You slam on the brakes of your car in a panic and skid a distance X on a straight, level road. If you had been traveling half as
fast under the same road conditions, you would have skidded a distance
1 answer:
Answer:
The required skidded distance would be one-forth of the distance skidded if the velocity of the car becomes half.
Explanation:
If '' be the velocity of the car where the brakes are slammed on and 'a' be the constant deceleration gained by the car before coming to stop after skidding a distance 'X' at time 't' where final velocity '
', then as can be seen from the figure,
If from any arbitrarily chosen point 'O', '' be the distance where the brakes are slammed on and '
' be the distance from the same point 'O' where the car stops, then
Now if the car would have been moving with a velocity '' and would have been skidding a distance 'Y', then from the above equation substituting the velocity we can have
So, if the car were moving at half as fast under the same road conditions, it would have skidded by a distance which is one-forth of the present skidded distance.
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Answer:
1020g
Explanation:
Volume of can=
Mass of can=80g=
1Kg=1000g
Density of lead=
By using
We have to find the mass of lead which shot can it carry without sinking in water.
Before sinking the can and lead inside it they are floating in the water.
Buoyancy force =
Where Density of water
Mass of can
Mass of lead
Volume of can
Substitute the values then we get
Hence, 1020 grams of lead shot can it carry without sinking water.
Answer:
a) I = 1,944 10⁻⁴⁶ Kg m²
, b) I = 7,915 10⁻⁵¹ Kg m²
Explanation:
a) The moment of inertia of point masses is
I = m r²
The nuclei have a very small size (10-15 m) so we can consider them punctual.
The distance to the center of mass passing through the middle of the two nuclei of equal mass
r = 0.6 10⁻¹⁰ m
The moment of total inertia
I = I₁ + I₂ = 2 I₀
I = 2 2.7 10⁻²⁶ (0.6 10⁻¹⁰)²
I = 1,944 10⁻⁴⁶ Kg m²
The kinetic energy of the rotation is
w = h / 2π
K = ½ I w²
K = ½ 1,944 10⁻⁴⁶ (h / 2π)² = ½ 1.944 10⁻⁴⁶ (6.63 10⁻³⁴ / 2π)²
K = 2.16 10⁻¹¹⁴ J (1eV / 1.6 10⁻¹⁹ J)
K = 2.16 10⁻⁹⁵ eV
B) the moment of inertia of the electron in the orbit, we can calculate it with the parallel axis theorem
I = + m R²
I = r² +
R²
Where R we calculate it by Pythagoras
R² = (0.6 10⁻¹⁰)² + (0.5 10⁻¹⁰) 2
R = √ (0.61 10⁻²⁰)
R = 0.78 10⁻¹⁰ m
I = 9.1 10⁻³¹ (0.5 10⁻¹⁰)² + 9.1 10⁻³¹ (0.78 10⁻¹⁰)²
I = (2,275 +5.54) 10⁻⁵¹
I = 7,915 10⁻⁵¹ Kg m²
The rotation energy of the electron
K = ½ I w²
If the angular velocity is the electrons outside the core, its kinetic energy is much lower by an order 10⁵, but the angular velocity of the electrons is much higher.
The correct answer is
<span>C) -10.7 m/s
In fact, the first rock is moving upward with velocity +4.5 m/s, while the second rock is moving downward with velocity -6.2 m/s, with respect to a fixed reference frame. In the reference frame of the first rock, instead, the second rock is moving with velocity equal to its velocity in the fixed frame minus the velocity of the reference frame of the first rock:
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<span>C) -10.7 m/s
In fact, the first rock is moving upward with velocity +4.5 m/s, while the second rock is moving downward with velocity -6.2 m/s, with respect to a fixed reference frame. In the reference frame of the first rock, instead, the second rock is moving with velocity equal to its velocity in the fixed frame minus the velocity of the reference frame of the first rock:
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