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tresset_1 [31]
3 years ago
11

You are stranded in a stationary boat. Your friend is on a dock, but the boat is just beyond his reach. There is a 5 kg anchor i

n the boat. You'd like to get the boat to move closer to the dock so your friend can rescue you. Select from the following list what effect each change will have on the position of the boat relative to the dock. A. The boat will move closer to the dock. B. The boat will move away from the dock. C. The position of the boat relative to the dock will not change.
Physics
1 answer:
fiasKO [112]3 years ago
8 0

Answer:

running away and launching the anchor that will give a greater speed towards the dock v₄.

Explanation:

To try to bring the boat closer to the dock, several cases can be carried out.

* move inside the ship so that the center of mass changes and since moving away you have a speed v, the ship will approach the dock at a speed v₂,

* Throw the anchor in the opposite direction to the dock so that using the conservation of the moment the boat moves towards it, it moves at a speed v₃

* A combination of the two processes running away and launching the anchor that will give a greater speed towards the dock v₄.

In all cases, the friction must be zero.

All other movements move the ship away from the dock

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A toy car has an initial acceleration of 2 m/s" across a horizontal surface after it is released from rest. After the car travel
Ganezh [65]

Answer:

Closed system, because the speed of the car is as expected in the case where an object has uniform acceleration for a time t

Explanation:

Here in the question it is mentioned that a toy car has an initial acceleration of 2m/s²  across a horizontal surface so we can say that it is acted upon by an external force

Assuming that the acceleration is constant and the reason for this assumption is there at the last

The major difference between an open system and closed system is in case of open system there will be transfer of matter and in case of closed system there will be no change in matter of the system

If acceleration is constant in case of closed system we can expect the speed of the car after a time t by using the formula

  s = u×t + 0·5×a×t²

where s is the distance travelled

t is the time taken to travel that distance

u is the initial velocity

a is the acceleration of that system

But in case of open system as there will be a change of mass there will be a change in velocity of the system so in this case we cannot expect the speed of the car after a time t

And if the acceleration is not constant then we cannot say that the toy car is an open system or closed system, that is why we are assuming that the acceleration of the toy car is constant

3 0
4 years ago
Read 2 more answers
At high speeds, a particular automobile is capable of an acceleration of about 0.540 m/s^2. At this rate, how long (in seconds)
Mama L [17]

Answer:

t = 6.68 seconds

Explanation:

The acceleration of the automobile, a=0.54\ m/s^2

Initial speed of the automobile, u = 91 km/hr = 25.27 m/s

Final speed of the automobile, v = 104 km/hr = 28.88 m/s

Let t is the time taken to accelerate from u to v. It can be calculated as the following formula as :

t=\dfrac{v-u}{a}

t=\dfrac{28.88-25.27}{0.54}

t = 6.68 seconds

So, the time taken by the automobile to accelerate from u to v is 6.68 seconds. Hence, this is the required solution.

5 0
3 years ago
When an electron in a certain excited energy level in a one-dimensional box of length 2.00 Å makes a transition to the ground st
Fiesta28 [93]

Answer:

Calculate the wavelength associated with an electron with energy 2000 eV.

Sol: E = 2000 eV = 2000 × 1.6 × 10–19 J

5 0
3 years ago
Jack and Jill have made up since the previous HW assignment, and are now playing on a 10 meter seesaw. Jill is sitting on one en
Airida [17]

Answer: 3 m.

Explanation:

Neglecting the mass of the seesaw, in order the seesaw to be balanced, the sum of the torques created by  gravity acting on both children  must be 0.

As we are asked to locate Jack at some distance from the fulcrum, we can take torques regarding the fulcrum, which is located at just in the middle of the length of the seesaw.

If we choose the counterclockwise direction as positive, we can write the torque equation as follows (assuming that Jill sits at the left end of the seesaw):

mJill* 5m -mJack* d = 0

60 kg*5 m -100 kg* d =0

Solving for d:

d = 3 m.

6 0
3 years ago
When landing after a spectacular somersault, a 35.0 kg gymnast decelerates by pushing straight down on the mat. calculate the fo
Sliva [168]

The deceleration experienced by the gymnast is the 9 times of the acceleration due to gravity.

Now from Newton`s  first law, the net force on gymnast,

F_{net} =F-W=ma

Here, W is the weight of the gymnast and a is the acceleration experienced by the gymnast (9\times g acceleration due to gravity)  

Therefore,

F= ma+W OR F=ma+mg=m(g+a)

Given m = 30 kg anda=9\times g=9\times 9.8 m/s^{2} =88.2 m/s^{2}

Substituting these values in above formula and calculate the force exerted by the gymnast,  

F=(40 kg) (88.2 m/s^{2} +9.8 m/s^{2} )

F=3.537\times10^{3}N

6 0
3 years ago
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