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svlad2
3 months ago
10

A fox locates rodents under the snow by the slight sounds they make. The fox then leaps straight into the air and burrows its no

se into the snow to catch its meal. If a fox jumps up to a height of 85 cm , calculate the speed at which the fox leaves the snow and the amount of time the fox is in the air. Ignore air resistance.

Physics
1 answer:
Ostrovityanka [3.2K]3 months ago
4 0

Response:

v = 4.08 m/s²

Clarification:

You might be interested in
The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.939 W/m2. The wave is incident u
Yuliya22 [3333]

Answer:

The total energy can be expressed as T = 169.02 \ J

Explanation:

The problem states that

The Poynting vector, which measures energy flux, equals k = 0.939 \ W/m^2

The rectangle's length is represented by l = 1.5 \ m

The width of the rectangle is w = 2.0 \ m

The duration considered is t = 1 \ minute = 60 \ s

Mathematically, the overall electromagnetic energy incident on the area is given by

T = k * A * t

where A denotes the area of the rectangle, calculated as

A= l * w

By plugging in the respective values

A= 2 * 1.5

A= 3 \ m^2

Again substituting values

T = 0.939 * 3 * 60

T = 169.02 \ J

5 0
3 months ago
A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along
Softa [3030]

Answer:

v = \frac{kQ}{a}  

Explanation:

We define the linear charge density as:

\lambda = \frac{Q}{L}

     Where L is the length of the rod, in this scenario the semicircle's length is L = πr

The potential at the center created by a differential element of charge is:

dv = \frac{kdq}{r}

          where k denotes Coulomb's constant

                     r signifies the distance from dq to the center of the circle

Thus.

v = \int_{}^{}\frac{kdq}{a}  

v = \frac{k}{a}\int_{}^{}dq

v = \frac{kQ}{a}     The potential at the semicircle's center

4 0
1 month ago
A locomotive is accelerating at 1.6 m/s2. it passes through a 20.0-m-wide crossing in a time of 2.4 s. after the locomotive leav
kicyunya [3294]

Response:

Once it has crossed, the locomotive requires 17.6 seconds to achieve a speed of 32 m/s.

Details:

  The locomotive's acceleration is 1.6 m/s^2

  The duration taken to pass the crossing is 2.4 seconds.

  We can apply the motion equation, v = u + at, where v represents final velocity, u indicates initial velocity, a denotes acceleration, and t signifies time.

  When the speed reaches 32 m/s, we have v = 32 m/s, u = 0 m/s, and a= 1.6 m/s^2.

   32 = 0 + 1.6 * t

    t = 20 seconds.

  Therefore, the locomotive attains a speed of 32 m/s after 20 seconds, and it passes the crossing in 2.4 seconds.

Thus, after clearing the crossing, it takes an additional 17.6 seconds to reach the speed of 32 m/s.

6 0
3 months ago
Read 2 more answers
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