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2 years ago

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Ball A was carried to the top of a hill in a straight line, while ball B was carried in a longer, zigzag path. At the top of the

hill, both balls were left at rest. A student draws a graph that depicts each ball's final potential energy. Which statement describes the final potential energy of the balls?(1 point)
Both balls have equal potential energy.

Neither ball has any potential energy.

Ball A has more potential energy.

Ball B has more potential energy.

2 answers:

lisov135 [29]2 years ago

6 0

Answer: 1. Both balls have equal potential energy.

2. Potential energy would decrease, while total mechanical energy would remain constant.

3. A slide

4. Mass is eliminated when equating gravitional potential energy with kinetic energy.

5. 1/2 mv^2 = 1/2 kx^2

Explanation:

I did the test by pure guess and failed but oh well, got the answers

marissa [1.9K]2 years ago

5 0

Answer: equal

Explanation:

Don’t know why honestly

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(b) The density of aluminum is 2.70 g/cm3. The thickness of a rectangular sheet of aluminum foil varies

vekshin1

Answer:

(i) $22.48 cm^3$

(ii) $1.5 mm$

Explanation:

Let t be the average thickness of the sheet.

Given that:

Density of the aluminum sheet is $2.70 g/cm^3$

Mass of sheet = 60.7 g

Length of sheet = 50.0 cm

Width of sheet = 30.0 cm

(i) Using, Density=Mass/Volume

$\Rightarrow \text{Volume}=\frac{\text{Mass}}{\text{Density}}$

$\Rightarrow \text{Volume}=\frac{60.7}{2.7}=22.48 cm^3$

Hence, the volume of the sheet is $22.48 cm^3$ .

(ii) Now, as this aluminum sheet is in the shape of a cuboid, so the volume of the sheet is

$\text{Volume}=\text{Length}\times\text{Width}\times\text{Thickness}$

$\Rightarrow 22.48=50\times 30 \times w$

$\Rightarrow w=\frac{22.48}{50\times 30}=0.015 cm$

Hence, the average thickness of the sheet is $1.5 mm$ .

6 0

2 years ago

Subatomic particles that do not possess any charge but provide mass to atoms are called

WARRIOR [948]

Answer:

Neutrons

Explanation:

Neutrons are subatomic particles that are electrically neutral and possess no charge in them.

6 0

2 years ago

A girl is running toward the front of a train at 10 m/s. If the train is going 75 m/s on the Southbound tracks, what is the spee

dimulka [17.4K]

We can take as positive direction the direction of the train, and as negative direction the direction of the girl. The speed of the train relative to the girl will be given by the difference between the two velocities:
$v' = v_t - v_g$
which means
$v'=75 m/s - (-10 m/s)=85 m/s$

5 0

2 years ago

Read 2 more answers

A student attaches a block to a vertical spring so that the block-spring system will oscillate if the block-spring system is rel

vodka [1.7K]

Answer:

Time period of the motion will remain the same while the amplitude of the motion will change

Explanation:

As we know that time period of oscillation of spring block system is given as

$T= 2\pi\sqrt{\frac{M}{k}}$

now we know that

M = mass of the object

k = spring constant

So here we know that the time period is independent of the gravity

while the maximum displacement of the spring from its mean position will depends on the gravity as

$mg = kx$

$x = \frac{mg}{k}$

so we can say that

Time period of the motion will remain the same while the amplitude of the motion will change

4 0

2 years ago

Suppose we replace the mass in the video with one that is four times heavier. How far from the free end must we place the pivot

Llana [10]

We must place the pivot to keep the meter stick in balance at 90 cm (10 cm from the weight) from the free end.

Answer: Option B

<u>Explanation:</u>

In initial stage, the meter stick’s mass and mass hanged in meter stick at one end are same. Refer figure 1, the mater stick’s weight acts at the stick’s mid-point.

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

$m \times g \times(x)+((m \times g)(x-50 \mathrm{cm}))=0$

$(m \times g \times x)-(50 \times m \times g)+(m \times g \times x)=0$

Taking out ‘mg’ as common and we get

$2 x-50=0$

$2 x=50$

$x=\frac{50}{2}=25 \mathrm{cm}$

Hence, the stick should be pivoted at a distance of,

$x^{\prime}=100 \mathrm{cm}-25 \mathrm{cm}=75 \mathrm{cm}$

So, the stick should be pivoted at a distance of 75 cm at the free end

Now, replace mass with another mass. i.e., four times the initial mass (as given)

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

$4 m g(x)+(m g)(x-50 c m)=0$

$4 m g x+m g x-50 m g=0$

Taking out ‘mg’ as common and we get

$5 x=50$

$x=\frac{50}{5}=10 \mathrm{cm}$

Hence, the stick should be pivoted at a distance of,

$x^{\prime}=100 \mathrm{cm}-10 \mathrm{cm}=10 \mathrm{cm}$

So, the stick should be pivoted at a distance of 10 cm from the free end.

Therefore, the option B is correct 90 cm (10 cm from the weight).

3 0

2 years ago