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3 years ago

(a) and (b) ??Thanks HELP d8 =1

Physics

1 answer:

garri49 [273]3 years ago

7 0

Explanation:

(a) You can solve this using kinematics or energy.

Using kinematics:

a = F/m = 90 N / 4 kg = 22.5 m/s²

v₀ = 0 m/s

Δx = 5 m

Find: v

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (22.5 m/s²) (5 m)

v = 15 m/s

Using energy:

W = ΔKE

Fd = ½ mv²

(90 N) (5 m) = ½ (4 kg) v²

v = 15 m/s

(b) ΔU = mg Δh

ΔU = (4 kg) (9.8 m/s) (12 m sin 40° − 15 m)

ΔU = -290 J

W = ΔU = -290 J

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3 years ago

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jeka57 [31]

(a) $4.14 rad/s^2$

The relationship beween centripetal acceleration and angular speed is

$a=\omega^2 r$

where

$\omega$ is the angular speed

r is the radius of the circular path

Here we gave

$a = 9g = 88.2 m/s^2$ is the centripetal acceleration

r = 5.15 m is the radius

Solving for $\omega$ , we find:

$\omega = \sqrt{\frac{a}{r}}=\sqrt{\frac{88.2 m/s^2}{5.15 m}}=4.14 rad/s^2$

(b) 21.3 m/s

The relationship between the linear speed and the angular speed is

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular speed

r is the radius of the circular path

In this problem we have

$\omega=4.14 rad/s$

r = 5.15 m

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3 years ago

Read 2 more answers

Which unit is used to measure resistance? a. volts b. charges c. amperes d. ohms

Ilia_Sergeevich [38]

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3 years ago

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4. A meter has a resistance of 100 Ω and gives a full scale deflection when it carries a current of 25 μA. (a) What resistor, Rx

frez [133]

Answer:

A=50mΩ

B≅50mΩ

Explanation:

A) To answer this question we have to use the Current Divider Rule. that rule says:

$Ix=.\frac{Req}{Rx} *Itotal$ (1)

Itotal represents the new maximun current, 50mA, Ix is the current going through the 100 ohms resistor, and Req. is the equivalent resitor.

We now have a set of two resistor in parallel, so:

$Req.=\frac{1}{\frac{1}{R1}+\frac{1}{R2} }$ (2)

where R1 is the resitor we have to calculate, and R2 is the 100 ohms resistor (25 uA).

substituting and rearranging (2)

$Req.=\frac{ 100*R1}{R1+100}$ (3)

Now substituting (3) in (1).

$25*10^{-6} =\frac{\frac{ 100*R1}{R1+100}}{100} *50*10^{-3}$

solving this, The value of R1 is: 50mΩ

This value of R1 will guaranty that the ammeter full reflection willl be at 50mA.

Given that R2 (100ohm) it too much bigger than 50mΩ, the equivalent resistor will tend to 50mΩ

If you substitude this values on (2) Req. will be 49.97 mΩ.

7 0

3 years ago

(a) and (b) ??Thanks HELP d8 =1​

(a) and (b) ??Thanks HELP d8 =1