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2 years ago

Help me pls really struggling with dat

Physics

1 answer:

pogonyaev2 years ago

7 0

So I’m gonna answer the first question on the second page:
For P the force is friction and for Q the force is forward thrust resultant force

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A 8.57-m ladder with a mass of 21.4 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upw

Andrei [34K]

Answer:

$1311.5\ \text{Nm}$

Explanation:

l = Length of ladder = 8.57 m

m = Mass of ladder = 21.4 kg

F = Force on ladder = 258 N

$\alpha$ = Angular acceleration = $1.63\ \text{rad/s}^2$

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Net torque is given by

$\tau=lf-\dfrac{l}{2}mg\\\Rightarrow \tau=8.57\times 258-\dfrac{8.57}{2}\times 21.4\times 9.81\\\Rightarrow \tau=1311.5\ \text{Nm}$

The net torque acting on the ladder is $1311.5\ \text{Nm}$ .

4 0

2 years ago

Three resistors of 4.0, 6.0, and 10.0 Ω are connected in parallel. If the combination is connected in series with a 12.0-V batte

Grace [21]

Answer:

0.536 A

Explanation:

From the question,

The combined resistance of the parallel connection is given as

1/Rt = 1/4+1/6+1/10

Rt = 120/15

Rt = 8 ohms.

Since the the combination is connected in series with a 10 ohms resistor,

Rt' = 8+10 = 18 ohms.

Current flowing through the series circuit

I = V/Rt'

V = 12 V, Rt' = 18 ohms

I = 12/18

I = 0.67 A.

Therefore voltage across the combined resistor

V' = 8(0.67)

v' = 5.36 V.

Since the resistors a connected in parallel, the same voltage drop across them.

Therefore, current through the 10 ohms

Ix = 5.36/10

Ix = 0.536 A

5 0

2 years ago

A force F (x) = (-5.0x2 + 7.0x) N acts on a particle. (a) How much work does the force do on the particle as it moves from x = 2

Archy [21]

Answer:

Part a)

Work done is given as

W = -121.5 J

Part b)

Potential energy of the system is given as

$U = - \frac{5x^3}{3} + \frac{7 x^2}{2}$

Explanation:

As we know that work done by variable force is given as

$W = \int F. dx$

here we know that

$F = - 5x^2 + 7 x$

so we have

$W = \int (-5 x^2 + 7x) dx$

so we have

$W = - 5x^3/3 + 7 x^2/2$

here we displace it from x = 2 to x = 5

so we will have

$W = -\frac{5}{3}(5^3 - 2^3) + \frac{7}{2}(5^2 - 2^2)$

$W = -195 + 73.5 = -121.5 J$

Part b)

As we know that change in Potential energy = work done

$U_f - U_i = -121.5$

$U - 0 = - \frac{5x^3}{3} + \frac{7 x^2}{2}$

$U = - \frac{5x^3}{3} + \frac{7 x^2}{2}$

3 0

2 years ago

_____ the actual brightness of a star as seen from earth

Flura [38]

Answer:

apparent

Explanation:

7 0

2 years ago

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Nady [450]

Fe/Fr I think hope this helps

8 0

3 years ago

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