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irina [24]
3 years ago
11

a car accelerates along a straight road from rest to 75km/h in 5.0s. What is the magnitude of its average acceleration? show up

solutions​
Physics
1 answer:
kykrilka [37]3 years ago
5 0

Answer:

4.186 m/s^2

Explanation:

First, convert km/hr to m/s:

(75 km/hr)(1000m/1km)(1hr/60min)(1min/60s) = 20.83 m/s

Then, divide 20.93 m/s by 5.0s

(20.93 m/s) / (5.0s) = 4.186 m/s^2

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When friction occurs, which of the following types of energy is produced?
sukhopar [10]

Answer:

heat energy

Explanation:

Friction causes the molecules on rubbing surfaces to move faster, so they have more energy. This gives them a higher temperature, and they feel warmer.

8 0
2 years ago
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A caterpillar climbs up a one-meter wall. For every 2 cm it climbs up, it slides down 1 cm. It takes 10 minutes for the caterpil
Brut [27]

100cm to go in 600 secs = 10mins

2 positive, 1 negative .... net 1 positive per step

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3 years ago
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A photon detector captures a photon with an energy of 4.29 ✕ 10−19 J. What is the wavelength, in nanometers, of the photon?
serious [3.7K]

Answer :  The wavelength of photon is, 4.63\times 10^{2}nm

Explanation : Given,

Energy of photon = 4.29\times 10^{-19}J

Formula used :

E=h\times \nu

As, \nu=\frac{c}{\lambda}

So, E=h\times \frac{c}{\lambda}

where,

\nu = frequency of photon

h = Planck's constant = 6.626\times 10^{-34}Js

\lambda = wavelength of photon  = ?

c = speed of light = 3\times 10^8m/s

Now put all the given values in the above formula, we get:

4.29\times 10^{-19}J=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{\lambda}

\lambda=4.63\times 10^{-7}m=4.63\times 10^{-7}\times 10^9nm=4.63\times 10^{2}nm

Conversion used : 1nm=10^{-9}m

Therefore, the wavelength of photon is, 4.63\times 10^{2}nm

6 0
3 years ago
You throw an 18.0 N rock into the air from ground level and observe that, when it is 15.0 m high, it is traveling upward at 17.0
stellarik [79]

Answer:

Vi = 24.14 m/s

Explanation:

If we apply Law of Conservation of Energy or Work-Energy Principle here, we get: (neglecting friction)

Loss in K.E of the Rock = Gain in P.E of the Rock

(1/2)(m)(Vi² - Vf²) = mgh

Vi² - Vf² = 2gh

Vi² = Vf² + 2gh

Vi = √(Vf² + 2gh)

where,

Vi = Rock's Speed as it left the ground = ?

Vf = Final Speed = 17 m/s

g = 9.8 m/s²

h = height of rock = 15 m

Therefore,

Vi = √[(17 m/s)² + 2(9.8 m/s²)(15 m)]

Vi = √583 m²/s²

<u>Vi = 24.14 m/s</u>

4 0
3 years ago
A ray at which angle would produce the most glare? A. 59 degrees C. 25 degrees D. 37 degrees
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<span>A. 59 degrees......
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