All categories

3 years ago

How does displacement, acceleration, time, and velocity affect motion?

Physics

2 answers:

Softa [21]3 years ago

6 0

Answer:

The relation between velocity and time is a simple one during uniformly accelerated, straight-line motion. The longer the acceleration, the greater the change in velocity. Change in velocity is directly proportional to time when acceleration is constant.

~Hoped this helped~

~Brainiliest?~

ASHA 777 [7]3 years ago

6 0

Answer:

We describe motion in terms of velocity and acceleration. Velocity: The rate of change of displacement of an object (displacement over elapsed time) is velocity. Velocity is a vector since it has both magnitude (called speed) and direction. ... Acceleration: The rate of change of velocity is acceleration.

Explanation:

Displacement is a vector which points from the initial position of an object to its final position. ... Instantaneous velocity, on the other hand, describes the motion of a body at one particular moment in time. Acceleration is a vector which shows the direction and magnitude of changes in velocity.

Displacement is the vector difference between the ending and starting positions of an object. Velocity is the rate at which displacement changes with time. ... The average velocity over some interval is the total displacement during that interval, divided by the time.

Hope this helps :)

You might be interested in

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

3 years ago

a bullet moving with a velocity of 100m/s pierce a block of wood and moves out with a velocityof 10 m/s.if the thickness of the

erma4kov [3.2K]

The emerging velocity of the bullet is 71 m/s.

The bullet of mass m moving with a velocity u has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed v when it emerges from the block.

If the block exerts a resistive force F on the bullet and the thickness of the block is x then, the work done by the resistive force is given by,

$W=Fx$

This is equal to the change in the bullet's kinetic energy.

$W=Fx=\frac{1}{2} m(u^2-v^2)......(1)$

If the thickness of the block is reduced by one-half, the bullet emerges out with a velocity v₁.

Assuming the same resistive forces to act on the bullet,

$F(\frac{x}{2} )=\frac{1}{2} m(u^2-v_1^2)......(2)$

Divide equation (2) by equation (1) and simplify for v₁.

$\frac{\frac{Fx}{2} }{Fx} =\frac{(u^2-v_1^2)}{(u^2-v^2)} \\\frac{100^2-v_1^2}{100^2-10^2} =\frac{1}{2} \\v_1^2=5050\\v_1=71.06 m/s$

Thus the speed of the bullet is 71 m/s

3 0

3 years ago

The French high-speed train travels at 300 km/h. How long

Agata [3.3K]

Answer:

given , v = 300 km/hr; distance d = 1500 km; then time t = d/v = 1500/300 = 5 hrs

Explanation:

4 0

3 years ago

Indiana jones (83.5 kg) is running 3.75 m/s when he jumps in a stationary 312 kg mine cart. what is their joint velocity afterwa

Lubov Fominskaja [6]

Answer:

.7917 m/s

Explanation:

This is a conservation of momentum question. You have an object initially at rest (cart) so that object is initially at 0 momentum. Indiana Jones is 83.5 kg and running 3.75 m/s so he starts with a momentum of 313.125 kg * m/s because momentum is equal to mass * velocity. Once the person jumps in the cart, the cart and the person can be considered one object and by conservation of momentum, the momentum of the Indiana-cart system is equal to 313.125 kg * m/s. By that, we can set that momentum equal to the combined mass * joint velocity. So 313.125 = (83.5kg + 312kg) * joint velocity. Then just solve for the velocity. The answer should be smaller than the intial velocity of the person of 3.75 m/s because the mine cart is HUGE at 312kg.

3 0

3 years ago

Two particles are moving along the x axis. Particle 1 has a mass m₁ and a velocity v₁ = +4.7 m/s. Particle 2 has a mass m₂ and a

nirvana33 [79]

Answer:

m₁ / m₂ = 1.3

Explanation:

We can work this problem with the moment, the system is formed by the two particles

The moment is conserved, to simulate the system the particles initially move with a moment and suppose a shock where the particular that, without speed, this determines that if you center, you should be stationary, which creates a moment equal to zero

p₀o = m₁ v₁ + m₂ v₂

pf = 0

m₁ v₁ + m₂ v₂ = 0

m₁ / m₂ = -v₂ / v₁

m₁ / m₂= - (-6.2) / 4.7

m₁ / m₂ = 1.3

Another way to solve this exercise is to use the mass center relationship

Xcm = 1/M (m₁ x₁ + m₂ x₂)

We derive from time

Vcm = 1/M (m₁ v₁ + m₂v₂)

As they say the velocity of the center of zero masses

0 = 1/M (m₁ v₁ + m₂v₂)

m₁ v₁ + m₂v₂ = 0

m₁ / m₂ = -v₂ / v₁

m₁ / m₂ = 1.3

4 0

3 years ago