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Eduardwww [97]
3 years ago
12

determinar el valor de la velocidad que lleva un cuerpo cuya masa es 3kg y su energia cinetica es de 400Joule

Physics
1 answer:
Korolek [52]3 years ago
6 0

Answer:

This is what they said in english if everyone was wondering

Explanation:

determine the value of the velocity carried by a body whose mass is 3kg and its kinetic energy is 400Joule

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At the equator earth rotates with a velocity of about 465 m/s.
Dafna1 [17]
The given velocity is 465 m/s.

Part a.
465 \,  \frac{m}{s} =(465 \times 10^{-3} \,  \frac{km}{s})*( 3600 \,  \frac{s}{h} ) = 1674 \,  \frac{km}{h}
Answer: 1674 km/h

Part b.
1674  \frac{km}{h} = (1674 \,  \frac{km}{h})*(24 \,  \frac{h}{day}  ) = 40176 \,  \frac{km}{day}
Answer: 40,176 km/day.

 
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3 years ago
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A box of mass 26 kg is initially at rest on a flat floor. The coefficient of kinetic friction between the box and the floor is 0
Kazeer [188]

Answer:

\Delta K = 52J

Explanation:

The change in kinetic energy will be simply the difference between the final and initial kinetic energies: \Delta K=K_f-K_i

We know that the formula for the kinetic energy for an object is:

K=\frac{mv^2}{2}

where <em>m </em>is the mass of the object and <em>v</em> its velocity.

For our case then we have:

\Delta K = K_f-K_i=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=\frac{m(v_f^2-v_i^2)}{2}

Which for our values is:

\Delta K = \frac{m(v_f^2-v_i^2)}{2} = \frac{(26Kg)((2m/s)^2-(0m/s)^2)}{2} = 52J

3 0
3 years ago
Hello everyone. This is a question about Dimensional Analysis and I came across this question but I am unable to wrap my head ar
omeli [17]

Answer:

2. [B] = [L]/[T] and [C] = [L]/[T]

Explanation:

I assume you mean this:

A = B² + 2B⁴/C²

Since you can't add numbers with different units (for example, you can't add seconds to meters), each term in the sum must have the same units as A.

B² = [L]²/[T]²

B = [L]/[T]

B⁴/C² = [L]²/[T]²

C²/B⁴ = [T]²/[L]²

C² = B⁴ [T]²/[L]²

C² = ([L]/[T])⁴ [T]²/[L]²

C² = [L]²/[T]²

C = [L]/[T]

Notice we ignore the 2 coefficient, which is unitless.

7 0
3 years ago
We are running late for school and we want to make our 0.5 kg tea (it’s at 90 C) colder. Let’s assume we can drink tea when it’s
PSYCHO15rus [73]

Answer:

x=0.154kg

Explanation:

(x*L)+(0.5kg*4200*50)+(x*4200*(-50)=0

(x*333 000J/kg*c)+(0.5kg*4200J/kg*C*(-40C))+(x*4200J/kg*C*50C)=0

6 0
2 years ago
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