The answer to this is Animal habitation
Answer:
Yes it will move and a= 4.19m/s^2
Explanation:
In order for the box to move it needs to overcome the maximum static friction force
Max Static Friction = μFn(normal force)
plug in givens
Max Static friction = 31.9226
Since 36.6>31.9226, the box will move
Mass= Wieght/g which is 45.8/9.8= 4.67kg
Fnet = Fapp-Fk
= 36.6-16.9918
=19.6082
=ma
Solve for a=4.19m/s^2
Answer:
The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s
Explanation:
Given that,
Mass of halfback = 98 kg
Speed of halfback= 4.2 m/s
Mass of corner back = 85 kg
Speed of corner back = 5.5 m/s
We need to calculate their mutual speed immediately after the touchdown-saving tackle
Using conservation of momentum
![m_{h}v_{h}+m_{c}v_{c}=m_{h+c}v_{h+c}](https://tex.z-dn.net/?f=m_%7Bh%7Dv_%7Bh%7D%2Bm_%7Bc%7Dv_%7Bc%7D%3Dm_%7Bh%2Bc%7Dv_%7Bh%2Bc%7D)
Where,
= mass of halfback
=mass of corner back
= velocity of halfback
= velocity of corner back
Put the value into the formula
![98\times4.2+85\times5.5=(98+85)\times v](https://tex.z-dn.net/?f=98%5Ctimes4.2%2B85%5Ctimes5.5%3D%2898%2B85%29%5Ctimes%20v)
![v=\dfrac{98\times4.2+85\times5.5}{98+85}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7B98%5Ctimes4.2%2B85%5Ctimes5.5%7D%7B98%2B85%7D)
![v=4.80\ m/s](https://tex.z-dn.net/?f=v%3D4.80%5C%20m%2Fs)
Hence, The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s
-- The resistance of the heater is (volts/current) = 5 ohms
-- The heating (RMS) value of a sinusoidal AC is V(peak)/√2 . For this particular alternator, V(peak)=100V, so the heating (RMS) equivalent is 70.71 V.
-- The heating power delivered to the electric heater is (E²/R).
Power = (100/√2)² / 5
Power = 5,000 / 5
<u>Power = 1,000 watts </u>
Answer:
the ans is D... good luck