What outside force slows down a parachute?

When performing a battery load test, what should the load be set to?

ASHA 777 [7]

Answer:

here you go

Explanation:

1. With a battery load tester, apply a load equal to one-half of the CCA rating of the battery for 15 seconds. ...

2. With a battery load tester, apply a load equal to one-half the vehicle's CCA specification for 15 seconds.

3. Disable the ignition and turn the engine over for 15 seconds with the starter motor.

3 0

3 years ago

The net electric flux through a cubic box with sides that are 22.0 cm long is 4950 N⋅m2/C . What charge is enclosed by the box?

Bond [772]

Answer:

43.83 nC

Explanation:

We know Q = ε₀Ψ where Q = charge enclosed by the cubic box, ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m and Ψ = net electric flux through cubic box = 4950 Nm²/C

So, substituting the values of the variables into the equation, we have

Q = ε₀Ψ

= 8.854 × 10⁻¹² F/m × 4950 Nm²/C

= 43827.3 × 10⁻¹² C = 4.38273 × 10⁻⁸ C

= 43.8273 × 10⁻⁹ C

= 43.8273 nC

≅ 43.83 nC

5 0

3 years ago

Write about the similarities and differences between kinetic and potential energy. Include specific, real world example in your

lord [1]

Answer:

Energy is transformed from potential to kinetic and vice versa

Explanation:

The energy is transformed from mechanical to kinetic energy when the object changes its position with respect to a reference point, where it loses height but increases its speed. When the object is at maximum height with respect to a reference point, it will have its maximum potential energy value. When the object passes through the reference point it will have potential energy equal to zero, but this energy will become kinetic energy.

The most characteristic and real example is that of a pendulum at one end, as can be seen in the attached image.

When the pendulum is located at the top end, as shown in Figure 1, at that point the maximum potential energy will be held. Then the pendulum is released and when it passes through the reference point and its height is zero, with respect to that point, all potential energy will have become kinetic energy in the same way at this point the maximum speed of the pendulum will be set.

8 0

3 years ago

A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?

Tcecarenko [31]

The weight of the meterstick is:
$W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N$
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
$d_1 = 0.50 m - 0.40 m=0.10 m$
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
$M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm$

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
$(mg) d_2 = 0.20 Nm$
from which we find the value of d2:
$d_2 = \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m$

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.

4 0

4 years ago

A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly

Mnenie [13.5K]

Answer:

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The acceleration of the tortoise is 0.28 m/s²

Explanation:

The equations that describe the position and velocity of the hare and the tortoise are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

To find the acceleration of the hare once it begins to slow down, we have to find how much time the hare traveled during the deceleration and what was its initial speed.

First, the hare moves with constant acceleration for 4.7 s. Then, its velocity at t = 4.7 s will be:

v = v0 + a · t (v0 = 0 because the hare starts form rest)

v = a · t = 0.9 m/s² · 4.7 s = 4.2 m/s

The distance traveled by the hare while accelerating can be calculated using the equation of the position:

x = x0 + v0 · t + 1/2 · a · t² (x0 = 0 and v0 = 0)

x = 1/2 · a · t² = 1/2 · 0.9 m/s² · (4.7)² = 9.9 m

Then, the hare runs at a constant speed of 4.2 m/s for 11.7 s. The distance traveled at constant speed will be:

x = v · t

x = 4.2 m/s · 11.7 s = 49.1 m

Then, the distance traveled by the hare while slowing down was:

Distance traveled while slowing down = 72 m - 49.1 m - 9.9 m = 13 m

Let´s find how much time it took the hare to come to stop, so we can calculate the acceleration. We know that when the position is 13 m, the velocity is 0.

v = v0 + a · t

0 = 4.2 m/s + a · t

-4.2 m/s / t = a

Replacing in the equation of the position:

x = v0 · t + 1/2 · a · t² (considering x0 as the point at which the hare started to slow down)

13 m = 4.2 m/s · t - 1/2 · 4.2 m/s / t · t²

13 m = 4.2 m/s · t - 2.1 m/s · t

13 m = 2.1 m/s · t

t = 13 m / 2.1 m/s

t = 6.2 s

Then, the acceleration of the hare while slowing down will be:

-v0/t = a

-4.2 m/s / 6.2 s = a

a = -0.68 m/s²

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The hare traveled 72 m in (6.2 s + 11.7 s + 4.7 s) 22.6 s. The tortoise reaches the final position of the hare at the same time, so, using the equation of the position we can calculate the acceleration of the tortoise:

x = x0 + v0 · t + 1/2 · a · t² (x0 = 0 and v0 = 0)

x = 1/2 · a · t²

72 m = 1/2 · a · (22.6 s)²

144 m / (22.6 s)² = a

a = 0.28 m/s²

The acceleration of the tortoise is 0.28 m/s²

6 0

3 years ago