- A horse cannot pull a cart and run in empty space.
- Passengers are thrown forward from their seats when a speeding bus stops suddenly.
- It is easier to pull a lawn mower than to push it.
- A cricketer moves his hands backwards while holding a catch.
An empty space is devoid of any such reaction force. Therefore, a horse cannot pull a cart and run in empty space.

The vertical component of this applied force acts upward. This reduces the effective weight of the mower. On the other hand, while pushing a lawn mower, a force at an angle $\theta$ is applied on it, as shown in the following figure.

In this case, the vertical component of the applied force acts in the direction of the weight of the mower. This increases the effective weight of the mower.
Since the effective weight of the lawn mower is lesser in the first case, pulling the lawn mower is easier than pushing it.
$\text{F}=\text{ma}=\frac{\text{m}\triangle\text{v}}{\triangle\text{t}}\ ....(\text{i})$
Where,
F = Stopping force experienced by the cricketer as he catches the ball
m = Mass of the ball
$\triangle\text{t}$ = Time of impact of the ball with the hand
It can be inferred from equation (i) that the impact force is inversely proportional to the impact time, i.e.,
$\frac{\text{F}\propto1}{\triangle\text{t}}\ ...(\text{ii})$
Equation (ii) shows that the force experienced by the cricketer decreases if the time of impact increases and vice versa.
While taking a catch, a cricketer moves his hand backward so as to increase the time of impact $(\triangle\text{t})$. This is turn results in the decrease in the stopping force, thereby preventing the hands of the cricketer from getting hurt.
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| Temperature | Pressure thermometer A | Pressure thermometer B |
| Triple-point of water | 1.250 × 105Pa | 0.200 × 105Pa |
| Normal melting point of sulphur | 1.797 × 105Pa | 0.287 × 105Pa |
What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?
What do you think is the reason behind the slight difference in answers of thermometers A and B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?