Describe rotational motion

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

jeka57 [31]

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

8 0

2 years ago

A child looked at

fgiga [73]

The answer to your question is A.

4 0

2 years ago

A straight wire of length 0.56 m carries a conventional current of 0.4 amperes. What is the magnitude of the magnetic field made

djverab [1.8K]

Answer:

The magnitude of the magnetic field made by current in the wire is 3.064 x 10⁻⁶ T.

Explanation:

Given;

length of the straight wire, L = 0.56 m

conventional current, I = 0.4 A

distance of magnetic field from the wire, r = 2.6 cm = 0.026 m

To determine magnitude of magnetic field made by current in the wire, we will apply Bio-Savart Law;

$B = \frac{\mu_o}{4\pi r} \frac{LI}{\sqrt{r^2 +(L/2)^2} } \\\\B = \frac{4\pi *10^{-7}}{4\pi *0.026} \frac{0.56*0.4}{\sqrt{(0.026)^2 +(0.56/2)^2} }\\\\B = 3.846*10^{-6}(0.7966)\\\\B = 3.064*10^{-6} \ T$

Therefore, the magnitude of the magnetic field made by current in the wire is 3.064 x 10⁻⁶ T.

3 0

3 years ago

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/ s. A I.O-kg stone is thrown

nadya68 [22]

(a) 296.6 m

The motion of the stone is the motion of a projectile, thrown with a horizontal speed of

$v_x = 15.0 m/s$

and with an initial vertical velocity of

$v_{y0} = -20.0 m/s$

where we have put a negative sign to indicate that the direction is downward.

The vertical position of the stone at time t is given by

$y(t) = h + v_{0y} t + \frac{1}{2}gt^2$ (1)

where

h is the initial height

g = -9.81 m/s^2 is the acceleration due to gravity

The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:

y(6.00 s) = 0

Substituting into eq.(1), we can solve to find the initial height of the stone, h:

$0 = h + v_{0y} y + \frac{1}{2}gt^2\\h = -v_{0y} y - \frac{1}{2}gt^2=-(-20.0 m/s)(6.00 s) - \frac{1}{2}(9.81 m/s^2)(6.00 s)^2=296.6 m$

(b) 176.6 m

The balloon is moving downward with a constant vertical speed of

$v_y = -20 m/s$

So the vertical position of the balloon after a time t is

$y(t) = h + v_y t$

and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:

$y(t) = 296.6 m + (-20.0 m)(6.00 s)=176.6 m$

(c) 198.2 m

In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.

The horizontal speed of the rock is

$v_x = 15.0 m/s$

So the horizontal distance travelled in t = 6.00 s is

$d_x = v_x t = (15.0 m/s)(6.00 s)=90 m$