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18 days ago

5

Alana is skateboarding at 19 km/h and throws a tennis ball at 11 km/h to her friend Oliver who is behind her leaning against a w

all. What is the speed of the tennis ball if someone was watching the exchange on the sidewalk?

1 answer:

Softa [2K]18 days ago

7 0

Answer: 11 km/h

Explanation:

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A copper sphere was moving at 25 m/s when it hit another object. This caused all of the KE to be converted into thermal energy f

kicyunya [2264]

Answer:

$\Delta T = 0.81 ^oC$

Explanation:

According to the principle of energy conservation

all kinetic energy will change into thermal energy to increase its temperature

$\frac{1}{2}mv^2 = ms\Delta T$

Next, divide both sides by the object's mass

$\frac{1}{2}v^2 = s\Delta T$

the resulting temperature change is expressed as

$\Delta T = \frac{v^2}{2s}$

$\Delta T = \frac{25^2}{2\times 387}$

$\Delta T = 0.81^oC$

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12 days ago

A 0.050-kg lump of clay moving horizontally at 12 m/s strikes and sticks to a stationary 0.15-kg cart that can move on a frictio

Ostrovityanka [2204]

Answer:

Explanation:

According to the parameters provided,

mass of the clay lump, m₁ = 0.05 kg

initial velocity of the lump, u₁ = 12 m/s

mass of the cart, m₂ = 0.15 kg

initial speed of the cart, u₂ = 0

As the clay adheres to the cart, we have an inelastic collision scenario. Let v represent the combined speed of both the cart and lump post-collision. Given that momentum is conserved, we have:

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$v=\dfrac{0.05\ kg\times 12\ m/s+0}{0.05\ kg+0.15\ kg}$

The resultant speed is v = 3 m/s.

Thus, the final speed of both cart and lump following the collision is 3 m/s. This concludes the solution.

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19 days ago

A person weighing 55 kg walks by applying 160 N of force on the ground, while pushing a 10-kg object. If the person accelerates

Sav [2226]

Answer:

I'm uncertain

Explanation:

since I didn't provide a correct answer, continue with my inquiries and you can use 'I'm uncertain' for 100 points.

6 0

19 days ago

A hard rubber rod with an electric potential energy of 5.2 × 10–3 J has a charge of 4.0 µC at the tip. What is the electric pote

Keith_Richards [2256]

1) The electric potential energy can be defined as the product of the electric potential and the associated charge:
$U=q V$
where
q refers to the charge
V denotes the electric potential

In this scenario, the charge on the rod is $q=4.0 \mu C = 4.0 \cdot 10^{-6}C$ , and the potential energy is $U=5.2 \cdot 10^{-3} J$ , thus we may rearrange the earlier formula to find the electric potential at the tip:
$V= \frac{U}{q}= \frac{5.2 \cdot 10^{-3}J}{4.0 \cdot 10^{-6} C}=1300 V=1.3 \cdot 10^3 V$

2) Using this same formula, if the charge changes to $q=2.0 \mu C = 2.0 \cdot 10^{-6} C$ , the resulting electric potential will be:
$V= \frac{U}{q}= \frac{5.2 \cdot 10^{-3} J}{2.0 \cdot 10^{-6}C}=2600 V = 2.6 \cdot 10^{3}V$

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10 days ago

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A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through

inna [2205]

Answer: The resulting speed is $\sqrt{2}v_{0}$ . Option (a) stands as the correct choice. Explanation: Given the context, the potential difference entails calculations linked to speed assessment. If instead accelerated through a different potential difference, the resulting speed will be computed accordingly.

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4 days ago