All categories

3 years ago

A cat weighing 7 kg chases a mouse at a speed of 4 m/s. What is the kinetic energy of the cat?

1 answer:

8 0

54J
You calculate kinetic energy by the equation.
KE= 1/2mv^2
m= mass
v= velocity
then we have
KE = 1/2 x 3 x 6^2
KE = 1/2 x 3 x 36 = 54J

You might be interested in

Which competition is the “ most important soccer competition in the world “ ?

Ganezh [65]

It would be B. The World Cup

Hope this helps

Have a great day/night

8 0

3 years ago

What is the resistance in a circuit that has a current of 0.75A and a voltage drop of 60V across the cell?( Please help) I will

Marrrta [24]

difference across the 12.0 Ω resistor is 1.20 V, calculate the emf of the cell. 45. Req = 13.5 Ω ... (b) I = V / Reff = 1.5 / 2 = 0.75 A.

3 0

3 years ago

Read 2 more answers

How deep under water would you need to be in order to be at double atomosphric pressure

nasty-shy [4]

You would need to be 33 feet deep.

4 0

2 years ago

Read 2 more answers

How will you relate the distance of the planets with their period of revolution

My name is Ann [436]

Answer:

${t}^{2} \alpha {r}^{2}$

In other words, the time period squared is directly proportional to the radius of their orbit cubed

Explanation:

By combing equations for gravitational force and centripetal force, you can create a formula for t or r.

Feel free to message me if you want more help

3 0

4 years ago

A very long, straight wire carries a current of 19.0 A in the +k direction. An electron 1.9 cm from the center of the wire in th

Anestetic [448]

(a) $1.03\cdot 10^{-16} N$ , -k direction

First of all, let's find the magnetic field produced by the wire at the location of the electron:

$B=\frac{\mu_0 I}{2 \pi r}$

where

I = 19.0 A is the current in the wire

r = 1.9 cm = 0.019 m is the distance of the electron from the wire

Substituting,

$B=\frac{(1.256\cdot 10^{-6})(19.0A)}{2 \pi (0.019 m)}=2\cdot 10^{-4} T$

and the direction is +j direction (tangent to a circle around the wire)

Now we can find the force on the electron by using:

$F=qvBsin \theta$

where

$q=1.6\cdot 10^{-19}C$ is the electron's charge

$v=3.23\cdot 10^6 m/s$ is the electron speed

$B=2\cdot 10^{-4} T$ is the magnetic field

$\theta$ is the angle between the direction of v and B

In this case, the electron is travelling away from the wire, while the magnetic field lines (B) form circular paths around the wire: this means that v and B are perpendicular, so $\theta=90^{\circ}, sin \theta=1$ . So, the force on the electron is

$F=(1.6\cdot 10^{-19}C)(3.23\cdot 10^6 m/s)(2\cdot 10^{-4} T)(1)=1.03\cdot 10^{-16} N$

The direction is given by the right hand rule:

- Index finger: direction of motion of the electron, +i direction (away from the wire)

- Middle finger: direction of magnetic field, +j direction (tangent to a circle around the wire)

- Thumb: direction of the force --> since the charge is negative, the sign must be reversed, so it means -k direction (anti-parallel to the current in the wire)

(b) $1.03\cdot 10^{-16} N$ , +i direction

The calculation of the magnetic field and of the force on the electron are exactly identical as before. The only thing that changes this time is the direction of the force. In fact we have:

- Index finger: direction of motion of the electron, +k direction (parallel to the current in the wire)

- Middle finger: direction of magnetic field, +j direction (tangent to a circle around the wire)

- Thumb: direction of the force --> since the charge is negative, the sign must be reversed, so it means +i direction (away from the wire)

(c) 0

In this case, the electron is moving tangent to a circle around the wire, in the +j direction. But this is exactly the same direction of the magnetic field: this means that v and B are parallel, so $\theta=0, sin \theta=0$ , therefore the force on the electron is zero.

6 0

3 years ago