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FinnZ [79.3K]
3 years ago
14

A truck pulls a caravan at 85 km/h [S]. When they hit a bump in the road the caravan detaches! If the caravan decelerates at a r

ate of 1.3 m/s² [N] along the road, how far it will travel before it stops?​
Physics
1 answer:
Brrunno [24]3 years ago
5 0

The caravan has an initial speed of

85 km/h × (1000 m/km) × (1/3600 h/s) ≈ 23.611 m/s

If it slows down at a constant rate, then it will cover a distance <em>x</em> such that

0² - (23.611 m/s)² = 2 (-1.3 m/s²) <em>x</em>

Solve for <em>x</em> :

<em>x</em> = -(23.611 m/s)² / (2 (-1.3 m/s²)) ≈ 214.417 m ≈ 210 m

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If a specimen was being viewed using a 20x objective lens and 10x ocular lens, what would be the total magnification
Paraphin [41]

Answer:

As Per Given Information

20x objective lens was used by specimen

10x ocular lens was also used by him.

we have to find the total magnification.

For calculating the total magnification we 'll simply do multiplication

Total Magnification = 20x × 10x

Total Magnification = 200x

So , the total magnification will be 200x .

6 0
2 years ago
An adiabatic nozzle has an inlet area of 1 m^2 and an outlet area of 0.25 m^2. Water enters the nozzle at a rate of 5 m^3/s and
svlad2 [7]

Answer:

v=20m/S

p=-37.5kPa

Explanation:

Hello! This exercise should be resolved in the next two steps

1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed

Q=VA

for he exitt

Q=flow=5m^3/s

A=area=0.25m^2

V=Speed

solving for V

V=\frac{Q}{A} \\V=\frac{5}{0.25} =20m/s

velocity at the exit=20m/s

for entry

V=\frac{5}{1} =5m/s

2.

To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

\frac{P1}{\alpha } +\frac{v1^2}{2g} =\frac{P2} {\alpha } +\frac{v2^2}{2g}

where

P=presure

α=9.810KN/m^3 specific weight for water

V=speed

g=gravity

solving for P1

(\frac{p1}{\alpha } +\frac{V1^2-V2^2}{2g})\alpha  =p2\\(\frac{150}{9.81 } +\frac{5^2-20^2}{2(9.81)})9.81  =p2\\P2=-37.5kPa

the pressure at exit is -37.5kPa

7 0
4 years ago
The magnetic field at point P due to a 2.0-A current flowing in a long, straight, thin wire is 8.0 μT. How far is point P from t
Tanzania [10]

Answer:

r = 0.05 m = 5 cm

Explanation:

Applying ampere's law to the wire, we get:

B = \frac{\mu_oI}{2\pi r}\\\\r =  \frac{\mu_oI}{2\pi B}

where,

r = distance of point P from wire = ?

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

I = current = 2 A

B = Magnetic Field = 8 μT = 8 x 10⁻⁶ T

Therefore,

r = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(2\ A)}{2\pi(8\ x\ 10^{-6}\ T)}\\\\

<u>r = 0.05 m = 5 cm</u>

8 0
3 years ago
A test charge of -1.4 x 10-7 coulombs experiences a force of 5.4 x 10-1 newtons. Calculate the magnitude of the electric field c
VARVARA [1.3K]

Answer:

3.86×10⁶ Newton/coulombs

Explaination:

Applying,

E = F/q....................... Equation 1

Where E = Electric Field, F  = Force, q = charge.

From the question,

Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs

Substitute these values into equation 1

E = 5.4×10⁻¹/ -1.4×10⁻⁷

E = -3.86×10⁶ Newtons/coulombs

Hence the magnitude of the electric field created by the

negative test charge is 3.86×10⁶ Newton/coulombs

5 0
3 years ago
Question 4 How much time does it take to walk 8 km north at a velocity of 3.8 km/h?​
IrinaVladis [17]

Given parameters:

Displacement = 8km

Velocity  = 3.8km/h

Unknown:

time  = ?

Solution:

Velocity is displacement divided by time.

  Velocity  = \frac{displacement}{time}  

      Displacement  = velocity x time

Input the parameters:

              8  = 3.8  x time

 Time  = \frac{8}{3.8}   = 2.1s

The time taken is 2.1s

6 0
3 years ago
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