Describe a situation in which an object's speed is constant but its velocity is changing.

1 answer:

5 0

1) If the object changes directions with the same speed, it will be changing its velocity, because velocity is a vector, which depends on both magnitude and direction. Speed is just magnitude regardless of direction.

2) Rotational motion.

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1 ) Starting from rest, a toy rocket accelerates at 12 m/sec/sec for exactly 4.0 seconds. It reaches 48 m/sec. Find the distance

dusya [7]

Displacement equals (Velocity times Time) plus half times the (acceleration times time squared). =. (48 * 4) + 1/2 * (12 *12^2) = 288meters

3 0

2 years ago

You pick up a 3.4-kg can of paint from the ground and lift it to a height of 1.8 m. (a) how much work do you do on the can of pa

MariettaO [177]

(a) For the work-energy theorem, the work done to lift the can of paint is equal to the gravitational potential energy gained by it, therefore it is equal to

$W=mg\Delta h$

where m=3.4 kg is the mass of the can, g=9.81 m/s^2 is the gravitational acceleration and $\Delta h=1.8 m$ is the variation of height. Substituting the numbers into the formula, we find

$W=(3.4 kg)(9.81 m/s^2)(1.8 m)=60.0 J$

(b) In this case, the work done is zero. In fact, we know from its definition that the work done on an object is equal to the product between the force applied F and the displacement:

$W=Fd$

However, in this case there is no displacement, so d=0 and W=0, therefore the work done to hold the can stationary is zero.

(c) In this case, the work done is negative, because the work to lower the can back to the ground is done by the force of gravity, which pushes downward. Its value is given by the same formula used in part (a):

$W=mg \Delta h=(3.4 kg)(9.81 m/s^2)(-1.8 m)=-60.0 J$

8 0

3 years ago

Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small

Vaselesa [24]

Answer:

The final kinetic energy of the Helium nucleus (alpha particle) after been scattered through an angle of 120° is

8.00 x 10-13J

Explanation:

In Rutherford Scattering experiment, the collision of the helium nucleus with the gold nucleus is an ELASTIC COLLISION. This means that the kinetic energy is conserved ( The same before and after the collision).

Thus, the final kinetic energy of the helium nucleus is the same as initial kinetic energy (8.00 x 10^-13Joules)

Although, the kinetic energy is converted to potential energy in Coulomb's law equation.

That is,

1/2(mv^2) = (K* q1q2)/r

Where m is the mass of helium nucleus, v is its colliding velocity, k is electrostatic constant, q1 is the charge on helium nucleus, q2 is the charge on gold nucleus, r is impact parameter

6 0

2 years ago

A string with a length of 4.00 m is held under a constant tension. The string has a linear mass density of \mu=0.000600~\text{kg

yulyashka [42]

Answer:

$T=245.76N$

Explanation:

We know that the frequency of the nth harmonic is given by $f_n=nf$ , where $f$ is the fundamental harmonic. Since we have the values of two consecutive frequencies, we can do: