The scientist who worked with uranium in the late 1800s was _____.1) Rutherford2) Curie3) Becquerel4) Bohr

Answer:

a. stay the same for very long

Explanation:

It is rare for any motion to stay the same for a very long time. The force applied on a body causes changes in the magnitude of motion.

Answer:

• Fundamental quantities are physical quantities that can't be expressed in any other forms by mathematical means while Derived quantities are physical quantities that can be expressed in terms of fundamental quantities by mathematical means.

Answer:

d. 60

Explanation:

If a body of mass 2 kg is moving with a velocity of 30 m/s, then

on doubling its velocity the momentum becomes

a 30 kgm/s

b 90 kgm/s

C 120 kgm/s

d 60 kgm/s

HALPLPLPPLL

Answer:

120

Explanation:

Answer:

a protractor

Explanation:

because protractors measure angles

Answer:

Explanation:

When discussing Newton's laws of motion, the most likely used terms when talking about Newton's third law of motion are action and reaction. According to this law, for every action there is an equal and opposite reaction

Answer: C. "action" and "reaction"

Explanation: Newtons 3rd law of motion: for every action, there is an equal and opposite reaction (action and reaction forces are described by this law).

Have a nice day. :)

The acceleration of Thomas is 0.233 m/s^2.

Acceleration is the rate of change of velocity. Thomas the Train chugs along at a velocity of 2 m/s.

Thomas needs to go faster so more coal is shoveled into his engine and he accelerates for 10 seconds until he is going 4.33 m/s.

We are to find the acceleration of Thomas.

The formula for acceleration is given as :

acceleration = (final velocity - initial velocity) / time

In the given problem, the initial velocity of Thomas, u = 2 m/s.

The final velocity of Thomas, v = 4.33 m/s The time for which Thomas accelerates, t = 10 s.

Therefore, the acceleration of Thomas will be given as:

a = (v - u) / ta = (4.33 - 2) / 10s => 2.33 / 10s => 0.233 m/s^2

Thus, the acceleration of Thomas is 0.233 m/s^2.

To summarize, the acceleration of Thomas is 0.233 m/s^2.

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The mass of the planet is  5.98 × 10^24 kg.

To calculate the mass of the planet, we can use Kepler's Third Law of Planetary Motion. This law states that the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit.

First, we need to convert the period of the satellite's orbit to seconds. We know that there are 60 minutes in an hour, so the period can be expressed as (6 × 60 + 12) minutes, which equals 372 minutes. Multiplying this by 60 seconds, we get a period of 22,320 seconds.

Next, we need to find the semi-major axis of the orbit. In a circular orbit, the semi-major axis is equal to the radius of the orbit. Therefore, the semi-major axis is 8.8 × 10^6 m.

Now, we can apply Kepler's Third Law to calculate the mass of the planet. The formula is T^2 = (4π^2/GM) × a^3, where T is the period of revolution, G is the gravitational constant, M is the mass of the planet, and a is the semi-major axis of the orbit.

Rearranging the formula, we can solve for the mass of the planet:
M = (4π^2/G) × a^3 / T^2

Plugging in the values, we get:
M = (4 × π^2 / 6.67430 × 10^-11) × (8.8 × 10^6)^3 / (22,320)^2

Evaluating this expression, we find that the mass of the planet is approximately 5.98 × 10^24 kg.

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The hypotenuse is 20.59. 10²+18² = C²

The hypotenuse of a right triangle is contrary to the proper angle. it is also the longest facet of the triangle. The hypotenuse of a triangle is defined as the longest side of a right triangle. opposite ninety ranges. it's far equal to the square root of the sum of squares of the opposite two aspects. square the lengths of the 2 sides referred to as a and b and add them collectively.

Take the square root of the end result to find the hypotenuse. The hypotenuse is the longest side the alternative facet is the facet contrary to the given attitude, and the adjacent facet is subsequent to the given attitude. We use special words to explain the perimeters of a proper triangle. The hypotenuse of a proper triangle is constantly contrary to the proper attitude. The sum of the angles of each triangle is a hundred and eighty°. In any triangle, the longest aspect is opposite the biggest angle and the shortest aspect is opposite the smallest perspective.

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Answer:

Power of 5 watts is used

Explanation:

Mechanical Work and Power

Mechanical work is the amount of energy transferred by a force.

Being F the force vector and s the displacement vector, the work is calculated as:

\(W=\vec F\cdot \vec s\)

If both the force and displacement are parallel, then we can use the equivalent scalar formula:

W=F.s

Power is the amount of energy converted per unit of time. The SI unit of power is the watt, equal to one joule per second.

The power can be calculated as:

\(\displaystyle P=\frac {W}{t}\)

Where W is the work and t is the time.

It's required to calculate the power used in t=30 seconds when W=150 Joules of work are completed. Substitute in the formula:

\(\displaystyle P=\frac {150}{30}\)

P = 5 Watt

Power of 5 watts is used

Answer:

λ = 5.734 x 10⁻⁷ m = 573.4 nm

Explanation:

The formula of the Young's Double Slit experiment is given as follows:

\(\Delta x = \frac{\lambda L}{d}\\\\\lambda = \frac{\Delta x d}{L}\)

where,

λ = wavelength = ?

L = distance between screen and slits = 8.61 m

d = slit spacing = 1.09 mm = 0.00109 m

Δx = distance between consecutive bright fringes = \(\frac{4.53\ cm}{10}\) = 0.00453 m

Therefore,

\(\lambda = \frac{(0.00453\ m)(0.00109\ m)}{8.61\ m}\)

λ = 5.734 x 10⁻⁷ m = 573.4 nm

The charge on the ball developed will be  1.78 x 10⁻⁷ Coulomb.

The field developed when a charge is moved. In this field, a charge experiences an electrostatic force of attraction or repulsion depending on the nature of charge,

Given is an electric field E = 100,000i N/C causes the 5.0 g point charge to hang at a 20° angle.

If T is the tension and θ is the angle, then

The force in  x direction Fx = T sinθ

Force in y direction Fy = T cosθ = mg

Dividing both the equation, we have

tan θ = qE /mg

Substitute the given values in the question, we have

q = 5 x10⁻³ x 9.81 x tan 20 /100000

q = 1.78 x 10⁻⁷ Coulomb.

Thus, the charge of ball is  1.78 x 10⁻⁷ Coulomb.

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Answer:

Explanation:

iphotosynthesis 2coal burning 3electric motor 4 electricity generator 5 waterturbines 6 solar bulb

The average force is given by the following formula:

where m is the mass of the car (1500 kg) and a its acceleration. To calculate the acceleration use the following expression:

where,

v: final speed = 28m/s

vo: initial speed = 18m/s

t: time of the acceleration = 4s

Replace the previous values into the formula for a and simplify:

Next, replace the previous value of a and the mass m into the expression for F and simplify:

Hence, the average force of the car is 3750N

The solution to the one dimensional Schrödinger equation for harmonic oscillator potential is given by:

ψn(x) = (1/√2n!) (mω/πh)1/4 exp(−mωx2/2h)Hn(√mωx/h)

where,

Generally, The Schrödinger equation is a fundamental equation in quantum mechanics which describes the time-dependent behavior of a system.

It is a partial differential equation that governs the behavior of a system's wave function, which describes the probability of a particle's location and momentum.

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At the lowest point on the Ferris wheel, there are two forces acting on the child: their weight of 430 N, and an upward centripetal/normal force with magnitude n; then the net force on the child is

∑ F = ma

n - 430 N = (430 N)/g • a

where m is the child's mass and a is their centripetal acceleration. The child has a linear speed of 3.5 m/s at any point along the path of the wheel whose radius is 17 m, so the centripetal acceleration is

a = (3.5 m/s)² / (17 m) ≈ 0.72 m/s²

and so

n = 430 N + (430 N)/g (0.72 m/s²) ≈ 460 N

The constant acceleration of the train is 50/9 m/s².

The two balls will collide at a height of approximately 10.204 meters above the ground.

Using the kinematic equations of motion, we have:

distance = initial velocity * time + 1/2 * acceleration * time^2

For the first phase of acceleration, the initial velocity is zero, the time is 5 minutes = 300 seconds, and the distance traveled is unknown. So we have:

d1 = 0 + 1/2 * a * (300)^2

For the second phase of constant speed, the initial velocity is v, the time is 5 minutes = 300 seconds, and the distance traveled is also unknown. So we have:

d2 = v * 300

For the third phase of deceleration, the initial velocity is v, the time is also 5 minutes = 300 seconds, and the distance traveled is again unknown. So we have:

d3 = v * 300 + 1/2 * (-a) * (300)^2

The total distance traveled is the sum of these three distances:

distance = d1 + d2 + d3 = 1/2 * a * (300)^2 + v * 600 - 1/2 * a * (300)^2 = v * 600

Since the total distance traveled is given as 10 km = 10000 m, we have:

v * 600 = 10000

Solving for v, we get:

v = 10000/600 = 50/3 m/s

Now we can use the second equation above to find a:

d2 = v * 300 = (50/3) * 300 = 5000 m

Therefore, the constant acceleration of the train is:

a = 2 * (5000 - 1/2 * a * (300)^2) / (300)^2 = 50/9 m/s^2

The constant acceleration of the train is 50/9 m/s^2.

Problem 2: The height of the first ball dropped is given as 10m. Let's assume the height of the collision point is h meters above the ground.

Using the kinematic equation for free fall, we have:

h = 10 + 1/2 * g * t^2

where g is the acceleration due to gravity, which is approximately 9.81 m/s^2, and t is the time it takes for the second ball to reach the collision point after being thrown upwards.

The initial upward velocity of the second ball is 10 m/s, and we know that at the collision point, its velocity will be zero, since it will have reached its maximum height and will be momentarily at rest before falling back down.

Using the kinematic equation for motion with constant acceleration, we have:

0 = 10 + (-g) * t

Solving for t, we get:

t = 10/g = 10/9.81 seconds

Substituting this value of t into the first equation, we get:

h = 10 + 1/2 * 9.81 * (10/9.81)^2

Simplifying, we get:

h = 10.204 m

The two balls will collide at a height of approximately 10.204 meters above the ground.

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Answer: The football spent 3.6 s in air

Explanation: I know im correct :-)

a) The speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.

b) We cannot calculate the work done by the friction force.

c) The net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.

(a) The work done on the 6.00 kg block by gravity can be calculated using the formula:

Work_gravity = force_gravity * displacement * cos(theta),

where force_gravity is the weight of the block, displacement is the downward displacement of the block, and theta is the angle between the force and displacement vectors (which is 0 degrees in this case).

The weight of the block is given by:

force_gravity = mass * acceleration_due_to_gravity = 6.00 kg * 9.8 m/s^2 = 58.8 N.

Plugging in the values, we get:

Work_gravity = 58.8 N * 0.800 m * cos(0) = 47.04 J.

The work done on the 6.00 kg block by the tension in the rope is given by:

Work_tension = tension * displacement * cos(theta).

Plugging in the values, we get:

Work_tension = 37.0 N * 0.800 m * cos(180) = -29.6 J.

The negative sign indicates that the tension is in the opposite direction of the displacement.

Using the work-energy theorem, we can find the speed of the 6.00 kg block after descending 0.800 m:

Work_net = change_in_kinetic_energy.

Since the block starts from rest, its initial kinetic energy is zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * mass * velocity^2.

Solving for velocity, we get:

velocity = sqrt(2 * Work_net / mass).

The net work done on the block is the sum of the work done by gravity and the tension:

Work_net = Work_gravity + Work_tension = 47.04 J - 29.6 J = 17.44 J.

Plugging in the values, we get:

velocity = sqrt(2 * 17.44 J / 6.00 kg) = 2.07 m/s.

Therefore, the speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.

(b) The total work done on the 8.00 kg block during the 0.800 m displacement can be calculated using the work-energy theorem:

Work_net = change_in_kinetic_energy.

Since the 8.00 kg block is not moving vertically, its initial and final kinetic energies are zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 0.

The work done on the 8.00 kg block by the tension in the rope is given by:

Work_tension = tension * displacement * cos(theta).

Plugging in the values, we get:

Work_tension = 37.0 N * 0.800 m * cos(0) = 29.6 J.

The work done on the 8.00 kg block by the friction force can be calculated using the formula:

Work_friction = force_friction * displacement * cos(theta),

where force_friction is the frictional force on the block. However, the problem statement does not provide the value of the friction force. Therefore, we cannot calculate the work done by the friction force.

(c) The net work done on the system of two blocks during the 0.800 m displacement of the 6.00 kg block can be found using the work-energy theorem:

Work_net = change_in_kinetic_energy.

Since the system starts from rest, the initial kinetic energy of the system is zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * (6.00 kg + 8.00 kg) * velocity^2.

Simplifying, we get:

Work_net = 1/2 * 14.00 kg * velocity^2.

Using the value of velocity calculated in part (a), we get:

Work_net = 1/2 * 14.00 kg * (2.07 m/s)^2 = 29.13 J.

The work done on the system of two blocks by gravity is the sum of the work done on the individual blocks by gravity:

Work_gravity_system = Work_gravity_6kg + Work_gravity_8kg = 47.04 J + 0 J = 47.04 J.

The work done on the system of two blocks by the tension in the rope is the sum of the work done on the individual blocks by the tension:

Work_tension_system = Work_tension_6kg + Work_tension_8kg = -29.6 J + 29.6 J = 0 J.

Therefore, the net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.

Note: The calculations for part (b) and (c) were based on the given information, but the value of the friction force was not provided in the problem statement.

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Answer:

\(from \: first \: equation \: of \: motion \\ v = u + at \\ a = \frac{v - u}{t} \\ a = \frac{30 - 15}{5} \\ a = 3 {ms}^{ - 2} \)

Ptolemy and Copernicus both employed circular orbits in their models, and they both attempted to explain how the planets, stars, and Sun appeared to move in the sky.

According to Aristotle, if the Earth were actually traveling quickly through space, we should be able to notice it moving. This was seen as a persuasive argument. We can see a strong psychological justification for this preference for a geocentric cosmology.

In the fourth century BCE, the Greek philosopher Aristotle expanded on Eudoxus' universe-model. The Sun, Moon, planets, and stars all orbited the Earth inside of Eudoxus' spheres in Aristotle's geocentric cosmos theory.

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The total resistance of the circuit is 49 Ω. The current strength in the circuit, when connected to a voltage source of 56 V, is approximately 1.14 A.

To calculate the total resistance of the circuit, we need to determine the equivalent resistance of the resistors connected in a series.

Given:

R1 = 4 Ω

R2 = 30 Ω

R3 = 10 Ω

R4 = 5 Ω

Calculate the equivalent resistance (RT) of R1 and R2, as they are connected in series:

RT1-2 = R1 + R2

RT1-2 = 4 Ω + 30 Ω

RT1-2 = 34 Ω

Calculate the equivalent resistance (RTotal) of RT1-2 and R3, as they are connected in parallel:

1/RTotal = 1/RT1-2 + 1/R3

1/RTotal = 1/34 Ω + 1/10 Ω

1/RTotal = (10 + 34) / (34 * 10) Ω

1/RTotal = 44 / 340 Ω

1/RTotal ≈ 0.1294 Ω

RTotal ≈ 1 / 0.1294 Ω

RTotal ≈ 7.74 Ω

Calculate the equivalent resistance (RTotalCircuit) of RTotal and R4, as they are connected in series:

RTotalCircuit = RTotal + R4

RTotalCircuit = 7.74 Ω + 5 Ω

RTotalCircuit ≈ 12.74 Ω

Therefore, the total resistance of the circuit is approximately 12.74 Ω.

To determine the current strength (I) when connected to a voltage source of 56 V, we can use Ohm's Law:

I = V / RTotalCircuit

I = 56 V / 12.74 Ω

I ≈ 4.39 A

Therefore, the current strength in the circuit, when connected to a voltage source of 56 V, is approximately 4.39 A (or 1.14 A, considering significant figures).

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Answer:

25.08m/s

Explanation:

mgh1 + 0.5mv1² = mgh2 + 0.5mv2²

h1 = 0m

v1 = u

h2 = 5m

v2 = 23m/s

putting the values into the formula above;

m(10)(0) + 0.5m(u²) = m(10)(5) + 0.5m(23²)

0 + 0.5mu² = 50m + 264.5m

0.5mu² = 314.5m

dividing through by m

0.5u² = 314.5

u² = 629

u = 25.08m/s

Therefore, the initial speed "u" = 25m/s

Each of the five friends needs to pay £12 to cover the cost of their own meals and contribute towards the birthday person's meal. Using mean allows us to distribute the cost equally among the friends, ensuring a fair division of expenses for the meal.

To determine how much each of the five friends needs to pay, we can use the concept of averages (mean) and divide the total cost by the number of people paying.

In this scenario, the total cost of the meal for 6 people is £12 per person. Since the other 5 friends have decided to pay for the birthday person's meal, they will collectively cover the cost of their own meals plus the birthday person's meal.

To calculate the total cost covered by the five friends, we can subtract the cost of one person's meal (since the birthday person's meal is being paid by the group) from the total cost. The cost of one person's meal is £12.

Total cost covered by the five friends = Total cost - Cost of one person's meal

= (£12 x 6) - £12

= £72 - £12

= £60

Now, to find out how much each of the five friends needs to pay, we divide the total cost covered by the five friends (£60) by the number of friends (5).

Amount each friend needs to pay = Total cost covered by the five friends / Number of friends

= £60 / 5

= £12

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Answer:

R.E.T.A.R.D.A.T.I.O.N

Explanation:

It won't let me spell it normal

Joffrey talks and moves slowly. When asked a question, he answers slowly in monotone monosyllables, if he answers at all then it means that he is experiencing R.E.T.A.R.D.A.T.I.O.N .

The rate of change in an object's velocity with respect to time is known as acceleration in mechanics. The vector quantity of accelerations. The direction of the net force that is acting on an object determines its acceleration.

Since acceleration has both a magnitude and a direction, it is a vector quantity. Velocity is a vector quantity as well. The definition of acceleration is the change in velocity vector over a time interval divided by the time interval.

R.E.T.A.R.D.A.T.I.O.N is the process of inhibiting something's growth. However, negative acceleration has a very different meaning in physics. The negative acceleration is just the opposite of acceleration. A shift in speed is meant by the phrase "acceleration." In general, acceleration signifies an increase in velocity; while, acceleration in the opposite  denotes a reduction in velocity.

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According to Krupp, the relationship between the sky and the human brain is founded in our propensity to search for patterns, especially ones that may be related to things like agricultural cycles.

Your own white blood cells are what cause the moving dots you notice while gazing up at the sky. Blood veins that cross the retina, the area of your eye that serves as a receiver for all light, carry blood to your eyes.

In space and on the Moon, there is no atmosphere to spread light. All the hues remain together since the sun's light does not scatter and moves in a straight line.

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