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4 years ago

8

Three resistors are connected into the section of a circuit described by the diagram. At which labeled point or points of the ci

rcuit could a fourth resistor be added in series with R1?

1 answer:

tangare [24]4 years ago

5 0

Answer:

Pont z only

Explanation:

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What is the difference between a chemical change and a physical change? *

slavikrds [6]

Answer:

I do belive that it is B hrs cn I an gn

7 0

3 years ago

Velocity v (m/s)

Sonja [21]

Explanation:

given solution

h=45m v^2=u^2+2gh

g=10m/s^2 v^2=0^2+2×10m/s^2×45m

vi=0 v^2=900m^2/s^2

v=30

3 0

3 years ago

A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio

Mariulka [41]

Answer:

Yes. Towards the center. 8210 N.

Explanation:

Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to $F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N$

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

$F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3$

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be $8.21\times 10^3~N$

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

5 0

3 years ago

The cockroach Periplaneta americana can detect a static electric field of magnitude 8.50 kN/C using their long antennae. If the

erica [24]

Answer:

0.235 nC

Explanation:

Given:

$E$ = the magnitude of electric field = $8.50\ kN/C =8.50\times 10^{3}\ N/C$

$F$ = the magnitude of electric force on each antenna = $2.00\ \mu N =2.00\times 10^{-6}\ N$

$q$ = The magnitude of charge on each antenna

Since the electric field is the electric force applied on a charged body of unit charge.

$\therefore E = \dfrac{F}{q}\\\Rightarrow q =\dfrac{F}{E}\\\Rightarrow q =\dfrac{2.00\times 10^{-6}\ N}{8.50\times 10^{3}\ N/C}\\\Rightarrow q =0.235\times 10^{-9}\ C\\\Rightarrow q =0.235\ nC$

Hence, the value of q is 0.235 nC.

4 0

3 years ago

Aaron's normal response time to apply the car brakes is 0.7 seconds. Aaron's response time doubles when he is tired. How far wil

Likurg_2 [28]

Aaron's car is moving at speed of 30 m/s

His reaction time is given as 0.7 s

but when he is tired the reaction time is doubled

Now we need to find the distance covered by his car when he is tired during the time when he react to apply brakes

So here since during this time speed is given as constant so we can say that distance covered can be product of speed and time

So here we can use

$d = v*t$

$d = 30 * 1.4$

$d = 42 m$

So the car will move to 42 m during the time when he apply brakes

3 0

3 years ago