Answer:
### Deceleration (a) =?

### a = –3.576 m/s²

### Time (t) =?

### t = 8.75 s

### Distance (s) =?

### s = 136.89 m

**A****.**The **time taken** for the **car** to **stop** is **8.75 s**

**B****.**The **distance travelled **when the **brakes** were **applied** till the **car stops** is **136.89 m**

**A**. Determination of the **time taken** for the **car** to **stop**.

**We'll**begin**by****calculating**the**deceleration**of the**car**

Initial velocity (u) = 70 mph = 0.447 × 70 = 31.29 m/s

Final velocity (v) = 30 mph = 0.447 × 30 = 13.41 m/s

Time (t) = 5 s

**Finally****,**we shall**determine**the**time taken**for the**car**to**stop**.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Thus, the **time taken** for the **car** to **stop** is **8.75 s**

**B****.**Determination of the **total distance travelled** when the **brakes** were **applied**.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Therefore, the **total distance travelled **by the **car** when the **brakes** were **applied** is **136.89 m**

Learn more: brainly.com/question/9163788

Answer:

**Answer:8.75 s,**

**136.89 m**

**Explanation:**

Given

Initial velocity

velocity after 5 s is

Therefore acceleration during these 5 s

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

(b)total distance traveled before stoppage

s=136.89 m

A toy car that is 0.12 m long is used to model the actions of an actual car thatis 6 m long. Which ratio shows the relationship between the sizes of themodel and the actual car?A: 5:1B: 1:50C: 50:1D: 1:5

Patricia places 1500g of fruit on a scale compressing it by 0.02m. What is the force constant of the spring on the scale. Hint: you need 2 equations to solve this.

A car with tires of radius 0.25 m come to a stop from 28.78 m/s (100 km/hr) in 50.0 m without any slipping of tires. Find: (a) the angular acceleration of the wheels; (b) number of revolutions made while coming to rest.

What form of braking is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the braketrail braking controlled braking threshold braking coasting

(a) How fast and in what direction must galaxy A be moving if an absorption line found at 550 nm (green) for a stationary galaxy is shifted to 450 nm (blue) for A? (b) How fast and in what direction is galaxy B moving if it shows the same line shifted to 700 nm (red)?

Patricia places 1500g of fruit on a scale compressing it by 0.02m. What is the force constant of the spring on the scale. Hint: you need 2 equations to solve this.

A car with tires of radius 0.25 m come to a stop from 28.78 m/s (100 km/hr) in 50.0 m without any slipping of tires. Find: (a) the angular acceleration of the wheels; (b) number of revolutions made while coming to rest.

What form of braking is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the braketrail braking controlled braking threshold braking coasting

(a) How fast and in what direction must galaxy A be moving if an absorption line found at 550 nm (green) for a stationary galaxy is shifted to 450 nm (blue) for A? (b) How fast and in what direction is galaxy B moving if it shows the same line shifted to 700 nm (red)?

The three given moon phases are:

Waxing Crescent Moon

Waning Crescent Moon

Full Moon

The things we need to match to the above three topicsare:

*visible near western horizon an hour after sunset

*rises about the same time the sun sets

*visible near eastern horizon just before sun rises

*occurs about 3 days before new moon (i know this is waningcrescent)

*visible due south at midnite

*occurs 14 days after new moon (i know this is full moon)

**Answer:**

**Explanation:**

*visible near western horizon an hour aftersunset **WAXING CRESCENT**

*rises about the same time the sun sets **FULLMOON**

*visible near eastern horizon just before sun rises **WANING CRESCENT**

*occurs about 3 days before new moon **WANING CRESCENT**

*visible due south at midnite ** FULL MOON**

*occurs 14 days after new moon ** FULL MOON**

b) decrease

The pressure decreases if the area increases. If the area decreases then the pressure increases.

**Answer:**

**Explanation:**

Using the diagram (see attachment) we extract the following position vectors:

Next step is to find unit vectors as follows:

Using the diagram we find the corresponding vectors Forces:

Equation of Equilibrium:

Comparing i , j and k components as follows:

Solving Above equation simultaneously we get:

1 cm = 100 m

1 mm = 100 cm

100 mm = 1 cm

1 m = 100 cm

**Answer:**

The last one

1m = 100 cm

**Explanation:**

If you do not trust me look it up

**We are given:**

The tuning fork vibrates at 15660 oscillations per minute

**Period of one back-and forth movement:**

the given data can be rewritten as:

1 minute / 15660 oscillations

60 seconds / 15660 oscillations *(1 minute = 60 seconds)*

*dividing the values*

**0.0038 seconds / Oscillation**

**Therefore, one back and forth vibration takes 0.0038 seconds**

wireequal to the strength of the Earth's magnetic field of about

5.0 x10^-5 T?

**Answer:**

The distance is 2 cm

**Solution:**

According to the question:

Magnetic field of Earth, B_{E} =

Current, I = 5.0 A

We know that the formula of magnetic field is given by:

where

d = distance from current carrying wire

Now,

d = 0.02 m 2 cm