7. MANSON'S FAVORITE SINGING GROUP WAS...

" when an earthquake hits, it causes terrible damage," a friend says. " That depends ," another friend answers . on what does it

Elis [28]

Your supposed to lay down on the floor but go up to the buildings roof and do it or get out of the building.

I am not sure though...

7 0

3 years ago

A rescue pilot drops a survival kit while her plane is flying at an ultitude of 2500m with a forward velocity of 95m. If the air

never [62]

Answer:

Approximately $2.1\; \rm km$ , assuming that $g = -9.8\; \rm m \cdot s^{-2}$ .

Explanation:

Let $t$ denote the time required for the package to reach the ground. Let $h(\text{initial})$ and $h(\text{final})$ denote the initial and final height of this package.

$\displaystyle h(\text{final}) = \frac{1}{2}\, g\, t^2 + h(\text{initial})$ .

For this package:

Initial height: $h(\text{initial}) = 2500\; \rm m$ .
Final height: $h(\text{final}) = 0\; \rm m$ (the package would be on the ground.)

Solve for $t$ , the time required for the package to reach the ground after being released.

$\displaystyle t^{2} = \frac{2\, (h(\text{final}) - h(\text{initial}))}{g}$ .

$\begin{aligned} t &= \sqrt{\frac{2\, (h(\text{final}) - h(\text{initial}))}{g} \\ &\approx \sqrt{\frac{2\times (0\; \rm m - 2500\; \rm m)}{(-9.8\; \rm m \cdot s^{-2})}} \approx 22.588\; \rm s\end{aligned}$ .

Assume that the air resistance on this package is negligible. The horizontal ("forward") velocity of this package would be constant (supposedly at $95\; \rm m \cdot s^{-1}$ .) From calculations above, the package would travel forward at that speed for about $22.588\; \rm s$ . That corresponds to approximately: $95\; \rm m \cdot s^{-1} \times 22.588\; \rm s \approx 2.1 \times 10^{3}\; \rm m = 2.1\; \rm km$ .

Hence, the package would land approximately $2.1\; \rm km$ in front of where the plane released the package.

5 0

3 years ago

To practice tactics box 5.2 working with objects in contact. a 1200-kg car pushes a 2100-kg truck that has a dead battery to the

Reptile [31]

Mass of the car = 1200 kg

Mass of the truck = 2100 kg

Total mass of car and truck = 2100 + 1200 = 3300 kg

Since, the car pushes the truck. Hence, they will move together and will have same acceleration.

Let the acceleration be a.

According to Newton's second law:

F(net) = ma

F = 4500 N

4500 = 3300 × a

$a = \frac{4500}{3300}$

a = 1.36 m/s^2

Let the force applied by the car on truck be F.

F = F(net) on the truck

F = ma

F = 2100 × 1.36

F = 2856 N

Hence, the force applied by the car on the truck is 2856 N

6 0

3 years ago

Read 2 more answers

What is the answer to this question?

icang [17]

Answer:

Explanation:

In a velocity/time (aka acceleration) graph, the slope of a line indicates the value of the acceleration in m/s/s. Acceleration is the change in velocity over the change in time. From 0 - 2 seconds, there is no change in velocity, so the acceleration during this interval is 0 (which is the same as the slope of the line). From 2 - 4 seconds, the slope of the line is -2, so the acceleration during the time interval from 2 to 4 seconds is -2 (negative because David is slowing down but is still going the same direction: to the right).

8 0

3 years ago

A tetrahedron has an equilateral triangle base with 25.0-cm-long edges and three equilateral triangle sides. The base is paralle

KATRIN_1 [288]

Answer:

electric flux through the three side = 2.35 N m²/C

Explanation:

given,

equilateral triangle of base = 25 cm

electric field strength = 260 N/C

Area of triangle = $\dfrac{\sqrt{3}}{4}a^2$

= $\dfrac{\sqrt{3}}{4} 0.25^2$

= 0.0271 m³

electric flux = E. A

= 260 × 0.0271

= 7.046 N m²/C

since, tetrahedron does not enclose any charge so, net flux through tetrahedron is zero.

electric flux through the three side = (electric flux through base)/3

= $\dfrac{7.046}{3}$

electric flux through the three side = 2.35 N m²/C

4 0

3 years ago