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s344n2d4d5 [400]
3 years ago
10

Under new homeland security measures boats must keep at least how many hours away from all military vessels and ships

Physics
1 answer:
crimeas [40]3 years ago
5 0
Do not approach within 100-yards, and slow to minimum speed within 500-yards of any U.S. naval vessel.
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The drag on a pitched baseball can be surprisingly large. Suppose a 145 g baseball with a diameter of 7.4 cm has an initial spee
kupik [55]

Answer:

<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2>v_f = 35.3 m/s</h2>

Explanation:

Part a)

As we know that drag force is given as

F = \frac{C_d \rho A v^2}{2}

C_d = 0.35

A = \frac{\pi d^2}{4}

A = \frac{\pi(0.074)^2}{4}

A = 4.3 \times 10^{-3} m^2

v = 40.2 m/s

so we have

F = \frac{0.35\times 1.2 (4.3 \times 10^{-3})(40.2)^2}{2}

F = 1.46 N

So acceleration of the ball is

a = \frac{F}{m}

a = \frac{1.46}{0.145}

a = 10.1 m/s^2

Part B)

As per kinematics we know that

v_f^2 - v_i^2 = 2 a d

v_f^2 - 40.2^2 = 2(-10.1)(18.4)

v_f = 35.3 m/s

4 0
3 years ago
In which one of the following situations is zero net work done? A) A ball rolls down an inclined plane. B) A physics student str
lidiya [134]

Answer:

Option D

Explanation:

The work done can be given by the mechanical energy used to do work, i.e., Kinetic energy and potential energy provided to do the work.

In all the cases, except option D, the energy provided to do the useful work is not zero and hence work done is not zero.  

In option D, the box is being pulled with constant velocity, making the acceleration zero and thus Kinetic energy of the system is zero. Hence work done in this case is zero.

6 0
3 years ago
Read 2 more answers
If the sun exploded would the solar system collaspe?
Sav [38]

Answer:

Yes, yes it would since we need light

Explanation:

5 0
3 years ago
A box weighs 100N and its base area of 2 m2. What pressure does it exert on the ground?
kotegsom [21]

Answer:

P = F/S = 100/2 =50 (N/m2)

5 0
3 years ago
34. A train, starting from rest, accelerates along the platform at a uniform rate of 0.6 m/s2. A passenger standing on the platf
ira [324]

Answer:

4.08 s

Explanation:

Let the passenger took "t" time to catch the train

so in this case the total distance moved by the train + 5 m = total distance moved by the passenger

so we will have

distance moved by train is given as

d_1 = \frac{1}{2}(0.6) t^2

also the distance moved by passenger

d_2 = \frac{1}{2}(1.2) t^2

so we will have

d_1 + 5 = d_2

0.3 t^2 + 5 = 0.6 t^2

0.3 t^2 = 5

t = 4.08 s

3 0
3 years ago
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