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The 68-kg crate is stationary when the force P is applied. Determine the resulting acceleration of the crate if (a) P = 0, (b) P

bogdanovich [222]

Explanation:

Mass of the crate, m = 68 kg

We need to find the resulting acceleration if :

(a) Force, P = 0

P = m a

⇒ a = 0

(b) P = 181 N

$a=\dfrac{P}{m}$

$a=\dfrac{181\ N}{68\ kg}$

$a=2.67\ m/s^2$

(c) P = 352 N

$a=\dfrac{P}{m}$

$a=\dfrac{352\ N}{68\ kg}$

$a=5.17\ m/s^2$

Hence, this is the required solution.

5 0

3 years ago

A jet liner must reach a speed of 82 m/s for takeoff. If the

SIZIF [17.4K]

Answer:

The acceleration that the jet liner that must have is 2.241 meters per square second.

Explanation:

Let suppose that the jet liner accelerates uniformly. From statement we know the initial ( $v_{o}$ ) and final speeds ( $v_{f}$ ), measured in meters per second, of the aircraft and likewise the runway length ( $d$ ), measured in meters. The following kinematic equation is used to calculate the minimum acceleration needed ( $a$ ), measured in meters per square second:

$a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot d}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $v_{f} = 82\,\frac{m}{s}$ and $d = 1500\,m$ , then the acceleration that the jet must have is:

$a = \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (1500\,m)}$

$a = 2.241\,\frac{m}{s^{2}}$

The acceleration that the jet liner that must have is 2.241 meters per square second.

3 0

2 years ago

How does an airbag help protect a passenger in a car accident?

Novosadov [1.4K]

Good question. It like a fail safe,if you hit a car tail gate going slightly fast it come out to protect you from hitting your on the the wheel or the mirror. See if you don't have a seat belt or it just don't fully stop you the airbag might help.Like concussions. I Hope This Help you :)

8 0

3 years ago

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The global positioning system (GPS) uses electromagnetic waves sent between computer devices on Earth and satellites in space to

zhenek [66]

Answer:

first plz mark brarinlest

Explanation:

5 0

2 years ago

Read 2 more answers

Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2