The answer is five because if you do 8×n=40 and if you count by five you would get the answer is 5
Answer:
The mass of the Al-duckie should be 30 kg.
Explanation:
We will use the first law of thermodynamics:
ΔU = m·Cv·ΔT
Since the specific heat of water is 4.185 J(gºC), the change in the water's internal energy would be:
ΔU = 100 kg · 4.185 J(gºC) · (42ºC - 38ºC) = 1674 KJ
Given that no heat is lost, all the internal energy that the water loses while cooling down will transfer to the duckie. So, if the duckie has ΔU = 1674 KJ and its final temperature is the desired 38 ºC, we can calculate its mass using the first law again:
![m=\frac{\Delta{U}}{Cv{\Delta{T}}}=\frac{1674}{0.9*[38-(-24)]}=30Kg](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B%5CDelta%7BU%7D%7D%7BCv%7B%5CDelta%7BT%7D%7D%7D%3D%5Cfrac%7B1674%7D%7B0.9%2A%5B38-%28-24%29%5D%7D%3D30Kg)
Answer:
Explanation:
Geologists record the seismic waves and study how they travel through Earth. The speed of these seismic waves and the paths they take reveal how the planet is put together. Using data from seismic waves, geologists have learned that Earth's interior is made up of several layers.
Answer:
Explanation:
Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.
1 ) Common acceleration a = force / total mass
= 234 / ( 25 +91 )
= 2.017 m s⁻².
2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂
= mass x acceleration
= 25 x 2.017
= 50.425 N
The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.
3 ) Maximum friction force that is possible between m₁ and m₂
= μ_s m₁g
= .79 x 25 x 9.8
= 193.55 N
Acceleration of m₁
= 193 .55 / 25
= 7.742 m s⁻²
This is the common acceleration in case of maximum tension required
So tension in rope
= ( 25 +91 ) x 7.742
= 898 N
4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁
= μ_k m₁g
= .62 x 25 x 9.8
= 151.9 N
Acceleration of m₁
= 151.9 / 25
= 6.076 m s⁻².
Answer:
Acceleration=4m/s²
Force applied =619.8N
Explanation:
Using equation of motion
V=u+at we have: u=o, v=50m/s
50= 0 + a×0.0121
a = 50/0.0121
a= 4m/s²
Neglecting resistance forces
F= ma, where a = v-u/t
F=m×(v-u)/t
F= 0.150 ×(50-0)/0.0121
F=7.5/0.0121
F= 619.8N
