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2 years ago

A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?

Physics

1 answer:

Tasya [4]2 years ago

3 0

<h3><u>Answer;</u></h3>

<em>Work = 125 joules </em>

<h3><u>Explanation and solution</u>;</h3>

Work is the product of force and the distance covered. Therefore, Work = force × distance.

Work is measured in joules.
Work is also a change in energy, such that work is done when energy changes, so when kinetic energy, or potential energy changes the there is work being done.

Thus; kinetic energy = work done

Kinetic energy = 1/2mv²

= 1/2 × 10× 5²

= 5 × 25

= 125 joules

Hence, work done is 125 joules.

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A thin copper rod 1.0 m long has a mass of 0.050 kg and is in a magnetic field of 0.10 t. What minimum current in the rod is nee

slamgirl [31]

Answer:

$i = 4.9 A$

Explanation:

Force on a current carrying rod due to magnetic field is given as

$F = iLB$

here we know that

$i =$ current in the rod

$B = 0.10 T$

$L = 1.0 m$

now magnetic force is balanced by the weight of the rod

so we will have

$iLB = mg$

$i(1.0)(0.10) = 0.05 \times 9.8$

$i = 4.9 A$

8 0

2 years ago

A carbon-dioxide laser emits infrared light with a wavelength of 10.6 μm. What is the length of a tube that will oscillate in th

alex41 [277]

Answer:

The length of a tube and number of rounds are 0.848 m and $1.77\times10^{8}\ trip\ per\ second$ .

Explanation:

Given that,

Wavelength $\lambda= 10.6\mu m$

m = 160000

We need to calculate the length

Using formula of wavelength

Laser tube behave like closed pipe

$m\dfrac{\lambda}{2}=L$

$L=160000\times\dfrac{10.6\times10^{-6}}{2}$

$L=0.848\ m$

Distance traveled by pulse of light in one back and fourth trip

$d=2L$

$d=2\times0.848$

$d=1.696\ m$

We need to calculate the time

Using formula for time

$t = \dfrac{d}{c}$

$t=\dfrac{1.696}{3\times10^{8}}$

$t=5.653\times10^{-9}\ s$

We need to calculate the number of round

Using formula of number of round

$N=\dfrac{1}{t}$

$N= \dfrac{1}{5.653\times10^{-9}}$

$N=1.77\times10^{8}\ trip\ per\ second$

Hence, The length of a tube and number of rounds are 0.848 m and $1.77\times10^{8}\ trip\ per\ second$ .

7 0

2 years ago

A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice

Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of $\sqrt{3} v_{e}$

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , $v_{i}=2 v_{e}$ . Here the radius is R, so find the escape velocity as follows,

$\frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0$

$\frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}$

$v_{e}^{2}=\frac{2 G M}{R}$

$v_{e}=\sqrt{\frac{2 G M}{R}}$

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as, $U_{i}=-\frac{G M m}{R}$

R tend to infinity when far away from the planet, so $v_{f}=0$

Then, kinetic energy at initial would be,

$k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}$

Similarly, kinetic energy at final would be,

$k_{f}=\frac{1}{2} m v_{f}^{2}$

Here, $v_{f}=\text { final velocity }$

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

$U_{i}+k_{i}=k_{f}+v_{f}$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}$

'm' and $\frac{1}{2}$ as common on both sides, so gets cancelled, we get as

$4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}$

We know, $v_{e}=\sqrt{\frac{2 G M}{R}}$ , it can be wriiten as $\left(v_{e}\right)^{2}=\frac{2 G M}{R}$ , we get

$4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}$

$v_{f}^{2}=3\left(v_{e}\right)^{2}$

Taking squares out, we get,

$v_{f}=\sqrt{3} v_{e}$

4 0

2 years ago

A father demonstrates projectile motion to his children by placing a pea on his fork's handle and rapidly depressing the curved

MariettaO [177]

Answer:

4.17 m/s

Explanation:

To solve this problem, let's start by analyzing the vertical motion of the pea.

The initial vertical velocity of the pea is

$u_y = u sin \theta = (7.39)(sin 69.0^{\circ})=6.90 m/s$

Now we can solve the problem by applying the suvat equation:

$v_y^2-u_y^2=2as$

where

$v_y$ is the vertical velocity when the pea hits the ceiling

$a=g=-9.8 m/s^2$ is the acceleration of gravity

s = 1.90 is the distance from the ceiling

Solving for $v_y$ ,

$v_y = \sqrt{u_y^2+2as}=\sqrt{(6.90)^2+2(-9.8)(1.90)}=3.22 m/s$

Instead, the horizontal velocity remains constant during the whole motion, and it is given by

$v_x = u cos \theta = (7.39)(cos 69.0^{\circ})=2.65 m/s$

Therefore, the speed of the pea when it hits the ceiling is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{2.65^2+3.22^2}=4.17 m/s$

5 0

2 years ago

Each shot of the laser gun most favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by th

mote1985 [20]

Answer:

U = 1794.005 × 10⁶ J

Explanation:

Data provided;

Capacitance of the original capacitor, C = 1.27 F

Potential difference applied to the original capacitor, V = 59.9 kV

= 59.9 × 10³ V

Now,

The Potential energy (U) for the capacitor is calculated as:

Potential energy of the original capacitor, U = $\frac{\textup{1}}{\textup{2}}$ × C × V²

on substituting the respective values, we get

U = $\frac{\textup{1}}{\textup{2}}$ × 1.27 × ( 59.9 × 10³ )²

U = 1794.005 × 10⁶ J

7 0

2 years ago