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andrezito [222]

2 years ago

9

The thermal energy of what system does change

1 answer:

Inessa05 [86]2 years ago

4 0

Answer:

there is the increase the temperature of cold body and decrease the temperature of hot body

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A 1000.0 kg truck accelerates from 20.0 m/s to 25.0 m/s over a distance of 300.0 m. What is the average net force on the truck?

choli [55]

Answer:

The average net force on the truck is 375 Newtons.

Explanation:

Using Newton's 3rd equation of motion, we have :

$v^{2} - u^{2} = 2$ ×a×s

where, v = final velocity = 25 m/s

u = initial velocity = 20 m/s

a = acceleration

s = distance traveled = 300 m

Using these values in the above equation, we get acceleration = 0.375 m/ $s^{2}$

Using Newton's second law, we have:

F=m×a

where m = mass = 1000 kg

a= acceleration = 0.375 m/ $s^{2}$

Putting values we have F=375 N

3 0

3 years ago

A flap of tissue that prevents blood from flowing back is a(n)________ .

ioda

A valve is a flap of tissue that prevents blood from flowing back

6 0

3 years ago

A magnetic B field of strength 0.9 T is perpendicular to a loop with an area of 3 m2. If the area of the loop is reduced to zero

Alexeev081 [22]

Answer:

The magnitude of induced emf is 5.4 V

Explanation:

Given:

Magnetic field $B = 0.9$ T

Area of loop $\Delta A = 3$ $m^{2}$

Time take to reduce loop to zero $\Delta t = 0.5$ sec

To find induced emf we use faraday's law,

Induced emf is given by,

$\epsilon = -\frac{\Delta \phi}{\Delta t}$

Here minus sign shows lenz law, for finding magnitude of emf we ignore it.

Where $\phi = B\Delta A$

Put the value of flux and find induced emf,

$\epsilon = \frac{B\Delta A}{\Delta t}$

$\epsilon = \frac{0.9 \times 3}{0.5}$

$\epsilon = 5.4$ V

Therefore, the magnitude of induced emf is 5.4 V

3 0

3 years ago

Two blocks of masses M 1 and M 2 are connected by a massless string that passes over a massless pulley as shown in the figure. M

stealth61 [152]

The mass M1 is 7.8 kg

Explanation:

Block M1 is hanging on the string while block M2 is on the frictionless ramp.

We have to write the equations of motion for the two blocks.

- For M1, the only two forces acting on it are the force of gravity $M_1 g$ (downward) and the tension in the string T (upward). So we can write

$M_1 g - T = M_1 a$

where

$M_1$ is the mass of the block

$g=9.8 m/s^2$ is the acceleration of gravity

$a$ is the acceleration of the system

- For M2, the only two forces acting on it are the tension in the string T (acting up along the ramp) and the component of the gravity acting down along the ramp, $M_2 g sin \theta$ . So the equation of motion is

$T-M_2 g sin \theta = M_2 a$

where

$M_2 = 13.5 kg$ is the mass of the 2nd block

$\theta=35.5^{\circ}$ is the angle of the ramp

In order for the two blocks to be in equilibrium, the acceleration must be zero:

$a=0$

So the two equations become:

$M_1 g - T=0\\T-M_2 g sin \theta = 0$

Isolating T from the 1st equation,

$T=M_1 g$

And substituting into the 2nd equation, we can find the value of the mass $M_1$ :

$M_1 g - M_2 g sin \theta = 0\\M_1 = M_2 sin \theta = (13.5)(sin 35.5^{\circ})=7.8 kg$

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

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brainly.com/question/2562700

#LearnwithBrainly

7 0

3 years ago

Determine the magnitude and direction of the force on a 200 m power line carrying a current of 5.0-A due west in a magnetic fiel

bogdanovich [222]

Answer:

Explanation:

l = 200 m

i = 5 A west

B = 6 micro Tesla direction 30 degree north of east

angle between length vector and the magnetic field vector = 180 - 30 = 150 degree

Write the length and the magnetic field in the vector form

$\overrightarrow{l}=- 200 \widehat{i}metre$

$\overrightarrow{B}= 6\times 10^{-6}\left ( Cos30\widehat{i}+Sin30\widehat{j} \right )Tesla$

$\overrightarrow{B}=\left ( 5.2\widehat{i}+3\widehat{j} \right )\times 10^{-6}Tesla$

$\overrightarrow{F} = i \overrightarrow{l}\times \overrightarrow{B}$

$\overrightarrow{F} = 5\times \left ( -200\widehat{i} \right )\times \left ( 5.2\widehat{i}+3\widehat{j} \right )\times 10^{-6}$

$\overrightarrow{F} =- 3\times 10^{-3}\widehat{k}Newton$

Thus, the magnitude of force is 3 x 10^-3 newton and it is directed towards negative z axis direction.

7 0

3 years ago