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vichka [17]
3 years ago
8

A car travels a distance of 320 km in 4 hours. What is your average speed in meters per second?

Physics
1 answer:
Andreas93 [3]3 years ago
4 0

Answer:

22.2 m/s

Explanation:

First, we need to convert km to m by multiplying by 1000. This means that the car traveled 320 000 meters.

Next, we convert hours to minutes by multiplying by 3600 (the number of seconds in an hour). This means that overall, the car traveled 320 000 m in 14 400 seconds.

The average speed can be found by using the equation \frac{distance}{time}. After substitution, this gives the fraction \frac{320 000}{14 400}, which reduces to 22 \frac{2}{9} m/s, or about 22.2 m/s.

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luda_lava [24]
A ) 
T = mB g + mB a
T + mA a - mA g sin 35° = (Mi) mA g cos 35°
------------------------------------------------------------
T = 2.7 · 9.81  + 2.7 a
T = 26.487 + 2.7 a
26.487 + 2.7 a + 2.7 a - 2.7 · 9.81 · 0.574 = 0.15 · 2.7 · 9.81 · 0.819
5.4 a + 26.487 - 15.2023 = 3.2539
5.4 a = 8.0296
a = 1.487 ≈ 1.5 m/s²
B )
T = 2,7 · 9.81 = 26.487 
26.487 - 15.2035 = (Mi) · 2.7 · 9.81 · 0.819
11.2835 = (Mi) · 21.69
(Mi) = 11.2835 : 21.69 = 0.52
4 0
3 years ago
Assume n1 and n2 are two adjacent energy levels of an atom. The emission of radiation with the longest wavelength would occur fo
Anni [7]

Answer:

Explanation:

If the energy levels of n1 and n2 are high, the difference of their energy level will tend to zero thereby making the energy of the emitted photons very low.

3 0
3 years ago
An object, initially at rest, is subject to an acceleration of 34 m/s^2. How long will it take for that object to reach 3400m ?
Norma-Jean [14]
Vf^2 = Vi^2 + 2ad
a= 34 m/s^2
Vi = 0 m/s
d = 3400m

Vf = 480.83 m/s

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t=480.83/34
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A singly charged ion of 7Li (an isotope of lithium which lost only one electron) has a mass of 1.16 ×10^-26 kg. It is accelerate
MaRussiya [10]

Explanation:

It is given that,

Mass of lithium, m=1.16\times 10^{-26}\ kg

It is accelerated through a potential difference, V = 224 V

Uniform magnetic field, B = 0.724 T

Applying the conservation of energy as :

\dfrac{1}{2}mv^2=qV

v=\sqrt{\dfrac{2qV}{m}}

q is the charge on an electron

v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 224\ V}{1.16\times 10^{-26}\ kg}}

v = 78608.58 m/s

v=7.86\times 10^4\ m/s

To find the radius of the ion's path in the magnetic field. The centripetal force is balanced by the magnetic force as :

qvB=\dfrac{mv^2}{r}

r=\dfrac{mv}{qB}

r=\dfrac{1.16\times 10^{-26}\ kg\times 7.86\times 10^4\ m/s}{1.6\times 10^{-19}\ C\times 0.724\ T}

r = 0.0078 meters

So, the radius of the path of the ion is 0.0078 meters. Hence, this is the required solution.

4 0
3 years ago
Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on
Natasha2012 [34]

Answer:

8.8\times 10^{-6} W/m^2

Explanation:

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1 nm=10^{-9} m

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1 m=100 cm

d=0.640 mm=0.64\times 10^{-3} m

1 mm=10^{-3} m

a=0.434 mm=0.434\times 10^{-3} m

y=0.830 mm=0.830\times 10^{-3} m

I_0=5.00\times 10^{-4}W/m^2

tan\theta=\frac{y}{R}

\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}

\theta=1.1\times 10^{-3}rad

tan\theta\approx \theta

Because \theta is small.

\phi=\frac{2\pi dsin\theta}{\lambda}

\sin\theta\approx \theta,

Therefore

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\phi=7.74 rad

\beta=\frac{2\pi a\theta}{\lambda}

\beta=5.3 rad

I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2

I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2

I=8.8\times 10^{-6} W/m^2

6 0
3 years ago
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