Most of the Earth's volcanoes occuWhen fossils or minerals form, certain unstable elements are sometimes locked into them. The proportion of these unstable elements gradually decreases over time as they decay into other materials in a predictable way. Scientists use the rate at which such unstable elements decay to determine when the fossils or minerals formed. The technique described above is known as as

Both balls will hit the ground simultaneously because the time it takes to fall is based on the height, not the initial velocity.

The time it takes for an object to fall to the ground depends only on its height and is independent of its initial velocity. In this case, both balls are dropped, so they have an initial velocity of 0 m/s. The ball dropped from 100 ft and the ball dropped from 75 ft will both hit the ground at the same time. This is because the only factor affecting the time it takes to fall is the height, not the initial velocity. Therefore, both balls will hit the ground simultaneously.

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The acceleration of an object in free fall on a planet that has twice the mass of the Earth and twice the Earth's radius is 4.9 m/s², half of the acceleration on Earth.

The acceleration of an object in free fall on a planet depends on the mass and radius of the planet. In this case, the planet has twice the mass of the Earth and twice the Earth's radius. To calculate the acceleration, we can use the formula for gravitational acceleration:

g = (G * M) / R^2

Where g is the acceleration, G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet.

Since the planet has twice the mass and twice the radius of the Earth, we can substitute those values into the formula:

g = (G * 2M) / (2R)^2 = (2 * G * M) / (4 * R^2) = G * M / (2 * R^2)

So the acceleration of an object in free fall on this planet is half of the acceleration on Earth, which is 4.9 m/s².

Answer:

a giant molecular cloud.

Explanation:

The centripetal acceleration of the ball will be 38.68 meter per sq.second.

The acceleration needed to move a body in a curved way is understood as centripetal acceleration.

The direction of centripetal acceleration is always in the path of the center of the course.

The given data in the problem;

r is the radius= 2.00m

frequency (f) = 0.7 rev/s

The angular velocity is found as;

[tex]\rm \omega = 2 \pi f \\\\ \omega = 2 \times 3.14 \times 0.7 \\\\ \omega = 4.398 \ rad/sec[/tex]

The centripetal acceleration is given by;

[tex]\rm a_c= \omega^2r \\\\\ a_c= (4.398)^2 \times 2.00 \\\\ a_c=38.68 \ m/sec^2[/tex]

Hence, the centripetal acceleration of the ball will be 38.68 meter per sq.second.

To learn more about the centripetal acceleration refer to the link;

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The correct answer to the question is gamma radiation . Therefore shielding that blocks gamma radiation, then it will most likely also block the other two types.

Radiation shielding is crucial to protect against the three types of radiation emitted by radioactive elements: alpha, beta, and gamma rays. Different materials are used for shielding based on the type of radiation and its penetrating power. For example, a sheet of paper can stop alpha particles, while thick lead or concrete is needed to absorb gamma rays.

Complete question:

You want to be shielded from all three types of nuclear radiation. If you find shielding that blocks ______ radiation, then it will most likely also block the other two types.

Answer:

Horizontal component: [tex]V_{x}=+5m/s[/tex]

Vertical component: [tex]V_{y}=+8.66m/s[/tex]

Explanation:

We have the launch speed [tex]V =+ 10m/s[/tex] and the angle of 60°, so the components of the speed  [tex]V_{x}[/tex] and  [tex]V_{y}[/tex] are those shown in the attached image.

[tex]V_{x}[/tex] which is the horizontal component of the velocity can be found by multiplying the cosine of the angle by the initial velocity:

[tex]V_{x}=+10m/s(cos60)\\V_{x}=+10m/s(0.5)\\V_{x}=+5m/s[/tex]

[tex]V_{y}[/tex] which is the verticalcomponent of the velocity is found by multiplying the sine of angle by the initial velocity

[tex]V_{y}=+10m/s(sin60)\\V_{y}=+10m/s(0.866)\\V_{y}=+8.66m/s[/tex]

In summary:

Horizontal component: [tex]V_{x}=+5m/s[/tex]

Vertical component: [tex]V_{y}=+8.66m/s[/tex]

Answer: The final velocity is 33.78 mi/h, so the driver did reduced his speed enough.

Explanation: The kinetic energy of an object can be calculated as:

K = (1/2)m*v^2

We know that the mass of the car is m=1100kg

and the initial velocity is 22m/s

The initial kinetic energy is:

K = (1/2)*1100*(22)^2 = 266,200 joules.

Now, if the kinetic energy decreases by 1.4x10^5 J, the new kinetic energy is:

K = 266,200j - 140,000j = 126,200j

So we now can find the new velocity in m/s.

126,200 = (1/2)*1100*v^2

126,200*2/1100 = v^2

229.45 = v^2

v = (229.45)^(1/2) = 15.1 m/s

We know that the limit is 35 mi/h, so we need to transform our result into miles per hour.

We know that in one hour, there are 3600 seconds, so the velocity per hour is:

15.1*3600 m/h = 54,360 m/h

and we know that one mile is 1609.34 meters, so we need to divide by 1609.34.

v = (54,360/1609.34) mi/h = 33.78 mi/h

this is less than the speed limit, so the driver reduced his speed enough.

After losing kinetic energy, the car's final velocity was approximately 18.99 m/s, exceeding the exit's speed limit of 15.64 m/s, hence the driver did not reduce their speed adequately.

To determine whether the driver reduced their speed enough when exiting the highway, we must calculate the car's speed after its kinetic energy decreases by 1.4×105J. The initial kinetic energy (KE) of the 1100-kg car traveling at 22 m/s can be calculated using the equation KE = ½ mv². Plugging in the values, we can find the initial kinetic energy:

KEinitial = ½ (1100 kg)(22 m/s)² = 5.28×105J

After losing 1.4×105J of energy, the remaining kinetic energy will be:

KEfinal = KEinitial - 1.4×105J = (5.28 - 1.4)×105J = 3.88×105J

We can solve for the final velocity (vfinal) using the remaining kinetic energy:

½ (1100 kg)vfinal² = 3.88×105J

vfinal = √((2×3.88×105J) / 1100 kg)

vfinal ≈ 18.99 m/s

To compare to the speed limit, we convert 35 mi/h to meters per second:

35 mi/h × 0.44704 (conversion factor) = 15.64 m/s

Since the final velocity of the car is 18.99 m/s, which is greater than the exit's speed limit of 15.64 m/s, the driver did not reduce their speed enough.