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4 years ago

What is the relationship between an object’s mass and its ability to pick

Physics

1 answer:

snow_tiger [21]4 years ago

8 0

Answer:

Force = m a

Explanation:

Force is the quantity that relates mass of an object with its accelerate.

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suppose you walk toward the rear of a moving train describe your motion as seen from a reference point on the train then describ

Alja [10]

From the travel car, it would look like you were running. If you were watching from the top of a tree, you would see the train moving.

5 0

4 years ago

PLEASE HELP! How much heat is absorbed by 57g iron skillet when its temperature rises from 11°c to 30°C?

Amiraneli [1.4K]

mass times specific heat times tem change.

57 x sh x (30-11)

convert to mks and look up sh iron

7 0

3 years ago

A pendulum consisting of a 0.5 kg mass tied to a 0.3 m string is set into oscillation at the same moment that a stone is dropped

lara31 [8.8K]

Answer:

2.72 cycles

Explanation:

First of all, let's find the time that the stone takes to reaches the ground. The stone moves by uniform accelerated motion with constant acceleration g=9.8 m/s^2, and it covers a distance of S=44.1 m, so the time taken is

$S=\frac{1}{2}at^2\\t=\sqrt{\frac{2S}{a}}=\sqrt{\frac{2(44.1m)}{9.8 m/s^2}}=3 s$

The period of the pendulum instead is given by:

$T=2 \pi \sqrt{\frac{L}{g}}=2 \pi \sqrt{\frac{0.3 m}{9.8 m/s^2}}=1.10 s$

Therefore, the number of oscillations that the pendulum goes through before the stone hits the ground is given by the time the stone takes to hit the ground divided by the period of the pendulum:

$N=\frac{t}{T}=\frac{3 s}{1.10 s}=2.72$

6 0

3 years ago

A small car of mass m and a large car of mass 4m drive along a highway at constant speeds VS and VL. They approach a curve of ra

Arte-miy333 [17]

Answer:

$v_S=\sqrt{2}v_L$

Explanation:

The acceleration experimented while taking a curve is the centripetal acceleration $a=\frac{v^2}{r}$ . Since $a_S=2a_L$ , we have that: $\frac{v_S^2}{r_S}=\frac{2v_L^2}{r_L}$

They take the same curve, so we have: $r_S=r_L=R$

Which means: $v_S^2=2v_L^2$

And finally we obtain: $v_S=\sqrt{2}v_L$

4 0

4 years ago

PLEASE HELP IM CONFUSED

Fittoniya [83]

Sound waves can be used in a similar way to "see" things. After turning on a sound source, we can look at the pattern of reflected sound waves that bounce back to us. Our own ears and brain don't process sound into mental pictures.

3 0

3 years ago