Answer:
yes
Explanation:
becausw yesnssjsdkwww
Suppose a car is traveling at 22.8 m/s, and the driver sees a traffic light turn red. After 0.404 s has elapsed (the reaction time), the driver applies the brakes, and the car decelerates at 9.00 m/s2. What is the stopping distance of the car, as measured from the point where the driver first notices the red light?
Answer:
38.09 m
Explanation:
We'll begin by calculating the distance travelled by the car during the reaction time. This can be obtained as follow:
Reaction time (tᵣ) = 0.404 s
Initial velocity (u) = 22.8 m/s,
Distance travelled during the reaction time (sᵣ) =?
sᵣ = utᵣ
sᵣ = 22.8 × 0.404
sᵣ = 9.21 m
Next, we shall determine the distance travelled by the car when the brake was applied. This can be obtained as follow:
Initial velocity (u) = 22.8 m/s
Acceleration (a) = –9 m/s² (since the car is decelerating)
Final velocity (v) = 0 m/s
Distance travelled when the brake was applied (s₆) =?
v² = u² + 2as₆
0² = 22.8² + (2 × –9 × s₆)
0 = 519.84 – 18s₆
Collect like terms
0 – 519.84 = –18s₆
–519.84 = –18s₆
Divide both side by –18
s₆ = –519.84 / –18
s₆ = 28.88 m
Finally, we shall determine the stopping distance of the car, as measured from the point where the driver first notices the red light. This can be obtained as follow:
Distance travelled during the reaction time (sᵣ) = 9.21 m
Distance travelled when the brake was applied (s₆) = 28.88 m
Stopping distance =?
Stopping distance = sᵣ + s₆
Stopping distance = 9.21 + 28.88
Stopping distance = 38.09 m
An 80 N rightward force is applied to a 10 kg object to accelerate it to the right.
The object encounters a friction force of 50 N.
net force = 30 N
mass = 8.16 kg
acceleration = 3.68 m/s²
Further explanationGiven
80 N force applied
mass of object = 10 kg
Friction force = 50 N
Required
Net force
mass
acceleration
Solution
net forceNet force = force applied(to the right) - friction force(to the left)
Net force = 80 - 50 = 30 N
massGravitational force(downward) : F = mg
m = F : g
m = 80 : 9.8
m = 8.16 kg
accelerationa = F net / m
a = 30 / 8.16
a = 3.68 m/s²
What inspired Ronald McNair to do science
Answer:
While working as a staff physicist at hughes Research Laboratories McNair learned that the National Aeronautics and Space Administration (NASA) was looking for scientist to join the shuttle program;)
On a level test track, a car with antilock brakes and 90% braking efficiency is determined to have a theoretical stopping distance (ignoring aerodynamic resistance) of 408 ft (after the brakes are applied) from 100 mi/h. The car is rear-wheel drive with a 110-inch wheelbase, weighs 3200 lb, and has a 50/50 weight distribution (front and back), a center of gravity that is 22 inches above the road surface, an engine that generates 300 ft-lb of torque, and overall gear reduction of 8.5 to 1 (in first gear), a wheel radius of 15 inches and a driveline efficiency of 95%. What is the maximum acceleration from the rest of this car on this test track
Answer:
a = 30.832 ft/s²
Explanation:
To solve this problem let's start by finding the braking acceleration using kinematics, where the distance is x = 408 ft, the initial velocity vo = 100 mi / h and the final velocity is zero v = 0
v² = v₀² - 2 a x
0 = v₀² - 2ax
a = [tex]\frac{v_o^2}{2x}[/tex]
Let's start by reducing the magnitudes to ft / s
v₀ = 100 mi / h (5280 foot / 1 mile) (1h / 3600 s) = 146.666 ft / s
let's calculate
a = [tex]\frac{146.66^2}{2 \ 408}[/tex]
a = 26.36 ft / s²
Let's call this acceleration a_effective, this acceleration is in the opposite direction to the speed of the vehicle.
Let's use a rule of three (direct proportions) to find the acceleration applied by the brake system (a1) which has an efficiency of 95%. or 0.95
a₁ = [tex]\frac{a_e}{0.95}[/tex]
Let's use another direct proportion rule If the acceleration of the brake system (a₁) for an applied acceleration (a) with an efficiency of 0.90
a = [tex]\frac{a_1}{0.90}[/tex]
we substitute
a = [tex]\frac{a_e}{0.95 \ 0.90}[/tex]
let's calculate
a = [tex]\frac{26.36}{ 0.95 \ 0.90}[/tex]
a = 30.832 ft/s²
This is the maximum relationship that the vehicle can have for when it brakes to stop at the given distance
What is the average speed of a car that travels 60 meters in 2
seconds?
Answer:
30 m/s
Explanation:
Speed is distance over time. 60 meters / 2 seconds, = 30 m/s.
How much force does it take to give a 70 kg object an acceleration of 20 mls2
Answer:
heyy
Explanation:how r uuu
help? its a short question
Answer:
i think its ancestor
Explanation:
sry if im wrong
Answer:
scientists compare organisms DNA to support the theory that all species share a common Ancestor.
Thomas knows that many machines transform electrical energy into other forms of energy
Answer:
Only the car transforms electrical energy into more than one form of energy.
Explanation:
The motion of the car is mechanical energy but it can also transform into another energy witch is electrical energy
The Moon has a smaller mass than the Earth. If you were to travel to the moon your weight would....
A Increase
B Decrease
C Stay the Same
D Vary with day and night
Answer:The Moon has a smaller mass than the Earth. If you were to travel to the moon your weight would..
Explanation: It would decrease.
The Moon has a smaller mass than the Earth. If you were to travel to the moon your weight would decrease because the acceleration due by gravity on the moon is less than the acceleration due to gravity on the earth, therefore the correct answer is option C.
What is gravity?It can be defined as the force by which a body attracts another body towards its center as the result of the gravitational pull of one body and another.
In comparison to the Earth, the Moon is less massive. Your weight would drop if you traveled to the moon because the acceleration caused by gravity there is lower than that caused by gravity here on Earth.
As a result of the less gravity on the moon, the weight would decrease.
Thus, the correct answer is option C.
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There is a very long straw of charge that is uniformly charged in electro static equilibrium. It has a charge per unit length of 4.0E-9 C/m (4.0 nC/m) and a radius of 0.5 m. What is the strength of the electric field a distance of 10.0 m from its center outside the straw
Answer:
2880 N/c
Explanation:
Given that:
Charge per unit length ; λ = 4 * 10^-9
radius, r = 10
Radius, R = 0.5m
Using the relation :
2λr / 4πE0R²
Columb's constant, k = 1/4πE0 =. 9* 10^9Nm²/C²
Hence, we have :
2λrk/ R²
(2 * 4 * 10^-9 * 10 * 9 * 10^9) / 0.5^2
(720 ÷ 0.25)
= 2880 N/c
If a 500-pound object is moved 200 feet how much work is being done?
a. 200 FT LB
b. 500 FT LB
c. 1000 FT LB
d. 100,000 FT LB
Answer:
D
Explanation:
Work = Distance x Mass
work done = 100,000 FT LB
What is work done ?
Work is done whenever a force moves something over a distance or The work done by a force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement.
Work done = force * displacement
given :
force = 500 pound
displacement = 200 feet
work done = 500 * 200 = 100,000 FT LB
correct option is d. 100,000 FT LB
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When the material in the mantle cools off near the surface then sinks
down towards the core and get heated again and rises back towards the
surface it is called?
Condensation
О
The water cycle
O
Convection currents
О
Apples and Bananas.
PLEASE HELP THIS IS URGENT ITS FOR A TEST
Answer:
convection currents
Explanation:
Did I hear correctly that the speed of light is different in deep space observation?
Answer:
Astronomers can learn about the elements in stars and galaxies by decoding the information in their spectral lines. There is a complicating factor in learning how to decode the message of starlight, however. If a star is moving toward or away from us, its lines will be in a slightly different place in the spectrum from where they would be in a star at rest. And most objects in the universe do have some motion relative to the Sun.
A warm hockey puck has a coefficient of restitution of 0.50, while a frozen hockey puck has a coefficient of restitution of only 0.35. In the NHL, the pucks to be used in games are kept frozen. During a game, the referee retrieves a puck from the cooler to restart play but is told by the equipment manager that several warm pucks were just put into the cooler. To check to make sure he has a game-ready puck, the referee drops the puck on its side from a height of 2 m. How high should the puck bounce if it is a frozen puck
Answer:
the required height is 0.2449 m only
Explanation:
Given the data in the question;
Initial height = 2m
so speed of the puck before hitting the ground will be;
u² = 2gh
Initial speed u_ball = √2gh
u_ball = √( 2 × 9.8 × 2 )
u_ball = √39.2
u_ball = 6.26 m/s
given that; FOR THE FROZEN PUCK, coefficient of restitution = 0.35 only
R = - (v_ball - v_ground / u_ball - u_ ground)
so
0.35 = - (v_ball - 0 / 6.26 - 0)
0.35 = -v_ball / - 6.26
-v_ball = 0.35 × (- 6.26)
-v_ball = -2.191 m/s
v_ball = 2.191 m/s
to get the height;
v² = 2gh
h = v² / 2g
we substitute
h = (2.191)² / 2×9.8
h = 4.800481 / 19.6
h = 0.2449 m
Therefore, the required height is 0.2449 m only
Which subatomic particle is NOT found in the nucleus of an atom? *
protons
neutrons
electrons
Answer:
Electrons
Explanation:
Only Protons and Neutrons are found in the nucleus
A clump of soft clay is thrown horizontally from 9.80 m above the ground with a speed of 20.0 m/s. Assume it sticks in place when it hits the ground At what time will the clay hit the ground
Answer:
Explanation:
The time to hit the ground will be same as time taken to fall from the height of 9.8 m with initial vertical velocity of zero .
Considering vertical displacement
initial velocity u = 0
displacement s = 9.8 m
acceleration a = g = 9.8 m /s²
time t = ?
s = ut + 1/2 g t²
9.8 = 0 + .5 x 9.8 x t²
t² = 2
t = √2 = 1.4 s
Peter, a 100 kg basketball player, lands on his feet after completing a slam dunk and then immediately jumps up again to celebrate his basket. When his feet first touch the floor after the dunk, his velocity is 5 m/s downward; when his feet leave the floor 0.50 s later, as he jumps back up, his velocity is 4 m/s upward. a. What is the impulse exerted on Peter during this 0.50 s
Answer:
Explanation:
Impulse = change in momentum
Initial momentum = mass x initial velocity = 100 x 5 = 500 kg m/s
final momentum = mass x final velocity = 100 x - 4 = -400 ( - ve sign due to reversal of direction )
change in momentum = final momentum - initial momentum
= - 400 - 500 = - 900 kg m/s .
As it is - ve , it acts upwards .
So magnitude of impulse on Perter = 900 kg m/s
Orion, also called the Hunter, has three stars that make up Orion's belt.
Which star is at the tip of the arrow? PLEASE HELP I NEED THIS FAST
A. Sirius
B. Betelgeuse
C. Rigel
D. Polaris
Answer - B. Betelguese.
I really hope this helps!!
a. Use the graph and the element made in question 2 to determine the mass of the star.
If an ocean wave has a wavelength of 2 m and a frequency of 1 wave/s, then its speed is m/s Enter the answer Check it CRATCHPAD Improve this questic 트
Answer:
2m/s
Explanation:
v=f×wavelength
v=2×1
=2m/s
I have a massive rock weighing 3,000 Newtons but I can only accelerate it to 500 m/s2 what is its mass?
Answer:
6 kg
Explanation:
F=ma
F is Force(newtons)
m is mass(kg)
a is acceleration(m/s^2)
Plug in the numbers
3000 = m(500)
divide both sides by 500 to cancel out the 500.
3000/500=6
6 = m
6kg is the mass
You are inside the Great Hall, 15 m from the north wall with the doors to the RMC, and centered between two open doors that are 3 m apart. Someone is blairing a 200 Hz tone outside the Great Hall so that it enters the doors as a plane wave. You hear a maximum intensity in your current position. As you walk along the direction of the wall with the doors (but maintain a distance 15 m from the wall), how far will you walk (in m) to hear a minimum in the sound intensity
Answer:
Δr = 0.425 m
Explanation:
This is a sound interference exercise, the expression for destructive interference is
Δr = (2n + 1) λ / 2
in this case the movement is in the same direction as the sound, therefore the movement is one-dimensional
let's use the relationship between the speed of sound and its frequency and wavelength
v = λ f
λ = v / f
the first minium occurs for n = 0
Δr = λ / 2
Δr = v / 2f
Δr = [tex]\frac{340}{2 \ 400}[/tex]
Δr = 0.425 m
this is the distance from the current position that we assume in the center of the room
a sensor light installed on the edge of a home can detect motion for a distance of 50 feet in front and with a range of motion of 200 degrees. what is the arc length of the area covered
Answer:
4363.3231 feets²
Explanation:
Given that :
Distance, r = 50 ft
θ = 200°
The arc length of area covered :
Arc length = θ/360° * πr²
Arc length = (200/360) * 50 ft ^2 * π
Arc length = 0.5555555 * 2500 * π
Arc length = 4363.3231 feets²
A 200-N force acts on a 10-kg object. The acceleration of the object is
A single, monochromatic indigo light source is shined through an etched, flat prism with a slit separation of .0250mm. The resulting interference pattern is viewed on a screen 1.25m away. The third maximum is found to be 6.6cm from the central maximum. What is the wavelength of the indigo light
Answer:
λ = 440 nm
Explanation:
The phenomenon of constructive interference is described by the expressions
d sin θ= m λ (1)
where d is the separation of the slits d = 0.0250 mm = 2.50 10⁻⁵ m, lam is the wavelength of the incident radiation and m is an integer indicating the order of interference
let's use trigonometry to find the angle
tan θ = y / L
where L is the distance to the screen L = 1.25 m
in general interference experiments angles are very small
tan θ = [tex]\frac{sin \ \theta }{cos \ \theta}[/tex]
ten θ = sin θ
substituting
sin θ = y / L
we substitute 1
d y / L = m λ
λ = [tex]\frac{ d \ y }{m \ L}[/tex]
in the exercise indicate
m = 3
y = 6.6 cm = 6.6 10-2 m
we calculate
λ = 2.50 10⁻⁵ 6.6 10⁻² /( 3 1.25)
λ = 4.4 10⁻⁷ m
let's reduce to nm
λ = 4.4 10⁻⁷ m (10⁹ nm / 1 m)
λ = 440 nm
A wheelbarrow is pushed 10 meters with a force of 75 N for 30 seconds. What is the work done?
How much power is needed?
Which of the following variables can be measured in joules?
A. momentum
B. Energy
C. Power
D. Work
Answer:
The variables that can be measured in joules are
B. Energy
D. Work
Hope it will help :)
20 points!!!! A 2,00ON steel rod that is 5 meters long is placed in a corner between the floor and a wall, and balanced at an angle using a cord attached to the wall The rod is balanced such that its top end is 2.38 meters away from the wall, The cord is 40 cm long, and it is attached to the wall at a height of 75 cm above the floor. The diagram to the right shows the situation If the lower end of the rod does not slip from the corner, what is the tension in the cord?
Answer:
WE NEED TO ADD ALL 40+2.38 +75+5
Explanation:
PLSE GIVE SOME POINTS DUDE
a 90 kilogram dog runs across the dog park at a speed of 6.5 meters per second. what is the magnitude and direction of the average force required to stop the dog in .85 seconds?
Answer:
am not sure about the answer
Explanation:
you need to find out the amount of force it's going in for example 10n or 100n then you need to times it the distance then devide by the time
Analyze the data to identify the mathematical relationship between the
amplitude and energy of a mechanical wave. If mechanical wave A has an
amplitude of 4 cm and mechanical wave B has an amplitude of 5 cm, what
will be the relationship between the energy carried by the two waves?
Amplitude
Energy
A. Wave A has about 1.25 times more energy than wave B.
ОО
B. Wave A has about 1.6 times more energy twan wave B.
C. Wave B has about 1.6 times more energy than wave A.
O D. Wave A has about 1.15 times more energy than wave B.
Answer:
Its C
Explanation:
Because I got it wrong for you
Wave B has about 1.6 times more energy than wave A.
What is energy?
Energy is the ability or capability to do tasks, such as the ability to move an item (of a certain mass) by exerting force. Energy can exist in many different forms, including electrical, mechanical, chemical, thermal, or nuclear, and it can change its form.
The amplitude and energy of a mechanical wave. If mechanical wave A has an amplitude of 4 cm and mechanical wave B has an amplitude of 5 cm wave B has about 1.6 times more energy than wave A.
Wave B has about 1.6 times more energy than wave A.
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