Answer:
s = 1.7 m
Explanation:
from the question we are given the following:
Mass of package (m) = 5 kg
mass of the asteriod (M) = 7.6 x 10^{20} kg
radius = 8 x 10^5 m
velocity of package (v) = 170 m/s
spring constant (k) = 2.8 N/m
compression (s) = ?
Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore
• Ei = Ef
• Ei = energy in the spring + gravitational potential energy of the system
• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}
• Ef = kinetic energy of the object
• Ef = \frac{1}{2}mv^{2}
• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}
• s =
s =
s = 1.7 m
Answer:
a. 5 × 10¹⁹ protons b. 2.05 × 10⁷ °C
Explanation:
Here is the complete question
A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is 0.42 A. (a) How many protons strike the target in 19 seconds? (b) Each proton has a kinetic energy of 6.0 x 10-12 J. Suppose the target is a 17-gram block of metal whose specific heat capacity is 860 J/(kg Co), and all the kinetic energy of the protons goes into heating it up. What is the change in temperature of the block at the end of 19 s?
Solution
a.
i = Q/t = ne/t
n = it/e where i = current = 0.42 A, n = number of protons, e = proton charge = 1.602 × 10⁻¹⁹ C and t = time = 19 s
So n = 0.42 A × 19 s/1.602 × 10⁻¹⁹ C
= 4.98 × 10¹⁹ protons
≅ 5 × 10¹⁹ protons
b
The total kinetic energy of the protons = heat change of target
total kinetic energy of the protons = n × kinetic energy per proton
= 5 × 10¹⁹ protons × 6.0 × 10⁻¹² J per proton
= 30 × 10⁷ J
heat change of target = Q = mcΔT ⇒ ΔT = Q/mc where m = mass of block = 17 g = 0.017 kg and c = specific heat capacity = 860 J/(kg °C)
ΔT = Q/mc = 30 × 10⁷ J/0.017 kg × 860 J/(kg °C)
= 30 × 10⁷/14.62
= 2.05 × 10⁷ °C
Answer:
the magnitude of acceleration will be 1.50m/s^2
Explanation:
To calculate your acceleration, you can use your formula that states that the net force on an object is equal to the mass of the object multiplied by the acceleration of the object. Fnet=ma
if you draw out this situation and label the forces you will have your vector towards the right with a magnitude of 20.0N and then your friction vector will be pointing to the left (in other words, in the negative direction) (opposing the direction of movement) with a magnitude of 5.00N, with the 10.0 kg box in the middle.
The net force will be calculated using F1+F2=Fnet where your F1=20.0N and F2= -5.00N (since it is towards the negative direction).
you will find that Fnet=15.0N
With that, plug in the values you know to calculate the acceleration of the block:
Fnet=ma
(15.0N)=(10.0kg)a from her you can divide both sides by 10 to isolate a:
1.50=a (and now make sure to label the units of your answer)
a=1.50m/s^2 (which is the typical unit for acceleration)
Both hits the ground <u>at the same time</u> because they have <u>same vertical acceleration</u>
<u></u>
<h3>What is vertical acceleration?</h3>
A vertical acceleration is typically one for which the direction of the vector is vertically upward, usually aligned with and opposite to the gravity vector. But this is a descriptive term, not a rigorous or technical term. A car may accelerate along a road and that would generally be assumed to be a horizontal.
The vector perpendicular to this direction, as perhaps a suspension motion over a bump, would be described as vertical even if it is not strictly vertical.
Note that acceleration is defined as the rate of change of the velocity vector. But the gravitation vector, ‘g’, generally vertically downward, is often denoted by what acceleration a mass in free fall (absent air resistance) would experience, i.e. the relationship between mass and weight.
Learn more about vertical acceleration
brainly.com/question/19528199
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Answer:
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Explanation:
Given that there are two force of 1 pound each at right angles to each other.
The from the vector law of addition:

where:
resultant force
be the two of the forces to be added.

