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2 years ago

When an object is turning around, is it also at rest at the point?

Physics

2 answers:

Ganezh [65]2 years ago

8 0

Answer:

yes

Explanation:

Depolarization

AysviL [449]2 years ago

8 0

When an object changes direction, it’s velocity goes to 0 at that point. So the object will also be considered at rest.

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In Challenge Example 11.9 (p. 280), after the explosion, suppose that the m1 fragment shot directly north at 12 m/s and the m3 f

Inga [223]

The question is incomplete. The mass of the object is 10 gram and travelling at a speed of 2 m/s.

Solution:

It is given that mass of object before explosion is,m = 10 g

Speed of object before explosion, v = 2 m/s

Let $m_1, m_2 \text{ and}\ m_3$ be the masses of the three fragments.

Let $v_1, v_2 \text{ and}\ v_3$ be the velocities of the three fragments.

Therefore, according to the law of conservation of momentum,

$mv=m_1v_1 +m_2v_2+m_3v_3$

$10 \times 2 \hat i=3 \times 12 \hat{j} + 3(v_{2x} \hat{i}+v_{2y} \hat{j})-4 \times 9 \hat{j}$

So the x- component of the velocity of the m2 fragment after the explosion is,

$3v_{2x} = 20$

∴ $v_{2x} = 6.67 \ m/s$

6 0

2 years ago

Which will ensure laboratory safety during the experiment? Check all that apply. using beaker tongs to handle the hot beaker rea

Aleks [24]

Answer: • using beaker tongs to handle the hot beaker.

• checking the beaker for chips prior to heating on the hot plate.

• Turning off the hot plate after use

Explanation:

The options that will ensure laboratory safety during the experiment will be:

• using beaker tongs to handle the hot beaker.

• checking the beaker for chips prior to heating on the hot plate.

• Turning off the hot plate after use.

We should note that the beaker tongs are simply used in the holding of the beakers that have hot liquids in them. Also, it s vital for the hot plate to be turned off after its use so as to prevent accident.

7 0

2 years ago

Read 2 more answers

After coming down a slope, a 60-kg skier is coasting northward on a level, snowy surface at a constant 15 m>s. Her 5.0-kg cat

Vinil7 [7]

To solve this exercise, it is necessary to apply the concepts of conservation of the moment especially in objects that experience an inelastic colposition.

They are expressed as,

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where,

$m_1$ = mass of the skier

$m_2$ = mass of the cat

$v_1$ = initial velocity of skier

$v_2$ = initial velocity of cat

$v_f$ = final velocity of both

Re-arrange to find V_f we have,

$V_f = \frac{m_1v_1+m_2v_2}{(m_1+m_2)}$

$V_f = \frac{(60)(15)+(5)(-3.8)}{(60+5)}$

$V_f = 13.55m/s$

Once the final velocity is found it is possible to calculate the change in kinetic energy, so

$\Delta KE = KE_i-KE_f$

$\Delta KE = \frac{1}{2}(m_1v_1^2+m_2v^2_2)-\frac{1}{2}(m_1+m_2)v_f^2$

$\Delta KE = \frac{1}{2}((60)(15)^2+(5)(-3.8)^2)-\frac{1}{2}(60+5)(13.55)^2$

$\Delta KE = 819.1J$

Therefore the amount of kinetic energy converted in to internal energy is 819J

3 0

2 years ago

Which statement below concerning the photoelectric effect is true? View Available Hint(s) Which statement below concerning the p

kakasveta [241]

There is a threshold frequency for each metal, and only light of a frequency higher than this threshold causes electrons to be emitted from the metal surface.

8 0

2 years ago

Very large forces are produced in joints when a person jumps from some height to the ground. Calculate the force produced if an

krok68 [10]

Answer:

A) 31 kJ

B) 1.92 KJ

C) 40 , 2.48

Explanation:

weight of person ( m ) = 79 kg

height of jump ( h ) = 0.510 m

Compression of joint material ( d ) = 1.30 cm ≈ 0.013 m

A) calculate the force

Fd = mgh

F = mgh / d

W = mg

F(net) = W + F = mg ( 1 + $\frac{h}{d} )$

= 79 * 9.81 ( 1 + (0.51 / 0.013) )

= 774.99 ( 40.231 ) ≈ 31 KJ

B) calculate the force when the stopping distance = 0.345 m

d = 0.345 m

Fd = mgh hence F = mgh / d

F(net) = W + F = mg ( 1 + $\frac{h}{d} )$

= 79 * 9.81 ( 1 + (0.51 / 0.345) )

= 774.99 ( 2.478 ) = 1.92 KJ

C) Ratio of force in part a with weight of person

= 31000 / ( 79 * 9.81 ) = 31000 / 774.99 = 40

Ratio of force in part b with weight of person

= 1920 / 774.99 = 2.48

4 0

2 years ago