At what constant rate of acceleration will a car starting from rest cover 18m in the first 3 seconds?

When ultra violets lights shine on glass what does it do to electrons in the glass structure?

andreev551 [17]

Answer:

No

Explanation:

7 0

2 years ago

A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. its speed is 40 m/s when i

Alenkinab [10]

Given that,

Initial velocity , Vi = 0

Final velocity , Vf = 40 m/s

Acceleration due to gravity , a = 9.81 m/s²

Distance can be calculated as,

2as = Vf² - Vi²

2 * 9.81 *s = 40² - 0²

s = 81.55 m

For half height, that is, s = 40.77m

Vf= ??

2as = Vf² - Vi²

2 * 9.81 * 40.77 = Vf² - 0²

Vf² = 800

Vf = 28.28 m/s

Therefore, speed of roller coaster when height is half of its starting point will be 28 m/s.

5 0

2 years ago

A T-shirt cannon can shoot a 0.085 kg T-shirt at nearly 30 m/s. The T-shirt cannon has a mass of 33 kg. If the initial net momen

IgorLugansk [536]

Answer:

Approximately $0.077\; {\rm m\cdot s^{-1}}$ (assuming that external forces on the cannon are negligible.)

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum $p$ of that object would be $p = m\, v$ .

Momentum of the t-shirt:

$\begin{aligned} p(\text{t-shirt}) &= m(\text{t-shirt}) \, v(\text{t-shirt}) \\ &= 0.085\; {\rm kg} \times 30\; {\rm m \cdot s^{-1}} \\ &= 2.55 \; {\rm kg \cdot m \cdot s^{-1}} \end{aligned}$ .

If there is no external force (gravity, friction, etc.) on this cannon, the total momentum of this system should be conserved. In other words, if $p(\text{cannon})$ denote the momentum of this cannon:

$p(\text{t-shirt}) + p(\text{cannon}) = 0$ .

$p(\text{cannon}) = -p(\text{t-shirt}) = -2.55\; {\rm kg \cdot m \cdot s^{-1}}$ .

Rewrite $p = m\, v$ to obtain $v = (p / m)$ . Since the mass of this cannon is $m(\text{cannon}) = 33\; {\rm kg}$ , the velocity of this cannon would be:

$\begin{aligned} v(\text{cannon}) &= \frac{p(\text{cannon})}{m(\text{cannon})} \\ &= \frac{-2.55\; {\rm kg \cdot m \cdot s^{-1}}}{33\; {\rm kg}} \\ &\approx 0.077\; {\rm m \cdot s^{-1}}\end{aligned}$ .