A dog is pulling on a leash while walking down

An ostrich can run at a speed of 43 mi/hr. How much ground can an ostrich cover if it runs at this speed for 15 minutes? (Hint:

Simora [160]

..... It would possibly she eenejjsjejeej 1.4

4 0

2 years ago

What direction does current flow from a battery in a series circuit?

telo118 [61]

What's now called "Conventional current" is thought of as the flow of positive charge, from the battery's positive terminal to its negative one.

But it turns out that positive charges don't flow. The physical flow of charge is the flow of electrons. They come out of the battery's negative terminal, and carry negative charge around the circuit to the battery's positive one.

6 0

2 years ago

Read 2 more answers

A straight segment of a current-carrying wire has a current element IL where I = 2.70 A and L = 3.20 cm i + 4.30 cm j. The segme

myrzilka [38]

The component of the force in negative z-direction is -0.144 N.

The given parameters;

current in the wire, I = 2.7 A
length of the wire, L = (3.2 i + 4.3j) cm
magnetic filed, B = 1.24 i

The force on the segment of the wire is calculated as follows;

$F = ILBsin(\theta)$

where;

θ is the angle wire and magnetic field

The force on the wire segment will be perpendicular in negative z-direction (applying right hand rule), so there won't be any x and y component of the force.

The angle between the wire and the magnetic field is calculated as follows;

$\theta = tan^{-1} (\frac{y}{x} )\\\\\theta = tan^{-1} (\frac{4.3}{3.2} )\\\\\theta = 53.3 \ ^0$

The magnitude of the wire length is calculated as follows;

$|l | = \sqrt{3.2^2 + 4.3^2} = 5.36 \ cm = 0.0536 \ m$

The component of the force in negative z-direction is calculated as;

$F_z = -ILB sin(\theta)\\\\F_z = -2.7 \times 0.0536 \times 1.24 \times sin(53.3)\\\\F_z = -0.144 \ N$

Thus, the component of the force in negative z-direction is -0.144 N.

Learn more here:brainly.com/question/22719779

6 0

2 years ago

Two resistors, R1=3.85 Ω and R2=6.47 Ω , are connected in series to a battery with an EMF of 24.0 V and negligible internal resi

Bond [772]

Answer:

(a) 2.33 A

(b) 15.075 V

Explanation:

From the question,

The total resistance (Rt) = R1+R2 = 3.85+6.47

R(t) = 10.32 ohms.

Applying ohm's law,

V = IR(t)..........equation 1

Where V = Emf of the battery, I = current flowing through the circuit, R(t) = combined resistance of both resistors.

Note: Since both resistors are connected in series, the current flowing through them is the same.

Therefore,

I = V/R(t)............. Equation 2

Given: V = 24 V, R(t) = 10.32 ohms

Substitute these values into equation 2

I = 24/10.32

I = 2.33 A.

Hence the current through R1 = 2.33 A.

V2 = IR2.............. Equation 3

V2 = 2.33(6.47)

V2 = 15.075 V

7 0

2 years ago

Consider a long rod of mass, m, and length, l, which is thin enough that its width can be ignored compared to its length. The ro

ehidna [41]

Answer:

Explanation:

The rod will act as pendulum for small oscillation .

Time period of oscillation

$T=2\pi\sqrt{\frac{l}{g} }$

angular frequency ω = 2π / T

= $\omega=\sqrt{\frac{g}{l} }$

b )

ω = 20( given )

velocity = ω r = ω l

Let the maximum angular displacement in terms of degree be θ .

1/2 m v ² = mgl ( 1 - cosθ ) ,

[ l-lcosθ is loss of height . we have applied law of conservation of mechanical energy .]

.5 ( ω l )² = gl( 1 - cos θ )

.5 ω² l = g ( 1 - cosθ )

1 - cosθ = .5 ω² l /g

cosθ = 1 - .5 ω² l /g

θ can be calculated , if value of l is given .

4 0

2 years ago