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alisha [4.7K]
3 years ago
8

What is the mass of 1.0 l of water in grams?

Physics
1 answer:
Maksim231197 [3]3 years ago
5 0
The mass of 1.0 l of water in grams is 1,000 g ;) 
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On a frictionless surface how much force is necessary to accelerate a 0.49 kg object to the left at 4.8 m/s2?
Alenkasestr [34]

Answer:

<h2>2.35 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

force = 0.49 × 4.8 = 2.352

We have the final answer as

<h3>2.35 N</h3>

Hope this helps you

8 0
3 years ago
If the archerfish spits its water 30 degrees from the horizontal aiming at an insect 1.2 m above the surface of the water, how f
Burka [1]

Answer:

The speed is  v =  9.8 \ m/s

Explanation:

From the question w are told that

    The angle  made is \theta  =  30^o

     The distance  above the surface of the water is  h_{max} = 1.2 \ m

     The  value of  g = 10 \  m/s^2

   

The maximum height attained by the fish is mathematically evaluate as

       h_{max} =  \frac{v^2 sin ^2 \theta }{2g }

Making v which is the speed of the fish the subject of the formula

      v =  \sqrt{ \frac{2gh_{max}}{ sin^2 \theta } }

  substituting values

     v =  \sqrt{ \frac{2*10 *1.2 }{ [sin (30)]^2  } }

     v =  9.8 \ m/s

3 0
3 years ago
You and a partner sit on the floor and stretch out a coiled spring to a length of 7.2 meters. You shake the coil so you
vekshin1

Answer:

Approximately 5.9\; {\rm m\cdot s^{-1}} (assuming that the partner is holding the other end of the coil stationary.)

Explanation:

In a standing wave, an antinode is a point that moves with maximal amplitude, while a node is a point that does not move at all. There is an antinode between every two adjacent nodes. Likewise, there is a node between every two adjacent antinodes.

The side of the spring that is being shaken moving with maximal amplitude. Hence, that point on this spring would also be an antinode. In contrast, the side of the spring that is held still (does not move at all) would be a node.

There would be a node between:

  • the antinode at the end of the spring that is being shaken, and
  • the antinode between the two ends of this spring.

Overall, the nodes and antinodes on this spring would be:

  • node at the end that is being held still,
  • antinode (as mentioned in the question),
  • node (inferred, not mentioned in the question), and
  • antinode at the end that is being shaken.

The distance between two adjacent nodes is equal to one-half (that is, (1/2)) the wavelength of the wave. The distance between a node and an adjacent antinode is one-quarter (that is, (1/4)) of the wavelength of the wave.

Thus, if the wavelength of the wave in this question is \lambda, the length of this spring would be:

\displaystyle \frac{1}{2}\, \lambda + \frac{1}{4}\, \lambda = \frac{3}{4}\, \lambda.

The question states that the length of this coiled spring is 7.2\; {\rm m}. In other words, (3/4) \, \lambda = 7.2\; {\rm m}. The wavelength of this wave would be (7.2\; {\rm m}) / (3/4) = 9.6\; {\rm m}.

The frequency f of this wave is the number of cycles in unit time:

\begin{aligned} f &= \frac{10}{16.3\; {\rm s}} \approx 0.613\; {\rm s^{-1}}\end{aligned}.

Hence, the speed v of this wave would be:

\begin{aligned} v &= \lambda\, f \\ &=9.6\; {\rm m} \times 0.613\; {\rm s^{-1}} \\ &\approx 5.9\; {\rm m \cdot s^{-1}}\end{aligned}.

3 0
2 years ago
during the course of songraphic exam, you notice lateral splaying of echoes in the far field. what can you do to improve the ima
horrorfan [7]

Answer:

lateral splaying of echoes in the far field can be improved by Increasing the maximum number of transmit focal zones and optimize their location.

8 0
3 years ago
Answer this plsss<br> A mobile phone operates at 900 MHz. <br> What wavelength does it use?
Aleks [24]

Answer:

The wavelength = 0.3333 meters at 900 MHz, therefore, = /4 = 0.08333 meters.

7 0
3 years ago
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