Answer:
b.it depends on the distance it falls
Answer:
1:2
Explanation:
It is given that,
Initial RMS AC voltage is 100 V and final RMS AC voltage is 200 V.
We need to find the ratio of the number of turns in the primary to the secondary for step up transformer.
For a transformer, ![\dfrac{V_1}{V_2}=\dfrac{N_1}{N_2}](https://tex.z-dn.net/?f=%5Cdfrac%7BV_1%7D%7BV_2%7D%3D%5Cdfrac%7BN_1%7D%7BN_2%7D)
So,
![\dfrac{N_1}{N_2}=\dfrac{100}{200}\\\\\dfrac{N_1}{N_2}=\dfrac{1}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7BN_1%7D%7BN_2%7D%3D%5Cdfrac%7B100%7D%7B200%7D%5C%5C%5C%5C%5Cdfrac%7BN_1%7D%7BN_2%7D%3D%5Cdfrac%7B1%7D%7B2%7D)
So, the ratio of the number of turns in the primary to the secondary is 1:2.
Answer:
60 kg m/s
Explanation:
Let
be the acceleration of the object.
As the acceleration of the object is constant, so
![a=\frac {v-u}{t}\cdots(i)](https://tex.z-dn.net/?f=a%3D%5Cfrac%20%7Bv-u%7D%7Bt%7D%5Ccdots%28i%29)
Given that applied force, F=6.00 N,
From Newton's second law, we have
,
[from equation (i)]
![\Rightarrow Ft=m(v-u)](https://tex.z-dn.net/?f=%5CRightarrow%20Ft%3Dm%28v-u%29)
![\Rightarrow Ft=mv-mu](https://tex.z-dn.net/?f=%5CRightarrow%20Ft%3Dmv-mu)
[given that time, t=10 s and F=6 N]
![\Rightarrow mv-mu=60 kg \;m/s](https://tex.z-dn.net/?f=%5CRightarrow%20mv-mu%3D60%20kg%20%5C%3Bm%2Fs)
Here mv is the final momentum of the object and mu is the initial momentum of the object.
So, the change in the momentum of the object is mv-mu.
Hence, the change in the momentum of the object is 60 kg m/s.
In order for the object to move upward, it needs an upward force
that's at least equal to its own weight.
Weight = (mass) x (gravity) = (35 kg) x (9.8 m/s²) = 343 N.
The engine thrust has to be more than 343 N.
Answer:
A) ![Q_a=74256\ J](https://tex.z-dn.net/?f=Q_a%3D74256%5C%20J)
B) ![Q=93562560\ J](https://tex.z-dn.net/?f=Q%3D93562560%5C%20J)
Explanation:
Given:
- temperature of air,
![T_a=-19+273=254\ K](https://tex.z-dn.net/?f=T_a%3D-19%2B273%3D254%5C%20K)
- temperature of lungs,
![T_l=37+273=310\ K](https://tex.z-dn.net/?f=T_l%3D37%2B273%3D310%5C%20K)
- specific Heat exchanged from the lungs ,
![c_l=0.47\ J.kg^{-1}.K^{-1}](https://tex.z-dn.net/?f=c_l%3D0.47%5C%20J.kg%5E%7B-1%7D.K%5E%7B-1%7D)
- specific heat of air,
![c_a=1020\ J.kg^{-1}.K^{-1}](https://tex.z-dn.net/?f=c_a%3D1020%5C%20J.kg%5E%7B-1%7D.K%5E%7B-1%7D)
- mass of 1 L air,
![m'=1.3\ kg](https://tex.z-dn.net/?f=m%27%3D1.3%5C%20kg)
- breath rate,
![b=21\ breath.min^{-1}](https://tex.z-dn.net/?f=b%3D21%5C%20breath.min%5E%7B-1%7D)
A)
Now,
amount of heat needed to warm the air of lungs to the body temperature:
![Q_a=m'.c_a.\Delta T](https://tex.z-dn.net/?f=Q_a%3Dm%27.c_a.%5CDelta%20T)
![Q_a=1.3\times1020\times (310-254)](https://tex.z-dn.net/?f=Q_a%3D1.3%5Ctimes1020%5Ctimes%20%28310-254%29)
![Q_a=74256\ J](https://tex.z-dn.net/?f=Q_a%3D74256%5C%20J)
B)
Amount of heat lost per hour:
<u>No. of breaths per hour:</u>
![B=b.60](https://tex.z-dn.net/?f=B%3Db.60)
![B=21\times 60](https://tex.z-dn.net/?f=B%3D21%5Ctimes%2060)
![B=1260](https://tex.z-dn.net/?f=B%3D1260)
<u>Now the total loss of energy in 1 hr.:</u>
![Q=Q_a.B](https://tex.z-dn.net/?f=Q%3DQ_a.B)
![Q=74256\times 1260](https://tex.z-dn.net/?f=Q%3D74256%5Ctimes%201260)
![Q=93562560\ J](https://tex.z-dn.net/?f=Q%3D93562560%5C%20J)