Answer:
true 知恵を yfyfhdftufvdughf
*Earth Science Question*
Uses an array of mirrors to allow the collection of more light*
A. Reflecting
B. Refracting
C. Multiple Mirror
D. CCD
WILL MARK BRAINLIEST PLS HELPPP -- Which of Newton’s Laws explains why the satellite would collide with the moon if gravity is “turned off?”
picture attached
Answer:
b
Explanation:
a 90 gram object moves around a circle whose diameter is 2.56 m at a constant speed off 16 m/s. what is the object acceleration? how much time will take to complete the around?
Answer:
[tex]a=200\ m/s^2[/tex] and t = 0.5 s
Explanation:
Given that,
Mass of the object, m = 90 gram = 0.09 kg
Diameter of the circle, d = 2.56 m
Radius, r = 1.28 m
Speed, v = 16 m/s
We need to find the object acceleration. On the circular path, acceleration is given by :
[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{(16)^2}{1.28}\\\\a=200\ m/s^2[/tex]
So, the object's acceleration is [tex]200\ m/s^2[/tex].
Let t is time it will take to complete the around. Speed is given by :
[tex]v=\dfrac{2\pi r}{t}\\\\t=\dfrac{2\pi r}{v}\\\\t=\dfrac{2\pi \times 1.28}{16}\\\\t=0.5\ s[/tex]
So, time to cover one complete round is 0.5 s.
Which statement best describe one of George Berkeley's arguments against materialism?
O A. There cannot be two different substances; my mind exists; therefore, matter does not exist.
OB. To be is to be perceived; we don't perceive matter; therefore, matter does not exist.
OC. I can only perceive my own mind; therefore, matter does not exist.
OD.
I can only perceive other minds; therefore, matter does not exist.
Reset
Next
Answer:
the answer is B
Explanation:
materialism is just wanting to always buy new things B does the opposite
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor
Complete Question
An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.
Answer:
The torque is [tex]\tau = 34.3 \ N\cdot m[/tex]
Explanation:
From the question we are told that
The mass of the steel ball is [tex]m = 3.0 \ kg[/tex]
The length of arm is [tex]l = 70 \ cm = 0.7 \ m[/tex]
The mass of the arm is [tex]m_a = 4.0 \ kg[/tex]
Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as
[tex]r = \frac{l}{2}[/tex]
=> [tex]r = \frac{ 0.7}{2}[/tex]
=> [tex]r = 0.35 \ m[/tex]
Generally the magnitude of torque about the athlete shoulder is mathematically represented as
[tex]\tau = m_a * g * r + m * g * L[/tex]
=> [tex]\tau = 4 * 9.8 * 0.35 + 3 * 9.8 * 0.70[/tex]
=> [tex]\tau = 34.3 \ N\cdot m[/tex]
Some help me on this please
What can be said for certain about the affect of an external net force on
an object? *
A.It changes the object's mass.
B.It changes the object's kinetic energy.
C.It changes the object's temperature.
D.It changes the object's momentum.
Answer:
Changes in the object's momentum (answer D)
Explanation:
A net force will cause an object to change its velocity, and that will affect the object's momentum, which is defined by the product of the object's mass times its velocity.
So, select the last option (D) in the given list.
Optical astronomers need a clear, dark sky to collect good data. Explain why radio astronomers have no problem observing in the UK where it is often very cloudy.
Answer:
In the clarification portion elsewhere here, the definition of the concern is mentioned.
Explanation:
So like optical telescopes capture light waves, introduce it to concentrate, enhance it, as well as make it usable through different instruments via study, so radio telescopes accumulate weak signal light waves, introduce that one to focus, enhance it, as well as make this information available during research. To research naturally produced radio illumination from stars, galaxies, dark matter, as well as other natural phenomena, we utilize telescopes.
Optical telescopes detect space-borne visible light. There are some drawbacks of optical telescopes mostly on the surface:
Mostly at night would they have been seen. Unless the weather gets cloudy, bad, or gloomy, they shouldn't be seen.Although radio telescopes monitor space-coming radio waves. Those other telescopes, when they are already typically very massive as well as costly, have such an improvement surrounded by optical telescopes. They should be included in poor weather and, when they travel through the surrounding air, the radio waves aren't obscured by clouds. Throughout the afternoon and also some at night, radio telescopes are sometimes used.
In a head-on collision, a ball of mass 0.3 kg travelling with velocity 2.8 m/s in the positive x-direction hits a stationary second ball of mass 0.4 kg. What is the velocity of the 0.3 kg ball after the collision? Assume collision is elastic.
Answer:
The final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.
Explanation:
Given;
mass of the first object, m₁ = 0.3 kg
initial velocity of the first ball, u₁ = 2.8 m/s
mass of the second ball, m₂ = 0.4 kg
initial velocity of the second ball, u₂ = 0
let the final velocity of the first ball, = v₁
let the final velocity of the second ball, = v₂
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(0.3 x 2.8) + (0.4 x 0) = 0.3v₁ + 0.4v₂
0.84 = 0.3v₁ + 0.4v₂
2.8 = v₁ + 1.333v₂ -------equation (1)
Apply one-direction velocity;
u₁ + v₁ = u₂ + v₂
2.8 + v₁ = 0 + v₂
v₂ = 2.8 + v₁
substitute the value of v₂ into equation (1)
2.8 = v₁ + 1.333v₂
2.8 = v₁ + 1.333(2.8 + v₁)
2.8 = v₁ + 3.732 + 1.333v₁
2.8 - 3.732 = v₁ + 1.333v₁
-0.932 = 2.333v₁
v₁ = -0.932 / 2.333
v₁ = -0.4 m/s
Therefore, the final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.
A bicycle starts at 2.5m/s and accelerates along a straight path to a speed of 12.5m/s in a time of 4.5 seconds. What is the bicyclist’s acceleration to the nearest tenth of a m/s^2 ?
Answer:
The bicyclist's acceleration is 2.2m/s^2
Explanation:
Given
[tex]u = 2.5m/s[/tex] ---- Initial Velocity
[tex]v = 12.5m/s[/tex] ---- Final Velocity
[tex]t = 4.5s[/tex] ---- Time
Required
Determine the acceleration
This will be solved using the first equation of motion
[tex]v = u + at[/tex]
Substitute values for v, u and a
[tex]12.5 = 2.5 + a * 4.5[/tex]
[tex]12.5 = 2.5 + 4.5a[/tex]
Collect Like Terms
[tex]4.5a = 12.5 - 2.5[/tex]
[tex]4.5a = 10.0[/tex]
Solve for a
[tex]a = 10.0/4.5[/tex]
[tex]a = 2.2m/s^2[/tex] ---- (approximated)
Hence, the bicyclist's acceleration is 2.2m/s^2
A 50-kg cart is on an incline of 30 degrees above the horizontal. The cart is at rest. The static coefficient of the slope is 0.85 and the kinetic coefficient is 0.50. What is the force due to friction on the cart?
The cart is at rest, so it is in equilibrium and there is no net force acting on it. The only forces acting on the cart are its weight (magnitude w), the normal force (mag. n), and the friction force (maximum mag. f ).
In the horizontal direction, we have
n cos(120º) + f cos(30º) = 0
-1/2 n + √3/2 f = 0
n = √3 f
and in the vertical,
n sin(120º) + f sin(30º) + (-w) = 0
n sin(120º) + f sin(30º) = (50 kg) (9.80 m/s²)
√3/2 n + 1/2 f = 490 N
Substitute n = √3 f and solve for f :
√3/2 (√3 f ) + 1/2 f = 490 N
2 f = 490 N
f = 245 N
(pointed up the incline)
50
1. Force = ? N mass = 65 kg acceleration = 25 m/s2 (Record your answer
by number only and don t.include units...)
Answer: el pepe
Explanation:
Answer:
8125 N.
Explanation:
F = M A
M is mass
A is acceleration
F = 65 X 25
F = 8125 N.
1. An object carries a charge of q = +4x10-8C.
How many electrons are needed to make it elec-
trically neutral?
Answer: i can help you
Explanation:
[tex]2.5[/tex]×[tex]10^{11}[/tex] electrons are needed to make an object neutral which is carrying a charge of [tex]q=+4[/tex]×[tex]10^{-8}[/tex]
What is a charge of an electron?The elementary charge (e) has a negative sign, and the electron's charge is equal to its magnitude. The charge of the electron is -1.602 x [tex]10^{-19}[/tex] coulombs (C), while the elementary charge is approximately 1.602 x [tex]10^{-19}[/tex] coulombs (C).Thus, -e can be used to represent the charge of an electron. The proton only possesses an e-charge despite being far more massive than the electron. The number of protons and electrons in neutral atoms is constant.The electron was unquestionably discovered by JJ Thomson. He only succeeded in obtaining the electron's charge-to-mass ratio, though, despite all the experiments he conducted on it. Robert Millikan holds the distinction of being the first to quantify the charge on an electron through his oil-drop experiment in 1909.To learn more about charge, refer to
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What is the ratio of a 0.7 in circle in centimeters.
hi people see bye 18 45 36
Answer:
bye
Explanation: