All categories

2 years ago

When is L larger than XL

Physics

2 answers:

polet [3.4K]2 years ago

8 0

-- In Roman Numerals, always ... 50 is larger than 40 .

-- In Algebra, whenever 'X' is less than ' 1 ' .

Brrunno [24]2 years ago

4 0

In roman numerals, nice riddle btw

You might be interested in

Is this true or false ethics deals with morals and values and can be measured and tested using the scientific method.

sammy [17]

This is false. Even though the first part is true, the second isn't because morals and values are simply widely accepted opinions that most humans have, and you can't test opinions with the scientific method, only facts!

3 0

2 years ago

Read 2 more answers

Any person for the job he applied Force moment or torque

VLD [36.1K]

Answer:

i think torque is the answer not confirmed

7 0

2 years ago

Which of the following has the largest momentum?

san4es73 [151]

Momentum = mass x velocity

Thus Option A is the correct answer

Momentum (dog) = 10 kg x (0.447 x 30) m/s

= 134.1 Kg m/s

Momentum ( bullet) = 0.02 kg x (0.447 x 800) m/s

= 7.152 Kg m/s

Momentum ( truck) = 0, as v = 0

tightrope has both low mass and low speed, thus its momentum will be low

4 0

3 years ago

When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0

lora16 [44]

Answer:

$d=0.165m$

Explanation:

Given

$m=0.25kg$ , $x_{1}=5cm*\frac{1m}{100cm}=0.05m$ , $x_{2}=4cm*\frac{1m}{100cm}=0.04m$ , $v=2\frac{rev}{s}$

The tension of the spring is

$F_{k}=K*x_{1}=m*g$

$K=\frac{m*g}{x_{1}}$

$K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m$

The force in the spring is equal to centripetal force so

$F_{c}=\frac{m*v^2}{r}$

$v=w*r=2\pi*r$

But Fc is also

Fc=KxΔr

$F_{c}=K*(r-x_{2})$

Replacing

$m*4\pi^2*r=K*(r-x_{2})$

$0.25kg*4\pi^2*r=49*(r-0.04m)$

$r=0.205m$

total distance is

$d=0.205-0.04=0.165m$

3 0

3 years ago

Một dây dẫn đặt trong không khí có dòng điện I = 12A chạy qua, được gấp thành hình vuông cạnh a = 10cm. Xác định vectơ cường độ

disa [49]

Answer:

The net magnetic field ta the center of square is $1.36\times10^{-4} T$ .

Explanation:

Current, I = 12 A , side ,a = 10 cm = 0.1 m

Let the magnetic field due to the one side is B.

The magnetic field is given by

$B = \frac{\mu o}{4\pi}\times \frac{I}{r}\times \left (Sin A +Sin B \right )\\\\B = 10^{-7}\times \frac{12}{0.05}\times \left ( sin 45 + sin 45 \right )\\\\B = 3.4\times 10^{-5} T$

Net magnetic field at the center of the square is

B' = 4 B

$B'= 4\times 3.4\times 10^{-5}\\\\B' = 1.36\times10^{-4} T$

4 0

3 years ago