What is the difference between constant and variable force ?

Cat walks 100 m East, then turns around and walks 25 m west. What is the cats displacement?

kvasek [131]

Answer:

Explanation:

If a cat walk 100m East, this means that it is walking in the positive x direction, the distance will therefore be +100m

If it turns around and walks 25 m west, the direction of movement is in the negative x direction i.e -25m

Taking the sum;

Displacement = +100m - 25m

Displacement of the cat = 75m

hence the cats displacement is 75m

8 0

3 years ago

A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t1 = 4.50 s, it is at

wariber [46]

Answer:

See it in the pic

Explanation:

See it in the pic

7 0

3 years ago

Any five physics problems

kow [346]

Explanation:

There are still some questions beyond the Standard Model of physics, such as the strong CP problem, neutrino mass, matter–antimatter asymmetry, and the nature of dark matter and dark energy.

7 0

2 years ago

Which option describes a phase change

anastassius [24]

Answer:

b. Gasloine evaporating into the air

I tryed to answer

6 0

3 years ago

Read 2 more answers

A block with mass m = 0.200 kg is placed against a compressed spring at the bottom of a ramp that is at an angle of 53.0∘ above

Murljashka [212]

Answer:

$0.566\; \rm N$ (assuming that while the block is moving, the friction on the block is constant.)

Explanation:

The mechanical energy of a system is the sum of its:

elastic potential energy,
gravitational potential energy, and
kinetic energy.

Friction does work on the block as the block moves up the ramp. The amount of energy that the block-string system has lost would be equal in size to the work that friction has done on the block. The size of the friction on the block could thus be computed.

Before this block was released, the block-spring system has no kinetic energy because there was no movement. Assume that the system has no gravitational potential energy at that moment, either. The only type of mechanical energy in this system at that moment would be elastic potential energy: $8\; \rm J$ according to the question.

The question states that the ramp is $53.0^\circ$ above horizontal. Therefore, after the block has traveled $3.00\; \rm m$ (along the ramp,) the height of this block would have increased by $\Delta h = 3.00\; \rm \sin\left(53^\circ\right) \approx 2.39591\; \rm m$ . Calculate the corresponding gain in gravitational potential energy:

$\begin{aligned}& m \cdot g \cdot \Delta h\\ &\approx 0.200\; \rm kg& \\\ &\quad\times 9.81\; \rm N \cdot kg^{-1} \\ &\quad \times 2.39591\; \rm m \\ &\approx 4.70077\; \rm J\end{aligned}$ .

On the other hand, the question states that the speed of the block ( $m = 0.200\; \rm kg$ ) at that moment is $v = 4.00\; \rm m \cdot s^{-1}$ . Calculate the corresponding kinetic energy:

$\begin{aligned}& \frac{1}{2}\, m \cdot v^{2} = 1.6\; \rm J \end{aligned}$ .

Because the block was already released, there should be no elastic potential energy in the spring.

Hence, the mechanical of the block-spring system would be approximately $4.70077\; \rm J+ 1.6\; \rm J \approx 6.30077\; \rm J$ .

Approximate the amount of mechanical energy that is lost:

$8\; \rm J - 6.30077\; \rm J \approx 1.69923\; \rm J$ .

In other words, when applied over $3\; \rm m$ , the friction on this block would do approximately $1.69923\; \rm J$ of work. Approximate the size of that friction:

$\begin{aligned}F &= \frac{W}{s} \\ &\approx \frac{1.69923\; \rm J}{3.00\; \rm m}\approx 0.566\; \rm N\end{aligned}$ .

3 0

3 years ago