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Alex777 [14]
3 years ago
8

Jimmy walks 3 miles east then 1 mile west. What is the distance Jimmy traveled?

Physics
1 answer:
AleksAgata [21]3 years ago
4 0

Answer:

2 miles

Explanation:

Jimmy walked forward 3 miles east (to the right)

Then, he backtracked and walked 1 mile west (to the left)

This can be written as:

3 - 1 = ?

3 - 1 = 2

Therefore, Jimmy traveled a total of 2 miles.

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On a hot day, you sit on the edge of a pool and dip your feet into the water, causing changes to occur that are related to the p
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Answer: d

Explanation:

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3 years ago
An unstoppable object is heading right toward an unmovable object. What's going to happen?
marshall27 [118]

Answer:

In my opinion the unstoppable object will hit the unmovable object and stop but the wheels will still be rolling and trying to move but can't.

<h3>Hope this helps.</h3><h3>Good luck ✅.</h3>
8 0
2 years ago
Arigid body must rotate about an axis in order for it to have angular momentum about that axis. True False
kompoz [17]

Answer:

False

Explanation:

Let's consider the definition of the angular momentum,

\vec{L} = I \vec{\omega}

where I = \int\limits_m r^2 dm = \lim_{n \to \infty} \sum\limits_{i=1}^n m_i r_i^2 is the moment of inertia for a rigid body. Now, this moment of inertia could change if we change the axis of rotation, because "r" is defined as the distance between the puntual mass and the nearest point on the axis of rotation, but still it's going to have some value. On the other hand,

\vec{\omega} = \frac{\vec{r} \times \vec{v}}{r^2} so \vec{\omega} \neq 0 unless \vec{r} ║  \vec{v}.

In conclusion, a rigid body could rotate about certain axis, generating an angular momentum, but if you choose another axis, there could be some parts of the rigid body rotating around the new axis, especially if there is a projection of the old axis in the new one.

7 0
3 years ago
A 5 kg toy is tied to a rope where the tension measures 150 N. What is the weight of the object?
elena-s [515]
Formula from physics to get the answer.
4 0
3 years ago
A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force cons
shepuryov [24]

Answer:

v=2.556m/s

Explanation:

From the conservation of mechanical energy

K_{E1}+U_1=K_{E2}+U_2

\frac{1}{2}m*v_1^2+\frac{1}{2}*K*x_1^2=\frac{1}{2}m*v_2^2+\frac{1}{2}*K*x_2^2

x_2=0.08m

v_1=0 m/s

Solve to velocity v2

m*v_2^2=k*x_1^2-k*x_2^2

v^2=\frac{k}{m}*(x_1^2-x_2^2)

v^2=\frac{5.5N/m}{0.24kg}*(0.54m^2-0.080^2)

v=\sqrt{6.54m^2/s^2}=2.556m/s

4 0
3 years ago
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