Answer: the higher the kinetic energy
Explanation:
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
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
Iron has a specific heat of o.45 J/g °C. Removing -1.16 E 2 J of energy lowered the temperature of iron from 89 °C to 26.41 °C. What was the mass of the iron?
Answer:
m = 4.11 grams
Explanation:
Given that,
The specific heat of Iron, c = 0.45 J/g°C
Heat removed, [tex]Q=1.16\times 10^2\ J[/tex]
Initial temperature, [tex]T_i=89^{\circ} C[/tex]
Final temperature, [tex]T_f=26.41^{\circ} C[/tex]
We need to find the mass of the iron. We know that the heat removed in terms of specific heat is given by :
[tex]Q=mc\Delta T\\\\m=\dfrac{Q}{c\Delta T}\\\\m=\dfrac{-1.16\times 10^2}{0.45\times (26.41-89)}\\\\m=4.11\ g[/tex]
So, the mass of the iron is 4.11 grams.
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
You are on the Pirates of the Caribbean attraction in the Magic Kingdom at Disney World. Your boat rides through a pirate battle, in which cannons on a ship and in a fort are firing at each other. While you are aware that the splashes in the water do not represent actual cannonballs, you begin to wonder about such battles in the days of the pirates. Sup-pose the fort and the ship are separated by 75.0 m. You see that the cannons in the fort are aimed so that their cannon-balls would be fired horizontally from a height of 7.00 m above the water.
(a) You wonder at what speed they must be fired in order to hit the ship before falling in the water.
(b) Then, you think about the sludge that must build up inside the barrel of a cannon. This sludge should slow down the cannonballs. A question occurs in your mind: if the can-nonballs can be fired at only 50.0% of the speed found ear-lier, is it possible to fire them upward at some angle to the horizontal so that they would reach the ship?
Answer:
a) v₀ₓ = 62.76 m / s, b) θ₁ = 17.6º, θ₂ = 67.0º
Explanation:
We can solve this exercise using the projectile launch ratios
a) Let's find the time it takes for the bullet to reach the water level
y = y₀ + v_{oy} t - ½ g t²
when it reaches the water its height is zero y = 0, as the bullet is fired horizontally its initial vertical velocity is zero
0 = y₀ + 0 - ½ g t²
t =[tex]\sqrt{2y_o/g}[/tex]
t = [tex]\sqrt{2 \ 7 /9.8}[/tex]
t = 1,195 s
now we can calculate the speed with the horizontal movement
x = v₀ₓ t
v₀ₓ = x / t
v₀ₓ = 75.0 / 1.195
v₀ₓ = 62.76 m / s
b) if the speed of the bullets is half of that found
v₀ = 62.76 / 2 = 31.38 m / s
let's write the expressions for the distance
x = v₀ cos θ t
y = y₀ + v_{oy} sin θ t - ½ g t²
t = [tex]\frac{x}{v_o \ cos \theta}[/tex]
we substitute
[tex]0 = y_o + v_o sin \theta \ \frac{x}{v_o \cos \thetay} - 1/2 g \ (\frac{x}{v_o \ cos \theta})^2[/tex]
[tex]0 = y_o + x tan \theta - \frac{1}{2} g \ \frac{x^2}{ v_o^2 \ cos^2 \theta}[/tex]
let's use the identified trigonometry
sec² θ = 1 + tan² θ
sec θ = 1 / cos θ
we substitute
[tex]0 = y_o + x tan \theta - \frac{g x^2}{2 v_o^2} ( 1 + tan^2 \theta)[/tex]
[tex]\frac{g x^2}{2v_o^2} tan^2 \theta - x tan \theta + \frac{gx^2}{2v_o^2} - y_o = 0[/tex]
we change variable
tan θ = H
[tex]\frac{gx^2}{2 v_o^2 } H^2 - x H + \frac{gx^2}{2v_o^2}-y_o =0[/tex]
we subtitle the values
[tex]\frac{9.8 \ 75^2}{2 \ 31.38^2} H^2 - 75 H + \frac{9.8 \ 75^2}{2 \ 31.38^2}-7 =0[/tex]
27.99 H² - 75 H + 20.99 = 0
H² - 2.679 H + 0.75 = 0
we solve the quadratic equation
H = [2.679 ± [tex]\sqrt{2.679^2 - 4 0.75}[/tex]] / 2
H = [2,679 ± 2,044] / 2
H₁ = 0.3175
H₂ = 2.3615
now we can find the angles
H₁ = tan θ₁
θ₁ = tan⁻¹ H₁
θ₁ = tan⁻¹ 0.3175
θ₁ = 17.6º
θ₂ = 67.0º
for these two angles the bullet hits the boat
How much force does it take to give a 70 kg object an acceleration of 20 mls2
Answer:
heyy
Explanation:how r uuu
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
what are the precautions to be taken while performing a rectangular glass prism experiment
Answer:
PRECAUTIONS
-The refracting faces of the glass prism should be smooth, transparent and without any air bubble or broken edge. ...
-Use a sharp pencil to draw boundary of the prism and rays of light.
-The alpins should have sharp tip and should be fixed exactly vertical to the plane of the paper.
Explanation:
Please give thanks to all my answers and please mark as brilliant and please follow me
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
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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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 :)
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
What is the symbol for carbonate ?
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
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
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
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.
Help please ! Ill give brainliest !! ☁️✨
6th grade science I mark as brainliest !
Answer:
first is Atoms
4) is True
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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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 point charge q1 = 2.40 uC is held stationary at the origin. A second point charge q2 = -4.30uC moves from the point x= .150 m, y= 0.0 m, to the point x = .250 m, y= 0.0m
a) what is the charge in potential energy of the pair of charges?
b) How much work is done by the electric force on q2
Answer:150M
Explanation:
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.
A 3" diameter germanium wafer that is 0.020" thick at 300K has 1.015 x 10^17 As atoms added to it. What is the resistivity of the wafer? Germanium has 4.42 x10^22 atoms/cc, electron and hole mobilities are 3900 and 1900 cm^2/(V*s). What is the resistivity of the Ge in ohm*microns?
Answer:
0.546 ohm / μm
Explanation:
Given that :
N = 1.015 * 10^17
Electron mobility, u = 3900
Hole mobility, h = 1900
Ng = 4.42 x10^22
q = 1.6*10^-19
Resistivity = 1/qNu
Resistivsity (R) = 1/(1.6*10^-19 * 1.015 * 10^17 * 3900)
= 0.01578880889 ohm /cm
Resistivity of germanium :
R = 1 / 2q * sqrt(Ng) * sqrt(u*h)
R = 1 / 2 * 1.6*10^-19 * sqrt(4.42 x10^22) * sqrt(3900*1900)
R = 1 /0.0001831
R = 5461.4964 ohm /cm
5461.4964 / 10000
0.546 ohm / μm
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
a. Use the graph and the element made in question 2 to determine the mass of the star.
A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is falling vertically at a constant velocity of 7.0 m/s downward with their parachute open (terminal velocity). A horizontal gust of wind helps push them towards the target. What must be their total speed if they want to just hit their target?
Answer:
Answer:
15.67 seconds
Explanation:
Using first equation of Motion
Final Velocity= Initial Velocity + (Acceleration * Time)
v= u + at
v=3
u=50
a= - 4 (negative acceleration or deceleration)
3= 50 +( -4 * t)
-47/-4 = t
Time = 15.67 seconds
We have that the speed must be at the speed below if they want to just hit their target
From the Question we are told that
Distance [tex]d=50m[/tex]
Height [tex]h=70m[/tex]
Constant Velocity [tex]v= 7.0 m/s[/tex]
Generally the equation for the time is mathematically given as
[tex]T=\frac{h}{v}\\\\T=\frac{70}{7}\\\\T=10s[/tex]
Therefore
The velocity required to make horizontal movement is
[tex]V=\frac{d}{T}\\\\V=\frac{50}{10}\\\\V=5m/s[/tex]
Given that
Velocity on the Vertical axis is
[tex]v_y=7m/s[/tex]
Velocity on the horizontal axis is
[tex]v_x=5m/s[/tex]
Therefore resultant speed
[tex]v_r=\sqrt{v_x^2+V_y^2}\\\\v_r=\sqrt{(5)^2+(7)^2}[/tex]
[tex]v_r=8.6023m/s[/tex]
In conclusion
[tex]v_r=8.6023m/s[/tex] must be their total speed if they want to just hit their target
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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!!
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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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.