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Oliga [24]
3 years ago
14

Parallel circuits are most likely to be used in which of the following?

Physics
2 answers:
lara31 [8.8K]3 years ago
6 0
They're most likely to be used in a house because if you turn on a switch in ur house the rest of them stay off they don't turn on
SIZIF [17.4K]3 years ago
3 0
The answer should be, D: a house
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You are packing for a trip to another star. During the journey, you will be traveling at 0.99c. You are trying to decide whether
Elenna [48]

Answer:

Do neither of these things ( c )

Explanation:

For length contraction : Is calculated considering the observer moving at a speed that is relative the object at rest applying this formula

L = (l) \sqrt{1 -\frac{v^{2} }{c^{2} } }

where l = Measured distance from object at rest, L =  contracted measured in relation to the observer , v = speed of clock , c = speed of light

you will do neither of these things because before you can make such decisions who have to view the object in this case yourself from a different frame from where you are currently are, if not your length and width will not change hence you can't make such conclusions/decisions .

7 0
3 years ago
On a highway, a car is driven 80. kilometers
nydimaria [60]

The average speed of the car for the entire trip can be calculate by using:

v=\frac{S}{t}

where S is the total distance covered by the car, and t is the total time taken.


The total distance travelled by the car is:

S=80 km+50 km+40 km=170 km

while the total time taken is:

t=1.00 h+0.50 h+0.50 h=2.00 h


so, the average speed of the car is:

v=\frac{S}{t}=\frac{170 km}{2.00 h}=85 km/h


so, the correct answer is (3) 85 km/h.

7 0
4 years ago
Read 2 more answers
On the kelvin scale what is the freezing point of water
Alla [95]
Freezing point of the water is known as 273 K

Hope this helps!
8 0
3 years ago
A pickup truck is traveling down the highway at a steady speed of 30.1 m/s. The truck has a drag coefficient of 0.45 and a cross
Sav [38]

Answer:

The energy that the truck lose to air resistance per hour is 87.47MJ

Explanation:

To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

F_D=\frac{1}{2}\rhoC_dAV^2

Our values are:

V=30.1m/s

C_d=0.45

A=3.3m^2

\rho=1.2kg/m^3

Replacing,

F_D=\frac{1}{2}(1.2)(0.45)(3.3)(30.1)^2

F_D=807.25N

We need calculate now the energy lost through a time T, then,

W = F_D d

But we know that d is equal to

d=vt

Where

v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,

W=(807.25N)(30.1m/s)(3600s/1hr)

W=87.47*10^6J (per hour)

Therefore the energy that the truck lose to air resistance per hour is 87.47MJ

4 0
3 years ago
Could someone please help me with this question?​
Elenna [48]
1 because the the mid night summer is dark
3 0
4 years ago
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