I need help with this answer

If an object falls from rest and lands 3.2 seconds later, how many<br> meters did it fall

DIA [1.3K]

Answer:

50.2 meters

Explanation:

When an object falls, it goes a distance d in time t according to the formula:

$d=\frac{1}{2}*g*t^{2}$

d is the distance in meter, g is the acceleration due to gravity with the value of 9.8 $m/s^{2}$ , t is the time in seconds

Therefore, $d = \frac{1}{2}*9.8*(3.2)^{2}$

d= 50.176 ≈ 50.2m

6 0

3 years ago

A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a very light rope that goes over an ideal pulley. If the

AleksAgata [21]

Answer:

acceleration = 2.4525‬ m/s²

Explanation:

Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²

Tension in the rope = T

Sol: m2 > m1

i) for downward motion of m2:

m2 a = m2 g - T

5 a = 5 × 9.81 m/s² - T

⇒ T = 49.05‬ m/s² - 5 a Eqn (a)‬

ii) for upward motion of m1

m a = T - m1 g

3 a = T - 3 × 9.8 m/s²

⇒ T = 3 a + 29.43‬ m/s² Eqn (b)

Equating Eqn (a) and(b)

49.05‬ m/s² - 5 a = T = 3 a + 29.43‬ m/s²

49.05‬ m/s² - 29.43‬ m/s² = 3 a + 5 a

19.62 m/s² = 8 a

⇒ a = 2.4525‬ m/s²

5 0

3 years ago

The maximum heart rate during exercise for a 40 year old is

irina1246 [14]

<span>Target HR Zone 50-85% would be </span>90-153 beats per minuet. The average maximum heart rate 100% is 180 beats per minuet.

4 0

3 years ago

In 2017, the company SpaceX became the first private company to send supplies to the International Space Station with a reusable

pav-90 [236]

Answer:

Approximately $3.98\; \rm m \cdot s^{-2}$ .

Assumption: air resistance on the rocket is negligible. Take $g = \rm 9.81\; m \cdot s^{-2}$ .

Explanation:

By Newton's Second Law of Motion, the acceleration of the rocket is proportional to the net force on it.

$\displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}}$ .

Note that in this case, the uppercase letter $\rm M$ in the units stands for "mega-", which is the same as $10^6$ times the unit that follows. For example, $\rm 1\; Mg = 10^6\; g$ , while $\rm 1\; MN = 10^6\; N$ .

Convert the mass of the rocket and the thrust of its engines to SI standard units:

The standard unit for mass is kilograms: $\displaystyle m = \rm 552\; Mg = 552 \times 10^6\; g \times \frac{1\; \rm kg}{10^3\; g} = 552 \times 10^3 \; kg$ .
The standard for forces (including thrust) is Newtons: $\text{Thrust} = \rm 7.61 \; MN = 7.61 \times 10^6\; N$ .

At launch, the velocity of the rocket would be pretty low. Hence, compared to thrust and weight, the air resistance on the rocket would be pretty negligible. The two main forces that contribute to the net force of the rocket would be:

Thrust (which is supposed to go upwards), and
Weight (downwards due to gravity.)

The thrust on the rocket is already known to be $\rm 7.61 \times 10^6\; N$ . Since the rocket is quite close to the ground, the gravitational acceleration on it should be approximately $9.81\; \rm m \cdot s^{-2} = 9.81 \; N \cdot kg^{-1}$ . Hence, the weight on the rocket would be approximately $9.81\; \rm N \cdot kg^{-1} \times 552 \times 10^3\; kg = 5.41412\times 10^6\; N$ .

The magnitude of the net force on the rocket would be

$\begin{aligned}&\text{Thrust} - \text{Weight} \\ &= 7.61 \times 10^6\; \rm N - 5.41412\times 10^6\; N \\ &\approx 2.19 \times 10^6\; \rm N\end{aligned}$ .

Apply the formula $\displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}}$ to find the net force on the rocket. To make sure that the output (acceleration) is in SI units (meters-per-second,) make sure that the inputs (net force and mass) are also in SI units (Newtons for net force and kilograms for mass.)

$\begin{aligned}\displaystyle &\text{Acceleration} \\ &= \frac{\text{Net Force}}{\text{Mass}} \\ &= \frac{2.19 \times 10^6\; \rm N}{552 \times 10^3\; \rm kg} \\ &\approx \rm 3.98\; \rm m \cdot s^{-2}\end{aligned}$ .

6 0

3 years ago

When you displace an object from its equilibrium position and the force pushing it back toward equilibrium is _____

hammer [34]

Answer

Linear

Explanation:

4 0

3 years ago