Answer:

**Answer:**

Δ = 84 Ω, = (40 ± 8) 10¹ Ω

**Explanation:**

The formula for parallel equivalent resistance is

1 / = ∑ 1 / Ri

In our case we use a resistance of each

R₁ = 500 ± 50 Ω

R₂ = 2000 ± 5%

This percentage equals

0.05 = ΔR₂ / R₂

ΔR₂ = 0.05 R₂

ΔR₂ = 0.05 2000 = 100 Ω

We write the resistance

R₂ = 2000 ± 100 Ω

We apply the initial formula

1 / = 1 / R₁ + 1 / R₂

1 / = 1/500 + 1/2000 = 0.0025

= 400 Ω

Let's look for the error (uncertainly) of Re

= R₁R₂ / (R₁ + R₂)

R’= R₁ + R₂

= R₁R₂ / R’

Let's look for the uncertainty of this equation

Δ / = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’

The uncertainty of a sum is

ΔR’= ΔR₁ + ΔR₂

We substitute the values

Δ / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

Δ / 400 = 0.1 + 0.05 + 0.06

Δ = 0.21 400

Δ = 84 Ω

Let's write the resistance value with the correct significant figures

= (40 ± 8) 10¹ Ω

You observe three carts moving to the right. Cart A moves to the right at nearly constant speed. Cart B moves to the right, gradually speeding up. Cart C moves to the right, gradually slowing down. Which cart or carts, if any, experience a net force to the right

A driver drives for 30.0 minutes at 80.0 km/h, then 45.0 minutes at 100 km/h. She then stops 30 minutes for lunch. She then travels for 30 minutes at 80 km/h. (a) Sketch a plot of her displacement versus time and speed versus time. (b) Calculate her average speed.

Sheba is texting and driving. She does not see the baby chinchilasitting in the road! If she was traveling at 35 m/s, and at best hercar can accelerate at -3.5 m/p/s, how far would it take her to stop?

What is the velocity at discharge if the nozzle of a hose measures 68 psi? 100.25 ft./sec 10.25 ft./sec 125.2 ft./sec 11.93 ft./sec

Help!!! Need answer ASAP.

A driver drives for 30.0 minutes at 80.0 km/h, then 45.0 minutes at 100 km/h. She then stops 30 minutes for lunch. She then travels for 30 minutes at 80 km/h. (a) Sketch a plot of her displacement versus time and speed versus time. (b) Calculate her average speed.

Sheba is texting and driving. She does not see the baby chinchilasitting in the road! If she was traveling at 35 m/s, and at best hercar can accelerate at -3.5 m/p/s, how far would it take her to stop?

What is the velocity at discharge if the nozzle of a hose measures 68 psi? 100.25 ft./sec 10.25 ft./sec 125.2 ft./sec 11.93 ft./sec

Help!!! Need answer ASAP.

False

A pressure antinode in a sound wave is not a region of high pressure, while a pressure node is not a region of low pressure.

The answer is false

The answer is false

A pressure antinode in a sound wave is indeed a region of high pressure, while a pressure node is a region of low pressure. These definitions hold** true for all types of waves.**

That's true. In terms of sound waves, a **pressure antinode** is a region of high pressure, while a **pressure node** is a region of low pressure. **This is true for all types of waves, **not only sound waves. In essence, a wave moves through a medium (in case of a sound wave, that medium is typically air) by creating areas of high and low pressure - the high pressure areas are called antinodes, and **the low pressure areas are called nodes.**

#SPJ2

3. Thu Tran, another Physics 2A student, grabs the ball and kicks it straight up to Hoai Nguyen, who is still up on the building rooftop. Assuming that the ball is kicked at 0.50 m above the ground and it goes on a vertical path, what is the minimum velocity required for the ball to make it to the building rooftop? Ignore air resistance. (Hint: the ball will pass the rooftop level with a higher speed...)

**Answer:**

1. 53.415 m

2. 32.373 m/s

3. 30.82 m/s

**Explanation:**

Let g = 9.81 m/s2. We can use the following equation of motion to calculate the distance traveled by the ball in 3.3s, and the velocity it achieved

1.

2.

3. If the ball is kicked at 0.5 m above the ground then the net distance between the ball and the roof top is

53.415 - 0.5 = 48.415 m

For the ball to at least make it to the roof top at speed v = 0 m/s. We can use the following equation of motion to calculate the minimum initial speed

where v = 0 m/s is the final velocity of the ball when it reaches the rooftop, is the initial velocity, is the distance traveled, g = -9.81 is the gravitational acceleration with direction opposite with velocity

**Answer:**

**200 **

**Explanation:**

The computation of the impedance of the circuit is shown below:

Provided that

RMS voltage = 120 v

Frequency = 60.0 Hz

RMS current = 0.600 A

Based on the above information, the formula to compute the impedance is

where,

And,

Now placing these above values to the formula

So, the impedance of the circuit is

= **200 **

**Answer:**

Part a)

Part b)

Part c)

Part d)

**Explanation:**

Part a)

When cabin is fully loaded and it is carried upwards at constant speed

then we will have

net tension force in the rope = mg

now it is partially counterbalanced by 400 kg weight

so net extra force required

now power required is given as

Part b)

When empty cabin is descending down with constant speed

so in that case the force balance is given as

now power required is

Part c)

If no counter weight is used here then for part a)

now power required is

Part d)

Now in part b) if friction force of 800 N act in opposite direction

then we have

now power is

Answer:

The position of the arrows will not be on the target i.e. outside the bull's eye, neither will they be close to one another (widely scattered).

Explanation:

Accuracy refers to the closeness of a measurement to an actual or accepted value while precision refers to the closeness of measurements to one another.

Using archery as an illustration of precision and accuracy, measurements (arrows) that are neither accurate not precise are those arrows that will be far away or outside the bull's eye region (target) of the board and also far apart from one another.

In a nutshell, the arrows will be distant from the bull's eye or target (not accurate) and also distant from one another (not precise).

The answers are :

1) when the sun, moon, and earth are in a line only

**2) when the gravitational forces of the Moon and the Sun are **

perpendicular to one**another with respect to the Earth. **

1) when the sun, moon, and earth are in a line only

perpendicular to one

**Answer:**

High and low tides are result of combined effect of gravitational pull of the sun and the moon. When the two align in a straight line, the range of tides is maximum. This happens on new moon and full moon day.

On the other hand, when the sun and the moon align at right angles, the effect of gravity is minimum and the range of the tides is minimum.