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2 years ago

Please help I'm stuck on this question

Physics

1 answer:

Afina-wow [57]2 years ago

8 0

Answer:

increase

decrease

Explanation:

using formula

Vt=mg/6πηr

so if m increases V increases

r is the denominator so if r increases V decreases

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A 150 N boy rides a 60 N bicycle a total of 200 m at a constant speed. The frictional force against the forward motion of the bi

soldier1979 [14.2K]

Answer:

W = 7000 J

Explanation:

To solve this problem we use that the speed of the bicycle is constant, therefore its acceleration is zero

F -fr = 0

F = fr

where F is the force applied by the child

Work is defined by

W = F. x

W = F x cos θ

in this case the child's force is parallel to the movement, therefore the angle is zero and cos 0 = 1

let's calculate

W = 35 200

W = 7000 J

8 0

3 years ago

A double nozzle lying in a horizontal x-y plane discharges water into the atmosphere at a rate of 0.5 m3 /s. Assume the water sp

Kisachek [45]

Answer:

The force is $F= 46.25kN$

Explanation:

The diagram for this question is shown on the first uploaded image

At Equilibrium the summation of the of force on the vertical axis is zero

i.e $\sum F_y =0$

=> $F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)$

$v_2$ is the is the speed of water at the nozzle which can be mathematically evaluated as

$v_2 = \frac{R}{A_n}$

substituting $0.5m^3/s$ for R and $\frac{\pi}{4}(12*\frac{1m}{100} )^2$ for $A_n$

$v_2 = \frac{0.5}{\frac{\pi}{4} * (12*\frac{1}{100} )^2 }$

$= 44.23 m/s$

$v_1$ is the is the speed of water at the pipe which can be mathematically evaluated as

$v_1 = \frac{R}{A_p}$

substituting $0.5m^3/s$ for R and $\frac{\pi}{4}(30*\frac{1m}{100} )^2$ for $A_p$

$v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }$

$= 7.07 m/s$

$\rho$ is he density of water with value $\rho =1000 kg /m^3$

Substituting values into the equation above

$F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)$

$= 21.99kN$

At Equilibrium the summation of the of force on the horizontal axis is zero

i.e $\sum F_x =0$

=> $F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)$

Since The speed at both A and B nozzle are the same then $v_2$ remains the same

Substituting values

$F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)$

=> $F_x = 40.69kN$

Hence the force acting on the flange bolts required to hold the nozzle in place is

$F = \sqrt{F_x^2 + F_y^2}$

$= \sqrt{40.69 ^2 + 21.99^2}$

$F= 46.25kN$

6 0

3 years ago

Read 2 more answers

To construct a solenoid, you wrap insulated wire uniformly around a plastic tube 7.1 cm in diameter and 57 cm in length. You wou

ASHA 777 [7]

Answer:

We need about 8769 meters of wire to produce a 2.6 kilogauss magnetic field.

Explanation:

Recall the formula for the magnetic field produced by a solenoid of length L. N turns, and running a current I:

$B=\mu_0\,\frac{N}{L} \,I$

So, in our case, where B = 2.6 KG = 0.26 Tesla; I is 3 amperes, and L = 0.57 m, we can find what is the number of turns needed;

$B=\mu_0\,\frac{N}{L} \,I\\0.26=4\,\pi\,10^{-7}\frac{N}{0.57} \,3\\N=\frac{0.26*0.57\,10^7}{12\,\pi} \\N=39311.27$

Therefore we need about 39312 turns of wire. Considering that each turn must have a length of $\pi\,D$ , where D is the diameter of the plastic cylindrical tube, then the total length of the wire must be:

$Length=39312\,(\pi\,D)=39312\,(\pi\,0.071)\approx 8768.66\,\,m$

We can round it to about 8769 meters.

5 0

3 years ago

An electron is emmited by an atomic nucleus in the process of...

Sliva [168]

I’m pretty sure it’s Radioactive decay

8 0

3 years ago

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A sprinter moving at 10 m/s slows down at a rate of 1.4 m/s2. how fast is the runner moving after 4 seconds?

AURORKA [14]

V=u+at
v=10+(-1.4 (4))
=10-5.6
=4.4m/s

3 0

3 years ago

Read 2 more answers

Please help I'm stuck on this question ​

Please help I'm stuck on this question