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3 years ago

Which are the upthrust forces?

Physics

2 answers:

anygoal [31]3 years ago

6 0

Explanation:

An object that is partly or completely submerged experiences a greater pressure on its bottom surface than on its top.

Bond [772]3 years ago

4 0

Answer:

upthrust force is an upward force exerted by a fluid that opposes the weight of a partially or fully immersed object

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An unbanked circular highway curve on level ground makes aturn

tino4ka555 [31]

Answer:

Explanation:

Given

Velocity of traffic $v=60\ mi/hr\approx 26.82\ m/s$

Maximum value of Centripetal force is one-tenth of weight

such that $(F_c)_{max}=\frac{mg}{10}$

to make a safe turn centripetal force with radius of curvature r is given by

$F_c=\frac{mv^2}{r}$

$(F_c)_{max}=\frac{mg}{10}=F_c=\frac{mv^2}{r}$

$\frac{mg}{10}=\frac{mv^2}{r}$

$r=\frac{10v^2}{g}$

$r=\frac{10\times 26.82^2}{9.8}$

$r=733.99\approx 734\ m$

4 0

4 years ago

How much work (MJ) must a car produce to drive 125 miles if an average force of 306 N must be maintained to overcome friction?

baherus [9]

Answer:

The right solution is "61.557 MJ". A further explanation is given below.

Explanation:

The given values are:

Force,

F = 306 N

Drive,

D = 125 miles,

i.e.,

= 201168

meters

As we know,

The work done will be:

= $F\times S$

On substituting the given values, we get

= $306\times 201168$

= $61557408 \ J$

On converting it in "MJ", we get

= $61.557\times 10^6 \ J$

= $61.557 \ MJ$

6 0

3 years ago

A self-driving car traveling along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a spe

Rasek [7]

Answer:

$56.5\ \text{s}$

$21.13\ \text{m/s}$

Explanation:

v = Final velocity

u = Initial velocity

a = Acceleration

t = Time

s = Displacement

Here the kinematic equations of motion are used

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{25-0}{2}\\\Rightarrow t=12.5\ \text{s}$

Time the car is at constant velocity is 39 s

Time the car is decelerating is 5 s

Total time the car is in motion is $12.5+39+5=56.5\ \text{s}$

Distance traveled

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{25^2-0}{2\times 2}\\\Rightarrow s=156.25\ \text{m}$

$s=vt\\\Rightarrow s=25\times 39\\\Rightarrow s=975\ \text{m}$

$v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-25}{5}\\\Rightarrow a=-5\ \text{m/s}^2$

$s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-25^2}{2\times -5}\\\Rightarrow s=62.5\ \text{m}$

The total displacement of the car is $156.25+975+62.5=1193.75\ \text{m}$

Average velocity is given by

$\dfrac{\text{Total displacement}}{\text{Total time}}=\dfrac{1193.75}{56.5}=21.13\ \text{m/s}$

The average velocity of the car is $21.13\ \text{m/s}$ .

6 0

3 years ago

I need help ASAP 8th GRADE SCIENCE

vaieri [72.5K]

Answer:

The answer is gravitational force

Explanation:

4 0

3 years ago

Read 2 more answers

A metal sample of mass M requires a power input P to just remain molten. When the heater is turned off, the metal solidifies in

vovangra [49]

Answer:

L = Pt/M

Explanation:

Power, P= Q/t = mL/t

we know that, (Q=m×l)

Now ⇒l= Pt/M

Thus l= Pt/M

4 0

3 years ago