All categories

3 years ago

Question 2:

Physics

2 answers:

timama [110]3 years ago

4 0

Answer:

we need the block

Explanation:

1×2 =4 lest 74 =345

Vesna [10]3 years ago

3 0

Explanation:

There are many students who can not get answers step by step and on time

So there are a wats up group where you can get help step by step and well explained by the trusted experts.

just join

post question

get answer

You might be interested in

20. A car battery with a 12-V emf and an internal resistance of 0.050 Ω is being charged with a current of 60 A. Note that in th

topjm [15]

Answer:

Part a)

V = 15 Volts

Part b)

P = 180 Watt

Part c)

Rate = 720 Watt

Explanation:

Part a)

When battery is in charging then the potential difference of the terminals of cell is given by

$\Delta V = EMF + iR$

here we know that

EMF = 12 volts

i = 60 A

r = 0.050 ohm

now we have

$\Delta V = 12 + (0.050)(60) = 15 Volts$

Part b)

Rate of thermal energy dissipated is the energy which is dissipating across the resistor

so here we have

$P = i^2 R$

$P = (60^2)(0.05)$

P = 180 Watt

Part c)

Rate at which Energy stored inside the cell is the rate of electrical energy that is converted into the chemical energy

$Rate = EMF \times i$

$Rate = (12)(60)$

$Rate = 720 Watt$

6 0

3 years ago

What happens when you combine purple and yellow?

Nana76 [90]

Umm You make brown lol

3 0

3 years ago

The chewing muscle, the masseter, is one of the strongest in the human body. It is attached to the mandible (lower jawbone) as s

kifflom [539]

Answer:

Explanation:

Check attachment

6 0

3 years ago

A hammer has a mass of 1 kg. What is its weight (i) on Earth (ii) on the

solong [7]

Given mass= 1kg

Weight on earth = mg(gravity of earth) = 9.8N

weight on moon = mg(gravity of moon)= 1.62N

weight on outer space mg(gravity outer space = 0) = 0N

4 0

2 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

8 0

1 year ago