Ask question

Ask question

All categories

3 years ago

7

cho hệ cơ học như hình vẽ hai đầu dây buộc hai vật có khối lượng tương ứng là m1=2kg và m2>m1 lấy g=10m/s sau 1s kể từ lúc bắ

t đầu chuyển dộng hệ vật đi được 50 cm tính m2 và sức căng của dây

1 answer:

Sever21 [200]3 years ago

7 0

xin lỗi không có sơ đồ vui lòng cho biết sơ đồ

You might be interested in

What determines the velocity of an object?

BARSIC [14]

I think it’s A not 100% sure

4 0

2 years ago

What is the guage pressure of an object with a mass of 30 kg and 5ft height?

wariber [46]

Answer: D=pg

Explanation:

?????

8 0

2 years ago

Solve these questions plz

8090 [49]

Answer:

For the First answer I cant answer it But I can help you :

The solid has constituent particles tightly packed and the lattice vibrations are carried out by them in their fixed position however oscillations take place about their mean position. These vibrations are increased as soon as there is increase in the temperature which eventually leads to the more chaotic motion of the constituents. At a fixed critical point of temperature, the bonds are broken and the constituent particles are spaced apart changing their phase into liquid. When more temperature is increased by gaining heat energy then the liquid changes into gas where the motion of constituent particles moving freely is dominant.

Explanation:

6 0

2 years ago

The flash unit in a camera uses a special circuit to "step up" the 3.0 V from the batteries to 360 V , which charges a capacitor

aksik [14]

Answer:

The capacitance of the capacitor is $1.85*10^{-5}F$

Explanation:

To solve this exercise it is necessary to apply the concepts related to Power and energy stored in a capacitor.

By definition we know that power is represented as

$P = \frac{E}{t}$

Where,

E= Energy

t = time

Solving to find the Energy we have,

$E = P*t$

Our values are:

$P = 1*10^5W$

$t = 12*10^{-6}s$

Then,

$E= (1*10^5)(12*10^{-6})$

$E = 1.2J$

With the energy found we can know calculate the Capacitance in a capacitor through the energy for capacitor equation, that is

$E=\frac{1}{2}CV^2$

Solving for C=

$C = \frac{1}{2}\frac{E}{V^2}$

$C = 2\frac{1.2}{(360^2-3^2)}$

$C = 1.85*10^{-5}F$

Therefore the capacitance of the capacitor is $1.85*10^{-5}F$

7 0

2 years ago

Baseball player a bunts the ball by hitting it in such a way that it acquires an initial velocity of 2.4 m/s parallel to the gro

LUCKY_DIMON [66]

Let $\mathbf r_A$ denote the position vector of the ball hit by player A. Then this vector has components

$\begin{cases}r_{Ax}=\left(2.4\,\frac{\mathrm m}{\mathrm s}\right)t\\r_{Ay}=1.2\,\mathrm m-\frac12gt^2\end{cases}$

where $g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}$ is the magnitude of the acceleration due to gravity. Use the vertical component $r_{Ay}$ to find the time at which ball A reaches the ground:

$1.2\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.49\,\mathrm s$

The horizontal position of the ball after 0.49 seconds is

$\left(2.4\,\dfrac{\mathrm m}{\mathrm s}\right)(0.49\,\mathrm s)=12\,\mathrm m$

So player B wants to apply a velocity such that the ball travels a distance of about 12 meters from where it is hit. The position vector $\mathbf r_B$ of the ball hit by player B has

$\begin{cases}r_{Bx}=v_0t\\r_{By}=1.6\,\mathrm m-\frac12gt^2\end{cases}$

Again, we solve for the time it takes the ball to reach the ground:

$1.6\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.57\,\mathrm s$

After this time, we expect a horizontal displacement of 12 meters, so that $v_0$ satisfies

$v_0(0.57\,\mathrm s)=12\,\mathrm m$

$\implies v_0=21\,\dfrac{\mathrm m}{\mathrm s}$

5 0

3 years ago