Which of the following is an example of a homogeneous mixture A. Bronze B. Zinc C. Copper D. Sand

A 5.00×105-kg subway train is brought to a stop from a speed of 0.500 m/s in 0.400 m by a large spring bumper at the end of its

Lesechka [4]

This can be calculated with the law of conservation of energy. The kinetic energy Ek of the train is completely transformed to potential energy of the spring Ep which stops the train:

Ek=Ep

(1/2)*m*v²=(1/2)*k*x², where m is mass of the train, v is the velocity of the train, k is the force constant of the spring and x is the path the train went while being stopped and also the amount of length the spring compressed.
We solve for k:

(1/2)*m*v²=(1/2)*k*x², 1/2 cancel out:

m*v²=k*x², we divide both sides by x² and get k:

k=(m*v²)/x²={(5*10^5)*(0.5²)}/(0.4²)=781250 N/m

So the force constant of the spring is: k=781250 N/m.

5 0

4 years ago

What is the mass of an object if a force of 17 N causes it to accelerate at 1.5 m/s/s?

DochEvi [55]

The answer is m=11.4g

4 0

3 years ago

A constant electric field with magnitude 1.50 ✕ 103 N/C is pointing in the positive x-direction. An electron is fired from x = −

romanna [79]

Answer:

The speed of electron is $1.5\times10^{7}\ m/s$

Explanation:

Given that,

Electric field $E=1.50\times10^{3}\ N/C$

Distance = -0.0200

The electron's speed has fallen by half when it reaches x = 0.190 m.

Potential energy $P.E=5.04\times10^{-17}\ J$

Change in potential energy $\Delta P.E=-9.60\times10^{-17}\ J$ as it goes x = 0.190 m to x = -0.210 m

We need to calculate the work done by the electric field

Using formula of work done

$W=-eE\Delta x$

Put the value into the formula

$W=-1.6\times10^{-19}\times1.50\times10^{3}\times(0.190-(-0.0200))$

$W=-5.04\times10^{-17}\ J$

We need to calculate the initial velocity

Using change in kinetic energy,

$\Delta K.E = \dfrac{1}{2}m(\dfrac{v}{2})^2+\dfrac{1}{2}mv^2$

$\Delta K.E=\dfrac{-3mv^2}{8}$

Now, using work energy theorem

$\Delta K.E=W$

$\Delta K.E=\Delta U$

So, $\Delta U=W$

Put the value in the equation

$\dfrac{-3mv^2}{8}=-5.04\times10^{-17}$

$v^2=\dfrac{8\times(5.04\times10^{-17})}{3m}$

Put the value of m

$v=\sqrt{\dfrac{8\times(5.04\times10^{-17})}{3\times9.1\times10^{-31}}}$

$v=1.21\times10^{7}\ m/s$

We need to calculate the change in potential energy

Using given potential energy

$\Delta U=-9.60\times10^{-17}-(-5.04\times10^{-17})$

$\Delta U=-4.56\times10^{-17}\ J$

We need to calculate the speed of electron

Using change in energy

$\Delta U=-W=-\Delta K.E$

$\Delta K.E=\Delta U$

$\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=4.56\times10^{-17}$

Put the value into the formula

$v_{f}=\sqrt{\dfrac{2\times4.56\times10^{-17}}{9.1\times10^{-31}}+(1.21\times10^{7})^2}$

$v_{f}=1.5\times10^{7}\ m/s$

Hence, The speed of electron is $1.5\times10^{7}\ m/s$

4 0

3 years ago

True or False

devlian [24]

Answer:

true

Explanation:

Hope it helped you brainiest plz

4 0

3 years ago

Read 2 more answers

Two points are on a disk turning at constant angular velocity. One point is on the rim and the other halfway between the rim and

DerKrebs [107]

Answer:

The point on the rim

Explanation:

All the points on the disk travels at the same angular speed $\omega$ , since they cover the same angular displacement in the same time. Instead, the tangential speed of a point on the disk is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the point from the centre of the disk

As we can see, the tangential speed is directly proportional to the distance from the centre: so the point on the rim, having a larger r than the point halway between the rim and the axis, will have a larger tangential speed, and therefore will travel a greater distance in a given time.

3 0

4 years ago