Answer: d. 8.25 m/s
Explanation:
We are given that Current= 5 m/s in j direction
Velocity= 8 m/s i + 3 m/s j
Now, we have to find Jada's speed with respect to the water.
First we find Jada's velocity with respect to water
v= (8 i + 3 j) - (5 j)
v= 8i - 2 j
To find the speed, we take the magnitude of this velocity vector we have
|v|= 
|v|=
= 8.246 m/s
which comes out to be around = 8.25 m/s
So option d is correct.
Well, the tension in the thread will probably quadruple, but the hanging body will continue to just hang there.
The question gives us no evidence that it is doing any oscillating, and there's no reason for it to start just because it suddenly got heavier.
Answer:
the total momentum is 8 .2 kg m/s in north direction.
Explanation:
given,
mass(m₁) 3.00 kg, moving north at v₁ = 3.00 m/s
mass(m₂) 4.00 kg, moving south at v₂ = 3.70 m/s
mass(m₃) 7.00 kg, moving north at v₃ = 2.00 m/s
north as the positive axis
south as the negative axis
now
total momentum = m₁v₁ + m₂ v₂ + m₃ v₃
total momentum = 3 x 3 - 4 x 3.7 + 7 x 2
= 9 - 14.8 + 14
= 8 .2 kg m/s
hence, the total momentum is 8 .2 kg m/s in north direction.
Answer:
Speed; v = 17 m/s
Explanation:
We are given;
Radius; r = 110m
Angle; θ = 15°
Now, we know that in circular motion,
v² = rg•tanθ
Thus,
v = √(rg•tanθ)
Where,
v is velocity
r is radius
g is acceleration due to gravity
θ is the angle
Thus,
v = √(rg•tanθ) = √(110 x 9.8•tan15)
v = √(288.85)
v = 17 m/s
Answer:
See the answers below.
Explanation:
The cost of energy can be calculated by multiplying each given value, a dimensional analysis must be taken into account in order to calculate the total value of the cost in Rs.
![Cost=0.350[kW]*12[\frac{hr}{1day}]*30[days]*4.5[\frac{Rs}{kW*hr} ]=567[Rs]](https://tex.z-dn.net/?f=Cost%3D0.350%5BkW%5D%2A12%5B%5Cfrac%7Bhr%7D%7B1day%7D%5D%2A30%5Bdays%5D%2A4.5%5B%5Cfrac%7BRs%7D%7BkW%2Ahr%7D%20%5D%3D567%5BRs%5D)
The fuse can be calculated by knowing the amperage.

where:
P = power = 350 [W]
V = voltage = 240 [V]
I = amperage [amp]
Now clearing I from the equation above:
![I=P/V\\I=350/240\\I=1.458[amp]](https://tex.z-dn.net/?f=I%3DP%2FV%5C%5CI%3D350%2F240%5C%5CI%3D1.458%5Bamp%5D)
The fuse should be larger than the current of the circuit, i.e. about 2 [amp]