Answer:
i dont know man
Explanation:
i dont know man
1. A train is moving north at 5 m/s on a straight track. The engine is causing it to accelerate northward at 2 m/s^2.
How far will it go before it is moving at 20 m/s?
A) 83
B) 43
C) 39
D) 94
E) 20
Answer:
It will go up to 93.75 m before it is moving at 20 m/s
Explanation:
As we know that
[tex]v^2 - u^2 = 2aS[/tex]
here v is the final speed i.e 20 m/s
u is the initial speed i.e 5 m/s
a is the acceleration due to gravity i.e 2 m/s^2
Substituting the given values in above equation, we get -
[tex]20^2 - 5^2 = 2*2*S\\S = 93.75[/tex]meters
what does loudness of a sound depend on?
Answer:
Amplitude
Explanation:
The loudness of a sound depends on the amplitude of vibration producing the sound
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I hope this helped!+*
Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54×107 light-years from Earth. If the lifetime of a human is taken to be 90.0 years, a spaceship would need to achieve some minimum speed vmin to deliver a living human being to this galaxy. How close to the speed of light would this minimum speed be? Express your answer as the difference between vmin and the speed of light c.
Answer:
[tex]0.0018833\ \text{m/s}[/tex]
Explanation:
[tex]d[/tex] = Distance of Andromeda Galaxy from Earth = [tex]2.54\times 10^7\ \text{ly}[/tex]
[tex]t[/tex] = Time taken = [tex]90\ \text{years}[/tex]
[tex]c[/tex] = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]
We have the relation
[tex]t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}[/tex]
[tex]c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}[/tex]
The required answer is [tex]0.0018833\ \text{m/s}[/tex].
A toy car rolls down a ramp. Which force causes the car to move
Answer:
Gravity
Explanation:
Gravity pulls things down to earth and it is a force
You are locked inside the train car and want to get it moving to draw attention to your plight. There is effectively no friction between the axle and the car, and the train is on horizontal tracks. To try and get the car moving with respect to the ground, you run and slam with all your force against the wall at the front. What happens with the car after you slammed against the wall of the car
Answer:
the car movves briefly as you ran, however, it stops again after you ran in to the wall
Explanation:
Our net total linear momentum was zero at the time both the train and the boy was at rest. As there is no pressure, the train and boy system's linear momentum can be conserved provided no external forces are working on it that might shift its momentum.
If the boy runs inside of the train with a velocity V1 in the forward direction, the train would have a velocity V2 in the reverse direction to V1 to conserve the system's momentum, resulting in Final momentum.
i.e
[tex]m \times V_1 + M \times V_2 = 0[/tex] --- (1)
here;
m = mass of the boy
M = mass of the train
Thus;
[tex]m \times V_1 =- M \times V_2[/tex] --- (2)
As the boy crashes into the train's wall, a pair of equal and opposing force F intervene on both the train and the boy. This force F causes the boy traveling with momentum m× V1 to come to a halt; its momentum remains zero. As the train moves with about the same momentum as the boy as seen in (2) and is subjected to the same force F, its momentum will be diminished to zero, and it will come to a halt.
A 4 kg box is at rest on a table. The static friction coefficient u, between the box and table is 0.30, and
the kinetic friction coefficient Hi is 0.10. Then, a 10 N horizontal force is applied to the box.
Answer:
The box will not move from its position.
Explanation:
First, we will calculate the static frictional force that is stopping the box to move from its position:
[tex]f = \mu R = \mu W=\mu mg[/tex]
where,
f = static frictional force = ?
μ = coefficient of static friction = 0.3
m = mass of box = 4 kg
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]f = (0.3)(4\ kg)(9.81\ m/s^2)\\f=11.77\ N[/tex]
Since the frictional force (11.77 N) is greater than the applied force (10 N).
Therefore, the box will not move from its position.
The classical theory of electromagnetism predicted that the energy of the electrons ejected should have been proportional to the intensity of the light.
a. True
b. False
Answer:
False
Explanation:
No, it is not true that energy of the electrons ejected should have been proportional to the intensity of the light. Perhaps the kinetic energy of the ejected electrons is independent of the intensity of the incident radiation. This is the very fact that classical theory of electromagnetism fails to explain in photoelectric effect. The kinetic energy of the electrons remains constant even if the amplitude of the incident light is increased.
The law of conservation of angular momentum states that if no external force acts on an object, then its angular momentum does not change. true or false
Answer:
the answer is false.
Explanation:
i took the test and it is false trust me!!!!!!!!!
A winch is capable of hauling a ton of bricks vertically two stories (6.35 m ) in 24.5 s .
If the winch’s motor is rated at 5.80 hp , determine its efficiency during raising the load.
Answer: 84 %
Explanation:
Philosophy: The Big Picture Unit 8
How does pragmatism differ from the utilitarianism of the previous era?
A. Utilitarianism did not consider rights and pragmatism did.
B. Pragmatism is concerned with function, utilitarianism with happiness.
C. Utilitarianism was created in England and pragmatism came from Germany.
D. Pragmatism applies to everyone, but utilitarianism is concerned with the upper class.
1.) Calculate the mass of a solid gold rectangular bar that has dimensions lwh = 4.30 cm ✕ 14.0 cm ✕ 27.0 cm. (The density of gold is 19.3 ✕ 103 kg/m3.)
kg
2.)A brass ring of diameter 10.00 cm at 17.3°C is heated and slipped over an aluminum rod of diameter 10.01 cm at 17.3°C. Assume the average coefficients of linear expansion are constant.
(a) To what temperature must the combination be cooled to separate the two metals?
(b) What if the aluminum rod were 10.06 cm in diameter?
Answer:
1) m = 0.3137 kg
2a)T_f = -181.7°C
2b) T_f = -1176.97°C
Explanation:
1) We are given;
Length; l = 4.30 cm = 0.043 m
Width; w = 14.0 cm = 0.014 m
height; h = 27.0 cm = 0.027 m
density of gold; ρ = 19.3 × 10³ kg/m³
Formula for the density is known as;
ρ = mass/volume
Thus;
m =ρV
m = 19.3 × 10³ × (lwh)
m = 19.3 × 10³ × (0.043 × 0.014 × 0.027)
m = 0.3137 kg
2a) We are given;
Diameter of brass; L_br = 10 cm
Diameter of aluminum; L_al = 10.01 cm
Now, to some for change in temperature we will use the formula;
L_f = L_i + αL_i(Δt)
Where α is coefficient of expansion.
Now, for the ring to be removed from the rod, the final diameter of the brass has to be same as the aluminium.
Thus;
L_f(brass) = L_f(aluminium)
From table attached, α_brass ≈ 19 × 10^(-6) /°C
Also, α_aluminium ≈ 24 × 10^(-6) /°C
Thus;
L_f(brass) = 10 + (19 × 10^(-6) × 10 × (Δt))
Similarly,
L_f(aluminium) = 10.01 + (24 × 10^(-6) × 10.01 × (Δt))
Since L_f(brass) = L_f(aluminium), then;
10 + (19 × 10^(-6) × 10 × (Δt)) = 10.01 + (24 × 10^(-6) × 10.01 × (Δt))
Rearranging, we have;
10.01 - 10 = (19 × 10^(-6) × 10 × (Δt)) - (24 × 10^(-6) × 10.01 × (Δt))
0.01 = Δt(-50.24 × 10^(-6))
Δt = 0.01/(-50.24 × 10^(-6))
Δt ≈ -199°C
Thus, temperature at which the combination must be cooled to separate the two metals is;
T_f = T_i + Δt
T_f = 17.3 + (-199)
T_f = -181.7°C
2b) Diameter of aluminum is now;
L_al = 10.06 cm
Thus;
10.06 - 10 = (19 × 10^(-6) × 10 × (Δt)) - (24 × 10^(-6) × 10.01 × (Δt))
0.06 = Δt(-50.24 × 10^(-6))
Δt = 0.06/(-50.24 × 10^(-6))
Δt = -1194.27°C
T_f = 17.3 + (-1194.27)
T_f = -1176.97°C
Describing a Wave
What does a wave carry?
Answer:
Waves carry energy from one place to another.
Explanation:
Because waves carry energy, some waves are used for communication, eg radio and television waves and mobile telephone signals.
What is the period of a wave with a speed of 20.0 m/s and a frequency of 10.0 Hz?
im confused hold on imma send you a link to the answerExplanation:
Light with a wavelength of 700 nm (7×〖10〗^(-7) m) is incident upon a double slit with a separation of 0.30 mm (3 x 10-4 m). A screen is located 1.5 m from the double slit. At what distance from the screen will the first bright fringe beyond the center fringe appear?
Answer:
[tex]0.0035\ \text{m}[/tex]
Explanation:
y = Distance from the center point
d = Separation between slits = 0.3 mm
D = Distance between slit and screen = 1.5 m
[tex]\lambda[/tex] = Wavelength = 700 nm
m = Order = 1
We have the relation
[tex]d\dfrac{y}{D}=m\lambda\\\Rightarrow y=\dfrac{Dm\lambda}{d}\\\Rightarrow y=\dfrac{1.5\times 1\times 700\times 10^{-9}}{0.3\times 10^{-3}}\\\Rightarrow y=0.0035\ \text{m}[/tex]
The distance from the screen at which the first bright fringe beyond the center fringe appear is [tex]0.0035\ \text{m}[/tex].
Which investigation BEST measures the gravitational force on a
toy car?
A. rolling the car down a steep ramp and measuring time
B. using a spring scale and measuring the weight of the car
C. pushing the car and measuring how far it travels before it stops
D. throwing the car in the air and measuring how far it goes before coming
down
Answer:
B : using a spring scale and measuring the weight of the car
Explanation:
A candle is placed 50 cm from a diverging lens with a focal length of 28. What is the image distance in cm.
Answer:
v = -17.94 cm
Explanation:
Given that,
The candle is placed at a distance of 50 cm, u = -50 cm
The focal length of the diverging lens, f = -28 cm (negative in case of a concave lens)
We need to find the image distance. We know that the lens formula is as follows:
[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(-28)}+\dfrac{1}{(-50)}\\\\v=-17.94\ cm[/tex]
So, the image distance is equal to 17.94 cm.
Consider a wheel (solid disk) of radius 1.12 m, mass 10 kg and moment of inertia 1 2 M R2 . The wheel rolls without slipping in a straight line in an uphill direction 37◦ above the horizontal. The wheel starts at angular speed 12.0536 rad/s but the rotation slows down as the wheel rolls uphill, and eventually the wheel comes to a stop and rolls back downhill. How far does the wheel roll in the uphill direction before it stops?
Answer:
d= 23.25 m
Explanation:
Assuming no other external forces acting on the disk, total mechanical energy must be conserved.Taking the initial height of the disk as the zero reference for the gravitational potential energy, initially. all the energy is kinetic.This kinetic energy is part translational kinetic energy, and part rotational kinetic energy, as follows:[tex]E_{o} = K_{transo} + K_{roto} (1)[/tex]
When the disk rolling uphill finally comes to an stop, its energy is completely gravitational potential energy, as follows:[tex]E_{f} = m*g*h (2)[/tex]
Since the angle with the horizontal of the track on which the disk is rolling, is 37º, we can express the height h in terms of the distance traveled d and the angle of 37º, as follows:[tex]h = d* sin 37 (3)[/tex]
Replacing (3) in (2):[tex]E_{f} = m*g* d * sin 37 (4)[/tex]
Since the wheel rolls without sleeping, this means that at any time there is a fixed relationship in the translational speed and the angular speed, as follows:[tex]v = \omega * R (5)[/tex]
For a solid disk, as mentioned in the question, the moment of inertia is just 1/2*M*R².The rotational kinetic energy of a rotating rigid body can be written as follows:[tex]K_{rot} = \frac{1}{2}* I * \omega^{2} (6)[/tex]
Replacing I from (6) and ω from (5), and remembering the definition of the translational kinetic energy, we can solve (1) in terms of v, m and r as follows:[tex]E_{o} = K_{transo} + K_{roto} = \frac{1}{2}* m* v^{2} +(\frac{1}{2}* \frac{1}{2}) *m*r^{2}*(\frac{v}{r}) ^{2} = \\ \frac{3}{4} * m * v^{2} (7)[/tex]
Since (4) and (7) must be equal each other, we can solve for d as follows:[tex]d =\frac{3}{4} * \frac{v^{2}}{g*sin37} = \frac{3}{4}*\frac{(\omega*r)^{2}}{g*sin 37} (8)[/tex]
Replacing by the values, we finally get:[tex]d =\frac{3}{4}*\frac{(\omega*r)^{2}}{g*sin 37} = \frac{3}{4} *\frac{(12.0536rad/sec*1.12m)^{2}}{9.8 m/s2*0.601} = 23. 25 m.[/tex]
Select the correct answer.
Each square dance begins with what?
A. Handshake
B. Dosado
C. Bow or curtsy
D. Promenade
Answer:
C
Explanation:
The men bow to the women and the women curtsy to the men.
(took on test and got it right)
what happens when water at 4° celsius is heated further?
Answer:
please give me brainlist and follow
Explanation:
4 degrees C turns out to be the temperature at which liquid water has the highest density. If you heat it or cool it, it will expand. ... Ice floats on top of lakes, preventing evaporation (and convection in the frozen layer), and lakes stay liquid underneath, allowing fish and other life to survive.
A coil of resistance 100ohm is placed in a magnetic field of 1mWb the coil has 100 turns and a galvanometer of 400ohm resistance is connected in series with it. find the average emf and the current if the coil is moved in one tenth of a second from the given field to a field of 2.0mWb
Answer:
C
Explanation:
C
HURRY IM TIMED
How can you make people feel inspired?
By leading them on an emotional journey through various states to inspiration
By talking about something that interests you
By proving yourself to be a trustworthy speaker
By making them laugh and feel comfortable
Answer:
By talking about something that interesto you’
sorry if wrong
Explanation:
10 POINTS! SPACE QUESTION!!
Answer; they are larger and made of rocky material
Which one of the following statements concerning the magnetic field inside (far from the surface) a long, current-carrying solenoid is true?
1) The magnetic field is zero.
2) The magnetic field is independent of the number of windings.
3) The magnetic field varies as 1/r as measured from the solenoid axis.
4) The magnetic field is independent of the current in the solenoid.
5) The magnetic field is non-zero and nearly uniform.
A running Marites launched the egg she
stole as she was about to be caught with
a velocity of 25 m/s in a direction making
an angle of 20° upward with the
horizontal
a) What is the maximum height reached by
the egg?
b) What is the total flight time (between
launch and touching the ground) of the
egg?
c) What is the horizontal range (maximum
* above ground) of the egg?
d) What is the magnitude of the velocity
of the egg just before it hits the ground?
Answer:
a) y = 3.73 m, b) t = 1.74 s, c) R = 40.99 m,
d) vₓ = 23.49 m/s, v_y = -8.5 m / s
Explanation:
This is a projectile launching exercise, we start by breaking down the initial velocity
sin θ = v_{oy} / v₀
cos θ = v₀ₓ / v₀
v_{oy} = v₀ sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 25 sin 20 = 8.55 m / s
v₀ₓ = 25 cos 20 = 23.49 m / s
a) when the egg reaches the maximum height its vertical speed is zero
v_y² = v_{oy}² - 2 g y
0 = v_[oy}² - 2g y
y = v_{oy}² / 2g
y = [tex]\frac{8.55^2}{2 \ 9.8 }[/tex]
y = 3.73 m
b) flight time
y = v_{oy} t - ½ g t²
the time of flight occurs when the body reaches the ground y = 0
0 = (v_{oy} - ½ g t) t
The results are
t₁ = 0s this time is for using the body star
v_{oy} - ½ g t = 0
t = [tex]\frac{2v_{oy}^2}{g}[/tex]
t = 2 8.55 / 9.8
t = 1.74 s
c) the range
R = v₀² sin 2θ / g
R = 25² sin (2 20) / 9.8
R = 40.99 m
d) speed at the point of arrival
horizontal speed is constant
vₓ = v₀ₓ = 23.49 m/s
vertical speed is
v_y = Iv_{oy} - g t
v_y = 8.55 - 9.8 1.74
v_y = -8.5 m / s
An infant's toy has a 120 g wooden animal hanging from a spring. If pulled down gently, the animal oscillates up and down with a period of 0.54 s . His older sister pulls the spring a bit more than intended. She pulls the animal 32 cm below its equilibrium position, then lets go. The animal flies upward and detaches from the spring right at the animal's equilibrium position. Part A If the animal does not hit anything on the way up, how far above its equilibrium position will it go
Answer:
the wooden animal will go 0.7068 m above its equilibrium
Explanation:
Given the data in the question;
mass of wooden animal m = 120 g = 0.12 kg
the animal oscillates up and down, T = 0.54 s
older sister pulls the animal 32 cm below its equilibrium position;
x = 32 cm = 0.32 m
g = 9.81 m/s²
We know that
k = mω²
where ω = 2π/T
So, k = m( 2π/T )²
we substitute
k = 0.12( 2π / 0.54 )²
k = 0.12 × (11.6355)²
k = 0.12 × 135.38486
k = 16.25 N/c
so Also,
kx²/2 = mgh
we solve for h
h = kx² / 2mg
we substitute
h = ( 16.25 × (0.32)²) / ( 2 × 0.12 × 9.81 )
h = 1.664 / 2.3544
h = 0.7068 m
Therefore, the wooden animal will go 0.7068 m above its equilibrium
A 35.0 g bullet strikes a 50 kg stationary piece of lumber and embeds itself in the wood. The piece of lumber and the bullet fly off together at 8.6 m/s. What was the speed of the ballot before it struck the lumbar? Define the bullet and the wood as a system
Answer:
12294.31 m/s
Explanation:
Momentum = (mass)(velocity)
Momentum before = Momentum after
(momentum of bullet)+(momentum of block)=(momentum of bullet and block)
0.035v+50(0)=(0.035+50)(8.6)
0.035v=430.301
v=12294.3142857m/s
9. Mr. Smith went skiing in Maine last weekend. He traveled 523 kilometers to Sugarloaf from
Leominster. His average speed was 109 km/hr. How long did it take Mr. Smith to hit the slopes?
Answer:
Time taken by Mr. smith = 4.80 hour (Approx.)
Explanation:
Given:
Distance travel by Mr. smith = 523 kilometer
Average speed of Mr. smith = 109 km/hr
Find;
Time taken by Mr. smith
Computation:
Time taken = Distance cover / Speed
Time taken by Mr. smith = Distance travel by Mr. smith / Average speed of Mr.
smith
Time taken by Mr. smith = 523 / 109
Time taken by Mr. smith = 4.798 hr
Time taken by Mr. smith = 4.80 hour (Approx.)
Momentum
Project: Egg Drop
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Answer:
get egg and try to make in not crack when it falls by exerting the momentum of the fall into something other than the egg ex. make a box full of bubble wrap and put your egg in it
Explanation:
What is the chemical formula for the molecule modeled?
Answer:
What is the chemical fórmula For the molecule modeled?
Explanation:
C6H12O2