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suter [353]
3 years ago
11

Use Kirchoff first law and second law to derise the expression for the total resistancs​

Physics
1 answer:
mestny [16]3 years ago
4 0

Answer:

Chupapi munyanyo

Explanation:

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Which one is made up of one or more molecules
zmey [24]
Ok so I already studied this and I realized it was water or also Oxygen
3 0
3 years ago
10. Which of the following is an acceleration?<br> a. 12 m/s2 down<br> b. 5 m/s up<br> c. 8N West
nadya68 [22]

Answer:

a. 12 m/s² down

Explanation:

Acceleration has units of length per time squared.  Acceleration is a vector, so it also has a direction.

3 0
3 years ago
Teeth are an example of which type of simple machine? lever wedge inclined plane pulley
andreev551 [17]

Answer: Teeth is an example of a wedge.

Explanation :

the machines that make our work easier are called simple machines. Some machines can be compound because they are a combination of more than two simple machines. For example, stapler.

Teeth are an example of a wedge. It is a simple machine which consists of two inclined planes. It is used to split apart objects.  

The mechanical advantage of a wedge is more than 1.

So, the correct option is (b) " Wedge".

3 0
3 years ago
Read 2 more answers
How much energy will be transferred to your eardrum while listening to this sound for 1.0 min?
Phoenix [80]

So E = 2x10^-3W/m^2*(π*(3.0x10^-3m)^2)*1min*60s... = 3.4x10^-6J

5 0
3 years ago
An automobile with a tangential speed of 54.1 km/h follows a circular road that has a radius of 41.6 m. The automobile has a mas
Viefleur [7K]

1) Available force of friction: 6174 N

2) No

Explanation:

1)

The magnitude of the frictional force between the car's tires and the pavement of the road is given by

F_f=\mu mg

where

\mu is the coefficient of friction

m is the mass of the car

g is the acceleration of gravity

For the car in this problem, we have:

\mu=0.500 (coefficient of friction)

m = 1260 kg (mass of the car)

g=9.8 m/s^2

Therefore, the force of friction is

F_f=(0.500)(1260)(9.8)=6174 N

2)

In order to mantain the car in circular motion, the force of friction must be at least equal to the centripetal force.

The centripetal force is given by

F=m\frac{v^2}{r}

where

m is the mass of the car

v is the tangential speed

r is the radius of the curve

In this problem, we have

m = 1260 kg

v=54.1 km/h =15.0 m/s is the tangential speed

r = 41.6 m is the radius of the curve

Therefore, the centripetal force is

F=(1260)\frac{15.0^2}{41.6}=6814 N

Therefore, the force of friction is not enough to keep the car in the curve, since F_f

4 0
3 years ago
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