Match the term to its correct definition:

I got the correct answer, but I don't know what I did. Why is the answer E?

MariettaO [177]

Explanation:

Displacement is the straight line distance from the starting position to the final position.

Person X walks halfway around circle. So her displacement is 100 m.

Person Y walks 3/4 of the way around. So his displacement is 50√2 m ≈ 70.7 m.

Person Z walks completely around the circle, so their displacement is 0 m.

Therefore:

Z < Y < X

6 0

2 years ago

A flatbed truck is carrying a 20.0 kg crate along a level road. the coefficient of static friction between the crate and the bed

Dahasolnce [82]

<span>3.92 m/s^2 Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so 0.4 * 9.8 m/s^2 = 3.92 m/s^2 If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so 20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N Multiply by the coefficient of static friction, giving 196 N * 0.4 = 78.4 N So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass 78.4 N / 20.0 kg = 78.4 kg*m/s^2 / 20.0 kg = 3.92 m/s^2 And you get the same result.</span>

6 0

3 years ago

A spring scale hung from the ceiling stretches by 6.4 cm when a 2.0 kg mass is hung from it. The 2.0 kg mass is removed and repl

dolphi86 [110]

In the first case, the force acting on the spring is the weight of the mass:
$F=mg=(2.0 kg)(9.81 m/s^2)=19.6N$
This force causes a stretching of $x=6.4 cm=0.064 m$ on the spring, so we can use these data to find the spring constant:
$k= \frac{F}{x}= \frac{19.6 N}{0.064 m}=306.3 N/m$

In the second case, the first mass is replaced with a second mass, whose weight is
$F=mg=(2.5 kg)(9.81 m/s^2)=24.5 N$
And since we know the spring constant, we can calculate the new elongation of the spring:
$x= \frac{F}{k}= \frac{24.5 N}{306.3 N/m}=0.080 m=8.0 cm$

5 0

3 years ago

What will be the final temperature if a 4.00 g silver ring at 41.0◦C if it gives off 18.0 J of heat to the surroundings? The spe

Eduardwww [97]

Answer:

Final temperature, $T_f=21.85^{\circ}$

Explanation:

Given that,

Mass of silver ring, m = 4 g

Initial temperature, $T_i=41^{\circ}C$

Heat released, Q = -18 J (as heat is released)

Specific heat capacity of silver, $c=0.235\ J/g\ C$

To find,

Final temperature

Solution,

The expression for the specific heat is given by :

$Q=mc\Delta T$

$Q=mc(T_f-T_i)$

$T_f=\dfrac{-Q}{mc}+T_i$

$T_f=\dfrac{-18}{4\times 0.235}+41$

$T_f=21.85^{\circ}$

So, the final temperature of silver is 21.85 degrees Celsius.

5 0

3 years ago

How many light years away is the sun from the middle of the Millky way

julsineya [31]

Answer:

The Milky Way is about 1,000,000,000,000,000,000 km (about 100,000 light years or about 30 kpc) across. The Sun does not lie near the center of our Galaxy. It lies about 8 kpc from the center on what is known as the Orion Arm of the Milky Way

4 0

3 years ago