Ask question

Ask question

All categories

3 years ago

5

A runner hopes to complete the 10,000m run in less than 30,0min, after running at constant speed for exactly 27.0 min , there ar

e still 1100m to go . The runner must then accelerate at 0.20 m/s2 for how many seconds in order to achieve the desired time ?

1 answer:

tankabanditka [31]3 years ago

3 0

Well I figured out at his current pace, he would finish the 1100 remaining meters in 200 seconds. However, he needs to complete it in 180 seconds. I'm not sure how to find out how long he has to accelerate at 0.20 m/s/s to complete it in 180 seconds.

You might be interested in

A hot iron ball of mass 200 g is cooled to a temperature of 22°C from 100°C. How much heat was

Nookie1986 [14]

Answer:

Q= -6900 J

Explanation:

use the formula Q=mC(T_2 - T_1) and sub in givens

Q=mC(T_2 - T_1)

Q= (200 g)(0.444 J/g°C)(22-100)

Q= -6900 J

The negative sign means heat is lost, which agrees with the decrease in temperature

6 0

3 years ago

In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

leonid [27]

Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by

$F_e = \dfrac{k|q_1||q_2|}{r^2}$

where,

$k$ = Coulomb's constant = $9\times 10^9\ \rm Nm^2/C^2.$
$r$ = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, $q_1 = +1.6\times 10^{-19}\ C.$

The charge on the electron, $q_2 = -1.6\times 10^{-19}\ C.$

These two are separated by the distance, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

$F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.$

Part (b):

The gravitational force of attraction between two objects of masses $m_1$ and $m_1$ respectively is given by

$F_g = \dfrac{Gm_1m_2}{r^2}.$

where,

$G$ = Universal Gravitational constant = $6.67\times 10^{-11}\ \rm Nm^2/kg^2.$
$r$ = distance of separation between the masses.

For the given system,

The mass of proton, $m_1 = 1.67\times 10^{-27}\ kg.$

The mass of the electron, $m_2 = 9.11\times 10^{-31}\ kg.$

Distance between the two, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

$F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.$

The ratio $\dfrac{F_e}{F_g}$ :

$\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.$

6 0

3 years ago

Calculate the kinetic energy of a moving 4 kg object traveling at a velocity of 3 m/a

Alina [70]

Answer:

Explanation:KE = 18 J

7 0

3 years ago

a machine with a mechanical advantage of 2.5 requires an input force of 120 newtons. What output force is applied by this machin

creativ13 [48]

If your machine has a mechanical advantage of 2.5, then WHATEVER force you apply to the input, the force at the output will be 2.5 times as great.

If you apply 1 newton to the machine's input, the output force is

(2.5 x 1 newton) = 2.5 newtons.

If you apply 120 newtons to the machine's input, the output force is

(2.5 x 120 newtons) = 300 newtons.

4 0

3 years ago

A solenoid of radius 2.0 mm contains 100 turns of wire uniformly distributed over a length of 5.0 cm. It is located in air and c

tankabanditka [31]

Answer:

The magnetic field strength inside the solenoid is $5.026\times10^{-3}\ T$ .

Explanation:

Given that,

Radius = 2.0 mm

Length = 5.0 cm

Current = 2.0 A

Number of turns = 100

(a). We need to calculate the magnetic field strength inside the solenoid

Using formula of the magnetic field strength

Using Ampere's Law

$B=\dfrac{\mu_{0}NI}{l}$

Where, N = Number of turns

I = current

l = length

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times100\times2.0}{5.0\times10^{-2}}$

$B=0.005026=5.026\times10^{-3}\ T$

(b). We draw the diagram

Hence, The magnetic field strength inside the solenoid is $5.026\times10^{-3}\ T$ .

4 0

3 years ago