Một vật rắn khối lượng 1 kg, chuyển động tịnh tiến ở độ cao 2 m so với mặt đất. Vật rắn có vận

In order to sail through the frozen Arctic Ocean, the most powerful icebreaker ever built was constructed in the former Soviet U

professor190 [17]

Answer:

955.36 seconds ≈ 16 minutes

Explanation:

Power(P) is the rate of doing work(W)

That is, P = W/t, where t is the time.

multipying both sides with 't' and dividing with 'P', we get: t=W/P

Here, W = 5.35 x 10^10 J and P = 5.6 x 10^7 W ( 1 W = 1 J/s).

Therefore , on dividing W with P, we get 955.36 seconds.

6 0

2 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

3 years ago

Object height 10cm is placed in front of plane mirror.The height of the image will be_______?

coldgirl [10]

Object height 10cm is placed in front of plane mirror. The height of the image will also be 10cm as the height of image is same as height of object in the case of plane mirror

7 0

2 years ago

A particle on a spring moves in simple harmonic motion along the x axis between turning points at x1 = 95 cm and x2 = 135 cm. (i

uranmaximum [27]

Answer:

(i) $x = 115\,cm$ , (ii) $x = 95\,cm$ , (iii) $x = 95\,cm$

Explanation:

(i) $x_{1}$ and $x_{2}$ represent the points where particle has a velocity of zero and spring reach maximum deformation, Given the absence of non-conservative force and by the Principle of Energy Conservation, the position where particle is at maximum speed is average of both extreme positions:

$x = 115\,cm$

(ii) Maximum accelerations is reached at $x_{1}$ and $x_{2}$ .

$x = 95\,cm$

(iii) Greatest net forces exerted on the particle are reached at $x_{1}$ and $x_{2}$ .

$x = 95\,cm$

8 0

2 years ago

Two ways that radio waves are used for transmitting information. Help me please.

Travka [436]

1. Stop transmitting start and stop sending photons. This is typical morse code like transmissions of information. By varying the pattern of starts stops you vary the information sent. Also used in digital communications over radios, modems, and pulse dialing over radio/phones.

2. Change the frequency. You use a pattern of shorter and longer wavelengths of radio waves higher and lower energies of individual photons to transmit information. This is frequency modulation

3 0

3 years ago