Answer:
4: toward which light rays converge but do not pass through
Explanation:
A 0.55 kg basketball moving 6.3 m/s to the right collides with a 0.06 kg tennis
ball moving 35 m/s to the left. After the collision, the tennis ball is moving
39.5 m/s to the right. What is the velocity of the basketball after the collision?
Assume an elastic collision occurred.
A. 1.8 m/s to the right
B. 14.4 m/s to the left
C. 14.4 m/s to the right D. 1.8 m/s to the left
Answer:
1.8 ms to the left
Explanation:
Answer:1.8 ms to the left
Explanation:
2. What are the units of mass?
kg
N
m/s
m/s^2
(kg)(m/s)
m
S
Answer:
Kg
Explanation:
The units of mass are gram (g) , kilograms (Kg)
But SI of mass accepted is Kilograms (Kg)
Answer:
kg
Explanation:
the s.i unit is kilogramA cylindrical 5.00-kg reel with a radius of 0.600 m and a frictionless axle, starts from rest and speeds up uniformly as a 3.00-kg bucket falls into a well, making a light rope unwind from the reel. The bucket starts from rest and falls for 4.00 s.
Required:
a. What is the linear acceleration of the falling bucket?
b. How far does it drop?
c. What is the angular acceleration of the reel?
true or false: the height from which cliff divers jump affects the velocity at which they will fall. velocity is speed in a certain direction.
Answer:
False
Explanation:
Because that is not true
Jennifer has taught her pet rat to run a maze. She thinks that the rat will
go faster if she puts its favorite treat at the end. She has the rat run the
maze ten times with the favorite treat and ten times with a regular food
pellet. She uses a stopwatch to measure how long it takes for the rat to get
to the reward.
O Replication
O Repetition
If galium-68 has a half life of 68.3 minutes, how much of a 10.0 mg sample is left after one half life. Two half lives? Three half lives?
If the mass of the original sample of Galium-68 is 10 mg, then . . .
== After one half-life, (1/2)(10 mg) = 5 mg remain
== After the second half-life, (1/2)(5 mg) = 2.5 mg remain
== After the third half-life, (1/2)(2.5 mg) = 1.25 mg remain
It doesn't matter how long the half-life is.
The amount of the 10.0mg sample of Galium-68 remaining after;
One half-life = 5mgTwo half-life = 2.5mgThree half-life = 1.25mgIrrespective of the half life of an element such as Galium-68; the half-life is the duration of time it takes for the element to remain half it's original mass.
In essence; Since the starting sample of galium-68 has a mass of 10.0mg.
After one half life;
The amount of Galium-68 remaining is; (10/2) mg
= 5mgAfter two half lifes;
The amount of Galium-68 remaining is; (5/2) mg
= 2.5mgAfter three half lifes;
The amount of Galium-68 remaining is; (2.5/2) mg
= 1.25 mgRead more;
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You pull a wagon with a force of 20 N. The wagon has a mass of 10 kg. What is the wagon's acceleration?
Answer:
The answer is 2 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
\(a = \frac{f}{m} \\ \)
where
f is the force
m is the mass
From the question
f = 20 N
m = 10 kg
We have
\(a = \frac{20}{10} \\ \)
We have the final answer as
2 m/s²Hope this helps you
Answer:.
Explanation:.
PLZ help 10 points!!! space question!
Answer:
B. They are smaller and made of rocky material
Explanation:
i think it's right??
The mediastinum does NOT contain which of the following?
A. Heart
B. Blood vessels
C. Lungs
D. Esophagus
125cm³ of a gas was collected at 15 °C and 755 mm of mercury pressure. Calculate the volume of the gas that will be collected at standard temperature and pressure
Answer:
119,2 см³
Explanation:
по формуле Клопейрона (P1×V1):T1=(P2×V2):T2
если из этой формулы найти V2, ответ будет равен примерно на 119,2 см³
What is the reactive force when a fish swims through water
The reactive force when a fish swims through water is the force of the water on the fish. This is an example of Newton’s third law of motion which states that for every action, there is an equal and opposite reaction. The active force is the fish pushing against the water, so the reactive force would be the reverse, the equal force of the water pushing back on the fish.
A bug starts out at rest, 3 m to the right of the origin. It then starts moving on a trip. After 1 s, the bug is seen at 9 m to the right of the origin, travelling at 5 m/s to the right. After 7 s (from the start of the trip), the bug is seen at 2 m to the left of the origin, travelling at 8 m/s to the right. a.) What is the position vector of the bug 1 s after the start of the trip
Answer:
d = 6 i^ m
Explanation:
This is an exercise in kinematics, where the position vector is the displacement of the body from one point to another.
In this case we are told that for t = 0 the worm is at x₀ = 3m and with velocity starts from zero, after t = 1 s it is at x₁ = 9m and with a velocity of
v₁ = 5 m / s
They ask what is the displacement for the time of 1 s
d = x₁ - x₀
d = 9 -3 m
d = 6 m
Bold indicates vector, displacement vector is
d = 6 i ^ m
The initial population of a town is 3000, and it grows with a doubling time of 10 years. What will the population be in 8 years?Please show step by step
Givens.
• The initial population is 3000. (P_0 = 3000)
,• The ratio is 2 because it's doubling. (r = 2)
,• The doubling time is 10 years. (u = 10)
,• The total number of years is 8, t = 8.
Represent the problem with an exponential function because population growth has an exponential behavior.
\(P(t)=P_0\cdot(r)^{\frac{t}{u}}\)Use the given values to find P(8).
\(P(8)=3000\cdot(2)^{\frac{8}{10}}\)Then,
\(\begin{gathered} P(8)=3000\cdot1.74110 \\ P(8)=5223.30 \end{gathered}\)Therefore, the population after 8 years is 5223.30.
15. A conductor has a surface charge density of 300nC/m^2. If the area of the conductor is
15m^2 what is the electric flux through the given area of the conductor (60= 8.85 x 10^-12C^2Nm?).
(a) 5.08 x10Nm²/C
(b) 2.08 x 10Nm2/C
(c) 1.08 x10
Nm²/C
(d) 0.08 x 10Nm2/C
E = sigma/2 * epselon naught
E = 300 * 10^-9/(2 * 8.854 * 10^-12)
E = 1.7 * 10^4 N/C
Assuming that the electric field is uniform all across the surface:
Then, Electric flux = E * A
Electrix flux = 1.7 * 10^4 * 15 = 2.5 x 10^5 Nm^2/C
It should be b)
The electric flux through the given area of the conductor is
2.26 x 10³ Nm²/C.
Surface charge density of the conductor, σ = 300 x 10⁻⁹ C/m²
Area of the conductor, A = 15 m²
The equation for surface charge density is,
σ = q/A where q is the charge in the conductor.
So,
Charge, q = σ A
q = 300 x 10⁻⁹/15
q = 2 x 10⁻⁸C
Therefore,
Electric flux through the conductor, Φ = q/ε₀
Φ = 2 x 10⁻⁸/8.85 x 10⁻¹²
Φ = 2.26 x 10³ Nm²/C
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A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A constant tangential force of 290 N applied to its edge causes the wheel to have an angular acceleration of 0.854 rad/s2.
(a) What is the moment of inertia of the wheel?
kg · m2
(b) What is the mass of the wheel?
kg
(c) If the wheel starts from rest, what is its angular velocity after 5.80 s have elapsed, assuming the force is acting during that time?
rad/s
Answer:
c
Explanation:
units c mark me brainless k
The achievement of lifting a rocket off the ground and into space can be explained by
Answer:
The achievement of lifting a rocket off the ground and into space can be explained by Newton's third law of motion. What is required for a rocket to lift off into space? Thrust is required for a rocket to lift off into space, ... An object that travels around another object in space is called a satellite.
Explanation:
A 8.15 kg mass oscillates up and down on a spring that has a force constant of 90 N/m.
(a) What is the angular frequency of this spring/mass system (in rad/s)?
(b) What is the period of this spring/mass system (in seconds)?
Answer:
(a) The angular frequency ($\omega$) of a spring/mass system with a force constant ($k$) and a mass ($m$) can be found using the formula:
\omega = \sqrt{\frac{k}{m}}
Plugging in the values given, we get:
\omega = \sqrt{\frac{90 N/m}{8.15 kg}} \approx 3.18 \text{ rad/s}
Therefore, the angular frequency of the spring/mass system is approximately 3.18 rad/s.
(b) The period ($T$) of a spring/mass system can be found using the formula:
T = \frac{2\pi}{\omega}
Plugging in the value of $\omega$ we found in part (a), we get:
T = \frac{2\pi}{3.18\text{ rad/s}} \approx 1.98 \text{ s}
Therefore, the period of the spring/mass system is approximately 1.98 s.
A cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 55.5 m above ground level, and the ball is fired with initial horizontal speed . The projectile lands at a distance D = 140 m from the cliff. Assume that the cannon is fired at time t = 0 and that the cannonball hits the ground at time . a. What is the value of ? b. What is the y position of the cannonball at the time c. Find the initial speed of the projectile.
a) The value of t u = 140/t`b.
b) The y position of the cannonball at the time t is 55.5 mc.
c) The initial speed of the projectile is 52.4 m/s.
Given that a cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 55.5 m above ground level, and the ball is fired with initial horizontal speed u. The projectile lands at a distance D = 140 m from the cliff. Assume that the cannon is fired at time t = 0 and that the cannonball hits the ground at time t.Now,We have to find the value of t, y position of the cannonball at the time t and the initial speed of the projectile.
a. To find the value of t:Here, we have to use the formula of distance
i.e.,S = ut + (1/2)gt², Where S = 140 m, u = u and g = 9.8 m/s².Hence,140 = u×t ………..(1)We know that, time taken by the cannonball to hit the ground can be calculated as,`(2H)/g`
Since the height of the cannon from the ground is 55.5m, the total height of the cannonball from the ground is
(2H) = 2 × 55.5
= 111 m`2H/g
= 111/9.8`
= 11.32653 s
From equation (1),u×t = 140u = 140/t
Therefore, `u = 140/t`b.
b)To find the y position of the cannonball at the time t:
Here, we have to use the formula of height i.e.,y = u×t – (1/2)gt²,
Where, y = height of the cannonball at time t, u = 140/t, t = time taken by the cannonball to hit the ground and g = 9.8 m/s².
We have already calculated the time taken by the cannonball to hit the ground in the previous step.`
y = 140 - (1/2) × 9.8 × t²`
On substituting the value of t as `t = 11.32653`,
we get,y = 140 - (1/2) × 9.8 × (11.32653)²= 55.5 mc.
c) To find the initial speed of the projectile:
To calculate the initial speed of the projectile, we need to use the formula of range of projectile
.i.e.,R = u²sin2θ/g
Where R = 140 m, g = 9.8 m/s², θ = 0° (horizontal)
u² = R × g/sin2θ
= 140 × 9.8/sin0°
= 2744m²/s²u
= \(\sqrt(2744m^2/s^2)\)
= 52.4 m/s
Hence, the initial speed of the projectile is 52.4 m/s.
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Greatest to least order
Answer:
Explanation:
FBEDAC
9. Rob Colley set a record in “pole-sitting" when he spent 42 days in a barrel at the
top of a flagpole with a height of 43 m. Supposed a friend wanting to deliver an ice-
cream sandwich to Colley throws the ice cream straight up with just enough speed
to reach the barrel. How long does it take the ice-cream sandwich to reach the
barrel?
Answer:
3 sec
Explanation:
T=squareroot((2*-43)/-9.81)
Which correctly describes latent heat?
A. The heat of molecules that are under pressure
B. The heat held inside of ice crystals colder than -2°C
C. The heat absorbed or lost by a substance while it's changing state
D. The heat used to change the temperature of a liquid
Option C. The heat absorbed or lost by a substance while it's changing state correctly describes latent heat
Latent heat is the heat absorbed or lost by a substance while it is changing state.
The latent heat is a type of heat that is transferred during phase change, i.e., while a substance undergoes a change of state.
For example, when ice melts into liquid water, or when liquid water evaporates into water vapor, heat is absorbed from the surroundings.
Latent heat is not associated with a temperature change; rather, it's associated with a change of state.
For instance, the temperature of water remains at 100°C while boiling.
When water is boiling, the latent heat of vaporization is absorbed and utilized to break the hydrogen bonds holding water molecules together to change water from the liquid phase to the gaseous phase.
When the water is boiling, adding more heat won't increase the water's temperature, instead, the extra heat will be absorbed to change the phase of water molecules.
Therefore, the correct answer to the given question is option C: The heat absorbed or lost by a substance while it is changing state.
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True or false. More people live in cities then they do in the country
Answer:
TrueExplanation:
That's my answer hope it helped
The density of a solid or liquid material divided by the density of water is called
Answer:
I believe the answer is specific gravity
Explanation:Hope this helps :)
On a warm summer day, a large mass of air (atmospheric pressure 1.01×105Pa) is heated by the ground to a temperature of 25.0 ∘C and then begins to rise through the cooler surrounding air. Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.70×104 Pa. Assume that air is an ideal gas, with γ=1.40. (This rate of cooling for dry, rising air, corresponding to roughly 1 ∘C per 100 m of altitude, is called the dry adiabatic lapse rate.)
The temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.70×10⁴ Pa is approximately 14.3°C.
Using the ideal gas law, we can write: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the absolute temperature. Since the mass of air is not changing, we can write: PV = constant.
Applying this to the situation where the air mass rises to a level where the pressure is 8.70×10⁴ Pa, we get:
(1.01×10⁵ Pa)×V = (nR/T1)×T1(8.70×10⁴ Pa)×V = (nR/T2)×T2Dividing the second equation by the first and using the fact that γ=Cp/Cv=1.40 for air, we get:
(T2/T1) = [(P2/P1)^(γ-1)/γ] = [(8.70×10⁴ Pa)/(1.01×10⁵ Pa)]^(1.4/1.4) = 0.813Solving for T2, we get:
T2 = T1×(P2/P1)^(γ-1)/γ = (25+273) K×0.813 ≈ 287.3 K ≈ 14.3°CThus, the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.70×10⁴ Pa is approximately 14.3°C.
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Part B
Enter into the table your calculated value for the spring constant, then play with different values of mk
until you get a close match to the motion. (Note: It will never be perfect. Remember that there are two
kinds of spring damping. Both are at work here, but we are not going to model both.) Once you're
satisfied with your model, record your model values in the table below.
The spring stiffness is quantified by the spring constant, or k. For various springs and materials, it varies.
What is Spring constant?The stiffer the spring is and the harder it is to stretch, the larger the spring constant.
Springs are pliable mechanical devices that regain their previous shape after deforming, i.e. after being stretched or compressed. They are an essential part of many different mechanical devices.
The well-known metal coil has evolved into an essential element in the modern world, appearing in everything from engines to appliances to tools to automobiles to medical equipment and even basic ball-point pens. The spring's ability to store mechanical energy accounts for its widespread use and applications.
Therefore, The spring stiffness is quantified by the spring constant, or k. For various springs and materials, it varies.
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A roller coaster is at a peak of 20m and has a mass of 900kg. What is the potential energy of the roller coaster?
O 100000 J
10000 J
O 9.8 J
O 176400 J
The potential energy of the roller coaster is 176,400 J (joules).
The potential energy of an object is given by the formula PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical position of the object.
In this case, the roller coaster is at a peak of 20m and has a mass of 900kg. The acceleration due to gravity, g, is approximately 9.8 \(m/s^2\).
Using the formula, we can calculate the potential energy:
PE = mgh
= (900 kg)(9.8 \(m/s^2\))(20 m)
= 176,400 J
Therefore, the potential energy of the roller coaster is 176,400 J (joules).
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The current sensitivity of a moving coil galvanometer increase with decrease in _________?
The current sensitivity of a moving coil galvanometer increases with a decrease in its resistance.
What is Current Sensitivity?
Current sensitivity is a measure of the responsiveness of an instrument to the current flowing through it. In particular, it is the deflection produced by a given amount of current passing through a device. The higher the current sensitivity, the smaller the amount of current required to produce a given deflection, and thus the more sensitive the instrument is to changes in current.
This is because the current sensitivity of a galvanometer is defined as the deflection produced per unit of current passed through it. Therefore, a higher current sensitivity means that a smaller current can produce a larger deflection, which can be achieved by decreasing the resistance of the galvanometer. This is because a lower resistance will allow a larger current to flow through the coil, which will result in a larger deflection of the coil.
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Answer:
Resistance
Explanation:
Sensitivity of a moving coil galvanometer can be increased by increasing the number of turns in the coil, the area of coil and magnetic field whereas by decreasing the couple per unit twist of the suspension.
Question 1
2 pts
Explain what causes a solution to be a strong acid.
Answer:
Cuanto más fuerte es el ácido, más rápido se disocia para generar H +start superscript, plus, end superscript. Por ejemplo, el ácido clorhídrico (HCl) se disocia completamente en iones hidrógeno y cloruro cuando se mezcla con agua, por lo que se considera un ácido fuerte.
Compare sound and earthquake waves
When materials vibrate, waves are created that travel through the substance, and this energy is what we hear as sound. Earthquakes are earth vibrations that cause the (potential) energy held within rocks to be released (as a result of their pressure-generating relative positions). Seismic waves are produced by earthquakes.
How do sound waves and earthquakes compare?
The waves lose energy as they move through the air with sound or through the ground with shaking during an earthquake. Therefore, a band can be heard louder close to the stage than farther away, and an earthquake can be felt more strongly close to the fault than farther away.
In actuality, sound in the air cannot match how quickly earthquake waves move. In rock, the compressional or "P" wave of an earthquake moves at the In actuality, sound in the air cannot match how quickly earthquake waves move. The speed of a P wave is typically 10,000 mph. The speed of sound through air is roughly 750 mph.
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A bowling ball falls freely from rest for 10 seconds before it hits the ground. Neglect air resistance.Is the average speed of the ball during the first 5 seconds greater than, less than, or the same as the average speed of the ball during the last 5 seconds?
Answer:
the average speed of the ball during the first 5 seconds is less than the average speed of the ball during the last 5 seconds
Explanation:
The average speed is defined as
\(v_{\text{avg}}=\frac{v_{i1}+v_{f1}}{2}\)where v_i = inital speed and v_f = final speed.
Now for the first five seconds, we know that the ball started from rest; therefore, v_i = 0, and so the average speed is
\(v_{\text{avg}}=\frac{0+v_{f1}}{2}\)\(\boxed{v_{\text{avg}}=\frac{v_{f1}}{2}}\)where v_f = final velocity at the end of the first 5 seconds.
Now the average velocity in the last 5 seconds is
\(v_{\text{avg}}=\frac{v_{i2}+v_{f2}}{2}\)Realising that v_i2 = v_f1 ( the initial velocity at the beginning of last 5 seconds = final velocity at the end of first 5 seconds) and that vf_2 > v_i2 gives us
\(undefined\)Answer:
100 m/s.
Explanation:
Reasoning: Speed increases at a rate of 10 m/s (actually 9.8 m/s) every second. Thus after 10 seconds, the speed is 10 x 10 = 100 m/s.