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stellarik [79]
3 years ago
11

An object with a lot of mass has ____________.

Physics
1 answer:
aleksley [76]3 years ago
3 0
C. a lot of inertia.............................. :-)
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Globalization is the process of
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a positively charged ion, due to a cosmic ray, is headed through earth's atmosphere toward the center of Earth. Due to Earth's m
Ganezh [65]

Answer:

East direction

Explanation:

Given that

Charge on the particle is positive.

Moving towards the center of earth .

We know that N(north ) pole in magnetic fields work as source of magnetic lines and S(South ) pole works and sink for magnetic lines.

Therefore due to the earth magnetic fields , the positive ions will deflect  towards  East direction.

Thus the answer will be East direction.

4 0
3 years ago
In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land
Marina CMI [18]

Answer:

(a) \vec{J}=176.4\frac{kg*m}{s} \hat{j}

(b) F=2125.30N

Explanation:

(a) According to the law of conservation of energy, the potential energy of the person at 0.40 m is equal to its kinetic energy before the colision with the floor:

\Delta U=\Delta K\\mgh=\frac{mv^2}{2}\\v=\sqrt{2gh}\\v=\sqrt{2(9.8\frac{m}{s^2})(0.40m)}\\v=2.8\frac{m}{s}

This is the initial velocity in the negative y-direction. Impulse is given by:

\vec{J}=\Delta \vec{p}\\\vec{J}=m\vec{v_f}-m\vec{v_i}\\\vec{J}=63kg(0 \hat{j})-63kg(-2.8\frac{m}{s} \hat{j})\\\vec{J}=176.4\frac{kg*m}{s} \hat{j}

(b) The average force is:

F=\frac{J}{\Delta t}\\F=\frac{176.4\frac{kg*m}{s}}{0.083s}\\F=2125.30N

6 0
3 years ago
The rope of a swing is 3.10 m long. Calculate the angle from the vertical at which a 76.0 kg man must begin to swing in order to
kakasveta [241]

Answer:

The angle from the vertical is 48.72°

Explanation:

Given :

Length of rope l = 3.10 m

Mass of man m = 76 Kg

Mass of car M = 1520 Kg

Velocity of car v = 1.01 \frac{m}{s}

According to conservation law,

Potential energy of man is converted to kinetic energy of car moving,

   \frac{1}{2} M v^{2} = mgh

We calculate height,

   h = \frac{M v^{2} }{2mg}

   h = \frac{1520(1.01)^{2} }{2\times 75 \times 9.8}                      ( ∵ g = 9.8 \frac{m}{s^{2} } )

   h = 1.05 m

This is the distance of the rope at the bottom,

So we take difference of it.

⇒ 3.10 - 1.05 = 2.05

We can calculate angle between them,

\cos \theta = \frac{2.05}{3.10}  = 0.6597

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Therefore, the angle from the vertical is 48.72°

8 0
3 years ago
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