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ddd [48]
3 years ago
7

How to calculate maximum height, when i know only V Pleeeaaassseee as fast as you can

Physics
1 answer:
Dahasolnce [82]3 years ago
8 0

Answer: was it this problem?

Explanation:

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What are light years actually? answer in brief
a_sh-v [17]
A light year is a distance. 
It's the distance that light travels in a year, through vacuum.
The distance is about 5.875 trillion miles. 
The nearest star outside our solar system is about 4.2 light years from us.

Sorry to have rambled on for so long.
6 0
3 years ago
What is the weight of a dog that has a mass of 47.0 kg?​
Nat2105 [25]

Answer:

while we often confuse mass with weight, 47 kg is 47 x 9.8 = 460.6 Newtons. 9.8 is acceleration of gravity in m/sec/sec

8 0
3 years ago
During a spring vacation, it rained on 13 days. When it rained in the morning, the afternoon was sunny. Every rainy afternoon wa
hammer [34]

Answer:

The holiday lasted for 18 days.

Explanation:

Let's assume the number of days are as following:

let the rain in the morning and lovely afternoon = X days

Clear morning and rain in the afternoon = Y days

No rain in the morning and in the afternoon = Z days

Now according to the question

Number of days with rain = X + Y = 13 days

Number of days with clear mornings = Y + Z = 11 days

Number of days with clear afternoons = X + Z = 12 days

Solving above 3 equations, we get

X = 7,  Y = 6 and  Z = 5

Now  total number of days on holiday = X+Y+Z = 7+6+5 =18

Hence, total number of days on holiday = 18 days.

8 0
3 years ago
Find the voltage drop (in volt) along a 93.4 meter long 10 gauge copper wire carrying acurrent of 72.5 A. The diameter of a 10 g
Sophie [7]

Answer: 5.41 V

Explanation:in order to explain this result we have to use the Ohm law given by:

ΔV=R*I where R is the resistance which is equal R= ρ*L/A . ρ is the resistivity, L the length of the wire and A is the cross section. I is the current.

Then we have

ΔV=ρ*L*I/A= 1.68 * 10^-8 Ωm*93.4 m*72.5A/2.1* 10^-5 m^2=5.41 V

3 0
3 years ago
A car covers 72 kilometers in the first hour of its journey. In the next hour, it covers 90 kilometers. What is the amount of wo
Nina [5.8K]

The work done is 2.8125 \times 10^{5} \mathrm{J}

Work Done = Change in Kinetic Energy (ΔKE)

<u>Explanation</u>

In first 1 hour it travels 72 km

So, Velocity = \frac{\text { distance }}{\text { time }}=\frac{72}{1} k m / h=72 \mathrm{km} / \mathrm{h}=\frac{72000}{3600} \mathrm{m} / \mathrm{s}=20 \mathrm{m} / \mathrm{s}

or, Initial Velocity (u) = 20 m/s

Similarly for the next hour it covers 90 km

So, Velocity = \frac{\text { distance }}{\text { time }}=\frac{90}{1} k m / h=90 \mathrm{km} / \mathrm{h}=\frac{90000}{3600} \mathrm{m} / \mathrm{s}=25 \mathrm{m} / \mathrm{s}

or, Final Velocity (v) = 20 m/s

Work done = Change in Kinetic Energy (ΔKE)

Work done = ΔKE = \frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}

ΔKE = \frac{1}{2} m\left(v^{2}-u^{2}\right)=\frac{1}{2} \times\left(2.5 \times 10^{3}\right) \times\left(25^{2}-20^{2}\right)

= \frac{2500 \times(625-400)}{2}=\frac{2500 \times 225}{2}=\frac{562500}{2}= 281250 joule  

= 2.8125 \times 10^{5} \mathrm{J}

4 0
3 years ago
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