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kipiarov [429]
2 years ago
5

A constant force is applied to an object, causing the object to accelerate at 10 m/s^2. What will the acceleration be if: a) The

force is halved? b) The object's mass is halved? c) The force and the object's mass are both halved?
Physics
1 answer:
Liula [17]2 years ago
5 0
What you need to know is that the force is

F=ma

The force is the product of mass and acceleration

this means that the acceleration is

a=F/m

a) The force is halved?
this means that f will be \frac{F}{2} now:

a=\frac{F}{2m}

So the accelaration will also he halved (it's the original acceleratation divided by 2)


 b) The object's mass is halved?
a=\frac{F}{m/2}=a=\frac{F2}{m}

which is the original acceleration times two!! so it will double


c) The force and the object's mass are both halved?
now we have

a=\frac{F/2}{m/2}=a=\frac{2F}{2m}=a=\frac{F}{m}

so they will cancel each other out and the acceleration will stay the same!











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Answer:

Explanation:

Given that, .

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t½ = 1 hr.

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After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

6 0
3 years ago
A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8 m/s^{2}

           L = length of the rod = 0.2 m

Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

              x = spring extension

The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

             = 0.1236 m

       k = 25 N/m

So, according to the law of conservation of energy

       \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}

      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

  \frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}

          v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

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Dehydration reactions link monomers together into polymers by releasing water, and hydrolysis breaks polymers into monomers using a water molecule. Monomers are just single unit molecules and polymers are chains of monomers.

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8 0
3 years ago
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lara31 [8.8K]

Answer:

The time taken to travel is, t = 12 minutes

Explanation:

Given data,

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The distance of travel, d = 12 km

The time taken for the travel is t = ?

The speed is defined as the distance divided by the time taken to travel. The formula for speed is,

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∴                                  t = d/v

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Hence, the time taken to travel is, t = 12 minutes.

5 0
3 years ago
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