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LUCKY_DIMON [66]
3 years ago
12

If it uses 5 watts, how much time would it take to do 100J of work?

Physics
1 answer:
Hoochie [10]3 years ago
7 0
POWER =ENERGY/TIME. making time the subject of the formula TIME=ENERGY/POWER. therefore, 100/5 =20seconds
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Two rockets are flying in the same direction and are side by side at the instant their retrorockets fire. Rocket A has an initia
Sedaia [141]

Answer:

a_2\ =\ -33.65\ m/s^2

Explanation:

Given,

For the first rocket,

  • Initial velocity of the first rocket A = u_1\ =\ 4600\ m/s.
  • Acceleration of the first rocket = a_1\ =\ -18\ m/s^2

For the second rocket,

  • Initial velocity of the second rocket B = u_2\ =\ 8200 m/s.
  • Displacement of both the rockets A and B = s = 0 m

Fro the first rocket,

Let 't' be the time taken by the first rocket A for whole the displacement

\therefore s\ =\ u_1t\ +\ \dfrac{1}{2}a_1t^2\\\Rightarrow 0\ =\ 4600t\ -\ 0.5\times 18t^2\\\Rightarrow t\ =\ \dfrac{4600}{0.5\times 18}\\\Rightarrow t\ =\ 511.11 sec

Let a_2 be the acceleration of the second rocket B for the same time interval

from the kinematics,

\therefore s\ =\ ut\ +\ \dfrac{1}{2}at^2\\\Rightarrow s\ =\ u_2t\ +\ \dfrac{1}{2}a_2t^2\\\Rightarrow a_2\ =\ \dfrac{2s\ -\ 2u_2t}{t^2}\\\Rightarrow a_2\ =\ \dfrac{0\ -\ 2u_2t}{t^2}\\

\Rightarrow a_2\ =\ -\dfrac{2u_2}{t}\\\Rightarrow a_2\ =\ -\dfrac{2\times 8600}{511.11}\\\Rightarrow a_2\ =\ -33.65\ m/s^2

Hence the acceleration of the second rocket B is -33.65\ m/s^2.

6 0
3 years ago
a block is pushed up a frictionless 40 incline. if the initial velocity after the push is 5.00 m/s. how far along the incline do
Lana71 [14]

Answer:

d=1.982m, t=1.019s

Explanation:

There are different approaches we can take to solve this problem. You could either solve this by using conservation of energy or by taking a kinematic approach. I'll solve this by using kinematics. So, the very first thing we need to do in order to solve this is do a drawing of the situation so we can analyze it better. (See attached picture).

So, since we are talking about an inclined plane, we can see that the force of gravity is being split into an x and y components if we incline the axis of coordinates. Taking this into account we can see that:

\sum F_{x}=ma_{x}

Since there is no friction in our system, then the only force acting upon the box is the force of gravity, or weight. Since we are taking the upwards direction as the positive direction of movement, we can say that the force of gravity is excerting a negative influence on our box, so this acceleration will be negative, so our sum of forces will now look like this:

-mg sin(40^{o})=ma

we can cancel the masses out so we can see that:

a=-g sin(40°)

a=-9.81m/s^{2} sin(40^{o})

We have now enough information to solve our problem.

we can take the following equation to find the distance the block travels up the incline:

x=\frac{V_{f}^{2}-V_{0}^{2}}{2a}

we know the final velocity must be zero, so we can use the provided data to solve our formula:

x=\frac{(0)^{2}-(5m/s)^{2}}{2(-9.81m/s^{2})sin 40^{o}}

which yields:

x=1.982m

In order to find the time it takes for the block to return to its original position we can use the following formula:

x=V_{0}t+\frac{1}{2}at^{2}

since x=0 is the starting point we can use that to solve our equation:

0=5t+\frac{1}{2}(-9.81sin 40^{o})t^{2}

which simplifies to:

0=5t-4.905t^{2}

which can now be solved for t

t(5-4.905t)=0

t=0                and          5-4.905t=0

t=0                 and         t=\frac{5}{4.905}=1.019

so the time it takes the block to return to its original position is

t= 1.019s

6 0
3 years ago
Which of the following has the most potential energy?
user100 [1]
Options A and C are correct but if you compare their masses, person has negligible mass compared to car. So a car at the top of the hill has the most potential energy. 
6 0
3 years ago
Trong thí nghiệm Y-âng về giao thoa ánh sáng : khoảng cách giữa hai khe là 0,5 mm ; khoảng cách từ mặt phẳng chứa hai khe đến mà
Natalija [7]

không có câu hỏi, trả lời như nào nhỉ?

8 0
3 years ago
Read 2 more answers
What happens when a colder drier air mass pushes against a warmer moister air mass
andrew-mc [135]
A cold<span> front </span>occurs when a cold air mass advances into a region occupied by a warm air mass<span>. If the boundary between the </span>cold<span> and </span>warm air masses<span> doesn't move, it is called a stationary front. This is due to the (usually) higher humidity in the </span>air<span> of </span>warm<span> fronts compared to that of </span>cold<span> fronts.

Hope this helps</span>
6 0
4 years ago
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