Answer:
for question no.4,
time =6 secs
acceleration due to gravity (g)= 9.8 m/s^2
velocity=gt
=6×9.8
=58.8m ans...
for question no .5,
time taken = 4seconds
aacceleration due to gravity (g)=9.8 m/s^2
distance=12gt
=12×9.8×4
=470.4meter answer...
That's right, the correct answer is
<span>A) The isotopes have a long half-life and only remain radioactive for a long time period
The half life of an isotope is the time it takes for the amount of the sample to reduce to half of its initial value. If an isotope has a long half-life, it means it takes a long time to reduce down to a significant level, so it will remain radioactive for a long time period.</span>
Answer:
31 m/s
Explanation:
As both the monkey and the darts are subjected to constant gravitational acceleration g = 9.8 m/s2 and both start from rest (vertically speaking). Their vertical position will always be the same. For the dart to hit the monkey, its horizontal position must be the same as the monkey's, which is unchanged before reaching the ground. Therefore, the time it takes for the dart to travel across 70 m must be less than the time it takes for the monkey to drop 25m to the ground. We can find it out using the following equation of motion
For the dart to takes less that 2.26 s to travel 70m, its horizontal speed must at least be 70 / 2.26 = 31 m/s
The work done by friction to move the sled is - 1,323 J.
<h3>
What is Coefficient of friction?</h3>
- The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them.
- Typically, it is represented by the Greek letter µ. In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
- The coefficient of friction has no dimensions because both F and N are measured in units of force (such as newtons or pounds). For both static and kinetic friction, the coefficient of friction has a range of values.
- When an object experiences static friction, the frictional force resists any applied force, causing the object to stay at rest until the static frictional force is removed. The frictional force opposes an object's motion in kinetic friction.
Solution:
Given that
Coefficient of friction (µ) = 0.10
Mass (m) = 90kg
distance covered (d) = 30m
We use the formula:
friction work = -µmgdcos∅
friction work = -0.100 × 90 kg × 9.8 m/s² × 30 m × cos 60°
friction work = - 1,323 J
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