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Thepotemich [5.8K]
2 years ago
15

7. An object can be accelerating even if its speed is

Physics
2 answers:
MatroZZZ [7]2 years ago
6 0

Answer:

D

Explanation:

If a car is moving very fast in a straight road or a highway in such a way that it has reached its limits as soon as it is going that fast it can break it limit

steposvetlana [31]2 years ago
4 0

Answer:

I think that the answer might be d

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Write the equation of a function h(t) that represents the amount of heat in joules required to heat the bar to a temperature of
bearhunter [10]
The initial temperature of the bar is 25. To get to the t temperature you need to add (t-25) degrees Celsius.

for 1 degree................... 7 Joules
      y given degree........  p Joules

p=7y

In our case y=(t-25) .

h(t) = 7(t-25) which is the final answer.

8 0
2 years ago
You are growing a little taller each year. What units could you use to measure how fast
vlada-n [284]

Answer:

inches and feet (or even centameters)

Explanation:

these are all common units used to measure height

6 0
2 years ago
Read 2 more answers
A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci
MA_775_DIABLO [31]

Answer:

v_B=3.78\times 10^5\ m/s

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is 6.63\times 10^{-27}\ kg

Speed of nucleus at A is v_A=6.2\times 10^5\ m/s

Potential at point A, V_A=1.5\times 10^3\ V

Potential at point B, V_B=4\times 10^3\ V

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s

So, the speed at point B is 3.78\times 10^5\ m/s.

7 0
3 years ago
Fill in the blanks for the following:
storchak [24]

Answer:

<em>a. 4.21 moles</em>

<em>b. 478.6 m/s</em>

<em>c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

Explanation:

Volume of container = 100.0 L

Temperature = 293 K

pressure = 1 atm = 1.01325 bar

number of moles n = ?

using the gas equation PV = nRT

n = PV/RT

R = 0.08206 L-atm-mol^{-1}K^{-1}

Therefore,

n = (1.01325 x 100)/(0.08206 x 293)

n = 101.325/24.04 = <em>4.21 moles</em>

The equation for root mean square velocity is

Vrms = \sqrt{\frac{3RT}{M} }

R = 8.314 J/mol-K

where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol

Vrms = \sqrt{\frac{3*8.314*293}{0.0319} }= <em>478.6 m/s</em>

<em>For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship</em>

\frac{Voxy}{Vnit} = \sqrt{\frac{Mnit}{Moxy} }

where

Voxy = root mean square velocity of oxygen = 478.6 m/s

Vnit = root mean square velocity of nitrogen = ?

Moxy = Molar mass of oxygen = 31.9 g/mol

Mnit = Molar mass of nitrogen = 14.00 g/mol

\frac{478.6}{Vnit} = \sqrt{\frac{14.0}{31.9} }

\frac{478.6}{Vnit} = 0.66

Vnit = 0.66 x 478.6 = <em>315.876 m/s</em>

<em>the root mean square velocity of the oxygen gas is </em>

<em>478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

6 0
2 years ago
A train travels 67 kilometers in 1 hours, and then 81 kilometers in 5 hours. What is its average speed?
Naddika [18.5K]
Answer: 24.7 km/h

Explanation:

1) Average speed definition and formula

The average speed is the total distance run divided by the time elapsed:

S = distance / time

2) Distance 1 = 67 km

3) Distance 2 = 81 km

4) Total distance traveled = 67 km + 81 km = 148 km

5) time 1 = 1 hour

6) time 2 = 5 hours

7) total time = time 1 + time 2 = 1 h + 5 h = 6 h

8) Average speed:

S = 148 km / 6 h = 24.7 km/h


4 0
2 years ago
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