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Bumek [7]
3 years ago
15

The​ time, t, required to drive a fixed distance varies inversely as the​ speed, r. It takes 9 hr at a speed of 20 ​km/h to driv

e a fixed distance. How long will it take to drive the same distance at a speed of 23 ​km/h?
Physics
1 answer:
Harlamova29_29 [7]3 years ago
6 0

Answer:

Time=7.84hrs

Explanation:

This is an inverse proportionality questions

Mathematically time t varies as 1/distance d

Hence T= dk/r

But speed r = distance/t

d= r*t= 9*20=180km/hr

We're k= constant of proportionality

At t=9hrs d=180km/hr

Hence k=r*t/d=20*9/1620=0.111

Finding t at at 1620km for 23km/hr

t=1620*0.111/23=7.84hrs

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shutvik [7]

Answer:

8.1 N/49 N=0.1653  which means 16.53% of the weight of the object on Earth.

Explanation:

On the Moon, where the gravitational constant is 1.62 \frac{m}{s^2}, the weight of the 5 kg object will be: weight_M=m*g_M = 5 kg * 1.62 \frac{m}{s^2} =8.1 N

Where the answer is in Newtons (N) since all quantities are given in the SI system.

On Earth, on the other hand, the weight of the object is:

weight_E=m*g_E= 5 kg* 9.8 \frac{m}{s^2} = 49N

Therefore the object's weight on the Moon compared to that on Earth will be:

8.1N/49N=0.1653

That is, 16.53% of the weight the object has on Earth.

5 0
3 years ago
I will fan and give 2 medals!!!!!!
Maksim231197 [3]
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I did this in class I got 100%

7 0
3 years ago
How many states of matter do you think are in the photo? Describe at least two.
laila [671]

Answer:

Three, the liquid in the man's cup, the stove is solid, and the air around them is a gas

Explanation:

3 0
3 years ago
If an object accelerates from rest, what will its velocity be after 1.3 s if it has a constant acceleration of 9.1 m/s^2?
HACTEHA [7]

\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time,  t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0  +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}

5 0
2 years ago
A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the
Alex17521 [72]

Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3  m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1       V = 1.50*10^-3  m^3       ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5

So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J

W = 3.12 J

Hope this helps!

4 0
3 years ago
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