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Ronch [10]
2 years ago
15

More friction causes more what? A. Drag B. Weight C. Lift D. Thrust

Physics
1 answer:
yaroslaw [1]2 years ago
6 0

Answer:

A. Drag

Explanation:

This is because, since friction tends to oppose the motion of an object, and drag affects an object moving in a fluid, the more the friction, the more the drag on the object.

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g When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 3.2 10-15 m before
AnnyKZ [126]

Answer:

Time, t=1.07\times 10^{-23}\ s

Explanation:

Given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 3.2\times 10^{-15}\ m before interacting.

Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,

t=\dfrac{d}{c}\\\\t=\dfrac{3.2\times 10^{-15}}{3\times 10^8}\\\\t=1.07\times 10^{-23}\ s

So, the time interval is 1.07\times 10^{-23}\ s

5 0
3 years ago
What is a lunar month
Ira Lisetskai [31]

Answer:

It's the duration between successive new moons. Also called a lunation or synodic month, it has a mean period of 29.53059 days (29 days 12 hours and 44 minutes).

7 0
3 years ago
Read 2 more answers
The 8.00-cm long second hand on a watch rotates smoothly.
dolphi86 [110]
The watch hand covers an angular displacement of 2π radians in 60 seconds.

ω = 2π/60
ω = 0.1 rad/s

v = ωr
v = 0.1 x 0.08
v = 8 x 10⁻³ m/s
4 0
3 years ago
How could you weaken the force of gravity between cars and the Earth?<br>**I WILL MARK BRAINLEST**​
scoundrel [369]

Answer:

The answer you have selected is correct.

Explanation:

Increase radius, force of gravity decreases

6 0
3 years ago
A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v
atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

y(x, t) = Acos(kx -wt)

The relationship between velocity and propagation constant is

v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\

v = 15.87m/s

Time taken, t = \frac{\lambda}{v}

= \frac{1.3}{15.87}\\\\=0.0819 sec

t = 0.0819s

2)

The velocity of transverse wave is given by

v = \sqrt{\frac{T}{\mu}}

v = \sqrt{\frac{W}{\frac{m}{\lambda}}}

mass of string is calculated thus

mg = 0.0125N

m = \frac{0.0125N}{9.8N/s}

m = 0.00128kg

\omega = \frac{v^2m}{\lambda}

\omega = \frac{(15.87^2)(0.00128)}{1.30}

\omega = 0.25N

3)

The propagation constant k is

k=\frac{2\pi}{\lambda}

hence

\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}

\lambda = 0.036 m

No of wavelengths, n is

n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\

n = 36

4)

The equation of wave travelling down the string is

y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]

without, unit\\\\y(x , t)= Acos[172x + 2730t]

7 0
3 years ago
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