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2 years ago

How fast must a 1000 kg car be moving to have a kinetic energy of 2.0*10^3

1 answer:

6 0

Kinetic Energy is defined by Ke=1/2mv^2. Plug in and solve for v.

2,000 = 1/2(1000)(v)^2
4=(v)^2
v=2 m/s

The car must move at 2 m/s to have a Ke of 2,000 Joules.

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2 years ago

Por una resistencia de 10 Ω fluyen 5A. ¿Cuál será la diferencia de potencial que se le debe aplicar a la resistencia?

ra1l [238]

Answer:

V = 50 volts

Explanation:

Given that,

Resistance, R = 10 ohms

Current, I = 5 A

We need to find the potential difference across the circuit. We know that,

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Put all the values,

V = 5 × 10

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2 years ago

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2 years ago

Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w

Tatiana [17]

Answer:

The average velocity is

$266\frac{m}{s},274\frac{m}{s}$ and $117\frac{m}{s}$ respectively.

Explanation:

Let's start writing the vertical position equation :

$L(t)=2t^{3}+t^{2}-5t+1$

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

$V_{avg}=\frac{Displacement}{Time}$ = Δx / Δt = $\frac{x2-x1}{t2-t1}$

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

$t2-t1=8s-5s=3s$

For the position variation we use the vertical position equation :

$x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m$

$x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m$

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

$\frac{798m}{3s}=266\frac{m}{s}$

For the second time interval :

t1 = 4 s → t2 = 9 s

$x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m$

$x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m$

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

$\frac{1370m}{5s}=274\frac{m}{s}$

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

$x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m$

$x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m$

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

$\frac{702m}{6s}=117\frac{m}{s}$

5 0

2 years ago

Consider two objects (Object 1 and Object 2) moving in the same direction on a frictionless surface. Object 1 moves with speed v

Semenov [28]

Answer:

A)Object 1 has the greater magnitude of its momentum.

B)The objects 2 have the greater kinetic energy.

Explanation:

For object 1 :

v₁ = v ,m₁ = 2 m

For object 2 :

$v_2=2\sqrt{v}$ ,m₂=m

We know that linear momentum given as

P = M V

M=Mass , V=Velocity

For object 1 :

P₁ =m₁ v₁

P₁ =2 m v

For object 2

$P_2=m_2v_2$

$P_2=2m\sqrt {v}$

We can say that object 1 have more momentum.

The kinetic energy

$KE_1=\dfrac{1}{2}m_1v_1^2$

$KE_1=\dfrac{1}{2}\times 2m\times v^2$

$KE_1=mv^2$

$KE_2=\dfrac{1}{2}m_2v_2$

$KE_2=\dfrac{1}{2}\times m\times 4v^2$

$KE_2=2mv^2$

Therefore both the object 2 have higher kinetic energy.

7 0

2 years ago