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My name is Ann [436]
2 years ago
6

A cruise ship travels across a river at 25 meters per minute. If the river is 6200 meters wide, how long

Physics
1 answer:
EleoNora [17]2 years ago
3 0

Answer:

248 minutes

Explanation:

6200/25=248

This means there is 248 25s in 6200

which means it will take 248 minutes to travel through the river

Also here's a neat trick:

The units for speed is meters/minute

The units for distance is meters

Dividing distance by speed will cancel out the meters and leave only the speed.

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A convex spherical mirror having a radius of curvature 18 cm (focal length = 1/2 radius of curvature for a spherical mirror) pro
tekilochka [14]

Answer:

distance between object and image =  18.9 cm

Explanation:

given data

radius of curvature = 18 cm

focal length = 1/2 radius of curvature

magnification = 40%

to find out

distance between object and image

solution

we know lens formula that is

1/f = 1/v + 1/u     ....................1

here f = 18 /2 and v and u is object and image distance

and we know m = 40% = 0.40

so 0.40 = -v / u

so here v = - 0.40 u

so from equation 1

1/f = 1/v + 1/u

2/18 = - 1/0.40u + 1/u

u = -13.5 cm   ..................2

and

v = -0.40 (- 13.5)

v = 5.4 cm     ......................3

so from equation 2 and 3

distance between object and image =  5.4 + 13.5

distance between object and image =  18.9 cm

6 0
3 years ago
A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele
evablogger [386]

Answer:

a)n= 3.125 x 10^{19 electrons.

b)J= 1.515 x 10^{6 A/m²

c)V_{d =1.114 x 10^{4m/s

d) see explanation

Explanation:

Current 'I' = 5A =>5C/s

diameter 'd'= 2.05 x 10^{-3 m

radius 'r' = d/2 => 1.025 x 10^{-3 m

no. of electrons 'n'= 8.5 x 10^{28}

a) the amount of electrons pass through the light bulb each second can be determined by:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

As we know that: Q= ne

where e is the charge of electron i.e 1.6 x 10^{-19C

n= Q/e => 5/ 1.6 x 10^{-19

n= 3.125 x 10^{19 electrons.

b)  the current density 'J' in the wire is given by

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x 10^{-3)²)

J= 1.515 x 10^{6 A/m²

c) The typical speed'V_{d' of an electron is given by:

V_{d = \frac{J}{n|q|}

    =1.515 x 10^{6 / 8.5 x 10^{28} x |-1.6 x 10^{-19|

V_{d =1.114 x 10^{4m/s

d) According to these equations,

J= I/A

V_{d = \frac{J}{n|q|} =\frac{I}{nA|q|}

If you were to use wire of twice the diameter, the current density and drift speed will change

Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.

Also drift velocity will decrease as it is inversely proportional to the area

 

5 0
2 years ago
Read 2 more answers
A cat with a mass of 4.50 kilograms sits on a ledge 0.800 meters above the ground. If it jumps to the ground, how much kinetic e
Slav-nsk [51]
Answer:

The cat will have <span>36J</span> of kinetic energy.

5 0
3 years ago
Read 2 more answers
A 60 KG skier’s racing down a course at 20 M/S what is the skiers momentum
Xelga [282]

Answer:1200kgm/s

Explanation:

Mass=60kg velocity=20m/s

Momentum=mass x velocity

Momentum=60 x 20

Momentum=1200kgm/s

6 0
3 years ago
PLS SHOW WORK !!!!!!! ASAP
Lynna [10]

Answer:

1- For the track B. The potential energy is the same for the two cars, but because of the slope of the track, the car B earn kinetic energy faster. The gravitation acceleration of the cars will be g•sinθ, and the angle of the track B will have a bigger value for sinθ

2- The conservation of energy applies because the roller coaster is a closed track. When a car climb the track, it earn GPE, which is given by mgh, when it get down in the track, it transform GPE in KE, which is given in 1/2mv².

3-

Position of car (m) GPE KE GPE + KE

top (30m) 60000 0 60000

bottom (0m) 0 60000 60000

halfway down (15m) 30000 30000 60000

three-quarters way down 15000 45000 60000

4 0
2 years ago
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