Ask question

Ask question

All categories

SVETLANKA909090 [29]

3 years ago

7

Spring has an unstretched length of 0.40 meters. The spring is stretched to a length of 0.60 meters when a 10.-newton weight is

hung motionless from one end. The spring constant of this spring is:

1 answer:

Nonamiya [84]3 years ago

7 0

Answer: The spring constant is 50N/m

Explanation: Please see the attachment below

You might be interested in

Why should you explore the potential roadblocks to your chosen career?

valina [46]

Answer: D

Explanation: D is the most reasonable answer because it's always good to plan ahead for anything, so if you were to plan ahead for future obstacles, then you can overcome them.

6 0

3 years ago

An electron is moving east in a uniform electric field of 1.47 {\rm N/C} directed to the west. At point A, the velocity of the e

zimovet [89]

Answer:

a) $v = 6.25\times 10^{5} m/s$

b) $v = 1.73\times 10^{4} m/s$

Explanation:

Given data:

Electric field = 1.47 N/C

velocity of electron is $4.55\times 10^5 m/s$

distance of point b from point A is 0.55 m

we know that acceleration of particle is given as

a) for electron

$a =\frac{q E}{m}$

$a = \frac{1.6\times 10^{-19} \times 1.47}{9.1\times 10^{-31}}$

$a = 2.58\times 10^{11} m/s^2$

from equation of motion we have

$v^2 = u^2 + 2aS$

$= 20.7025 \times 10^{10} + 2\times 2.58\times 10^{11} \times 0.355$

$v = 6.25\times 10^{5} m/s$

b) for proton

$a = \frac{1.6\times 10^{-19} \times -1.47}{1.6\times 10^{-27}}$

$a = -1.41\times 10^{8} m/s^2$

from equation of motion we have

$v^2 = u^2 + 2aS$

$= 3.8025 \times 10^{8} - 2\times 1.41\times 10^{8} \times 0.355$

$v = 1.73\times 10^{4} m/s$

3 0

3 years ago

A small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k

rewona [7]

The kinetic energy of the small ball before the collision is

   KE = (1/2) (mass) (speed)²

   = (1/2) (2 kg) (1.5 m/s)

   =    (1 kg) (2.25 m²/s²)

   =    2.25 joules.

Now is a good time to review the Law of Conservation of Energy:

   Energy is never created or destroyed.
   If it seems that some energy disappeared,
   it actually had to go somewhere.
   And if it seems like some energy magically appeared,
   it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision. The large ball
and the small ball will just have to share the same 2.25 joules.

3 0

3 years ago

1. An asphalt block has a mass of 90 kg and a volume of 0.075 m. Determines the density of the asphalt.

Kisachek [45]

Answer:

1. Density = 1200[kg/m^3]; 2. Volume= 0.005775[m^3], mass= 15.59[kg]

Explanation:

1. We know that the density is defined by the following expression.

$Density = \frac{mass}{volume} \\where:\\mass=90[kg]\\volume=0.075[m^{3} ]\\density=\frac{90}{0.075} \\density=1200[\frac{kg}{m^{3} }]$

2. First we need to convert the units to meters.

wide = 35[cm] = 35/100 = 0.35[m]

long = 11 [dm] = 11 decimeters = 11/10 = 1.1[m]

Thick = 15[mm] = 15/1000 = 0.015[m]

Now we can find the density using the expression for the density.

$density= \frac{mass}{volume} \\where:\\volume = wide*long*thick\\volume=0.35*1.1*0.015 = 0.005775[m^3]\\\\mass= density*volume = 2700*0.005775 = 15.59[kg]$

7 0

3 years ago

I got part c right but idk why the other parts are wrong HELP!

dedylja [7]

a) The impulse is 76.5 Ns

b) The average force is 546.4 N

c) The final speed is 31.5 m/s

Explanation:

a)

The impulse exerted on an object is defined as

$J=\int F\Delta t$

where

F is the magnitude of the force exerted on the object

$\Delta t$ is the time interval during which the force is applied

If we consider a graph of the force applied vs time, it follows that the impulse exerted is equal to the area under the graph.

Therefore, in this problem, we can calculate the impulse by computing the area under the graph. We have a trapezium, whose bases are

$B=0.14-0 = 0.14s\\b=8-5=3s$

and whose height is

$h=900 N$

Therefore, the area (and the impulse) is

$J=\frac{(B+b)h}{2}=\frac{(0.14+0.03)(900)}{2}=76.5 Ns$

b)

In this problem, the force applied is not constant. However, we can rewrite the impulse also as

$J=F_{avg} \Delta t$

where

$F_{avg}$ is the average force exerted during the whole time $\Delta t$

In this problem we have

J = 76.5 Ns is the impulse (calculated in part a)

$\Delta t = 0.14 s$ is the time interval

Solving for the average force, we find

$\Delta t = \frac{J}{F_{avg}}=\frac{76.5}{0.14}=546.4 N$

c)

According to the impulse theorem, the impulse exerted on an object is equal to the change in momentum of the object:

$J=\Delta p = m(v-u)$

where

m is the mass of the object

v is the final velocity

u is the initial velocity

In this problem, we have

J = 76.5 Ns

m = 3.0 kg is the mass

u = 6.0 m/s is the initial velocity

Solving for v, we find the final velocity (and speed):

$v=u+\frac{J}{m}=6.0+\frac{76.5}{3}=31.5 m/s$

Learn more about impulse and momentum:

brainly.com/question/9484203

#LearnwithBrainly

6 0

3 years ago