Distance = (speed) x (time)
Distance = (70 mi/hr) x (0.5 hr)
Distance = 35 miles
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
What can be described as a technology that uses high-frequency sound waves to create images?
A. ultrasound
B. echolocation
C. analog signal
D. digital signal
Answer:
Ultrasound can be described as a technology that uses high-frequency sound waves to create images.
Explanation:
Ultrasound can be described as a technology that uses high-frequency sound waves to create images.
What is ultrasound?Ultrasound imaging uses sound waves to produce pictures of the inside of the body. It helps diagnose the causes of pain, swelling and infection in the body's internal organs and to examine an unborn child (fetus) in pregnant women.
In infants, doctors commonly use ultrasound to evaluate the brain, hips, and spine. . It also helps guide biopsies, diagnose heart conditions, and assess damage after a heart attack. Ultrasound is safe, non-invasive, and does not use radiation.
This procedure requires little to no special preparation. Your doctor will tell you how to prepare, including whether you should not eat or drink beforehand. Leave jewellery at home and wear loose, comfortable clothing. You may need to change into a gown.
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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
1.) True or False. Scientists have drilled through the crust to reach the mantle.
Answer:
False
Explanation:
We have never made it to the mantle.
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]
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Answer:
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Explanation:
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.
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.