All categories

2 years ago

Frick and Frack are standing back-to-back, leaning on each other, but not moving. If Frick weighs

Physics

2 answers:

mylen [45]2 years ago

7 0

Answer:

Frick is pushing harder

Explanation:

if Frack weighs more and he was pushing harder they would be moving, but if Frick pushes harder then they wont move

Lana71 [14]2 years ago

4 0

Answer:

they are both pushing with the SAME force

Explanation:

Since no one is moving, the system of the two friends is NOT accelerated, which means that the net force on it is zero. then, the force applied by each is of the SAME magnitude but in opposite directions.

Therefore they are both pushing with the SAME force.

You might be interested in

I cant solve this problem, and our teacher said that this would be in the test we'll have tomorrow, can someone help me?

Ad libitum [116K]

Answer:

d = 11.1 m

Explanation:

Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:

$\frac{1}{2} m {v}^{2} = mgh$

Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:

$\sin(15) = \frac{h}{d} \\ or \: h = d \sin(15)$

Plugging this into the energy conservation equation and cancelling m, we get

${v}^{2} = 2gd \sin(15)$

Solving for d,

$d = \frac{ {v}^{2} }{2g \sin(15) } = \frac{ {(7.5 \: \frac{m}{s}) }^{2} }{2(9.8 \: \frac{m}{ {s}^{2} })(0.259)} \\ = 11.1 \: m$

3 0

3 years ago

Name one metal and one non-metal from the periodic table? <br><br> Y’all I need help<br> Thank you!

Stella [2.4K]

Answer:

Zinc and Neon!

Explanation:

zinc is an element that is a transition metal and neon is a noble gas

8 0

3 years ago

2. Determine the current in a 120-watt bulb plugged into a 120-volt outlet.

Sergio039 [100]

Answer:

1 ampere

Explanation:

Power = VI

V = voltage

I = current

Given

Power = 120 watts

V = 120 volts

Therefore

120 = 120 x I

Divide both sides by 120

120/120 = 120/120 x I

1 = I

I = 1 ampere

8 0

3 years ago

61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

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

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

6 0

3 years ago

A proton and an alpha particle are momentarily at rest at adistance r from each other. They then begin to move apart.Find the sp

Arte-miy333 [17]

Answer:

The unknown quantities are:

E and F

The final velocity of the proton is:

√(8/3) k e^2/(m*r)

Explanation:

Hello!

We can solve this problem using conservation of energy and momentum.

Since both particles are at rest at the beginning, the initial energy and momentum are:

Ei = k (q1q2)/r

pi = 0

where k is the coulomb constant (= 8.987×10⁹ N·m²/C²)

and q1 = e and q2 = 2e

When the distance between the particles doubles, the energy and momentum are:

Ef = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

pf = m1v1 + m2v2

with m1 = m, m2 = 4m, v1=vf_p, v2 = vf_alpha

The conservation momentum states that:

pi = pf

Therefore:

m1v1 + m2v2 = 0

That is:

v2 = (1/4) v1

The conservation of energy states that:

Ei = Ef

Therefore:

k (q1q2)/r = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

Replacing

m1 = m, m2 = 4m, q1 = e, q2 = 2e

and v2 = (1/4)v1

We get:

(1/2)mv1^2 = k e^2/r + (1/2)4m(v1/4)^2 = k e^2/r + (1/8)mv1^2

(3/8) mv1^2 = k e^2/r

v1^2 = (8/3) k e^2/(m*r)

3 0

3 years ago