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3 years ago

Unscramble this caybnetruopergs * starts with "T" ends with "S" Please help me

1 answer:

8 0

Typebars
Turbocars
Teacups
Teapoys
Tourneys
Toecaps
Trepangs
Tabourers
Trouncers
Troupers
Terraces
Trances
Turbans
Toupees
Toucans
Sorry if this is not what you want me to do and I just threw a bunch of words at you.

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the mass of a lump of gold is constant everywhere but its weight isnt explain both in weight and mass

DiKsa [7]

That is very true, but what is the question asking you.

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3 years ago

Read 2 more answers

A motorboat maintained a constant speed of 15 miles per hour relative to the water in going 10 miles upstream and then returning

Lesechka [4]

Answer:

speed of current is 5 mile/hr

Explanation:

GIVEN DATA:

speed of motorboat = 15 miles/hr relative with water

let c is speed of current

15-c is speed of boat at upstream

15+c is speed of boat at downstream

we know that

travel time=distance/speed

$\frac{10}{15-c} +\frac{10}{15+c} = 1.5$

150+10c+150-10c=1.5(15-c)(15+c)

300=1.5(225-c^2)

300=337.5-1.5c^2

200=225-c^2

c^2=25

c = 5

so speed of current is 5 mile/hr

6 0

3 years ago

A combination lock has a 1.3-cm-diameter knob that is part of the dial you turn to unlock the lock. To turn that knob, you twist

galina1969 [7]

Answer:

0.04225 Nm

Explanation:

N = Force applied = 5 N

$\mu$ = Coefficient of static friction = 0.65

d = Diameter of knob = 1.3 cm

r = Radius of knob = $\frac{d}{2}=\frac{1.3}{2}=0.65\ cm$

Force is given by

$F=N\mu\\\Rightarrow F=5\times 0.65\\\Rightarrow F=3.25\ N$

When we multiply force and radius we get torque

Torque on thumb

$\tau_t=F\times r\\\Rightarrow \tau_t=3.25\times 0.0065\\\Rightarrow \tau_t=0.021125\ Nm$

Torque on forefinger

$\tau_f=F\times r\\\Rightarrow \tau_f=3.25\times 0.0065\\\Rightarrow \tau_f=0.021125\ Nm$

The total torque is given by

$\tau=\tau_t+\tau_f\\\Rightarrow \tau=0.021125+0.021125\\\Rightarrow \tau=0.04225\ Nm$

The most torque that exerted on the knob is 0.04225 Nm

4 0

3 years ago

Pwease help me with thesse three

DiKsa [7]

Answer:

1. Spring

2. 23.5

3. Summer

7 0

3 years ago

At what height h above the ground does the projectile have a speed of 0.5v?

dedylja [7]

Answer:height above ground at which projectile have velocity

0.5v is (0.0375v^2)

Explanation:

Using Vf = Vi - gt

Where Vf is final velocity

Vi is initial velocity

g is the acceleration due to gravity

t is the time taken

So, 0.5v = v - gt

t = 0.05v

Therefore height h = vt - 0.5gt^2

Subtitute t

h = 0.05v^2 - 0.0125v^2

h = 0.0375v^2

3 0

3 years ago