If the mass of an object is 44 kilograms and its velocity is 10 meters per second east, how much Kinetic Energy does it have?

Answer: 760 cW/microgram

2 significant figures

Explanation: solution attached:

Answer:

760 [tex]\frac{cW}{micrograms}[/tex]

Explanation:

You have to know that:

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

So:

[tex]x=\frac{c*b}{a}[/tex]

In this case, you have [tex]\frac{760 GW}{1 kg}[/tex]

In this case, the rule of three is applied as follows: if 1 gigawatt [GW] = 100000000000 centiwatt [cW], 7.600 GW how many cW are they?

[tex]\frac{7.600 GW}{kg} *\frac{100000000000 cW}{1 GW} *\frac{1 kg}{1000000000 micrograms}=760 \frac{cW}{micrograms}[/tex]

Answer:

Original length = 2.97 m

Explanation:

Let the original length of the pendulum be 'L' m

Given:

Acceleration due to gravity (g) = 9.8 m/s²

Original time period of the pendulum (T) = 3.45 s

Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, [tex]L_1=L-1[/tex]

New time period of the pendulum is, [tex]T_1=2.81\ s[/tex]

We know that, the time period of a simple pendulum of length 'L' is given as:

[tex]T=2\pi\sqrt{\frac{L}{g}}[/tex]-------------- (1)

So, for the new length, the time period is given as:

[tex]T_1=2\pi\sqrt{\frac{L_1}{g}}[/tex]------------ (2)

Squaring both the equations and then dividing them, we get:

[tex]\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1[/tex]

Now, plug in the given values and calculate 'L'. This gives,

[tex]L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m[/tex]

Therefore, the original length of the simple pendulum is 2.97 m

The period of a simple pendulum can be calculated with the formula T = 2π√(L/g). When the length of the pendulum is shortened by 1.0m, the period becomes 2.81 seconds. By solving a system of equations related to the period and the length of the pendulum, we can get the original length and the acceleration.

The subject matter relates to physics, specifically kinematics, and the topic is the period of a simple pendulum. The formula to calculate the period of a simple pendulum is T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. We know that when the length of the pendulum was shortened by 1.0m, its period became 2.81 seconds. We can plug these values into our formula and solve for the original length (L).

To find the original length, we first let L0 be the original length and L1 be the shortened length. Therefore, L1 = L0 - 1m. Now we have two equations, T0 = 2π√(L0/g) and T1 = 2π√(L1/g). Plug in the known values and solve the system of equations, you can get the original length and the acceleration. Note, that the typical value for acceleration due to gravity on Earth is g = 9.8 m/s².

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Final answer:

To find the final speed of the roller coaster at the second hill, we can use the conservation of energy equation. Substituting the given values into the equation, we find that the final speed of the roller coaster at the bottom of the first hill is approximately 14.14 m/s. Assuming negligible friction, this will be the speed of the roller coaster at the top of the second hill.

Explanation:

To find the final speed of the roller coaster at the second hill, we can use conservation of energy. At the top of the first hill, the roller coaster has gravitational potential energy which is converted into kinetic energy at the bottom of the first hill.

Using the conservation of energy equation: mgh = (1/2)mv², where m is the mass of the roller coaster, g is the acceleration due to gravity, h is the height of the hill, and v is the final velocity, we can solve for v. Plugging in the values, we have (150 kg)(9.8 m/s²)(50 m) = (1/2)(150 kg)v².

Solving this equation, we find that the final speed of the roller coaster at the bottom of the first hill is approximately 14.14 m/s. Since the work done by frictional forces is negligible, we can assume that the roller coaster will maintain this speed as it goes up the second hill.

Answer

B

Explanation:

i took that same test i'm pretty sure it's B

Nuclear decay forms new B) A radioactive particle

There are many types of nuclear decay:-

Alpha decay

beta decay

gamma decay

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Answer:

weathering

Explanation:

A) Displacement after 6.0 s 43.2 m uphill.

B) Displacement after 9.0 s 43.2 m uphill.

Explanation:

A car moving upwards in a hill is [tex]12 ms^{-1}[/tex].

Its uniform backward acceleration is [tex]-1.6ms^{-2}[/tex]. (since backward acceleration is a negative acceleration, it is mentioned in negative)

We need to find the displacement of the car after some time.

Using the equation of the motion formula, we know can identify the displacement.

D=[tex]vt+\frac{1}{2} at^2[/tex].

a) Displacement after 6.0 seconds,

D = [tex]12(6.0)+\frac{1}{2}(-1.6)(6.0)^2[/tex].

=[tex]72+\frac{1}{2} (36)(-1.6).[/tex]

=[tex]72+\frac{1}{2}(-57.6).[/tex]

=72-28.8.

D=43.2 m.

b) Displacement after 9.0 seconds,

D= [tex]12(9.0)+\frac{1}{2}(-1.6)(9.0)^2[/tex].

=[tex]108+\frac{1}{2} (81)(-1.6).[/tex]

=[tex]108+\frac{1}{2}(-129.6).[/tex]

= 108-64.8.

D=43.2 m.

The car's displacement after 6.0 s is 14.4 m and after 9.0 s is 43.2 m.

To find the displacement of the car after a given time, we can use the equation:
Displacement (d) = Initial velocity (v) * time (t) + (1/2) * acceleration (a) * time^2

a. After 6.0 s:
Initial velocity (v) = 12 m/s
Acceleration (a) = -1.6 m/s^2 (negative because it's a backward acceleration)
Substituting the values into the equation:
d = (12 m/s) * (6.0 s) + (1/2) * (-1.6 m/s^2) * (6.0 s)^2 = 72 m - 57.6 m = 14.4 m

b. After 9.0 s:
Using the same equation and substituting the new time, we can calculate the displacement:
d = (12 m/s) * (9.0 s) + (1/2) * (-1.6 m/s^2) * (9.0 s)^2 = 108 m - 64.8 m = 43.2 m

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Answer:

F = 3.20 N

Explanation:

Given:

Work done by child = 80.2 j

Distance that the car moves = 25.0 m

We need to find the force acting on the car.

Solution:

Using work done formula as.

[tex]W = F\times d[/tex]

Where:

W = Work done by any object.

F = Force (push or pull)

d = distance that the object moves.

Substitute [tex]W = 80.2\ J\ and\ d =25.0\ m[/tex] in work done formula.

[tex]80.2 = F\times 25[/tex]

[tex]F=\frac{80.2}{25}[/tex]

F = 3.20 N

Therefore, force acting on the car F = 3.20 N

Answer:

They have not erupted for a long time

Explanation:

They have not erupted for a long time is correct because dormant volcanoes are volcanoes that are active and have erupted once in the last 10,000 years. Because they are active, they will erupt again in the future.

Answer:

D

Explanation:

the answer is d because gravitational force is what allows them to rotate

hope this was helpful

Answer:

18 mV

Explanation:

The maximum emf is:

ε = N B A ω

where N is the number of turns in the coil,

B is the strength of the magnetic field,

A is the area of the coil,

and ω is the angular velocity.

The number of turns N is equal to the length of the wire divided by the circumference of the coil.

N = L / (2πr)

Area of the coil is:

A = πr²

The angular velocity in rad/s is:

ω = rpm × 2π / 60 s

Therefore:

ε = (L / (2πr)) B (πr²) (rpm × 2π / 60 s)

ε = L B r (rpm × π / 60 s)

Plugging in values:

ε = (1.6 m) (0.071 T) (0.032 m) (93 × π / 60 s)

ε = 0.018 Tm²/s

ε = 18 mV

To find the maximum emf for a rotating coil in a magnetic field, use Faraday's law. The given wire length, rotation speed, and magnetic field strength can be used to calculate the emf after determining the number of turns and the coil area. The calculation results in an approximate maximum emf of 18 mV.

To find the maximum emf induced in a rotating coil within a magnetic field, we can use Faraday's law of electromagnetic induction which states that the emf (ε) induced in a coil is directly proportional to the rate of change of magnetic flux through the coil. Mathematically, ε = NABωsin(ωt+φ), where N is the number of turns, A is the area of the coil, B is the magnetic field strength, ω is the angular velocity, and ωt+φ is the phase angle. In this scenario, the coil is rotating at 93 RPM (revolutions per minute), which we can convert to radians per second (rad/s) since 1 revolution is equal to 2π radians: 93 RPM * 2π rad/rev * 1 min/60 s = 9.735 rad/s. The area (A) of the coil with radius (r = 3.2 cm = 0.032 m) can be found using the formula for the area of a circle: A = πr².

As the wire length is 1.6 m, we need to calculate the number of turns (N) by dividing the length of the wire (L = 1.6 m) by the circumference of the coil (C = 2πr), yielding N = L/C. Finally, to find the peak or maximum emf, we consider that sin(ωt+φ) is 1 at maximum. Plugging in the values, we get the maximum emf:

N = 1.6 m / (2π * 0.032 m) ≈ 7.958 turns (approx. 8 turns since a coil cannot have a fraction of a turn).
A = π * (0.032 m)² ≈ 0.0032 m².
ε(max) ≈ 8 turns * 0.0032 m² * 9.735 rad/s * 7.1 x 10⁻² T.

Calculating this givesapproximately 18 mV as the maximum emf, which should be provided in millivolts (mV) and rounded to two significant figures according to the question, resulting in 18 mV.