Science homework help

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.320 with the floor. If

mart [117]

Answer: 29.50 m

Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:

f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).

f=m.a=μ*N= m*a= μ*m*g= m*a

then

a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2

With this value we can determine the short distance to stop the train

as follows:

x= vo*t- (a/2)* t^2

Vf=0= vo-a*t then t=vo/a

Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m

5 0

3 years ago

A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed

Vladimir [108]

Answer:

we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level

for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.

Explanation:

A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed of the rock at 2m on the way down compare with its speed at 2m on the way up?

It decreases in speed on its way down and increases in speed on its way down.

it decreases in speed on its way up because the the vertical motion is against the earths gravitational pull on an object to the earth's center

.It increases in speed on his way down because its under the influence of gravity

from newton's equation of motion we can check by

using V^2=u^2+2as

we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level

for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.

5 0

3 years ago

A mass m is tied to an ideal spring with force constant k and rests on a frictionless surface. The mass moves along the x axis.

7nadin3 [17]

Answer: $x=\frac{x_m}{\sqrt{2}}$

Explanation:

Given

initially mass is stretched to $x_m$

Let k be the spring Constant of spring

Therefore Total Mechanical Energy is $\frac{kx_m^2}{2}$

Position at which kinetic Energy is equal to Elastic Potential Energy

$K=\frac{mv^2}{2}$

$U=\frac{kx^2}{2}$

it is given

$k=U$

thus $2U=\frac{kx_m^2}{2}$

$2\times \frac{kx^2}{2}=\frac{kx_m^2}{2}$

$2x^2=x_m^2$

$x=\frac{x_m}{\sqrt{2}}$

3 0

3 years ago

A box of mass 12 kg is at rest on a flat floor. The coefficient of static friction between the box and floor is 0.42. What is th

vladimir2022 [97]

As we know that friction force on box is given by

$F_s = \mu_s N$

here we know that

$N = mg$

here we have

m = 12 kg

$\mu_s = 0.42$

so now we have

$N = 12(9.8) = 117.6 N$

now we will have

$F_s = 0.42(12)(9.8)$

$F_s = 49.4 N$

so it required minimum 49 N(approx) force to move the block

5 0

3 years ago

Read 2 more answers

PLEASE HELP 15 POINTS The electric field around a positive charge is shown in the diagram. Describe the nature of these lines. P

DENIUS [597]

The field lines spread apart as we move away from the charge, and they point away from the charge

Explanation:

The electric field produced by a single-point positive charge is a radial field, whose strength is given by the equation

$E=k\frac{Q}{r^2}$

where

k is the Coulomb's constant

Q is the magnitude of the charge

r is the distance from the charge at which the field is calculated

There are two pieces of information given by the field lines shown in the graph:

The spacing between the lines gives an indication of the strength of the field: the closer to each other they are, the stronger the field. In this case, as we move away from the charge, the spacing between the lines increases, and this means that the field becomes weaker (in fact, it follows an inverse square law, $E\propto \frac{1}{r^2}$
The direction of the lines gives the direction of the electric field, which points away from the central charge. This is because the direction of the electric field corresponds to the direction of the force that a positive test charge would feel when immersed in the electric field: in this case, if we place a positive test charge in this field, then it would get repelled away from the central charge (remember that the electric force between two positive charges is repulsive), and therefore, the direction of the electric field is away from the central charge.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

8 0

3 years ago