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3 years ago

A car travels 8km in 7 minutes. Find the speed of the car.

Physics

1 answer:

AleksandrR [38]3 years ago

3 0

Answer:

42.6083 mi/h

Explanation:

Given: A car travels 8km in 7 minutes.

To find: Find the speed of the car.

Formula: $Speed = \frac{Distance}{Time}$

Solution: Since the formula for the speed of an object (which is the car) is speed = distance ÷ time, divide the distance (8km) by the time (7min)

Speed = 42.6083 miles per hour

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A triply charged ion with velocity 7.00 × 10^6 m/s moves in a path of radius 36.0 cm in a magnetic field of 0.55 T in a mass spe

saw5 [17]

Answer:

Mass of ion will be $22\times 10^{-13}kg$

Explanation:

We have given ion is triply charged that is $q=3\times 1.6\times 10^{-19}=4.8\times 10^{-19}C$

Radius r = 36 cm = 0.36 m

Velocity of the electron $v=7\times 10^6m/sec$

Magnetic field B = 0.55 T

We know that radius of the path is given by $r=\frac{mv}{qB}$

$m=\frac{rqB}{v}=\frac{0.36\times 4.8\times 10^{-19}\times 7\times 10^6}{0.55}=22\times 10^{-13}kg$

6 0

3 years ago

In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu

Ronch [10]

Answer:

$i_2=3.61\ A$

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

$q, q_1, q_2$ = charge of the capacitor in any time $t, t_1, t_2$

$q_o$ = initial charge of the capacitor

$\omega$ =angular frequency of the circuit

$i, i_1, i_2$ = current through the circuit in any time $t, t_1, t_2$

The charge in an LC circuit is given by

$q(t) = q_0 \, cos (\omega t )$

The current is the derivative of the charge

$\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).$

We are given

$q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}$

It means that

$q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]$

$i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]$

From eq 1:

$\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}$

From eq 2:

$\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}$

Squaring and adding the last two equations, and knowing that

$sin^2x+cos^2x=1$

$\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1$

Operating

$\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2$

Solving for $q_o$

$\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}$

Now we know the value of $q_0$ , we repeat the procedure of eq 1 and eq 2, but now at the second time $t_2$ , and solve for $i_2$

$\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2$

Solving for $i_2$

$\displaystyle i_2=w\sqrt{q_o^2-q_2^2}$

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

$\displaystyle f=\frac{1}{4\pi}\ KHz$

$w=2\pi f=500\ rad/s$

$\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}$

$q_0=0.01166\ c$

Finally

$\displaystyle i_2=500\sqrt{0.01166^2-.006^2}$

$i_2=5\ A$

3 0

3 years ago

Hooke's law describes a certain light spring of unstretched length 31.8 cm. when one end is attached to the top of a doorframe a

lubasha [3.4K]

Answer:

Explanation:

extension in the spring = 40.4 - 31.8 = 8.6 cm = 8.6 x 10⁻² m .

kx = mg

k is spring constant , x is extension , m is mass

k x 8.6 x 10⁻² = 7.52 x 9.8

k = 856.93 N/m

= 857 x 10⁻³ KN /m

b ) Both side is pulled by force of 188 N .

Tension in spring = 188N

kx = T

856.93 x = 188

x = .219.38 m

= 21.938 cm

= 21.9 cm .

length of spring = 31.8 + 21.9

= 53.7 cm .

6 0

2 years ago

Calculate the force needed to bring a 1,019-kg car to rest from a speed of 29 m/s in a distance of 118 m

son4ous [18]

Acceleration = ▵v/▵t
Time = d/v
Fisrt calculate time : ( 118/29 ) = 4 seconds
Then calculate acceleration
A = 29/4 = 7.25 m/s²
Now the force.
Force = mass * acceleration.
F= 1,019 * 7.25
F= 7,387 N

8 0

3 years ago

What is the formula for the moment of inertia of the person/single particle rotating in a circle? (Give these values with a subs

Ann [662]

Moment of inertia of single particle rotating in circle is I1 = 1/2 (m*r^2)

The value of the moment of inertia when the person is on the edge of the merry-go-round is I2=1/3 (m*L^2)

Moment of Inertia refers to:

the quantity expressed by the body resisting angular acceleration.
It the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.

The moment of inertia of single particle rotating in a circle I1 = 1/2 (m*r^2)

here We note that the,

In the formula, r being the distance from the point particle to the axis of rotation and m being the mass of disk.

The value of the moment of inertia when the person is on the edge of the merry-go-round is determined with parallel-axis theorem:

I(edge) = I (center of mass) + md^2

d be the distance from an axis through the object’s center of mass to a new axis.

I2(edge) = 1/3 (m*L^2)

learn more about moment of Inertia here:

<u>brainly.com/question/14226368</u>

#SPJ4

7 0

2 years ago