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VLD [36.1K]
3 years ago
9

Which statements describe acceleration? Check all that apply. Negative acceleration occurs when an object slows down in the posi

tive direction. Negative acceleration occurs when an object slows down in the negative direction. Negative acceleration occurs when an object speeds up in the negative direction. Positive acceleration occurs when an object speeds up in the positive direction. Positive acceleration occurs when an object speeds up in the negative direction. Positive acceleration occurs when an object slows down in the negative direction
Physics
1 answer:
garik1379 [7]3 years ago
7 0

Answer:

Negative acceleration occurs when the acceleration vector points to the left.

1. Object slowing down in the positive direction.

2. Object speeding up in the  negative direction.

Following six statements:

1. T

2. F

3. T

4. T

5. F

6. T

Check direction of acceleration vector.

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Why is cathode positive in leclanche cell??
laila [671]

Answer:

Because the zinc is reluctant

Explanation:

A leclanche cell contains a conducting solution (electrolyte) of ammonium chloride, a cathode (positive terminal) of carbon, a depolarizer of manganese dioxide (oxidizer), and an anode (negative terminal) of zinc (reductant).

As the Zn2+ ions move away from the anode, leaving their electrons on its surface,

Zn → Zn2+ + 2e−

the anode becomes more negatively charged than the cathode. When the cell is connected to an external electrical circuit, the excess electrons on the zinc anode flow through the circuit to the carbon rod, the movement of electrons forming an electric current.

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Which of the following is NOT true about sweat? Sweat helps cool the body. Sweat glands are located all over the body. Sweat hel
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A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta
jeyben [28]

Answer:

V_0

Explanation:

Given that, the range covered by the sphere, M, when released by the robot from the height, H, with the horizontal speed V_0 is D as shown in the figure.

The initial velocity in the vertical direction is 0.

Let g be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. V_0 will remain constant throughout the projectile motion.

So, if the time of flight is t, then

D=V_0t\; \cdots (i)

Now, from the equation of motion

s=ut+\frac 1 2 at^2\;\cdots (ii)

Where s is the displacement is the direction of force, u is the initial velocity, a is the constant acceleration and t is time.

Here, s= -H, u=0, and a=-g (negative sign is for taking the sigh convention positive in +y direction as shown in the figure.)

So, from equation (ii),

-H=0\times t +\frac 1 2 (-g)t^2

\Rightarrow H=\frac 1 2 gt^2

\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)

Similarly, for the launched height 2H, the new time of flight, t', is

t'=\sqrt {\frac {4H}{g}}

From equation (iii), we have

\Rightarrow t'=\sqrt 2 t\;\cdots (iv)

Now, the spheres may be launched at speed V_0 or 2V_0.

Let, the distance covered in the x-direction be D_1 for V_0 and D_2 for 2V_0, we have

D_1=V_0t'

D_1=V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_1=\sqrt 2 D [from equation (i)]

\Rightarrow D_1=1.41 D (approximately)

This is in the 3 points range as given in the figure.

Similarly, D_2=2V_0t'

D_2=2V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_2=2\sqrt 2 D [from equation (i)]

\Rightarrow D_2=2.82 D (approximately)

This is out of range, so there is no point for 2V_0.

Hence, students must choose the speed V_0 to launch the sphere to get the maximum number of points.

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The nucleus of an atom is composed of unchanged particles as well as of the following
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