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sattari [20]
3 years ago
13

A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball wh

en it reaches its highest point
Physics
1 answer:
xenn [34]3 years ago
3 0

Answer:

The velocity of the ball at it's highest point is zero.

Explanation:

When we toss a ball upwards as we know that it's motion is an uniformly accelerated motion with acceleration = 9.81m/s^{2} in the downward direction.

Hence as we throw the ball upwards it's speed will go on decreasing as it gains height. The ball will stop gaining height when it's velocity becomes zero and the ball starts to descend again towards earth.

If the ball is tossed at an angle with the horizontal in that case it's velocity at the top most point will not be zero but in our case since the ball is tossed vertically it's velocity at the topmost point will become zero.

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The correct answer is Relational Database

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Which of the following statements about the force on a charged particle due to a magnetic field are not valid?
Vinil7 [7]
The correct answer is "None of the above; all of these statements are valid." All the statements namely, it depends on the particle's charge, it depends on the strength of the external magnetic field, it depends on the particle's velocity, and it acts at right angles to the direction of the particle's motion are all valid. Thank you for posting your question. I hope this answer helped you. Let me know if you need more help. 
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A 6.00-μf parallel-plate capacitor has charges of 40.0 μc on its plates. how much potential energy is stored in this capacitor?
dedylja [7]

Thew energy stored in a capacitor of capacitance C and voltage between the plates V is

E=\frac{1}{2} CV^2=\frac{1}{2C} Q^2.

Substituting numerical value

E=\frac{1}{2*6*10^{-6}} (40*10^{-6})^2\\ E=133.33\; \mu J

7 0
3 years ago
Can someone help me wit hthis science question?
klemol [59]

Answer:

see explanation  

Explanation:

There is an increasing demand for materials and natural resources from a growing global population, especially those in more economically developed countries. The world's resources are being used up more quickly. The consumption of resources is spread unequally between MEDCs (more economically developed countries), who use more resources, and LEDCs (less economically developed countries), who use less.

The gap between the rich and poor is more evident when the resources are shared so unevenly and unfairly and natural resources like materials and natural energy cannot reach the demand of the people which can have consequences and be very difficult to manage. Having a lack of these materials in a country can result in prices going up for them, and the industry could be harder to work in because of a lack of materials.

6 0
3 years ago
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

5 0
2 years ago
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