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2 years ago

This table shows the weights of four different objects that are sitting on the

Physics

1 answer:

bekas [8.4K]2 years ago

6 0

I think that it would be A

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Only answer if you would send your eggplant

Mariulka [41]

Answer:

yup

Explanation:

8 0

3 years ago

A man pushes a 60.8 kg crate across a rough surface with an applied force of 125 N and at a CONSTANT SPEED.

Sedbober [7]

Answer: Colby we both dumb if we need brainly lol

Explanation:

8 0

3 years ago

A solenoid used to produce magnetic fields for research purposes is 2.1 m long, with an inner radius of 28 cm and 1000 turns of

Veseljchak [2.6K]

Answer:

I = 2172.46 A

Explanation:

Given that,

The length of a solenoid, l = 2.1 m

The inner radius of the solenoid, r = 28 cm = 0.28 m

The number of turns in the wire, N = 1000

The magnetic field in the solenoid, B = 1.3 T

We need to find the current carried by it. We know that, the magnetic field in a solenoid is given by :

$B=\mu_o nI\\\\or\\\\B=\mu_o \dfrac{N}{L}I\\\\I=\dfrac{BL}{\mu_o N}$

Put all the values,

$I=\dfrac{1.3\times 2.1}{4\pi \times 10^{-7}\times 1000}\\\\I=2172.46\ A$

So, it carry current of 2172.46 A.

7 0

3 years ago

A delivery truck leaves a warehouse and travels 2.60 km north. The truck makes a left turn and travels 1.25 km west before makin

yulyashka [42]

Answer:

4.19 km and 107.35 degrees north of east

Explanation:

So in the end, the truck is (2.6 + 1.4 = 4km) north and 1.25 km west from the warehouse. We can use the Pythagorean formula to calculate the magnitude and direction α of the truck displacement from the warehouse:

$s = \sqrt{s_n^2 + s_w^2} = \sqrt{4^2 + 1.25^2} = \sqrt{16 + 1.5625} = \sqrt{17.5625} = 4.19$ km

$tan\alpha = \frac{s_n}{s_w} = \frac{4}{1.25} = 3.2$

$\alpha = tan^{-1}3.2 = 1.27 rad \approx 72.65 degrees$ north or west or (180 - 72.65) = 107.35 degrees north of east

3 0

2 years ago

Two resistors R1 = 3 Ω and R2 = 6 Ω are connected in parallel. What is the net resistance in the circuit?

gtnhenbr [62]

Answer:

"2Ω" is the net resistance in the circuit.

Explanation:

The given resistors are:

R1 = 3Ω

R2 = 6Ω

The net resistance will be:

⇒ $\frac{1}{R_{net}} =\frac{1}{R_1} +\frac{1}{R_2}$

On substituting the values, we get

⇒ $\frac{1}{R_{net}} =\frac{1}{3} +\frac{1}{6}$

On taking L.C.M, we get

⇒ $\frac{1}{R_{net}} =\frac{2+1}{6}$

⇒ $\frac{1}{R_{net}} =\frac{3}{6}$

⇒ $\frac{1}{R_{net}} =\frac{1}{2}$

On applying cross-multiplication, we get

⇒ $R_{net}=2 \Omega$

3 0

2 years ago