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goldfiish [28.3K]
3 years ago
9

Speed differs from velocity in the same way that __________ differs from displacement.

Physics
2 answers:
Taya2010 [7]3 years ago
6 0

Speed differs from velocity in the same way that DISTANCE differs from displacement.

lyudmila [28]3 years ago
5 0

Speed differs from velocity in the same way that distance differs from displacement.

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An objects speed is 2.40 m/s, and its momentum is 120.0 kg m/s. what is the mass of the object?
jarptica [38.1K]
Momentum is defined as 

P = m*v

where m is the mass of the object
and v is the speed of the object

We find m.............> m = P/v = 120 [kg.m/s] / 2.4 [m/s] = 50 kg

The mass is 50 kg
4 0
4 years ago
At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r
Oksana_A [137]

Answer:

<u>For proton:</u>

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

<u>For electron:</u>

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s    

Explanation:

<u>For a proton we have</u>:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.                

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

W = \Delta Vq    (1)                                        

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C    

V: is the potential    

Also, the work is equal to:

W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}     (2)      

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

v_{a}: is the velocity in the point a

v_{b}: is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

\Delta Vq = \frac{1}{2}mv_{a}^{2}

(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}

v_{a} = 2.9 \cdot 10^{4} m/s

<u>For an electron we have</u>:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}

Since v_{a} = 0 (starts from rest) and m_{e} = 9.1x10⁻³¹ kg (electron's mass), we have:

(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}    

v_{b} = 1.2 \cdot 10^{6} m/s

I hope it helps you!

6 0
4 years ago
In a game of pool, why will the balls eventually stop after a collision?
Aneli [31]

As a ball rolls ... even if there were no air resistance ... the ball
has to push down and climb over every one of those little fibers
that stick up from the felt fabric on the table.  As it does that, each
little fiber brushes against the ball, and robs a tiny tiny bit of its
kinetic energy. 

7 0
3 years ago
A 37-cm-long wire of linear density 18 g/m vibrating at its second mode, excites the third vibrational mode of a tube of length
lora16 [44]

Answer:

T = 4.42 10⁴ N

Explanation:

this is a problem of standing waves, let's start with the open tube, to calculate the wavelength

        λ = 4L / n                 n = 1, 3, 5, ...    (2n-1)

How the third resonance is excited

       m = 3

       L = 192 cm = 1.92 m

       λ = 4 1.92 / 3

       λ = 2.56 m

As in the resonant processes, the frequency is maintained until you look for the frequency in this tube, with the speed ratio

      v = λ f

      f = v / λ

      f = 343 / 2.56

      f = 133.98 Hz

       

Now he works with the rope, which oscillates in its second mode m = 2 and has a length of L = 37 cm = 0.37 m

The expression for standing waves on a string is

           λ = 2L / n

           λ = 2 0.37 / 2

           λ = 0.37 m

The speed of the wave is

          v = λ f

As we have some resonance processes between the string and the tube the frequency is the same

          v = 0.37 133.98

          v = 49.57 m / s

Let's use the relationship of the speed of the wave with the properties of the string

              v = √ T /μ

              T = v² μ

              T = 49.57²   18

              T = 4.42 10⁴ N

4 0
3 years ago
Read 2 more answers
A 53 g ice cube at −30◦C is dropped into a container of water at 0◦C. How much water freezes onto the ice? The specific heat of
vladimir2022 [97]

For A 53 g ice cube at −30◦C is dropped into a container of water at 0◦C, the amount of water that freezes onto the ice?  is mathematically given as

x = 9.93 g

<h3>What is the amount of water that freezes onto the ice?</h3>

Where

Energy received = energy given out

Generally, the amount of water is mathematically given as

(53)(0.5)(30) = (80)(x)

Therefore

x = (49)(0.5)(16)/(80)

x = 9.93 g

In conclusion, the mass of water

x = 9.93 g

Read more about  mass

brainly.com/question/15959704

4 0
2 years ago
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