Answer:
3.94 s
Explanation:
Take down to be positive. Given:
Δy = 76.0 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
76.0 m = (0 m/s) t + ½ (9.8 m/s²) t²
t = 3.94 s
A radio station can be heard if the receiver is tuned to a frequency of 6x10^5 Hz. What is the wavelength of the radio waves
5 * 10² m
Explanation:We are Given:
Frequency of the wave = 6 * 10⁵ Hz
Wavelength of the wave:
We know the relation:
c = λν
[where λ (Lambda) is the wavelength ,ν (nu) is the frequency and c is the speed of light ]
3 * 10⁸ m/s= (6 * 10⁵ )* λ [replacing known values]
λ = [tex]\frac{3 * 10^{8}}{6 * 10^{5}}[/tex] [dividing both sides by 6 * 10⁵]
λ = 1/2 * 10³
λ = 1/2 * 10 * 10² [10³ can be rewritten as 10 * 10²]
λ = 5 * 10² m
Therefore, the wavelength of the wave is 5 * 10² m
A radio wave transmits 38.5 W/m2 of power per unit area. A flat surface of area A is perpendicular to the direction of propagation of the wave. Assuming the surface is a perfect absorber, calculate the radiation pressure on it.
Answer:
[tex]P=2.57\times 10^{-7}\ N/m^2[/tex]
Explanation:
Given that,
A radio wave transmits 38.5 W/m² of power per unit area.
A flat surface of area A is perpendicular to the direction of propagation of the wave.
We need to find the radiation pressure on it. It is given by the formula as follows :
[tex]P=\dfrac{2I}{c}[/tex]
Where
c is speed of light
Putting all the values, we get :
[tex]P=\dfrac{2\times 38.5}{3\times 10^8}\\\\=2.57\times 10^{-7}\ N/m^2[/tex]
So, the radiation pressure is [tex]2.57\times 10^{-7}\ N/m^2[/tex].
Sandra pays $11 for 2.75 pounds of cheese. What is the cost per pound?
$0.40
x
$0.44
$4.00
$4.40
Hi
Answer:
$4.00
Explanation:
The cost per pound is $4.00, $11 divided by 2.75 is 4.
Answer:
Th cost per pound is $4.00.
Explanation:
11/2.75= 4
Two identical wind-up cars A and B are released. Car B has a 2 kilogram weight strapped to the back of the car. Which will have the greatest average speed towards the end of the motion?
A)Car A
B)They will both have an average speed of zero.
C)They will have the same average speed.
D)Car B
E)There is not enough information to answer.
Answer:
Car A would have a better average speed
Explanation:
added weight to a object that is self propelled will be slower than a identical object with no added weight
A ball is thrown 24 m/s into the air. How high does it go?
556.4 m
0 m
29.4 m
-556.4 m
Answer:
option c is correct
Explanation:
we know that
2as=vf^2-vi^2
vf=24 m/s
vi= 0 m/s
a=g= 9.8 m/s^2
s=vf^2-vi^2/2a
s=(24)²-(0)²/2*9.8
s=576/19.6
s=29.4 m
therefore option c is correct
It is more difficult to climb a vertical staircasethan a slanted one give reason
Answer:
The reason is similar to the reason why it is difficult to roll an object on a surface with a positive incline than rolling it on the ground
The more the path becomes vertical, the more force we have to apply to oppose the force of gravity
but when we are moving horizontally, we don't have to move against the gravity and hence, it is less difficult than going vertically upwards
An 7.40 kg block drops straight down from a height of 0.83 m, striking a platform spring having a force constant of 9.50 102 N/m. Find the maximum compression of the spring.
Answer:
0.25 m.
Explanation:
mass of the block = 7.40 kg, height = 0.83 m, force constant of the spring = 9.50 x [tex]10^{2}[/tex] N/m.
The maximum compression on the spring can be determined by;
Potential energy stored in the spring = [tex]\frac{1}{2}[/tex] K[tex]x^{2}[/tex]
But, potential energy = mgh
So that,
mgh = [tex]\frac{1}{2}[/tex] K[tex]x^{2}[/tex]
7.4 x 9.8 x 0.83 = 9.50 x [tex]10^{2}[/tex] x [tex]x^{2}[/tex]
60.1916 = 9.50 x [tex]10^{2}[/tex] x [tex]x^{2}[/tex]
[tex]x^{2}[/tex]= [tex]\frac{60.1916}{9.50*10^{2} }[/tex]
= 0.06336
x = 0.2517
x = 0.25 m
The maximum compression of the spring is 0.25 m.
What would happen if there is more male hyenas than female hyenas in a population?
Choices:
Male hyenas will compete to mate with the females.
Some male hyenas will die.
Male hyenas for wait for more females to join the population.
Answer:
Option 1
Explanation:
I always see animals do that
4. A substance has a density of 0.79 g/cm'. It is soluble in water. List all the possibilities of what it might be How could you determine the actual identity?
Answer:
See explanation
Explanation:
Given that the density of the unknown substance is 0.79 g/cm3 and is soluble in water, the possible substances it could be are;
i) t-butanol
ii) ethanol
iii) 2-propanol
iv) acetone
However, the actual identity of the unknown substance can be obtained by carrying out a boiling point test. The four substances listed above have different boiling points. Hence the boiling point of the unknown substance ultimately discloses its identity.
An elastic conducting material is stretched into a circular loop of 13.6 cm radius. It is placed with its plane perpendicular to a uniform 0.871 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 73.9 cm/s. What emf is induced in volts in the loop at that instant?
Answer: The Answer is attached to the image below
(A star if you answer this question) A school bus is traveling at 11.1 m/s and has a
momentum of 152,625 kgm/s. What is the mass of the bus?
[tex]\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}[/tex]
Actually Welcome to the Concept of the Kinematics in real world.
So, as given here, we have to find the Mass of the bus from the given momentum, so we get as,
P = m * V
momentum = mass * velocity
here, P= 152625 kgm/s and v= 11.1 m/s
so substituting we get as,
m = 152625 ÷ 11.1 => 13,750 kg
hence,the mass of the bus is 13,750 kg.
Three moles of a monatomic ideal gas are heated at a constant volume of 1.20 m3. The amount of heat added is 5.22x10^3 J.(a) What is the change in the temperature of the gas?________ ? K(b) Find the change in its internal energy.________ ? J(c) Determine the change in pressure.________ ? Pa
Answer:
A) 140 k
b ) 5.22 *10^3 J
c) 2910 Pa
Explanation:
Volume of Monatomic ideal gas = 1.20 m^3
heat added ( Q ) = 5.22*10^3 J
number of moles (n) = 3
A ) calculate the change in temp of the gas
since the volume of gas is constant no work is said to be done
heat capacity of an Ideal monoatomic gas ( Q ) = n.(3/2).RΔT
make ΔT subject of the equation
ΔT = Q / n.(3/2).R
= (5.22*10^3 ) / 3( 3/2 ) * (8.3144 J/mol.k )
= 140 K
B) Calculate the change in its internal energy
ΔU = Q this is because no work is done
therefore the change in internal energy = 5.22 * 10^3 J
C ) calculate the change in pressure
applying ideal gas equation
P = nRT/V
therefore ; Δ P = ( n*R*ΔT/V )
= ( 3 * 8.3144 * 140 ) / 1.20
= 2910 Pa
A) The change in the temperature of the gas is; ΔT = 139.5 K
B) The change in internal energy of the gas is; ΔU = 5.22 × 10³ J
C) The change in pressure of the gas is; ΔP = 2899.5 Pa
We are given;
Volume of Monatomic ideal gas; V = 1.2 m³
Amount of heat added; Q = 5.22 × 10³ J
number of moles; n = 3
A) To calculate the change in temperature of the monatomic idea gas, we will use formula;
Q = ³/₂nRΔT
Where R is a constant = 8.314 J/mol.K
ΔT is the change in temperature
Making ΔT the subject of the formula;
ΔT = ²/₃(Q/(nR))
ΔT = ²/₃(5.22 × 10³)/(3 × 8.314)
ΔT = 139.5 K
B) Due to the fact that no work was done, then from first law of thermodynamics, we can say that;
ΔU = Q
Thus;
change in internal energy; ΔU = 5.22 × 10³ J
C) The change in pressure will be calculated from the formula;
ΔP = (n*R*ΔT)/V
ΔP = (3 * 8.314 * 139.5)/1.2
ΔP = 2899.5 Pa
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A plane mirror is placed to the right of an object. The image formed by the mirror will be a
real image that appears to be on the right of the mirror.
real image that appears to be on the left of the mirror.
virtual image that appears to be on the right of the mirror.
virtual image that appears to be on the left of the mirror.
Hamish is studying what happens when he sends a sound wave through different mediums, and he records his data in a table.
A 2-column table with 4 rows titled Hamish's Waves. The first column labeled Wave has entries 1, 2, 3, 4. The second column labeled Information has entries liquid, solid, gas, liquid.
Which statement could made about the data collected in Hamish’s table?
Wave 1 will move the fastest.
Wave 2 will move the slowest.
Wave 3 will move the slowest.
Wave 4 will move the fastest.
What is common between transverse waves and longitudinal waves?
Both include an amplitude, crest, and rarefactions
Both move faster at higher temperatures
Both move slower through densely packed molecules
Both include a wavelength from compression to compression
An angle of refraction is the angle between the refracted ray and the
incident ray.
normal.
medium.
boundary.
Answer:
A plane mirror is placed to the right of an object. The image formed by the mirror will be a virtual image that appears to be on the left of the mirror.
Explanation:
a current of 200 mA through a conductor converts 40 joules of electrical energy into heat in 30 seconds determine the p
otential drop across the conductor
Answer:
ou have I=200mA, E=40J, t=30s, and you want to find the voltage drop.
First, you should know that P=V⋅I , so V=PI
Second, you have the amount of energy converted in a certain amount of time, so E=P⋅t
So, find the power and use it to find the voltage drop.
this works , but i thought energy was defined by W = P * t whitch would then be P = W/t
A model of a helicopter rotor has four blades, each 3.4 m in length from the central shaft to the tip of the blade. The model is rotated in a wind tunnel at 550 rev/min. What is the radial acceleration of the blade tip, expressed as a multiple of the acceleration g due to gravity?
A. (5.72 × 104)g
B. (6.23 × 102)g
C. (1.15 × 103)g
D. (2.25 × 103)g
which statement accurately describes the relationship between force and momentum?
A. As the mass of an object increases its momentum increases, and it takes more force to change its motion.
B. As the velocity of an object increases, its momentum decreases and it takes less force to change its motion.
C. As the mass of an object increases its momentum decreases and it takes less force to change it motion
D. As the velocity of an object decreases its momentum increases and it takes more force to change its motion
Answer:
A
Explanation:
just did it
As the mass of an object increases its momentum increases, and it takes more force to change its motion. So, option A.
What is meant by momentum ?Mass in motion is quantified by momentum, which is the measure of the amount of mass in motion.
Here,
Momentum of an object, which is under motion can be defined as the product of the mass and velocity of the object.
Momentum, P = mv
According to Newton's second law, the force is defined as the rate of change of momentum, or the momentum per unit time.
F = dP/dt
So, force is proportional to the amount of momentum imparted on the object.
Therefore, if the mass or velocity of the object increases, it will eventually cause the momentum to be increased and as a result, the force required to exert on the object will increase.
Hence,
As the mass of an object increases its momentum increases, and it takes more force to change its motion. So, option A.
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Dolphins rely on echolocation to be able to survive in the ocean. In a 20 °C ocean, a dolphin produces an ultrasonic sound with a frequency of 125 kHz. What is the wavelength of this sound, in meters?While remaining stationary, the dolphin emits a sound pulse and receives an echo after 0.220 s. How far away, in meters, is the reflecting object from the dolphin?
Answer:
wavelength = 0.01 m
distance = 162.8 m
Explanation:
Given that;
Speed of sound in water = 1,480 meters per second
Frequency of ultrasound = 125KHZ
From=
v=λf
v= speed of sound
λ= wavelength of sound
f= frequency of sound
λ= 1,480 ms-1/125 * 10^3 Hz
λ= 0.01 m
From
v = 2x/t
where;
v= velocity of sound in water
x= distance traveled
t = time taken
x = vt/2
x = 1,480 ms-1 * 0.220 s/2
x= 162.8 m
1. A spring extends by 10 cm when a mass of 100 g is attached to it. What is the spring constant? (Calculate your answer in N/m)10N/m
2. What will be the extension of this spring if the load is a) 4N and b) 75 g?
Answer:
1) k = 10 [N/m]
2) a-) x = 0.4 [m]
b) x = 0.075 [m]
Explanation:
To be able to solve this type of problems that include springs we must use Hooke's law, which relates the force to the deformed length of the spring and in the same way to the spring coefficient.
F = k*x
where:
F = force [N] (units of Newtons]
k = spring constant [N/m]
x = distance = 10 [cm] = 0.1 [m]
Now, the weight is equal to the product of the mass by the gravity
W = m*g = F
where:
m = mass = 100 [g] = 0.1 [kg]
g = gravity acceleration = 10 [m/s²]
F = 0.1*10 = 1 [N]
Now clearing k
k = F/x
k = 1/0.1
k = 10 [N/m]
2)
a ) if the force is 4 [N]
clearing x
x = F/k
x = 4/10
x = 0.4 [m]
m = 75 [g] = 0.075 [kg]
W = m*g = F
F = 0.075*10 = 0.75 [N]
x = .75/10
x = 0.075 [m]
A block slides down an inclined plane from rest. Initially the block is at 4.5m above the ground. Find the speed of the block when it is 1.5m above the ground. 1) 7.7m/s 2) 9.4m/s 3) 5.4m/s 4) 3.2m/s
Since, no external force is acting , so the system is in equilibrium .
Initial total energy = Final total energy
[tex]mg(4.5) = mg(1.5) + \dfrac{mv^2}{2}\\\\\dfrac{v^2}{2}=3\times g \\\\v^2=3\times 9.8\times 2\\\\v = \sqrt{58.8}\ m/s\\\\v = 7.67 \ m/s[/tex] ( Here , g = acceleration due to gravity = 9.8 m/s² )
Therefore, option 1) is correct.
Hence, this is the required solution.
A car’s brakes decelerate it at a rate of -2.40 m/s2. If the car is originally travelling at 13 m/s and comes to a stop, then how far, in meters, will the car travel during that time?
Answer:
Approximately [tex]35.2\; \rm m[/tex].
Explanation:
Given:
Initial velocity: [tex]u = 13\; \rm m \cdot s^{-1}[/tex].
Acceleration: [tex]a = -2.40\; \rm m \cdot s^{-2}[/tex] (negative because the car is slowing down.)
Implied:
Final velocity: [tex]v = 0\; \rm m \cdot s^{-1}[/tex] (because the car would come to a stop.)
Required:
Displacement, [tex]x[/tex].
Not required:
Time taken, [tex]t[/tex].
Because the time taken for this car to come to a full stop is not required, apply the SUVAT equation that does not involve time:
[tex]\begin{aligned} x &= \frac{v^2 - u^2}{2\, a} \\ &= \frac{{\left(0\; \rm m \cdot s^{-1}\right)}^2 - {\left(13\; \rm m \cdot s^{-1}\right)}^2}{2\times \left(-2.40\; \rm m\cdot s^{-2}\right)} \approx 35.2\; \rm m \end{aligned}[/tex].
In other words, this car would travel approximately [tex]35.2\; \rm m[/tex] before coming to a stop.
If a rock is dropped from the top of a tower at the front of and it takes 3.6 seconds to hit the ground. Calculate the final velocity of the penny in m/s.
Answer:
36 m/s
Explanation:
t = 3.6s
u = 0m/s
a = +g = 10m/s²
v = ?
using,
v = u + at
v = 0 + 10(3.6)
v = 36 m/s
In the winter sport of curling, players give a 20 kg stone a push across a sheet of ice. The Slone moves approximately 40 m before coming to rest. The final position of the stone, in principle, onlyndepends on the initial speed at which it is launched and the force of friction between the ice and the stone, but team members can use brooms to sweep the ice in front of the stone to adjust its speed and trajectory a bit; they must do this without touching the stone. Judicious sweeping can lengthen the travel of the stone by 3 m.1. A curler pushes a stone to a speed of 3.0 m/s over a time of 2.0 s. Ignoring the force of friction, how much force must the curler apply to the stone to bring it op to speed?A. 3.0 NB. 15 NC. 30 N
D. 150 N2The sweepers in a curling competition adjust the trajectory of the slope byA. Decreasing the coefficient of friction between the stone and the ice.
B. Increasing the coefficient of friction between the stone and the ice.C. Changing friction from kinetic to static.D. Changing friction from static to kinetic.3. Suppose the stone is launched with a speed of 3 m/s and travel s 40 m before coming to rest. What is the approximate magnitude of the friction force on the stone?A. 0 NB. 2 NC. 20 ND. 200 N4. Suppose the stone's mass is increased to 40 kg, but it is launched at the same 3 m/s. Which one of the following is true?A. The stone would now travel a longer distance before coming to rest.B. The stone would now travel a shorter distance before coming to rest.C. The coefficient of friction would now be greater.D. The force of friction would now be greater.
Answer:82. Since you have a distance and a force, then the easiest principle to use is energy, i.e. work.
The work done by friction is F * d. This work cancels out the kinetic energy of the stone (1/2)mv^2
Fd = (1/2)mv^2
F = (1/2)mv^2/d.
Plug in m = 20 kg, v = 3 m/sec, d = 40 m.
83. With more mass, the kinetic energy is higher now. The work needed is higher. W = F * d and F is the same.
Explanation:Hope I helped :)
How many significant figures are in 0.0067?
Answer:
2
Explanation:
there are 2 significant figures in there
Is Geothermal Energy renewable? Why or why not? Use in your own words.
Answer:
Yes, geothermal energy is a renewable energy resource because the water can be heated by pumping it through the rocks.
1. A stone of mass 0.8 kg is attached to a 0.9 m long string. The string will break if the tension exceeds 60 N. The stone is whirled in a horizontal circle on a frictionless tabletop while the other end of the string remains fixed. What is the maximum speed the stone can attain without breaking the string?
A. 8.22 m/s
B. 7.30 m/s
C. 9.34 m/s
D. 7.76 m/s
2. A highway curve with radius 900 ft is banked so that a car traveling at 55 mph will not skid sideways even in the absence of friction. At what angle should the curve be banked to prevent skidding?
A. 14.6°
B. 18.9°
C. 10.9°
D. 12.7°
3. A button will remain on a horizontal platform rotating at 40 rev/min as long as it is no more than 0.15 m from the axis. How far from the axis can the button be placed without slipping if the platform rotates at 60 rev/min?
A. 0.365 m
B. 0.338 m
C. 0.225 m
D. 0.294 m
4. What is the tension in a cord 10 m long if a mass of 5 kg is attached to it and is being spun around in a circle at a speed of 8 m/s?
A. 67 N
B. 28 N
C. 32 N
D. 50 N
5. A 0.5 kg mass attached to a string 2 m long is whirled around in a horizontal circle at a speed of 5 m/s. What is the centripetal acceleration of the mass?
A. 11.3 m/s2
B. 12.5 m/s2
C. 5.9 m/s2
D. 10.2 m/s2
6. Find the maximum speed with which a car can round a curve that has a radius of 80 m without slipping if the road is unbanked and the coefficient of friction between the road and the tires is 0.81.
A. 44.3 m/s
B. 20.8 m/s
C. 25.2 m/s
D. 30.6 m/s
Answer:
1. A. 8.22. m/s
Explanation:
When a potential difference of 10 V is placed across a certain solid cylindrical resistor, the current through it is 2 A. If the diameter of this resistor is now tripled, the current will be:______.A) 18 A.
B) 2/3 A.
C) 3 A.
D) 2/9 A.
E) 2 A.
Answer:
sorry I wish I could it help you
Need help ASAP..please help
Answer:
option 3
Explanation:
can i get brainliest
In principle, when you fire a rifle, the recoil should push you backward. How big a push will it give? Let's find out by doing a calculation in a very artificial situation. Suppose a man standing on frictionless ice fires a rifle horizontally. The mass of the man together with the rifle is 70 kg, and the mass of the bullet is 10 g. If the bullet leaves the muzzle at a speed of 500 m/s, what is the final speed of the man?
Answer:
Explanation:
m1v1=m2v2
m1=70 kg
m2=10 g=0.01 kg
v2=500 m/s
m1v1=m2v2
v1=m2v2/m1
v1=0.01*500/70
v1=0.07
Measurements of the radioactivity of a certain isotope tell you that the decay rate decreases from 8255 decays per minute to 3110 decays per minute over a period of 4.50 days.
What is the half-life (T1/2) of this isotope?
Answer:half-life (T1/2) of this isotope =
Explanation:
The number of nuclei of any radioactive substance at a given time is expressed by
Nt = N0e⁻kt
Nt=decay of material at a time t, =3110 decays per minute
N=decays at t=0, 8255 decays per minute
k=constant
Nt=N0e−kt
3110= 8255 e⁻k(4.50)
3110/ 8255=e−k(4.50)
0.3767 = e−k(4.50)
In 0.3767 = -k (4.50)
0.976=-4.5k
k=0.976/4.5
=0.2159
Also we know that t 1/2= time that it takes half the original material to decay.it is related to the rate constant by
T₁/₂=ln 2 / k
Therefore half-life (T1/2) of this isotope
T₁/₂=ln 2/0.2159
T₁/₂=3.12 days
What is the energy contained in a 0.950 m3 volume near the Earth's surface due to radiant energy from the Sun?