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arsen [322]
3 years ago
8

The -34?ffrfefefffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff

Physics
1 answer:
Soloha48 [4]3 years ago
7 0

Answer:

owo

Explanation:

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In the absence of air, why does the horizontal component of velocity for a projectile remain constant while the vertical compone
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Because there is no horizontal force on the object, and we all know what happens to velocity in a direction where there is no force.
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3 years ago
A 4.3 kg steel ball and 6.5 m cord of negligible mass make up a simple pendulum that can pivot without friction about the point
Alekssandra [29.7K]

Answer

given,

mass of steel ball, M = 4.3 kg

length of the chord, L = 6.5 m

mass of the block, m = 4.3 Kg

coefficient of friction, μ = 0.9

acceleration due to gravity, g = 9.81 m/s²

here the potential energy of the bob is converted into kinetic energy

m g L = \dfrac{1}{2} mv^2

v= \sqrt{2gL}

v= \sqrt{2\times 9.8\times 6.5}

      v = 11.29 m/s

As the collision is elastic the velocity of the block is same as that of bob.

now,

work done by the friction force = kinetic energy of the block

f . d = \dfrac{1}{2} mv^2

\mu m g. d = \dfrac{1}{2} mv^2

d=\dfrac{v^2}{2\mu g}

d=\dfrac{11.29^2}{2\times 0.9 \times 9.8}

    d = 7.23 m

the distance traveled by the block will be equal to 7.23 m.

8 0
3 years ago
A 1kg cart slams into a stationary 1kg cart at 2 m/s. The carts stick together and move forward at a speed of 1 m/sl. Determine
finlep [7]

Answer:

No, it is not conserved

Explanation:

Let's calculate the total kinetic energy before the collision and compare it with the total kinetic energy after the collision.

The total kinetic energy before the collision is:

K_i = K_1 + K_2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2=\frac{1}{2}(1 kg)(2 m/s)^2+\frac{1}{2}(1 kg)(0)^2=2 J

where m1 = m2 = 1 kg are the masses of the two carts, v1=2 m/s is the speed of the first cart, and where v2=0 is the speed of the second cart, which is zero because it is stationary.

After the collision, the two carts stick together with same speed v=1 m/s; their total kinetic energy is

K_f = \frac{1}{2}(m_1+m_2)v^2=\frac{1}{2}(1 kg+1kg)(1 m/s)^2=1 J

So, we see that the kinetic energy was not conserved, because the initial kinetic energy was 2 J while the final kinetic energy is 1 J. This means that this is an inelastic collision, in which only the total momentum is conserved. This loss of kinetic energy does not violate the law of conservation of energy: in fact, the energy lost has simply been converted into another form of energy, such as heat, during the collision.

3 0
3 years ago
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