Newton's second law states that $F=m \times a$ (Force is mass times acceleration). Out of the given options, “a 10 kg ball thrown with a 50 Newton force”, an example have the greatest acceleration.

Explanation: 

To do the calculation, we already have the formula given derived from Newton’s second law of motion. To know the acceleration, we can simply modify the formula as,

$\text { Force }=\text { mass } \times \text { acceleration }$

$\text {acceleration}=\frac{\text {Force}}{\text {mass}}$

For the 10kg ball threw in 50N, we have mass = 10 kg and Force = 50 N

Acceleration = $\frac{50}{10}=5 \mathrm{m} / \mathrm{s}^{2}$

Similarly for 1 kg ball threw in 0.5N, substituting the values, we get ,

Acceleration = $\frac{0.5}{1} = 0.5 \mathrm{m} / \mathrm{s}^{2}$

For launching 50kg student by catapult of 100N,

acceleration = $\frac{100}{50}=2 m / s^{2}$

For accelerating 500 kg car in 1000N engine,

acceleration = $\frac{1000}{500} = 2 m / s^{2}$

8 0

3 years ago

The rigid beam is supported by the three suspender bars. bars ab and ef are made of aluminum and bar cd is made of steel. if eac

faltersainse [42]

Answer:

Pmax = 67.5 KN

Explanation:

We need to calculate the maximum allowable value of P for both aluminum and steel bars.

FOR STEEL BARS:

Since,

(σallow)st = (Pmax)st/A

where,

(σallow)st = maximum allowable stress of steel bar = 200 MPa = 2 x 10⁸ Pa

A = Cross-sectional area of steel bar = 450 mm² = 0.45 x 10⁻³ m²

(Pmax)st = Maximum allowable force for steel bar = ?

Therefore,

2 x 10⁸ Pa = (Pmax)st/0.45 x 10⁻³ m²

(Pmax)st = (2 x 10⁸ Pa)(0.45 x 10⁻³ m²)

(Pmax)st = 9 x 10⁴ N = 90 KN

FOR Aluminum BARS:

Since,

(σallow)al = (Pmax)al/A

where,

(σallow)al = maximum allowable stress of Al bar = 150 MPa = 1.5 x 10⁸ Pa

A = Cross-sectional area of Aluminum bar = 450 mm² = 0.45 x 10⁻³ m²

(Pmax)al = Maximum allowable force for Aluminum bar = ?

Therefore,

1.5 x 10⁸ Pa = (Pmax)al/0.45 x 10⁻³ m²

(Pmax)al = (1.5 x 10⁸ Pa)(0.45 x 10⁻³ m²)

(Pmax)al = 6.75 x 10⁴ N = 67.5 KN

Since,

(Pmax)al < (Pmax)st

Therefore,

The maximum allowable force will be:

Pmax = (Pmax)al

Pmax = 67.5 KN

3 0

4 years ago