All categories

3 years ago

A 15.0 kg bucket is lowered vertically by a rope in which there is 188 N of tension at a given instant

Physics

1 answer:

saw5 [17]3 years ago

6 0

Answer: 2.72 m/s²

Explanation:

F = m(g + a)

188 = 15.0(9.81 + a)

188 = 147.15 + 15.0a

40.85 = 15.0a

a = 2.7233333... ≈ 2.72 m/s²

This is an upward acceleration as the force being applied is greater than that needed to maintain constant velocity. (147 N)

So even though the bucket is being lowered, it's rate of descent is decreasing with time. If the acceleration continues for long enough, the bucket will begin to rise.

You might be interested in

suppose that a large cargo truck needs to cross a bridge. the truck is 30 m long and 3.2 m wide. the cargo exerts a force nof 54

Tamiku [17]

<span>No, because the truck applies more pressure than the bridge can support.</span>

7 0

3 years ago

Look at the circuit diagram. Which of these components is part of the circuit?

Roman55 [17]

Answer:

b ac power source

Explanation:

ieififuffucjcjcicogore8e8e9e9e9e9ew0ww0w0w

7 0

3 years ago

What is the ability to complete extended periods of physical activity?

Galina-37 [17]

Answer:

D) WELLNESS

Explanation:

3 0

3 years ago

Read 2 more answers

Plz help 20 points!!

vazorg [7]

Answer:

1-d

2- a

3-e

4-b

5-c

This is the correct sequence

5 0

2 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago