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3 years ago

Find the net force and acceleration. 10 points. Will give brainliest.

Physics

2 answers:

Svetllana [295]3 years ago

5 0

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the FRICTION.

Force by Xavier = 25 N

Force of friction in the opposite direction of Xavier = f = umg

where, u = coefficient of kinetic friction.

so we here get as ,

Net force = 16 N

and acceleration = 2.13 m/s^2

Afina-wow [57]3 years ago

3 0

Answer:

a) 16 N

b) 2.13 m/s²

Explanation:

Draw a free body diagram of the tv stand. There are four forces:

Weight force mg pulling down,

Normal force N pushing up,

Friction force Nμ pushing left,

and applied force P pulling right.

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

The net force in the x direction is:

∑F = P − Nμ

∑F = P − mgμ

∑F = 25 N − (7.5 kg) (10 m/s²) (0.12)

∑F = 16 N

Net force equals mass times acceleration:

∑F = ma

16 N = (7.5 kg) a

a = 2.13 m/s²

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3 years ago

Two resistances, R1 and R2, are connected in series across a 9-V battery. The current increases by 0.450 A when R2 is removed, l

Rina8888 [55]

Answer:

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

Explanation:

Since the resistors R1 and R2 are connected in series, the current flowing through them when the 9 V battery is applied is 9/R1 + R2.

When the current increases by 0.450 A wen only R1 is in the circuit, the current is

9/R1 + R2 + 0.450 A = 9/R1 (1)

When the current increases by 0.225 A when only R2 is in the circuit, the current is

9/R1 + R2 + 0.225 A = 9/R2 (2)

equation (1) - (2) equals

9(1/R1 - 1/R2) = 0.450 A - 0.225

9(1/R1 - 1/R2) = 0.125

(1/R1 - 1/R2) = 0.125 A/9 = 0.0138

1/R1 = 0.0138 + 1/R2

R1 = R2/(1 + 0.0138R2) (3)

From (1)

9/R1 - 9/R1 + R2 = 0.450 A

9R2/[R1(R1 + R2)] = 0.450 A

R2/[R1(R1 + R2)] = 0.450 A/9 = 0.5

R2/[R1(R1 + R2)] = 0.5 (4)

From (3) R2/R1 = (1 + 0.0138R2) and from (4) R2/R1 = 0.5(R1 + R2). So,

(1 + 0.0138R2) = 0.5(R1 + R2)

0.5R1 + 0.5R2 = 1 + 0.0138R2

0.5R1 = 1 + 0.0138R2 - 0.5R2

0.5R1 = 1 - 0.4862R2 (5)

Substituting (3) into (5) we have

0.5R2/(1 + 0.0138R2) = 1 - 0.4862R2

R2 = (1 + 0.0138R2)(1 - 0.4862R2)

R2 = 1 - 0.4724R2 - 0.0067R2²

Collecting like terms, we have

0.0067R2² + 0.4724R2 + R2 - 1 = 0

0.0067R2² + 1.4724R2 - 1 = 0

Using the quadratic formula,

$R_{2} = \frac{-1.4724 +/-\sqrt{(1.4724)^{2} - 4 X 0.0067 X -1} }{2 X 0.0067} \\= \frac{-1.4724 +/-\sqrt{2.1680 + 0.0268} }{0.0268}\\= \frac{-1.4724 +/-\sqrt{2.1948} }{0.0268}\\= \frac{-1.4724 +/- 1.4815 }{0.0268}\\= \frac{-1.4724 + 1.4815 }{0.0268} or \frac{-1.4724 - 1.4815 }{0.0268}\\= \frac{0.0091 }{0.0268} or \frac{-2.9539}{0.0268}\\= 0.340 or -110.22$

We choose the positive answer.

So R2 = 0.340 Ω

From (5)

R1 = 0.5 - 0.9931R2

= 0.5 - 0.9931 × 0.340

= 0.5 - 0.338

= 0.162 Ω

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

5 0

3 years ago