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irina1246 [14]
3 years ago
7

A ball rolls from x = - 5m to x = 0m in 1 second . What was its average velocity? (Units = m/s) Don't forget: velocities and dis

placements to the right are +, to the left are -
Physics
1 answer:
katrin2010 [14]3 years ago
8 0

Answer:

+ 5 m/s

Explanation:

change in displacement = ΔX=final position - initial position

                                            ΔX = 0-(-5) =0+5 =+ 5 m

average velocity =  ΔX/t

                             = +5/1

                             = + 5 m/s

positive sign shows that ball rolls towards right

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93 cm3 liquid has a mass of 77 g. When calculating its density what is the appropriate number of significant figures
lubasha [3.4K]

Answer:

828 kg/m³ or 0.828 g/cm³

Explanation:

Applying,

D = m/V............. Equation 1

Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.

From the question,

Given: m = 77 g , V = 93 cm³

Substitute these values into equation 1

D = 77/93

D = 0.828 g/cm³

Converting to kg/m³

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6 0
3 years ago
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Vanyuwa [196]
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3 years ago
You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t
Sladkaya [172]

Answer:

0.767

Explanation:

The work done on the truck by the frictional drag force is given by

W=-Fd

where

F is the magnitude of the frictional force

d = 38.0 m is the maximum displacement allowed for the truck

The negative sign is due to the fact that the force of friction is opposite to the motion of the truck

The force of friction can also be written as:

F=\mu mg

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\mu is the coefficient of kinetic friction between the truck and the lane

m is the mass of the truck

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So we can rewrite the work done as

W=-\mu mg d (1)

According to the work-energy theorem, the work done by friction is equal to the change in kinetic energy of the truck:

W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (2)

where

v = 0 is the final velocity of the truck

u = 23.9 m/s is the initial velocity of the truck

By combining (1) and (2) we get

-\frac{1}{2}mu^2 = -\mu mg d

And solving for \mu, we find the minimum coefficient of kinetic friction able to stop the truck in a distance d:

\mu = \frac{u^2}{2gd}=\frac{23.9^2}{2(9.8)(38.0)}=0.767

7 0
3 years ago
Car A, moving in a straight line at a constant speed of 20. meters per second, is initially 200 meters behind car B, moving in t
MA_775_DIABLO [31]
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For Car B: y= 15x      Car B moves 15 meters every second and starts at our basis point

Set the two equations equal to one another to find the time x at which they meet:
20x - 200 = 15x
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D , since Voltage is one joule per coulomb
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