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pogonyaev
3 years ago
5

How are average speed and velocity related to each other for an object in uniform motion?

Physics
1 answer:
olya-2409 [2.1K]3 years ago
5 0

Speed is the size (absolute value) of the velocity.

Speed is a scalar.  Velocity is a vector.
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Can u help me with C
Vladimir79 [104]
I believe it would be 2m/s.
6 0
3 years ago
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Calculate the de Broglie wavelength of (a) a mass of 1.0 g traveling at 1.0 m s−1 , (b) the same, traveling at 1.00 × 105 km s−1
lesantik [10]

Answer:

a)\lambda=6.63\times10^{-31}m

b)\lambda=6.63\times10^{-39}m

c)\lambda=9.97\times10^{-11}m

d)\lambda=4.03\times10^{-36}m

e)λ=∞

Explanation:

De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

\lambda=\frac{h}{P}=\frac{h}{mv}

with h the Plank's constant (6.63\times10^{-34}\frac{m^{2}kg}{s}) and P the momentum of the object that is mass (m) times velocity (v).

a)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}kg*1.0)}

\lambda=6.63\times10^{-31}m

b)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}*(1.00\times10^{8}))}

\lambda=6.63\times10^{-39}m

c)\lambda=\frac{6.63\times10^{-34}}{(6.65\times10^{-27}*1000)}

\lambda=9.97\times10^{-11}m

d)\lambda=\frac{6.63\times10^{-34}}{(74*2.22)}

\lambda=4.03\times10^{-36}m

e) \lambda=\frac{6.63\times10^{-34}}{(74*0)}

λ=∞

6 0
3 years ago
What do we call the movement of electric charge?<br> A. Power B. Voltage C. Resistance D. Current
Alex Ar [27]
D. Current

Hope I helped!
4 0
2 years ago
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39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th
tatuchka [14]

<u>Answer:</u> The final temperature of the solution is 80.14^oC

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, \text{heat}_{absorbed}=\text{heat}_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 39 g

m_2 = mass of coffee = 166 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 83^oC

c_1 = specific heat of aluminium = 0.904J/g^oC

c_2 = specific heat of coffee= 4.1801J/g^oC

Putting all the values in equation 1, we get:

39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]

T_{final}=80.14^oC

Hence, the final temperature of the solution is 80.14^oC

4 0
3 years ago
Brainliest if correct Question 1 of 10
Art [367]

Answer:

B

Explanation:

hope this helps you

4 0
2 years ago
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