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3 years ago

13

You are traveling at 100 km/hr to make it to work on time. You have to be to work in .25 hr or you will be late. You have 16 km

to travel. Will you make it on time?

1 answer:

alexgriva [62]3 years ago

4 0

Answer: Yes

Explanation:

Velocity $V$ is defined as the distance traveled $d$ in a specific time $t$ :

$V=\frac{d}{t}$

If you are traveling at $V=100 km/h$ a distance $d=16 m$ , then the time it will take you to be at work is:

$t=\frac{d}{V}=\frac{16 km}{100 km/h}$

$t=0.16 h$

This means you will make it on time, because this time is less than 0.25 h.

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2 years ago

"Two long parallel wires 24.0 cm apart carry currents of 3.0 A and 8.0 A in the same direction. How far from the wire carrying 3

avanturin [10]

Answer:

6.5454 m

Explanation:

Let the distance from the wire carrying 3 A current is x

Then the distance from the the carrying current 8 A is 24-x

We know that magnetic field due to long wire is given by $B=\frac{\mu _0i}{2\pi r}$

It is given that magnetic field is zero at some distance so

$\frac{\mu _0i_1}{2\pi x}=\frac{\mu _0i_2}{2\pi (24-x)}$

Here $i_1=3\ A \ and\ i_2=8\ A$

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3 0

3 years ago

A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th

Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(4)}{2}(2(0)+4)$

$F = 25000 lb$

b) Force on deep end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(10)}{2} (2(0)+10)$

$F = 187500 lb$

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

$F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)$

$x_{1} = 0\\h_{s} = 4ft$

$F = \frac{(62.5)(100)(2)}{2}(2(0)+4)$

$F =25000lb$

2) Force on the triangular part:

$F = \frac{pg(l.h)}{6} (3x_{1} +2h)$

here

h = h(d) - h(s)

h = 10-4

h = 6ft

$x_{1} = 4ft\\$

$F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))$

$F = 150000 lb$

now add both of these forces,

F = 25000lb + 150000lb

F = 175000lb

d) Force on the bottom:

$F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s}) }{2}$

$F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}$

$F = 2187937.5 lb$

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3 years ago

Why do clothes often cling together after tumbling in a clothes dryer?

nadya68 [22]

Answer:

Explained

Explanation:

The electrons get rubbed off some items onto others. This causes an excess or electron on one item and deficiency of electron on the other. Therefore an electrostatic force of attraction is produced and hence clothes often cling together after tumbling in a clothes dryer

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3 years ago