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2 years ago

Is the formula for velocity the same as speed or different?

Physics

1 answer:

a_sh-v [17]2 years ago

5 0

Answer:

always same

Explanation:

velocity and speed are same upto some extend but velocity is vector while speed is scalar quantity

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What is the resistance of a 600 W kettle that draws a current of 5.0 A? please answer with steps

lukranit [14]

P=I^2 *R

600 =5.0^2 *R

R=24

Answer: 24 ohms

I hope it’s correcttttttt...

5 1

2 years ago

A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are

igor_vitrenko [27]

Answer:

0.231 N

Explanation:

To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

$\alpha = \frac{\theta}{t} = \frac{6.37}{9.87} = 0.6454 rad/s^2$

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

$I = \frac{mL^2}{3} = \frac{1.27*0.847^2}{3} = 0.303kgm^2$

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

$T = I\alpha = 0.303*0.6454 = 0.196 Nm$

So the force acting on the other end to generate this torque mush be:

$F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N$

4 0

3 years ago

PLEASE PLEASE HELPPPPPP!!!!!!!’

Ostrovityanka [42]

The answer is B

explanation

3 0

2 years ago

If, for a given velocity, the maximum range is at a projection angle of 45, then there must be equal ranges for angles above and

tiny-mole [99]

Explanation:

The range R of a projectile is given by

$R = \frac{v_0^2}{g} \sin 2\theta$

The maximum range $R_{max}$ occurs when $\sin 2\theta = 1\:\text{or}\:\theta = 45°$ . Let $\alpha$ be the angle above or below 45°. Now let's look at the ranges brought about by these angle differences.

Case 1: Angle above 45°

We can write the range as

$R_+ = \dfrac{v_0^2}{g} \sin 2(45° + \alpha)= \dfrac{v_0^2}{g} \sin (90° + 2\alpha)$

$\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} (\sin 90° \cos 2\alpha + \cos 90° \sin 2\alpha)$

$\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} \cos 2\alpha\:\:\:\:\:(1)$

Case 2: Angle below 45°

We can write the range as

$R_- = \dfrac{v_0^2}{g} \sin 2(45° - \alpha)= \dfrac{v_0^2}{g} \sin (90° - 2\alpha)$

$\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} (\sin 90° \cos 2\alpha - \cos 90° \sin 2\alpha)$

$\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} \cos 2\alpha\:\:\:\:\:(2)$

Note that the equations (1) and (2) are identical. Therefore, the ranges are equal if they differ from 45° by the same amount.

3 0

2 years ago

Thiết kế mạch khếch đại công suất loại B? Tính toán và giải thích

taurus [48]

Answer:

sorry i dont know ur language

4 0

2 years ago