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malfutka [58]
3 years ago
15

What is the conversion factor between km/h and mi/h

Physics
1 answer:
Aloiza [94]3 years ago
5 0
<span> 1 mi = 1.609 km
so X mi/hr = 1.609 * X km/hr hope this helps!!</span>
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What is the mass of a stone moving at a speed of 15 m/s and having a monument at 7.1 kg meters per second
adell [148]

Answer:

<h3>The answer is 0.47 kg</h3>

Explanation:

The mass of the object given it's momentum and velocity can be found by using the formula

m =  \frac{p}{v}  \\

where

p is the momentum

v is the velocity

We have

m =  \frac{7.1}{15}  \\  = 0.4733333...

We have the final answer as

<h3>0.47 kg</h3>

Hope this helps you

4 0
3 years ago
Hello people ~
lozanna [386]

Answer:

It’s called a conservative field.

Explanation:

I think it’s going to be the conservative field because in the question it talks about how it is able to become possible to define potential at a point in an electric field because electric field.

3 0
2 years ago
Read 2 more answers
A car is traveling with 90km/hr and another car with 20m/s in opposite direction. Calculate the relative velocity.
Slav-nsk [51]
90 km/h is 25 m/s
the relative velocity when cars are traveling in opposite directions is the sum of the two
25+20= 45 m/s
5 0
3 years ago
When you trace the outline of your palm how do you find its area​
Tcecarenko [31]

Answer:

you count the squares or messure it

Explanation:

you can raw equal squares about 1 cm wide if possible all equal and count the squares eg theres 10 squares (small hand) so that would be 10cm squared

7 0
3 years ago
Read 2 more answers
A 1300 kg steel beam is supported by two ropes. (Figure
Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

  • the net horizontal force acting on the beam is

R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

  • the net vertical force acting on the beam is

R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

8 0
1 year ago
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