speed of sound is the final velocity [tex] v=1100[/tex]
and initial velocity $u=600$ , $a=10$
$v=u+at$
$t=\dfrac{v-u}{a}= \dfrac{500}{10}$
$\implies t=50 h$
A glacier advances at 7.4 x 10^-6 cm/s
How far will it move in 4.2 y?
Answer in units of cm.
cm/s.
Answer:
236.8 cm
Explanation:
1 yr = 365 days = (365×24×60×60) secs
= 32 × 10^6
distace = velocity × time
= 7.4 × 10^-6 × 32 × 10^6
= 236.8 cm
The glacier will move 976.8 cm when it advances at 7.4 7.4×10⁻⁶ cm/s in 4.2 y
From the question above when we talk about how far, is still the same as distance.
Applying the equation of velocity,
V = d/t................... Equation 1
Where V = velocity, d = distance, t = time
make d the subject of equation 1
d = Vt................ Equation 2
From the question,
Given: V = 7.4×10⁻⁶ cm/s, t = 4.2 year = (4.2×3.154×10⁷) seconds = 1.32×10⁸ seconds
Substitute these values into equation 2
d = (7.4×10⁻⁶)(1.32×10⁸)
d = 9.768×10² cm
d = 976.8 cm
Hence the glacier will advances 976.8 cm
Learn more about distance here: https://brainly.com/question/22491931
-
Speed is a scalar, a quantity that is
described by
alone.
Answer:
True
Explanation:
Thats is true since it cannot be a vector because speed only gives a magnitude, not a direction as well.
when you turn on music on your phone, what energy transformation takes place ?
Answer: A Battery contains chemical energy which transfers electrical energy into the phone, powering it and also transferring Light energy, Sound energy and Heat energy.
Explanation:
Determine the magnitude of the force that must be applied to a wooden block whose weight has a magnitude of 8N, so that it acquires an acceleration of 0.5 m / s²
Answer:
F = 0.408 N
Explanation:
It is given that,
Weight of a wooden block, W = 8 N
Weight, W = mg
m is mass of wooden block
[tex]m=\dfrac{W}{g}\\\\m=\dfrac{8\ N}{9.8\ m/s^2}\\\\m=0.816\ kg[/tex]
Acceleration of the block, a = 0.5 m/s²
Force, F = ma
[tex]F=0.816\ kg\times 0.5\ m/s^2\\\\F=0.408\ N[/tex]
So, the magnitude of force applied to the wooden block is 0.408 N.
Two mass m1 and m2 lie on a frictionless surface. Between the two masses is a compressed spring, with spring constant k. The system is held in place. At time t=0 the blocks are released. The blocks move off in opposite directions with velocities v1 and v2. how much was the spring compressed?
Answer:
The spring was compressed the following amount:
[tex]\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }[/tex]
Explanation:
Use conservation of energy between initial and final state, considering that the surface id frictionless, and there is no loss in thermal energy due to friction. the total initial energy is the potential energy of the compressed spring (by an amount [tex]\Delta x[/tex]), and the total final energy is the addition of the kinetic energies of both masses:
[tex]E_i=\frac{1}{2} k\,(\Delta x)^2\\\\E_f=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2[/tex]
[tex]E_i=E_f\\[/tex]
[tex]\frac{1}{2} k\,(\Delta x)^2=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2\\k\,(\Delta x)^2=m_1\,v_1^2+ m_2\,v_2^2\\(\Delta x)^2=\frac{m_1\,v_1^2+ m_2\,v_2^2}{k} \\\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }[/tex]
An object accelerates to a velocity of 230 m/s over a time of 2.5 s. The acceleration it experienced was 42 m/s2. What was its initial velocity?
Answer:
230 = x + 105
x= 125
Explanation:
v = v0 + at
White light is spread out into its spectral components by a diffraction grating. If the grating has 1,970 grooves per centimeter, at what angle (in degrees) does red light of wavelength 640 nm appear in first order? (Assume that the light is incident normally on the gratings.)
Answer;
7.238°
Explanation
From question we know that the grating has 1,970 grooves per centimeter, we can convert to from (cm) to (nm) for unit consistency
The slit separation is expressed below
d=1cm/1970
d=0.0005076
=5076nm
Then the angle (in degrees) that the red light of wavelength 640 nm appear in first order can be calculated using expression below
Sin(x)= mλ/d
Where λ= wavelength=640 nm
d=slit separation
Sin(x)= mλ/d
Sin(x)=(1×640)/5076
=0.126
sin-1(0.126)
X= 7.238°
Therefore,the angle (in degrees) that the red light of wavelength 640 nm appear in first order is 7.238°
How is heat transferred during thermal conduction? Through direct contact of particles Through electromagnetic waves Through motion of hot and cold liquid Through the rising of a warm gas
Answer:
Through direct contact of particlesExplanation:
Heat is transferred in the following ways listed below
1. Conduction(Through direct contact of particles )
2. Convection(Through motion of hot and cold liquid )
3. Radiation(Through electromagnetic waves Through the rising of a warm gas )
1. Conduction: This has to do with the transfer of heat from one medium to another having direct contact with each other, the medium can either be solid or liquids
2. Convection: Has to do with the transfer of heat energy via molecules of the liquids in motion with different or varying temperature gradients
3. Radiation: Mostly occurs in gases, it is the transfer of energy through waves
d. Write the symbol for the nucleus that completes each nuclear equation. (1 point each)
Answer:
Ra, 88
Ru, 44
Cr, 24
Explanation:
i
the remaining mass number is 224(228 - 4), and it's atomic number is 88(90 - 2). The element is Radium, Ra
ii
the remaining mass number is 98(98 - 0), and it's atomic number is 44(45 - 1). The element is Ruthenium, Ru
iii
The remaining mass number is 52(52 - 0), and it's atomic number is 24(23 - -1). The element is Chromium, Cr
How large is theta? Please help!!!!
Answer:
30°
Explanation:
The angle between the normal and the vertical is the same as the angle between the incline and the horizontal. We can show this with geometry.
A ball with a mass of 0.5 kg is attached to one end of a light rod that is 0.5 m long. The other end of the rod is loosely pinned at a frictionless pivot. The rod is raised until it is vertical, with the ball above the pivot. The rod is released and the ball moves in a vertical circle. The tension in the rod as the ball moves through the bottom of the circle is closest to:
Answer:
The tension in the rod as the ball moves through the bottom circle is 9.8 N
Explanation:
When the ball is released from rest, the centripetal force equals the weight of the ball. So mv²/r = mg where m = mass of ball = 0.5 kg, v = speed of ball, r = radius of vertical circle = length of rod = 0.5 m and g = acceleration due to gravity = 9.8 m/s²
v = √gr = √9.8 m/s² × 0.5 m = √4.9 = 2.21 m/s
Now at the bottom of the circle T - mg = mv²/r where T = tension in the rod
T = m(g + v²/r)
= m(g + (√gr)²/r)
= m(g+ gr/r)
= m(g + g)
= 2mg
= 2 × 0.5 kg × 9.8 m/s²
= 9.8 N
So, the tension in the rod as the ball moves through the bottom circle is 9.8 N
within a conventional light bulb, which of the following would not be considered a state of matter
a. the glass shell of the bulb
b. the metal base of the bulb
c. the air inside the bulb
d. the light from the bulb
Answer:
D
Explanation:
because light has no space nor mass
Se suelta un cuerpo desde la azotea de un edificio de 180m de alto. Luego de 3s se lanza verticalmente hacia abajo otro cuerpo. Si ambos llegan simultáneamente al piso, calcular la rapidez con la que fue lanzado el segundo cuerpo (g=10m/s2).
Answer:CcCcccccc
Explanation:
Rfsgqh
Matter is anything that has mass and takes up
space.
Which of the following is an example of
matter?
A. ear phones
B. music
C. sunlight
D. heat
Answer: ear phones
Explanation:
You can physically hold ear phones, but you can't hold music, sunlight, or heat.
Mr. Franklin uses a machine that has an efficiency of 80%. The machine applies a force of 800 N through a distance of 20 m. Mr. Franklin applies a force of 50 N on the machine. Through what distance does Mr. Franklin apply the force? 4 m 16 m 260 m 400 m
Answer:
The distance Mr Franklin applied the force is 400 m
Explanation:
The efficiency of the machine = 80%
The force applied by the machine = 800 N
The distance over which the force is applied by the machine = 20 m
The force applied by Mr. Franklin = 50 N
The work done by the machine = 800 × 20 = 16000 J
The ideal work at 100% efficiency = X
80% of X = 16000 J
X = 16000 N/0.8 = 20,000 J
Therefore, we have have;
Work = Force × Distance;
20,000 J = 50 N × Distance Mr Franklin applied the force
Distance Mr Franklin applied the force = 20,000 J/(50 N) = 400 m
The distance Mr Franklin applied the force = 400 m.
Answer:
the awnser in 400m
Explanation:
did the test on edge 2020 and got it right
what is acceleration produced by a force of 12 newton exerted on an object of mass 3kg
Answer:
a=F/m
a=12N/3kg (here newton can be written as kgm/s^2 so kg will be cancelled)
a=4m/s^2
Explanation:
What is the approximate pressure of a storage cylinder of recovered R-404A that does not contain any non-condensable impurities and is stored in a room where the temperature is 80°F?A. 68 psigB. 72 psigC. 174psigD. 288 psig
Answer:
D. 288 psig
Explanation:
The storage cylinder of recovered R-404A is a mixture of R-125, R-143 and R-134. The refrigeration sector has low temperatures for installation. The pressure of cylinder at 80 F will be 288 psig. The pure refrigerants have inside a container have saturation temperature which is equal to ambient temperature.
Each corner of a right-angled triangle is occupied by identical point charges "A", "B", and "C" respectively. Draw a sketch of this arrangement. "A" exerts force F on "B". An equal force F is exerted by "C" on "B" (/_ ABC= 90 degrees). Determine an expression for the net force on "B".
Answer:
Fnet = F√2
Fnet = kq²/r² √2
Explanation:
A exerts a force F on B, and C exerts an equal force F on B perpendicular to that. The net force can be found with Pythagorean theorem:
Fnet = √(F² + F²)
Fnet = F√2
The force between two charges particles is:
F = k q₁ q₂ / r²
where
k is Coulomb's constant, q₁ and q₂ are the charges, and r is the distance between the charges.
If we say the charge of each particle is q, then:
F = kq²/r²
Substituting:
Fnet = kq²/r² √2
A student uses a spring scale attached to a textbook to compare the static and kinetic friction between the textbook and the top of a lab table. If the scale measures 1,580 g while the student is pulling the sliding book along the table, which reading on the scale could have been possible at the moment the student overcame the static friction?
Answer:1860g
Explanation:The force would have to be greater than 1580g, which is when the student is pulling the sliding book with the spring scale
I need help. can I get some
Explanation:
a) Acceleration is the slope of the velocity vs time graph. When velocity vs time is a straight line, that means the acceleration is a constant. F = ma, so the resultant force is also constant.
b) As air resistance increases, the acceleration begins to decrease.
c) A falling object stops accelerating when it reaches terminal velocity.
d) F = ma, so when the acceleration is 0, the resultant force is also 0.
If you have a 200g sample of a radioactive isotope that has a half life of 50 years, hc
much will you have left at the end of 200 years?
a. 12.5 g
b. 100 g
c. 50 g
d. 25 g
Answer:
The correct option is;
a. 12.5 g
Explanation:
The given parameters are;
The initial mass of the radioactive substance = 200 g
The half life of the radioactive substance = 50 years
The time duration for the disintegration = 200 years
The formula for half life is given as follows;
[tex]N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}}[/tex]
Where:
N(t) = Quantity of the remaining substance
N₀ = Initial radioactive substance quantity = 200 g
t = Time duration = 200 years
[tex]t_{1/2}[/tex] = Half life of the radioactive substance 50 years
Therefore, we have;
[tex]N(t) =200 \times \left (\dfrac{1}{2} \right )^{\dfrac{200}{50} } =200 \times \left (\dfrac{1}{2} \right )^4 = 12.5 \ g[/tex]
Therefore, at the end of 200 years, the quantity left = 12.5 g.
Guys, can you please answer these three questions. I'm having trouble doing them.
Answer:
17) a) 30 kg·m/s
b) 30 kg·m/s
c) 0.3 m/s.
18) 2.4 m/s
19) 360,000 N
The different make of car is safer because the force acting on it is 28,800 N, resulting much less impact
Explanation:
17) By conservation of linear momentum principle, we have;
m₁·v₁ + m₂·v₂ = m₁·v₁' + m₂·v₂'
Where;
m₁ = The mass of the astronaut = 100 kg
v₁ = The initial velocity of the astronaut = 0 m/s
m₂ = The mass of the equipment box = 2.5 kg
v₂ = The initial velocity of the equipment box = 0 m/s
v₁' = The final velocity of the astronaut
v₂' = The final velocity of the equipment box = 12 m/s
The final momentum of the equipment box = Mass of the equipment box × Final velocity
a) The final momentum of the equipment box = 2.5 kg × 12 m/s = 30 kg·m/s
b) Given that the box and the astronaut will move in opposite directions, we have;
m₁·0 + m₂·0 = -m₁·v₁' + m₂·v₂'
m₁·v₁' = m₂·v₂'
Therefore, the momentum of the astronaut will be equal to the momentum of the equipment box = 30 kg·m/s
c) The velocity of the astronaut is given by th following relation;
100×0 + 2.5×0 = -100×v₁' + 2.5×12
100×v₁' = 2.5×12
v₁' = 2.5×12/100 = 30/100 = 0.3 m/s
The final velocity of the astronaut = 0.3 m/s.
18) By conservation of linear momentum principle, we have;
m₁·v₁ + m₂·v₂ = m₁·v₁' + m₂·v₂'
Where;
m₁ = The mass of the railway truck= 4000 kg
v₁ = The initial velocity of the railway truck= 8 m/s
m₂ = The mass of the stationary truck = 6000 kg
v₂ = The initial velocity of the stationary truck = 0 m/s
v₁' = The velocity of the two trucks immediately after collision
Therefore, we have;
4000 kg × 6 m/s + 6000 × 0 = (6000 + 4000) × v₁' = 10,000·v₁'
24,000 kg·m/s = 10,000·v₁'
v₁' = 24,000/10,000 = 2.4 m/s
The velocity of the two trucks immediately after collision is = 2.4 m/s
19) The change in kinetic energy of the car due to the collision = 1/2·m·(v₂² - v₁²)
The initial velocity of the car, v₁ = 12 m/s
The final velocity, v₂ = 0 m/s
The mass of the car, m= 600 kg
The time in which the car is brought to rest = 2 seconds
Initial time at point of collision, t₁ = 0 s
Time after collision, t₂ = 0.02 s
By Newton's second law of motion, we have
The force F acting on the car = Rate of change in momentum produced by the force
F = m × dv/dt
Force = m × (v₂ - v₁)/(t₂ - t₁) = 600×(0-12)/(0.02 - 0) = -360,000 kg·m/s²
The force acting on the car is equal and opposite to the force of the car = 360,000 N
b) For the different make of the car, we have;
m × (v₂ - v₁)/(t₂ - t₁) = 600×(0-12)/(0.25 - 0) = -28,800 kg·m/s². = -28,800 N
Therefore, the different make of car is safer because the force acting on it is 28,800 N, resulting much less impact.
WILLL GIVE 5 STARS BRAINIEST AND THANKS AND 20 POINTS EACH ANSWER!!!!
A diver performs a dive from a 10 m-platform. She
jumps vertically upwards at 3 m s initially.
Find the time she takes to reach the water surface
Answer:
The time she takes to reach the water from when she jumps off the platform is 1.71 s
Explanation:
According to the equations of motion, we have;
v = u - g·t
v² = u² - 2·g·s
s₁ = u₁·t₁ + 1/2·g₁·t₁²
The given parameters are;
The height of the platform (assumption: above the water) = 10 m
The velocity with which she jumps, u = 3 m/s
The acceleration due to gravity, g = 9.81 m/s²
The height to which she jumps, s, is found as follows;
v² = u² - 2·g·s
At maximum height, v = 0, which gives;
0 = 3² - 2×9.81×s
2×9.81×s = 3² = 9
s = 9/(2×9.81) = 0.4587 m
s = 0.4587 m
The time to maximum height, t, is found as follows;
v = u - g·t
0 = 3 - 9.81×t
9.81×t = 3
t = 3/9.81 = 0.3058 s
The total distance, s₁ from maximum height to the water surface = s + 10 = 0.4587 + 10 = 10.4587 m = 10.46 m
The time to reach the water from maximum height, t₁, is found as follows;
s₁ = u₁·t₁ + 1/2·g₁·t₁²
Where;
s₁ = The total distance from maximum height to the water surface = 10.46 m
u₁ = The initial velocity, this time from the maximum height = 0 m/s
g₁ = The acceleration due to gravity, g (positive this time as the diver is accelerating down) = 9.81 m/s²
t₁ = The time to reach the water from maximum height
Substituting the values gives;
s₁ = u₁·t₁ + 1/2·g₁·t₁²
10.46 = 0·t₁ + 1/2·9.81·t₁²
t₁²= 10.46/(1/2×9.81) = 2.13 s²
t₁ = √2.13 = 1.46 s
Total time = t₁ + t = 1.46 + 0.3058 = 1.7066 ≈ 1.71 s.
Therefore, the time she takes to reach the water from when she jumps off the platform = 1.71 s.
An average skater averages 11 m/s over the first 5 seconds of a race. find the average speed required over next 10 seconds to average 12 m/s overall.
Answer:
usa
Explanation:
A round steel plate with a diameter of 20 cm is supplied with heat in such a way that its temperature changes from 22 ° C to 150 ° C. How much does its area increase?
Answer:
1.05×10⁻⁴ m²
Explanation:
ΔA = 2 A₀ α ΔT,
where ΔA is the change in area,
A₀ is the initial area,
α is the linear expansion coefficient,
and ΔT is the change in temperature.
A₀ = π (0.20 m / 2)² = 0.0314 m².
For steel, α = 13×10⁻⁶ /°C.
ΔT = 150°C − 22°C = 128°C.
ΔA = 2 (0.0314 m²) (13×10⁻⁶ /°C) (128°C)
ΔA = 1.05×10⁻⁴ m²
Which is
easier
Push
or pull.
when force is at angle
Answer:
pull
Explanation:
because when you push there is one component of force that adds to tge weight of body and hence there is more friction. Whwn you pull the vertical component of force is against the weight of body and hence friction is low.
What is the atomic number of hydrogen? A. 26 B. 14 C.35 D. 1
Answer:
1
Explanation:
Hydrogen is the first element in the periodic table so its atomic number is 1.
If forces acting on an object are unbalanced, the object could experience a change in ________.
Answer:
velocity
Explanation:
Unbalanced forces on an object cause it to accelerate. Acceleration is a change in velocity over time.
Answer:
a
Explanation:
none
Why is it important to wear loose-fitting clothing when exercising?
It isn't best to wear loose fitting clothing when exercising, as you may not feel too comfortable. It depends on the person. However there are a few advantages.
• Tight clothing may not be comfortable as well
• Looser clothing allows easy evaporation of sweat
Then again when it says " loose - fitting " it may mean not too baggy. Baggy clothing can have an impact on your physical activity.
Answer All of the above
For people on a p e x
How use of SI units in daily life is important .
Answer:
SI units are the physical units of measurement. SI units include kilogram, meter, second, ampere, kelvin, candela, and mole etcetera.
SI units are very important for daily life while measuring daily life commodities, such as:
Use of kilograms and grams while buying or selling sugar or other food items.Use of kilometers and meters while covering distances.Use of meters or inch by carpenters while estimating the lengths for furniture.Use of hours, seconds and minutes to measure time duration.Use of ampere and volt to measure current and voltage respectively by electricians.Without SI units, it would be difficult to meausre the amount of object required or any other measurement object.
