All categories

3 years ago

When does the velocity and speed of a moving body become identical?

Physics

2 answers:

raketka [301]3 years ago

8 0

Velocity and speed of a moving body become identical when it tends to move in a straight line.

Nat2105 [25]3 years ago

8 0

Velocity and speed are never identical. Velocity is a vector quantity. It takes two numbers to describe a velocity. One number for its size the other number for its direction. Speed is a scalar quantity. It only has size and only needs one number to describe it. The size of a vector may be identical to a speed. But the two quantities are never completely identical.

You might be interested in

A 0.5-kilogram apple falls from a height of 2 meters to 1.50 meters. Ignoring frictional effects, what is the kinetic energy of

spin [16.1K]

The final kinetic energy of the ball is 2.45 J

Explanation:

We can solve this problem by using the law of conservation of energy.

In absence of frictional effect, the mechanical energy of the apple must be conserved during the fall. So we can write:

$U_i +K_i = U_f + K_f$

where :

$U_i$ is the initial potential energy, at the top

$K_i$ is the initial kinetic energy, at the top

$U_f$ is the final potential energy, at the bottom

$K_f$ is the final kinetic energy, at the bottom

By explicing the potential energy, we can rewrite the equation as:

$mgh_i + K_i = mgh_f + K_f$

where:

m = 0.5 kg is the mass of the apple

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 2 m$ is the initial height

$h_f=1.50 m$ is the final height

The initial kinetic energy is zero, since the ball starts from rest:

$K_i = 0$

Therefore we can solve the equation for $K_f$ , the final kinetic energy of the ball:

$K_f = mg(h_i-h_f)=(0.5)(9.8)(2-1.50)=2.45 J$

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

3 0

3 years ago

Read 2 more answers

The atomic nucleus contains two subatomic particles, the proton and the neutron. Atoms of different elements have different numb

9966 [12]

The answer is option A .

7 0

3 years ago

Read 2 more answers

Please answer will give brainliest

laila [671]

3b. No

4a. 50N

I hope this helps

8 0

2 years ago

A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of

34kurt

Answer:

The centripetal acceleration of the runner is $1.73\ m/s^2$ .

Explanation:

Given that,

A runner completes the 200 m dash in 24.0 s and runs at constant speed throughout the race. We need to find the centripetal acceleration as he runs the curved portion of the track. We know that the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

v is the velocity of runner

$v=\dfrac{200\ m}{24\ s}\\\\v=8.34\ m/s$

Centripetal acceleration,

$a=\dfrac{(8.34)^2}{40}\\\\a=1.73\ m/s^2$

So, the centripetal acceleration of the runner is $1.73\ m/s^2$ . Hence, this is the required solution.

5 0

3 years ago

A roller coaster car is traveling at a constant 3 m/s when it reaches a downward slope. On the slope, the car accelerates at a c

kirill [66]

Answer:

Velocity of the car at the bottom of the slope: approximately $20.3\; \rm m \cdot s^{-2}$ .

It would take approximately $3.9\; \rm s$ for the car to travel from the top of the slope to the bottom.

Explanation:

The time of the travel needs to be found. Hence, make use of the SUVAT equation that does not include time.

Let $v$ denote the final velocity of the car.
Let $u$ denote the initial velocity of the car.
Let $a$ denote the acceleration of the car.
Let $x$ denote the distance that this car travelled.

$v^2 - u^2 = 2\, a\cdot x$ .

Given:

$u = 3\; \rm m \cdot s^{-1}$ .
$a = 4.5\; \rm m \cdot s^{-2}$ .
$x = 45\; \rm m$ .

Rearrange the equation $v^2 - u^2 = 2\, a\cdot x$ and solve for $v$ :

$\begin{aligned}v &= \sqrt{2\, a \cdot x + u^2} \\ &= \sqrt{2 \times 4.5\; \rm m \cdot s^{-2} \times 45\; \rm m + \left(3\; \rm m \cdot s^{-1}\right)^{2}} \\ &\approx 20.3\; \rm m \cdot s^{-1}\end{aligned}$ .

Calculate the time required for reaching this speed from $u = 3\; \rm m \cdot s^{-1}$ at $a = 4.5\; \rm m \cdot s^{-2}$ :

$\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{20.3\; \rm m \cdot s^{-1} - 3\; \rm m \cdot s^{-1}}{4.5\; \rm m \cdot s^{-2}} \approx 3.9\; \rm m \cdot s^{-1}\end{aligned}$ .

3 0

3 years ago