Hi hi someone talk to meeeee

The 6 kg block is then released and accelerates to the right, toward the 4 kg block. The surface is rough and the coefficient of

KatRina [158]

Answer:

v =3.41 m/s

Explanation:

given,

mass of block 1 = 6 Kg

mass of another block 2 = 4 Kg

coefficient of friction = 0.3

Assuming 6 Kg block is attached to the spring of spring constant 350 N/m

and distance between the two block is equal to 0.5 m

using formula

$U = \dfrac{1}{2}kx^2$

$U = \dfrac{1}{2}\times 350 \times 0.5^2$

U = 43.75 J

using conservation of energy

KE = U - f.d

where f is the frictional force acting

$\dfrac{1}{2}mv^2 = 43.75- \mu m g d$

$\dfrac{1}{2}\times 6 \times v^2 = 43.75- 0.3\times 6 \times 9.8 \times 0.5$

$v= \sqrt{11.643}$

v =3.41 m/s

4 0

3 years ago

Give two examples of human activity that affect the earth systems

joja [24]

Humans affect the eath by burning fossil fuels

4 0

3 years ago

How does size influence the appearance of a star? Give an example in your response.

vlabodo [156]

The size of the star defines wither of not it is a dwarf star, or a red giant.

5 0

4 years ago

If half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration it can achieve o

Scilla [17]

Answer:

Maximum acceleration will be equal to $3.43m/sec^2$

Explanation:

We have given coefficient of kinetic friction $\mu _k=0.7$

And coefficient of static friction $\mu _s=1$

Acceleration due to gravity $g=9.8m/sec^2$

When truck moves maximum force will be equal to $F=\mu _kmg$

It is given that half of the weight is supported by its drive wheels

So force required $=\frac{\mu _kmg}{2}$

From newtons law maximum acceleration will be equal to $a=\frac{\frac{\mu _kmg}{2}}{m}=\frac{\mu _kg}{2}=\frac{0.7\times 9.8}{2}=3.43m/sec^2$

8 0

3 years ago

Two girls,(masses m1 and m2) are on roller skates and stand at rest, close to each other and face-to-face. Girl 1 pushes squarel

Arturiano [62]

Answer:

$\vec{v}_1 = -\frac{\vec{v}_2m_2}{m_1}$

Explanation:

The center of mass of the system (two girls) is constant, as the velocity of the center of mass of the system is also constant.

$\vec{v}_{cm} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}$

<u>The initial velocity of the system is zero, since both girls are at rest.</u> So the velocity of the total system at any point should be zero as well.

$0 = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}\\\vec{v}_1 = -\frac{\vec{v}_2m_2}{m_1}$

This is true, because there is no friction between the girls and the ground. Otherwise, the velocity of the center of mass wouldn't be constant.

5 0

3 years ago