, as the wind blows, but the problem doesn't tell us when to calculate the resultant force (if during the acceleration or after) + we don't have the mass of the jet.
The maximum speed with which an 1000 kg car makes a 180-degree turn is 10 m/s. The radius of the circle through which the car is turning is 24 m. Determine the force of friction and the coefficient of friction acting upon
the car.
American Football originated from which two sports?
A) Soccer and Rugby
B) Rugby and Basketball
C) Cricket and Soccer
D) Soccer and Volleyball
Answer:
b rugby and basketball
Explanation:
basket ball was made by a white man
American Football originated from Soccer and Rugby. Hence, option (A) is correct.
What is American Football?American football, sometimes known as gridiron, is a team sport played by two teams of eleven players on a rectangular field with goalposts at each end. It is known simply as football in the United States and Canada. The defense, the team without the ball, seeks to halt the offense's movement and seize control of the ball for themselves.
The offence, the team in possession of the oval-shaped football, attempts to advance down the field by running with the ball or throwing it. The offence is awarded a new set of four downs to resume the drive if they successfully gain at least 10 yards in four downs or plays. If they fail, they forfeit the ball to the defense.
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f. Protons and neutrons
2. TRUE or FALSE: An object that is positively charged contains all protons and no electrons
3. TRUE or FALSE: An object that is negatively charged could contain only electrons with me
accompanying protons
4. TRUE or FALSE: An object that is electrically neutral contains only neutrons.
An object which has more electron than proton is negatively charged, otherwise positively charged. Every statement is false
What is atom?Atom is the smallest unit of the matter consist of the positive charged nucleus and the electrons which moves around it. The atom can not be divided further.
The atom of a matter is made by three elements-
1) Neutron-Neutron is the element of atom, which has zero charge.2) Proton-Proton is the element of atom, which has positive charge.3) Electron-Electron is the element of atom, which has negative charge.For the Protons and neutrons, lets check all the statement wheather they are true or false.
2. An object that is positively charged contains all protons and no electrons- An object which is positively charged, has more number of proton than electron. This is a false statement.3. An object that is negatively charged could contain only electrons with me accompanying protons- An object, which is negatively charged, has more number of electron than proton. This is a false statement.4. An object that is electrically neutral contains only neutrons-The number of electron and proton is equal in an electrically neutral object. This is a false statement.Thus, every statement is false as a object which has more electron than proton is negatively charged, otherwise positively charged.
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Two satellites are in orbit around a planet. Satellite 1 has a mass of mm and an orbital radius of 2r2r. Satellite has a mass of 2m2m and an orbital radius of rr. If the magnitude of the gravitational force the planet exerts on Satellite 1 is FF, the magnitude of the gravitational force the planet exerts on Satellite 2 is
a) F
b) 1/2F
c) 2F
d) 1/4F
The magnitude of the gravitational force the planet exerts on Satellite 2 is 1/2F.
We recall that from Newton's law of universal gravitation, the gravitational force is given by;
F =G m1m2/r^2
m1 and m2 are the respective masses
r is the distance of separation.
Let the mass of the planet be M and the mass of the satellite be m. The distance of separation is r.
For satellite 1;
F = GmM/(2r)^2
F = GmM/4r^2
For satellite 2;
F2= G2mM/r^2
Hence, the magnitude of the gravitational force the planet exerts on Satellite 2 is 1/2F.
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A 70 kg man is running up the stairs which is 3m high in 2s.(a)How much work is done by the man?(b)What is the power exerted by the man? (Use g = 10ms 2)
Explanation:
m = 70 kg
s = 3m
t =2s
g = 10 m/s²
(a)How much work is done by the man?
W = Fs
= mg X s
= 70 x 3 x 10
= 210 x 10
= 2100 Joule
(b)What is the power exerted by the man?
P = W/t
= 2100/2
P = 1050 Watt
Cats are groomed often by their owners to remove hair. What logical inference can be made based on this statement?
Answer:
its keeps the fur straight.....im kinda confused with the question
Explanation:
The average value of the load between A and B is 6.0 N. The spring has an unstretched length
increases from 4.0 cm to 6.0 cm.
The change in the length of the spring as it stretches from 4.0 cm to 6.0 cm is 2.0 cm.
The change in length of the spring can be calculated by subtracting the initial unstretched length (4.0 cm) from the final stretched length (6.0 cm).
Change in length = Final length - Initial length
Change in length = 6.0 cm - 4.0 cm
Change in length = 2.0 cm
Therefore, the change in the length of the spring as it stretches from 4.0 cm to 6.0 cm is 2.0 cm. This means the spring elongates by 2.0 cm under the applied load between points A and B.
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The complete question is:
The average value of the load between A and B is 6.0 N. The spring has an unstretched length increases from 4.0 cm to 6.0 cm. What is the change in the length of the spring as it stretches from 4.0 cm to 6.0 cm?
A. 4.0 cmB. 1.0 cmC. 2.0 cmD. 6.0 cmAn airplane accelerates from rest down a runway at 9.50 m/s2 for 29.3 seconds when it
takes off. What is the distance traveled while taking off?
Answer:
Explanation:
s = ½at²
s = ½(9.50)(29.3²) = 4,077.8275
s = 4080 m
when rounded to the three significant digits of the question numerals.
Velocity v (m/s)
11/A stone is dropped from the top of a 45 m high building, How
fast will it be moving when it reaches the ground? And what
lits velocity be?
Explanation:
given solution
h=45m v^2=u^2+2gh
g=10m/s^2 v^2=0^2+2×10m/s^2×45m
vi=0 v^2=900m^2/s^2
v=30As a result of friction, the angular speed of a wheel changes with time according to dθ/dt = ω0e^−σt where ω0 and σ are constants. The angular speed changes from 3.70 rad/s at t = 0 to 2.00 rad/s at t = 8.60 s.
a. Use this information to determine σ and ω0.
σ = _______s−1
ωo = ______rad/s
b. Determine the magnitude of the angular acceleration at t = 3.00 s.
______rad/s2
c. Determine the number of revolutions the wheel makes in the first 2.50 s
_______rev
d. Determine the number of revolutions it makes before coming to rest.
_______rev
Hi there!
a.
We can use the initial conditions to solve for w₀.
It is given that:
[tex]\frac{d\theta}{dt} = w_0e^{-\sigma t}[/tex]
We are given that at t = 0, ω = 3.7 rad/sec. We can plug this into the equation:
[tex]\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}[/tex]
Now, we can solve for sigma using the other given condition:
[tex]2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}[/tex]
b.
The angular acceleration is the DERIVATIVE of the angular velocity function, so:
[tex]\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}[/tex]
c.
The angular displacement is the INTEGRAL of the angular velocity function.
[tex]\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]
[tex]\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.[/tex]
[tex]\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}[/tex]
[tex]\theta = 8.471 rad[/tex]
Convert this to rev:
[tex]8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}[/tex]
d.
We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.
[tex]0 = 3.7e^{-0.0714t}\\\\t = \infty[/tex]
Evaluate the improper integral:
[tex]\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]
[tex]\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.[/tex]
[tex]\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad[/tex]
Convert to rev:
[tex]51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}[/tex]
Which statement is true for a series circuit
Answer: they have one path to flow
Explanation: share the same current
Blaine and her sister are identical twins riding roller coasters at Kinetic Kars. They each ride the
roller coaster on their own once. Next time, they ride the roller coaster together. On which ride
do you think they have the most kinetic energy? Explain your answer using information from class
activities.
identify the origins of breakdown when using a spectrum analyzer
Four regions of the electromagnetic spectrum that astronomers use when observing objects in the space are the following enumerated answers.
1. First is Ultraviolet
2. Next is Infrared
3. Then the radio
4. Lastly the Visible lights.
These are the answers to the problem.
Two rods of mass m, length L are stuck together to form an X shape and spun around the center. What is the rotational inertia